for Class XI & XII, Engineering Entrance and other ...
Transcript of for Class XI & XII, Engineering Entrance and other ...
for Class XI amp XII Engineering Entranceand other Competitive Exams
Mathematics at a Glance
Sanjay MishraB Tech (IIT-Varanasi)
ISBN 978-93-325-2206-0
Copyright copy 2015 Pearson India Education Services Pvt Ltd Published by Pearson India Education Services Pvt Ltd CIN U72200TN2005PTC057128 formerly known as TutorVista Global Pvt Ltd licensee of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisherrsquos prior written consent This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material in this eBook at any time
Head Office A-8 (A) 7th Floor Knowledge Boulevard Sector 62 Noida 201 309 URegistered Office Module G4 Ground Floor Elnet Software City TS-140 Block 2 Salai Taramani Chennai 600 113 Tamil Nadu India Fax 080-30461003 Phone 080-30461060 wwwpearsoncoin Email companysecretaryindiapearsoncom
eISBN 978-93-325-3736-1
Contents
Preface ivAcknowledgements v
1 Foundation of Mathematics 11-128 2 Exponential Logarithm 229-236 3 Sequence and Progression 337-347 4 Inequality 448-454 5 Theory of Equation 555-563 6 Permutation and Combination 664-678 7 Binomial Theorem 779-783 8 Infinite Series 884-886 9 Trigonometric Ratios and Identities 987-997 10 Trigonometric Equation 1098-10109 11 Properties of Triangle 11110-11120 12 Inverse Trigonometric Function 12121-12131 13 Properties of Triangle 13132-13139 14 Straight Line and Pair of Straight Line 14140-14151 15 Circle and Family of Circle 15152-15161 16 Parabola 16162-16172 17 Ellipse 17173-17179 18 Hyperbola 18180-18188 19 Complex Number 19189-19211 20 Sets and Relation 20212-20225 21 Functions 21226-21254 22 Limit Continuity and Differentiability 22255-22272 23 Method of Differentiation 23273-23277 24 Application of Derivatives 24278-24304 25 Indefinite Integration 25305-25321 26 Definite Integration and Area Under the Curve 26322-26336 27 Differential Equation 27337-27350 28 Vectors 28351-28365 29 Three Dimensional Geometry 29366-29381 30 Probability 30382-30391 31 Matrices and Determinants 31392-31411 32 Statistics 32412-32419
Any presentation or work on Mathematics must be conceived as an art rather than a text This is where this work holds it differently During my school days and throughout my long teaching career I realized that most of the JEE aspirants feel the need of a book that may provide them with rapid revision of all the concepts they learned and their important applications throughout their two years long time of preparation I prefer to call it Mathematics at a Glance The present book is written with sole objective of that The entire syllabus of Mathematics for AIEEE JEE Mains and JEE Advanced has been presented in an unprecedented format The reader ought to have the following pre requisites before going through it
(i) HeShe must have ample knowledge of high school Mathematics (ii) Must have conceptualtheoretical knowledge behind the various mathematical thoughts presented (iii) Must be confident enough that heshe is not the father of Mathematics and if not comfortable with
any concept or text we shall be thankful to have your valuable advice
As the name of this work suggests that it has been designed to help during revision It must be kept in mind that the motive of the text is to provide a recapitulation of the entire mathematics that you have studied in your mainstream syllabus While going through the book if you want detailed analysis of any thought or idea you must go for
ldquoFundamentals of Mathematics---By Sanjay MishrardquoAll the suggestions for improvement are welcome and shall be greatfully acknowledged
mdashSanjay Mishra
Preface
I am really grateful to ldquoPearson Educationrdquo for showing their faith in me and for providing me an opportunity to transform my yearning my years-long teaching experience and knowledge into the present rapid revision book ldquoMathematics at a Glancerdquo I would like to thank all teachers and my friends for their valuable criticism support and advice that was really helpful to carve out this work I wish to thank my parents and all my family members for their patience and support in bringing out this book and contributing their valuable share of time for this cause I extend my special thanks to my team including my assistant teachers Rakesh Gupta Parinika Mishra managers and computer operators for their hard work and dedication in completing this task
mdashSanjay Mishra
Acknowledgements
Chapter 1Foundation oF MatheMatiCs
MatheMatical Reasoning
11 INTRODUCTION
Mathematics is a pure application of brains To crack mathematical problems an analytical approach is required
12 PRE-REQUISITES
Flush out your thoughts of maintaining algorithms for mathematical problemsTry to connect the text and work in this chapter from high-school mathematics and make conclusive
analysis of applying basic principles of mathematics
121 Greek Words (Symbols)
Symbol Meaning Symbol Meaning Symbol Meaning
α Alpha β Beta g Gammaδ D Delta isin ε Epsilon ξ Zeta
η Eta θ Theta i Iotaκ Kappa λ Lambda micro Muv Nu ξ Xi o Omicronπ Pi ρ Rho σ sum Sigmaτ Tau υ Upsilon f Phiχ Chi ψ Psi ω Omega
13 UNDERSTANDING THE LANGUAGE OF MATHEMATICS
Well Obviously mathematics is no language by itself but as remarked by Albert Einstein ldquoMathematics is the language in which god has written the universerdquo
12 Mathematics at a Glance
131 Mathematical Symbols
Symbol Meaning Symbol Meaning Symbol Meaning
Therefore int Single Integration D Triangle
∵ Because Since int int Double Integration rArr Implies
Such that Σ Sigma N The set of natural numbers So as a Proportionate to hArr Implies and is implied by Ratio f Function Z or I The set of integers
Proportion infin Infinity Q The set of rational numbers= Equal to _ Line bracket ℝ The set of real numbersne Not equal to () Small bracket |x| Absolute value of xgt Greater than Mid bracket ie ie (that is)lt Less than [] Large bracket eg example gratia (for example)
ge Greater than or equal to
isin Belongs to QED Quod erat demonstrandum
le Less than or equal to
notin Does not belong to nsub Is not a subset of
∢ Not less than sub Is a subset of cup Universal setnth root cup Union of sets ~ Similar toCube root cap Intersection of sets iff If and only if
ang Angle A times B Cartesian product of A and B
|| Parallel
^ Perpendicular A ndash B Difference of two sets A and B
f Null Set (phi)
Congruent to forall For all cap Arc$ There exists
14 STATEMENTS AND MATHEMATICAL STATEMENTS
141 Statement
It is a sentence which is complete in itself and explains its meanings completely eg Delhi is the capital of India
142 Mathematical StatementsA given statement is mathematical if either it is true or it is false but not both
143 Scientific StatementA given sentence will qualify as a scientific statement even if it may be true conditionally eg mass can be neither created nor destroyed
Foundation of Mathematics 13
15 CLASSIFICATION OF MATHEMATICAL STATEMENTS
1 axiom Mathematical statements which are accepted as truth without any formal proof given for it eg Equals added to equals are equals
2 Definition Mathematical statement which is used to explain the meaning of certain words used in the subject
Eg ldquoThe integers other than plusmn1 and 0 which are divisible by either one or by themselves are called prime integersrdquo
3 Theorems A mathematical statement which is accepted as lsquotruthrsquo only when a formal proof is given for it like summation of interior angles of a triangle is 180 degree is a theorem
151 ConjecturesIn mathematics a conjecture is an unproven proposition that appears correct For example every even integer greater than two can be expressed as a sum of two primes
152 Mathematical Reasoning
Reasoning is a process of logical steps that enables us to arrive at a conclusion In mathematics there are two types of reasoning These are as follows 1 inductive Reasoning Like that in mathematical induction 2 Deductive Reasoning Series of steps to deduct one mathematical statement from the other and
their proof which will be discussed in the text
16 WORKING ON MATHEMATICAL STATEMENTS
161 Negation of a Statement
The denial of a statement is called its negation To negate a statement we can use phrases like ldquoIt is falserdquo ldquois notrdquo Rita is at home rArr Rita is not at home
162 Compounding of Statements
Compounding of statements is defined as combining two or more component statements using the connecting words like lsquoandrsquo and lsquoorrsquo etc The new statement formed is called a compound statement
Compounding with OR
p x is odd prime numberq x is perfect square of integer
x is a odd prime or a perfect square integer
Compounding with AND
p 2 is a prime numberq 2 is an even number
2 is a prime and even natural number
NoteOR be inclusive or exclusive depending both conditions are simultaneously possible or not respectively
14 Mathematics at a Glance
17 IMPLICATION OF A STATEMENT
If two statement p and q are connected by the group of words lsquoifhellip thenhelliprsquo the resulting compound statement
lsquoif p then qrsquo is called lsquoconditional implicationsrsquo of p and q is written in symbolic form as lsquop rarr qrsquo (read as lsquop implies qrsquo)
eg p the pressure increases q the volume increasesThen implication of the statements p and q is given by p rarr q if the pressure increases then the
volume decreases
171 Converse of a Statement
it is given by p rArr q means q rArr pIf a integer n is even then n2 is divisible by 4 Converse is ldquoIf n2 is divisible by 4 then n must be evenrdquo
172 Contra Positive of a Statement p rArr q is ~q rArr ~p
If a triangle has two equal sides then it is isosceles triangle Its contrapositive is lsquoif a triangle is not isosceles then it has no two sides equalrsquo
18 TRUTH VALUE
The truth (T) or falsity (F) of any statement is called its truth value Eg every mathematical statement is either true or false Truth value of a true statement is (T) and that of a false statement is (F)
Given below in the table are Venn Diagrams and truth tables of various mathematical and logical operations
Operation Venn Diagram Truth Table And p q p and qp ^ q
T T TT F FF T FF F F
Or p q p or qp or qT T TT F TF T TF F F
Foundation of Mathematics 15
Operation Venn Diagram Truth Table Negation p ~p
T FF T
Implies and is Implied by
p q p rarr q q rarr p (prarrq) ^ (qrarrp)
T T T T TT F F T FF T T F FF F T T T
Implication p q p rarr q
T T TT F FF T TF F T
19 QUANTIFIERS
These are phrases like ldquothere exists $rdquo ldquofor all forallrdquo less than greater than etc For example there exist a polygon having its all sides equal is known as a regular polygon
191 Proofs in MathematicsWe can prove a mathematical statement in various ways which are categorized as straightforward Mathod of exhaustion Mathematical induction Using counter example Contradiction and Contrapositive statements
192 What is a Mathematical AssumptionA mathematical statement which is assumed to be true until a contradiction is achieved An assumed statement may prove to be false at a later stage of mathematical analysis
nuMbeR systeMWell Life without numbers is unpredictable Numbers have been used since ages to facilitate our transac-tions regarding trade exchange or other mathematical purposes Number system has successfully replaced the Bartar system of exchange In this text we will discuss the number system followed by mathematical analysis of real world problems Our present number system is known as Indo-Arabic number system
110 SET OF NATURAL NUMBERS
ℕ = x x is counting number) Counting numbers are called lsquonatural numbersrsquo and their set is denoted as = 1 2 3 4 5
16 Mathematics at a Glance
If 0 is not included in the set of natural numbers then we obtain whole numbers (W) W = 0 1 2 3
1101 Algebraic Properties of Natural Numbers
They are associative and commutative ie for all a b c in the set of natural numbersassociative law a + (b + c) = (a + b) + c a(bc) = (ab)ccommutative law a + b = b + a ab = ba
The cancellation law holds for natural numbers If a b c are natural numbers a + c = b + crArr a = b ac = bc rArr a = b (c is not equal to zero)
Distribution of multiplication over addition a(b + c) = ac + bc Order properties (i) law of trichomy Given any two natural numbers a and b exactly one of the following
holds a gt b or a lt b or a = b (ii) transitivity For each triplet of natural numbers a b c a gt b and b gt c implies that a gt c (iii) Monotone Property for addition and Multiplication For each triplet of natural
numbers a gt b rArr a + c gt b + c and ac gt bc existence of additive and multiplicative identity Zero is an additive identity element and 1 is
a multiplicative identity element existence of additive and multiplicative inverse For every integer x there always exists its
negative ndashx which when added to x makes additive identity Multiplicative inverse of x is an element which when multiplied to x makes multiplicative identity 1
111 SET OF INTEGERS
When negatives of natural numbers are included in a set of whole numbers then a set of integers is formed ℤ = ndash4 ndash3 ndash2 ndash1 0 1 2 3 4
112 GEOMETRICAL REPRESENTATION OF INTEGERS
Greek Mathematicians invented Geometrical method of representing numbers on a line known as lsquonumber linersquo In this method a point is marked as zero (0) and with respect to zero the numbers are located in order of their magnitude The distance of number (x) from zero represents its magnitude (|x|)
1121 Properties of Integers (a) It is closed commutative associative and distributive for addition subtraction and multiplication (b) Zero is the identity element for addition and 1 is the identity for multiplication
Foundation of Mathematics 17
(c) Additive inverse of x is equal to ndashx Q x+ (ndashx) = 0 (d) Multiplicative inverse of x is 1x provided x ne 0 as x 1x = 1 (e) Cancellation law holds for addition as well as multiplication (f) Property of order forall x y isinℤ either x gt y or x = y or x lt y Also known as law of trichomy
113 DIVISION ALGORITHM
Given are two integers a and b such that a gt b and b gt 0 then there exist two integers q and r such that a = bq + r where a dividend b divisor q quotient r remainder
Properties The remainder r is a non-negative integer which is less than the divisor b 0 le r lt b where r = 0 1 2 3 4 b ndash 1 If the remainder r = 0 then a = bq Then a is called completely divisible by b (ie multiple of b) and b and q are called factors of a
1131 Even and Odd Integers (a) Set of even integer = x x = 2k where k isin ℤ (b) Set of odd integers = x x = 2k + 1 where k isin ℤ
1132 Prime Integer
An integer x (other than 0 ndash1 and 1) is called prime iff it has only positive divisors as 1 and itself eg 2 3 5 7 etc
11321 Properties
An integer other than 0 ndash1 and 1 which are non-primes are called composite numbers A composite integer has atleast three factors
1 ndash1 0 are neither prime nor composite Twin Primes A pair of primes is said to be twin primes if they differ by 2 ie 3 5 and 11 13 etc Co-Primes A pair of integers is said to be co-primes if they have no common positive divisor except
1 eg 8 5 and 12 35 If p is prime and greater than or equal to 5 then p is either 6k + 1 or 6k ndash 1 but converse is not
necessarily true If p is prime and greater than 5 then p2 ndash 1 is always divisible by 24
114 FACTORIAL NOTATION
Factorial of r is denoted as r and is defined as product of first r natural numbers ie r = 1 2 3 4hellip (r ndash 1)reg 1 = 1 2 = 2 3 = 6 4 = 24 5 = 120 6 = 720 7 = 5040
Product of any r consecutive integer is always divisible by r
18 Mathematics at a Glance
1141 Related TheoremsTheorem 1 xn ndash yn is divisible by (x ndash y) forall x isin ℕ since putting x = y makes expression xn ndash yn = yn ndash yn = 0 Therefore x ndash y must be factor in the above expression
Theorem 2 xn ndash yn is divisible by (x + y) forall odd natural numbers n Since putting x = ndashy makes expression xn + yn = yn + (ndashy)n = yn + (ndash1)n yn = yn ndash yn = 0 Thus x + y must be factor in the above expression (xn + yn) = (x + y)(xnndash1 ndash xnndash2y + xnndash3y2 ndashhellip+ (ndash1)nndash1 ynndash1)
Theorem 3 Given n isin ℕ and p and p is prime such that ldquon is co-prime to prdquo then np ndashn is always divisible by pFermatrsquos Theorem n = 2 and p = 5 rArr 5|25 ndash 2 rArr 5|24 ndash 1
corollary 1 np ndash n is also divisible by n and (n ndash 1)corollary 2 np ndash n is divisible by n(n ndash 1) Since n and (n ndash 1) are always co-primecorollary 3 npndash1 ndash 1 is always divisible by p
Theorem 4 (fundamental theorem of arithmetic) A natural number N can be expressed as product of non-negative exponent of primes N = pa qb rc sd hellip where p q r s are primes and a b c d are whole numbers eg 1800 = 23325270
Theorem 5 (Wilsonrsquos theorem) if p is a prime number then 1 + (p ndash 1) is divisible by p ie 16 + 1 is divisible b
1142 Divisors and Their PropertyA natural number x = pa qb rg is called divisor of N = pa qb rc rArr N is completely divisible by x
hArr all the prime factors of x are present in NhArr 0 le α le a 0 le b le b 0 le g le c where a b g are whole numbers Set of all divisors of N is given as x x = pα qb rg where 0 le α le a 0 le b le b 0 le g le c
1143 Number of Divisorsn (αbg) 0 le α le a 0 le b le b 0 le g le c = number of ways the integers a b g can take values applying the above restrictions = (a + 1)(b + 1)(c + 1)
sum of Divisor of n = pa qb rc (1 + p + p2 ++ pa) (1 + q + q2 ++ qb) (1 + r +r2 ++ rc)
improper Divisors of N = pa qb rc when a = b = g = 0 rArr x = 1 this is divisor of every integer and a = a b = b and g = c then x becomes number N itself These two are called lsquoimproper divisorrsquo The number of proper divisors of N = (a + 1)(b + 1)(c + 1) ndash 2
If p = 2 then number of even divisors = a(b + 1)(c + 1) Number of odd divisors = (b + 1)(c + 1)Number and sum of divisors of N divisible by a natural number 1 1 1a b cy p q r=
Let x = pa qb rg be such divisors ∵ 1 1 1a b cy | x p q r | p q rα β γrArrrArr a1 le α le a and b2 le b le b and c1 le g le c rArr Number of such divisors = (a ndash a1 + 1) (b ndash b1 + 1) (c ndash c1 + 1)
Sum of such divisors Sy = 1 1 1 1 1 1a a 1 b b 1 c c 1a b cyS (p p p )(q q q )(r r r )+ + += + + + + + + + + +
= 1 1 1a a b b c c2 2y(1 p p p )(1 q q q )(1 r r )minus minus minus+ + + + + + + + + + +
= 1 1 1a a 1 b b 1 c c 1p 1 q 1 r 1y
p 1 q 1 r 1
minus + minus + minus + minus minus minus
minus minus minus
Foundation of Mathematics 19
Notes
1 The number of ways of resolving n into two factors is + + +1
( a 1)( b 1)( c 1)2
when n is not a perfect
square and + + + +1
( a 1)( b 1)( c 1) 12
when n is a perfect square
2 Every number n has two improper divisors 1 and n itself and the remaining divisors are called proper divisors Eg number of proper divisors of 108 is 10
1144 Least Common Multiple (LCM)
LCM of set of numbers is the smallest number (integerrational) which is completely divisible by each of them ie x is said to be LCM of y and z iff y divides x z also divides x and x is least positive of all such numbers Eg LCM of 6 4 9 is 36
Let x and y be two given integer x = paqβrgsd and 1 1 1 1y p q r sα β γ δ= where p q r are primes
If z is LCM of x and y then 1 1 1 1max( ) max( ) max( ) max( )z p q r sα α β β γ γ δ δ=
LCM LCM (a and c)a cand
b d HCF (b and d)=
1145 Greatest Common Divisor (GCD)Highest Common Factor (HCF)
HCF of a given set of numbers is the largest number which divides each of the given numbers HCF of y and z is also denoted as (y z) Therefore x is said to be GCD of y and z if x divides both y and z and x is largest of such numbers So clearly every common divisor of y and z also divides x and x ne 0
Eg HCF of 12 and 64 is 4 GCD of 6 and 35 is 1 (co-prime)
HFC HCF (a and c)a cand
b d LCM (b and d)=
Method to find hcF For two given integers x and y
Method 1 Consider their prime factors 1 1 1 2 2 2x p q r and y p q r α β γ α β γ= =If z is HCF of x and y rArr zx and zy z contains the least power for each corresponding prime factor rArr 1 2 1 2 1 2min min min z (x y) p q r α α β β γ γ= =
1146 Decimal Representation of Number
given a natural number x abcde= where e d c b a are respectively digits occupying unit tenrsquos hundredth thousandth ten thousandth places So the numerical value of x is defined as lsquosum of products of digits multiplied by their corresponding place valuesrsquo
th th th
4 3 2 1 0
tens place unit placetenthousand thousand hundredvalue valueplace value place value place value
x = a 10 + b 10 + c 10 + d 10 + e 10
minus minus minus
times times times times times
Theorem If an integer x is divided by 10 the reminder is a digit at the unit place of x
Proof = = + + + + = +4 3 2x abcde a(10 ) b(10 ) c(10 ) d(10) e 10m e rArr Remainder is e
110 Mathematics at a Glance
Theorem The remainder if an integer x is divided by 5 is e 0 e 4e 5 5 e 9
le le minus le le
where e is are unit place
digit of the number 4 3 2x abcde a(10 ) b(10 ) c(10 ) d(10) e= = + + + +
= a(104) + b(103) + c(102) + d(10) + e = 5m + e 0 le e le 9
0
5m e 0 e 4 5m e 0 e 45m 5 e 5 5 e 9 5m (e 5) 5 e 9
+ le le + le le = = + + minus le le + minus le le
1147 Periodic Properties of Integers
Theorem 1 Unit digit of nth power of an integer having zero at its unit place is zero
rArr n1 1 1(abc0) (a b c 0)=
Theorem 2 Unit digit of nth power of an integer having one at its unit place is one
rArr n1 1 1(abc1) (a b c 1)=
Theorem 3 Unit digit of nth power of an integer having two at its unit place is described as follows
rArr cn1 1 1(abc2) (a b c 2)= if n = 4k + 1 ie n
1 1 1(abc2) (a b c 4)= if n = 4k + 2
ie n1 1 1(abc2) (a b c 8)= if n = 4k + 3 ie n
1 1 1(abc2) (a b c 6)= if n = 4k
Theorem 4 Unit digit of nth power of an integer having three at its unit place is described as follows
rArr n1 1 1(abc3) (a b c 3)= if n = 4k + 1 ie n
1 1 1(abc3) (a b c 9)= if n = 4k + 2
rArr ie n1 1 1(abc3) (a b c 7)= if n = 4k + 3 ie n
1 1 1(abc3) (a b c 1)= if n = 4k
Theorem 5 Unit digit of nth power of an integer having four at its unit place is described as follows
rArr n1 1 1(abc4) (a b c 4)= if n = 2k + 1 ie n
1 1 1(abc4) (a b c 6)= if n = 2k
Theorem 6 Unit digit of nth power of an integer having five at its unit place has five at unit place
rArr n1 1 1(abc5) (a b c 5)= if n isin ℕ
Theorem 7 Unit digit of nth power of an integer having six at its unit place has six at unit place
rArr n1 1 1(abc6) (a b c 6)= if n isin ℕ
Theorem 8 Unit digit of nth power of an integer having seven at its unit place is described as follows
ie n1 1 1(abc7) (a b c 7)= if n = 4k + 1 ie n
1 1 1(abc7) (a b c 9)= if n = 4k + 2
ie n1 1 1(abc7) (a b c 3)= if n = 4k + 3 ie n
1 1 1(abc7) (a b c 1)= if n = 4k
Theorem 9 Unit digit of nth power of an integer having eight at its unit place is described as follows
ie n1 1 1(abc8) (a b c 8)= if n = 4k + 1 ie n
1 1 1(abc8) (a b c 4)= if n = 4k + 2
ie n1 1 1(abc8) (a b c 2)= if 4k + 3 n
1 1 1ie (abc8) (a b c 6) if n 4k= =
Theorem 10 Unit digit of nth power of an integer having nine at its unit place is described as followsn
1 1 1ie (abc9) (a b c 9) if n 2k 1= = + n1 1 1ie (abc9) (a b c 1) if n 2k= =
Foundation of Mathematics 111
115 TESTS OF DIVISIBILITy
1 Divisibility by 2 A number N is divisible by 2 if and only if its last digit is divisible by 2 (ie even) 2 Divisibility by 3 A number N is divisible by 3 if and only if the sums of all digits are divisible by 3 3 Divisibility by 4 A number N is divisible by 4 if its units digit plus twice its tenrsquos digit is divisible by 4 4 Divisibility by 5 A number N is divisible by 5 if and only if its last digit is divisible by 5 (ie if it
ends in 0 or 5) 5 Divisibility by 6 A number N is divisible by 6 if and only if its unitsrsquos digit is even and the sum of
its digits are divisible by 3 6 Divisibility by 7 A number N is divisible by 7 if and only if 3 times unitrsquos digit + 2 times tenrsquos digit ndash 1
times hundredrsquos digit ndash 3 times thousandrsquos digit -2 times ten thousandrsquos digit + 1 times hundred thousandrsquos digit is divisible by 7 ie 3(a0) + 2(a1) ndash 1(a2) ndash 3(a3) ndash 2(a4) + 1(a5) + 3(a6) + is divisible by 7
ie If there are more digits present in the sequence of multipliers 3 2 ndash 1 ndash 3 ndash 2 1 is repeated as often necessary
7 Divisibility by 8 A number N is divisible by 8 if and only if its unitrsquos digit + 2times tenrsquos digit + 4 times hundredrsquos digit is divisible
8 Divisibility by 9 A number N is divisible by 9 if and only if the sum of its digits is divisible by 9 9 Divisibility by 10 A number N is divisible by 10 if and only if the last digit is 0 10 Divisibility by 11 N is divisible by 11 if and only if the difference between the sum of the digits in
the odd places (starting from the right) and the sum of the digits in the even places (starting from the right) is a multiple of 11 eg 1221 123321 2783 etc
12 Divisibility by 13 A number N is divisible by 13 if and only if 10 times unitsrsquos digit ndash 4 times tenrsquos digit ndash 1 times hundredrsquos digit + 3 times thousandrsquos digit + 4 times ten thousandrsquos digit + 1 times hundred thousandrsquos digit is divisible by 13 (If there are more digits present the sequence of multipliers 10 ndash4 ndash1 3 4 1 is repeated as often as necessary)
116 RATIONAL (ℚ) AND IRRATIONAL NUMBERS (ℚprime)
A number x in the form pq where p and q are integers and q is not equal to 0 is called rational and
otherwise it is called irrational numbers ( or ) eg 1 3 5 0 25 1016 107 are rational while radic2 radic3 radic5hellip radicx x is not a perfect square of rational are irrationals Pie (p) is ratio of circumference of any circle to the diameter of the same circle It is an irrational number approximately equal to rational numbers 227 or 314
euler number (e) 1 1 1e 1 27 e 81 2 3
= + + + + infinrArr lt lt
1161 Properties of Rational and Irrational Numbers
If a number x in decimal form is written as x cdepqr= then
th
2 1 0 1 2 3
tens place unit place first decimal Second decimal third decimalhundredvalue value place value place value place valueplace value
x c 10 d 10 e 10 p 10 q 10 r 10minus minus minus
minus minus minusminus
= times + times + times + times + times + times
All terminating decimals are rational eg 4
abcdeabcde10
= = = 1 2 n1 2 3 n n
ax x xx ax x x x10
112 Mathematics at a Glance
If a rational pq (in lowest term) is terminating decimal then q = 2m5n ie q must not contain any prime factor other than 2 or 5
Non-terminating but repeating decimals are also rationals eg y = xab ab ab helliphellip y xabrArr = helliphellip(i)
If number of repeating digits be n then multiply both side by 10n ie 210 y xabab= helliphellip(ii)
Subtracting (i) from (ii) we get xab xy99minus
= (which is a rational number)
Non-terminating and non-repeating decimals are irrationals 271354921275718 hellip (no periodic re-occurrence up to micro)
Set of rational numbers is countable while set of irrational numbers is uncountable
1162 nth Root of a NumberA real number y is called nth root of real number x where n is a natural number (n ge 2) Iff yn = x When n = 2 then it is called as square root and for n = 3 known as cube root All the numbers other than zero have more than one nth roots eg both 2 and ndash2 are square root of 4
1163 Principal nth RootThe principal nth root of a real number x (having atleast one n-th root) is that nth root which has its sign same as that of x It is denoted by a radical symbol n x
The positive integer n is known as the index of the radical symbol Usually we omit the index from the radical sign if index n = 2 and write as x
eg 27 336 68 2
= = and 5 4243 ( 3) 16 2minus = minus = whereas 4 16minus is a non-real number since fourth
power of no real number can be ndash16 which is negative
1164 Properties of nth Root
(i) Every positive real number x has exactly two real nth roots when n is a positive even natural number
(n = 2m) denoted by 2m 2mx and xminus are two real fourth roots of 256 eg 4 4256 4 256 4= minus = minus
(ii) Every real number x has only one real nth roots when n is a positive odd natural number
(n= 2m + 1) denoted by 2m 1 x+ eg 3 3125 5 125 5= = minus
(iii) nth root of a negative real number x is non-real when n is an even integer Eg 424 16minus minus has
no real values 1minus is a non-real number symbolized as i (iota)
(iv) Zero is only real number which has only one nth root and n 0 0= (v) Integers such as 1 4 9 16 25 and 49 are called perfect squares because they have integer
square roots (vi) Integers such as 1 8 27 64 are called perfect cubes as they have integer cube roots
square roots If b is the square root of a where a is the non-negative real number then b when squared must become equal to a
rArr b2 = a rArr b2 ndash a = 0 rArr (b ndash radica) (b + radica) = 0rArr b ndash radica = 0 or b + radica = 0 rArr b = radica (positive) or b = ndashradica (negative)
Foundation of Mathematics 113
11641 Properties of Square Roots
(i) Zero has only one square root ie zero (ii) Every positive real number (except zero) has two square roots One of them is positive (called as
principle square root denoted as radica) and the other is negative denoted as (ndashradica) (iii) Magnitude of real number x denoted as |x| and defined as the quantity of x is
2
x if x 0| x | x 0 if x 0
x if x 0
gt= = =minus lt
1165 Algebraic Structure of and
closure law For addition and subtraction multiplication commutative law For addition and multiplication associative law For both addition and multiplication Distributive law For addition and subtraction operation
Notes
(i) Zero is the identity element for addition and 1 is the identity for multiplication
Q x + 0 = x and 1
x 1x
times =
forall x isin ℚ x ne 0
(ii) Additive inverse of x = p | q is equal to ndashx Q x + (ndashx) = 0
(iii) Multiplicative inverse of = =p 1
xq x
provided x ne 0 as 1x 1
x=
cancellation law holds for addition as well as multiplication 1 2 1 3 2 3
1 2 1 3 2 3
x x x x x xx x x x x x+ = + rArr =
= rArr =
provided x1 ne 0 Property of order forall x y isin ℚ either x gt y or x = y or x lt y Also known as law of trichotomy Union of set of rationals and set of irrationals is called set of real numbers ℝ
117 SURDS AND THEIR CONJUGATES
Sum of a rational and an irrational number is always irrational and called as surd denoted by s
rational irrationalpart of s part of s
s a b= + where b is not a perfect square of the rational number
For every surd s there exist element s s a b= minus where s a b= + called as conjugate of s
Rationalization of denominator of an irrational number2
2 2
s a b (a b)(a b) a b 2a bs a b a b(a b) (a b)(a b)
+ + + += = = +
minus minusminus minus +
114 Mathematics at a Glance
118 REAL NUMBERS SySTEM
Union of set of Rationales and set of Irrationals is called set of Real numbers (ℝ) = cup
Properties
(i) Square of real numbers is always non-negative If x isin ℝ rArr x2 ge 0 (ii) Between any two real numbers there are infinitely many real numbers (iii) Magnitude of real number x is denoted as |x| and defined as the quantity of x
ie 2
x if x 0| x | x 0 if x 0
x if x 0
gt= = =minus lt
(iv) They are represented on a straight line called as real number line in order of their magnitude such that distance of the number of x from zero is equal to magnitude of x (|x|)
(v) A real number line is infinitely dense and continuous line ie between two any two number (how so ever closed they are) there lies infinitely real number
1181 Concept of IntervalAs the set of all real numbers lying between two unequal real numbers a and b can never be expressed in roster form therefore these are expressed in set builder form using the concept of intervals
open interval Denoted as (a b) x isin (a b) = x a lt x lt b x isin ℝ ie end points are not included
closed interval [a b] x isin [a b] = x a le x le b xisin ℝ the end points are included
semi-open interval x isin (a b] rArr a lt x le b and x isin [a b) rArr a lt x le b
1182 Intersection and Union of Two or More Intervals To find the intersection or union of two or more intervals locate each interval over the same real number line and for intersection take the interval which is common to both and for union locate the interval which includes the numbers of all the interval considered
119 MATHEMATICAL INDUCTION
Mathematical induction is a mathematical tool by which we can prove the correctness of any mathematical statement or proposition It works on the principle that results for higher integers are induced from the results for lower integers
Foundation of Mathematics 115
Statement Working RuleFirst principle of mathematical induction
The set of statements P(n) n isin N is true for each natural number n ge m is provided thatP(m) is trueP(k) is true for n = k (where k ge m)rArr P(n) is true for
n = k + 1
Let there be a proposition or a mathematical statement namely P(n) involving a natural num-ber n In order to prove that P(n) is true for all natural numbers n ge m we proceed as followsVerify that P(m) is trueAssume that P(k) is true (where k ge m)Prove that P(k + 1) is trueOnce step ndash (c) is completed after (a) and (b) we are through ie P(n) is true for all natural numbers n ge m
Second principle of mathematical induction
The set of statements P(n) n isin N is true for each natural number n ge m provided thatP(m) and P(m + 1) are true P(n) is true for n le k (where k ge m)rArr P(n) is true for
n = k + 1This is also called extended principle of Mathematical Induction
Verify that P(n) is true for n = m n = m + 1Assume that P(n) is true for n le k (where k ge m)Prove that P(n) is true for n = k + 1Once rule (c) is completed after (a) and (b) we are through That is P(n) is true for all natural numbers n ge m This method is to be used when P(n) can be expressed as a combination of P(n -1) and P(n - 2) In case P(n) turns out to be a combination of P(n -1) P(n - 2) and P(n -3) we can verify for n = m + 2 also in Rule(a)
1191 Ratio and ProportionRatio and proportions are algebraic operations which are operated on one or more variables as
Ratio It is a rational between two quantities that tells us what multiplepart one quantity is of the other Therefore if x and y are two quantities of the same kind then their ratio is x y which may be denoted by xy (This may be an integer or fraction)
1 A ratio may be represented in a number of ways eg x mx nxy my ny= = = where m nare
non-zero numbers 2 To compare two or more ratios always reduce them to a common denominator
3 Ratio of two fractions may be represented as the ratio of two integers eg x z xy xu y u zu yz
= or xu yz
4 Ratios are compounded by taking their product ie x z v xzv y u w yuw
=
5 Duplicatetriplicate ratio If x y is any ratio then its duplicate ratio is x2 y2 triplicate ratio is x3 y3 etc If xy is any ratio then its sub-duplicate ratio is x12 y12 sub triplicate ratio is x13 y13 etc
ProportionWhen two ratios ab and cd are equal then the four quantities composing them are said to be propor-tional If abcd are proportional then ab = cd and it is written as ab = cd or ab c d 1 lsquoarsquo and lsquodrsquo are known as extremes whereas lsquob and crsquo are called as means 2 Product of extremes = product of means
116 Mathematics at a Glance
1192 Some Important Applications of Proportion
If four a b c d are proportional then many other useful proportions can be derived using various laws of fraction which are extremely useful in mathematical calculations and simplifications
invertando If a b = c d then b a = d calternando If a b = c d then a c = b d
componendo If ab = cd then a b c d
b d+ +
=
∵ a cb d= adding 1 from both sides a c1 1
b d+ = + rArr
a b c db d+ +
=
Dividendo If a b = c d then a b c db dminus minus
=
∵ a cb d= subtracting one to both sides a c1 1
b dminus = minus rArr
a b c db dminus minus
=
componendo and dividendo If a b = c d then applying both componendo and dividendo operations
together we get a b c da b c d+ +
=minus minus
If a c eb d f= = (say = l) then
1nn n n
n n n
xa yc zexb yd zf
+ + + +
1193 Linear EqualitiesAn expression of the form y = ax + b where a and b isin ℝ is called a linear polynomial function of x y and set of points (x y) satisfying the above relations if plotted on the xy plane a straight line is obtained An equation of the form ax + by + c = 0 is termed as linear equation in x and y
solving simultaneous linear equations in two unknowns
To solve a pair of linear equation a1x + b1y = c1 (i)a2x + b2y = c2 (ii)
The following three approaches are adopted
1194 Method of ComparisonFrom both equations find the value of any one variable say y in terms of other ie x
1 1 2 2
1 2
c a x c a xyb bminus minus
= = rArr 1 2 1 2
1 2 1 2
c c a a xb b b b
minus = minus
rArr 2 1 1 2
1 2 2 1
b c b ca b a b
minusminus
and similarly we get 2 1 1 2
1 2 1 2
a c a cyb a a b
minus=
minus
1195 Method of SubstitutionTo solve equations (i) and (ii) substitute the value of y from equation (i) to (ii) get x and y then can also be
obtained 1 12 2 2
1
c a xa x b cb
minus+ =
a2b1x + b2c1 ndash b2a1x = b1c2
rArr (a2b1 ndash a1b2)x = b1c2 ndash b2c1 rArr 1 2 2 1
2 1 1 2
b c b cxa b a b
minus=
minus and so we get 1 2 2 1
1 2 2 1
a c a cya b a b
minus=
minus
Foundation of Mathematics 117
1196 Method of Eliminationa1x + b1y = c1 (i)a2x + b2y = c2 (ii)
Multiplying equation (i) by a2 and equation (ii) by a1 and subtracting x gets eliminated
a1a2x + b1a2y = a1c1 (iii)a1a2x + a1b2y = a1c2 (iv)
Subtracting equation (iii) and (iv) 2 1 1 2
2 1 1 2
a c a cya b a b
minus=
minus and thus 1 2 2 1
2 1 1 2
b c b cxa b a b
minus=
minus
11961 Method of cross-multiplication
It is a very useful method for solving pair of linear equations in two or three variables Given two equations a1x + b1y + c1 helliphellip (i)a2x + b2y + c2z helliphellip (ii)
Dividing both equations by z and replacing 0x xz= and 0
y yz= we get
a1x0 + b1y0 + c1 helliphellip (iii)
a2x0 + b2y0 + c2z helliphellip (iv) Solving by any of the above mentioned three elementary methods we get
2 1 1 2 2 1 1 20 0
2 1 1 2 2 1 1 2
b c b c b c b cx xx xa b a b z a b a b z
minus minus= = = =
minus minus
that can be symmetrically expressed as 1 2 2 1 1 2 2 1 1 2 2 1
x y zb c b c c a c a a b a b
= =minus minus minus
Thus we can conclude that the set of solution of above pair of equation can always be expressed by the ratio x y z in terms of coefficients of the equations
step (1) Express the coefficients of x y z beginning with y in cyclic order as shown in the figure and take the product of the coefficients indicated by arrows
step (2) The product formed by descending arrows is considered positive and those by ascending arrows is taken negative
step (3) So we get x y z (b1c2 ndash b2c1) (c1a2 ndash c2a1) (a1b2 ndash a2b1)
FunDaMentals oF inequality
120 INTRODUCTION
The concept of inequality finds its origin from the property of order of real numbers An inequation is marked by the use of logical operations such as lt gt le ge ne etc An inequation can have one or more than one variables ax + by + c ge 0
inequation An inequation is a statement involving sign of inequality ie lt gt le ge ≮ ≯ ne
118 Mathematics at a Glance
1201 Classification of InequalityInequalities are of four types
If a ndash b gt 0 rArr a gt b (read a greater than b)If a ndash b ge 0 rArr a ge b (read a greater than or equal to b)If a ndash b lt 0 rArr a lt b (read a is less than b)If a ndash b le 0 rArr a le b (read a is less than or equal to b)
linear inequality Inequality having variables in one degree eg 2x + 3y gt 5 x ndash 2y + 3z = 5 etc
solution of inequality The values of unknown variable which satisfies the given inequation are called solutions of inequality eg x = 2 y = 4 is a particular solution of inequality 2x + 3y gt 5
12011 Basic properties of inequality and laws
(i) transition property If a gt b and b gt c rArr a gt c (ii) law of trichotomy If x and y are two real numbers then exactly one of the three statements
hold ie x gt y or x lt y or x = y (iii) If a gt b then a + c gt b + c and a ndash c gt b ndash c forall c isin ℝ (iv) If x lt y lt 0 rArr |x| gt |y| (Larger the number smaller the magnitude) (v) If x gt y gt 0 rArr |x| gt |y| (Larger the number larger the magnitude) (vi) If a gt b then ac gt bc forall c gt 0 (sign of inequality does not change on multiplying by positive
real number) (vii) If a gt b then ac lt bc forall c lt 0 (sign of inequality gets reverse when multiplied both sides by negative
real number)
(viii) If a gt b then a b for c 0c cgt gt and a b for c 0
c clt lt
(ix) If a cb dge then ad ge bc if b and d same sign
(x) If a cb dge then ad le bc if b and d are opposite signs
(xi) law of addition If a1 gt b1 and a2 gt b2hellip and an gt bn rArr (a1+a2+hellip+ an) gt (b1+ b2 +hellip+ bn) (xii) law of Multiplication If a1 gt b1 gt 0 and a2 gt b2 gt 0hellip and an gt bn gt 0 rArr (a1a2a3hellipan) gt (b1b2b3hellipbn) (xiii) laws of reciprocal
(a) If 0 lt a lt b then 1 1a bgt (b) If b lt a lt 0 then 1 1
b agt
(c) If x isin [a b] then
gt lt minusinfin cup infin lt gt = isin infin = gt = minusinfin = lt =
1 1 for a b 0 or a b 0b a
1 1 for a 0 b 0 not defined at x 0a b1
x 1 for a 0 b 0 not defined at x 0b
1 for b 0 a 0not defined at x 0a
Foundation of Mathematics 119
(xiv) laws of squares or positive even powers
2 2
2 2
2 2
a b if both ab 0agtb a b If |a| = |b|
a b If ab lt 0
gt ge
rArr = lt
If a and b have opposite sign and a gt b ie a gt 0 and b lt 0 then
2 2
2 2
2 2
a b iff |a | | b |a b a b iff |a | | b |
a b iff |a | | b |
gt gt
gt rArr = = lt lt
This law can be extended for any even natural power (2n)
If x isin [a b] then
2 2
2 22 2
2
2
[a b ] for a b 0[b a ] for a b 0
x x[0a ] for a b and ab 0
[0b ] for b a andab 0
gt
ltisin isingt lt
gt lt
similar is the case for x2n n isinℕ
(xv) law of square root If a and b both are non-negative and 2n 2n
a ba b
a b
gtgt rArr gt
(xvi) laws of cubes or positive odd powers If x isin [a b] then x3 isin [a3 b3] similarly x2n+1 isin [a2n+1 b2n+1] for n isin ℕ
(xvii) law of cube root a gt b rArr a3 gt b3 and a13 gt b13 forall a b isinℝ a lt b rArr a3 lt b3 and a13 lt b13 forall a b isin ℝ this law can be extended for any odd natural power (2n+ 1) and odd root
(xviii) laws of exponential inequality (a) If 0 lt a lt 1 and r isin ℝ+ then 0 lt ar lt 1 lt andashr (b) If a gt 1 and r isin ℝ+ then ar gt 1 gt andashr gt 0(c) For a gt 1 ax gt ay for x gt y and x y isin ℝ(d) For 0 lt a lt 1 ax lt ay for x gt y and x y isin ℝ(e) For a bisin (0 1) or a b isin (1infin) if a gt b then ax lt bx for x lt 0 and ax gt bx for x gt 0(f) For a isin (0 1) and b isin (1 infin) ax gt bx for x lt 0 and ax lt bx for x gt 0
(xix) laws of logarithmic inequality (a) x ge y hArr logax ge logay for a gt 1 (b) x ge y hArr logax le logay for 0 lt a lt 1(c) ax ge y rArr x ge logay for a gt 1 (d) ax ge y rArr x le logay for 0 lt a lt 1
RemarkAbove two results follow from the fact that logarithmic and exponential function to the base a gt 1 are increasing function and when base lies between 0 and 1 then they become decreasing function
(xx) inequalities containing modulus functions(a) |x| lt a hArr ndasha lt x lt a where a gt 0 ie x isin (-a a)(b) |x| le a hArr ndasha le x le a where a gt 0 ie x isin [-a a](c) |x| gt a hArr x lt ndasha or x gt a ie x isin (ndashinfin ndasha] cup (a infin) (d) |x| ge a hArr x le ndasha or x ge a ie x isin (ndashinfinndasha] cup [a infin)(e) a lt |x| lt b hArr x isin (ndashb b) for a le 0(f) a lt |x| lt b hArr x isin (ndashb ndasha] cup [a b) for a gt 0
120 Mathematics at a Glance
(xxi) triangle inequality | |x| ndash |y| | le |x plusmn y| le |x| + |y| forall x y isin ℝ Further(a) |x + y| = |x| + |y| for xy ge 0 (b) |x + y| lt |x| + |y| for xy lt 0(c) |x ndash y| = |x| + |y| for xy le 0 (d) |x ndash y| lt |x| + |y| for xy gt 0(e) | |x| ndash |y| | = |x + y| for xy le 0 (f) | |x| ndash |y| | lt |x + y| for xy gt 0(g) | |x| ndash |y| | lt |x ndashy| for xy lt 0 (h) | |x| ndash |y| | = |x ndashy| for xy ge 0
12012 Solutions of linear in equations in two variables
1 by graphical method Let L equiv ax + by + c = 0 be a line then by = ndashax ndash c Since the P point satisfies the equation of the line aa + bb + c = 0 From the given diagram we interpret that g gt b bg gt bb for b gt 0 rArr aa + bg gt aa + bb rArr aa + bg + c gt aa + bb + c rArr aa + bg + c gt 0 Thus all the points lying in the half plane II above the line ax + by + c = 0 satisfies the
inequality ax + by + c gt 0 Similarly in case b lt 0 we can prove that the point satisfying ax + by + c gt 0 lies in the
half plane I Hence we infer that all points satisfying ax + by + c gt 0 lies in one of the half plane II or I
according as b gt 0 or b lt 0 and conversely Thus the straight line ax + by + c = 0 divides the whole x ndashy plane into three regions (a) For b lt 0 (i) R1 = (a b) aa + bb + c = 0 (ii) R2 = (a b) aa + bb + c lt 0 (iii) R3 = (a b) aa + bb + c gt 0 (b) For b gt 0
2 short-cut method step i Consider the equation from the Inequality step ii Draw the straight line representing the Equation step iii Consider a Point P (a b) (not on the line) and find the sign of
linear expression for P (ab) step iV Check whether it satisfies the inequality or not If it satisfies
then the inequality represents the half plane which contains the point and shade the region
step V Otherwise the inequality represents that half plane which does not contain the point within it
For convenience the point (0 0) is preferred step Vi The set R1 is a straight line while the sets R2 and R3 are called open half planes The set
R1 cup R3 represent the points whose co-ordinates satisfy ax + by + c ge 0 (b gt 0) and R1 cup R2 represent the points whose co-ordinates satisfying ax + by + c le 0 (b gt 0) Here R2 is the solution region of inequality ax + by + c lt 0 b gt 0 and R3 is the solution region of inequality ax + by + c gt 0 b gt 0
+ndashndashndashndash ndash ndashndash ndash ndashndash ndash ndash ndash ndash ndashndash ndash
++++++++++++++++
ndash ndash ndashndash ndash ndashndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash
+++++++++++++++++
ndashndashndashndash ndashndashndash ndash ndashndash ndash ndash ndash ndash ndashndash ndash
+++++++++++++++++
Foundation of Mathematics 121
1202 Rational Algebraic Inequalities
type 1 P(x) P(x) P(x) P(x)0 0 0 0Q(x) Q(x) Q(x) Q(x)
gt lt ge le P(x) Q(x) are polynomials
step 1 Factor P(x) and Q(x) into linear factors
step 2
(i) P(x) 0 P(x)Q(x) 0 P(x) 0Q(x) 0 or P(x) 0Q(x) 0Q(x)
gt rArr gt rArr gt gt lt lt
(ii) P(x) 0 P(x)Q(x) 0 P(x) 0 Q(x) 0 or P(x) 0Q(x) 0Q(x)
lt rArr lt rArr gt lt lt gt
(iii) P(x)Q(x) 0P(x) 0 P(x) 0 Q(x) 0 or P(x) 0 Q(x) 0
Q(x) 0Q(x)gege rArr rArr ge gt le lt ne
(iv) P(x)Q(x) 0P(x) 0 P(x) 0 Q(x) 0or P(x) 0 Q(x) 0
Q(x) 0Q(x)lele rArr rArr ge lt le gt ne
step 3 For solving the above inequalities formed eg P(x) Q(x) gt 0 use wavy curve method or solution set is given by x P(x) gt 0 Q(x) gt 0 cup x P(x) lt 0 Q(x) lt 0
type ii For solving inequality of the form P(x) R(x)Q(x) S(x)
lt
step 1 P(x) R(x) R(x) P(x)0or 0Q(x) S(x) S(x) Q(x)
minus lt minus gt
rArr P(x)S(x) R(x)Q(x) R(x)Q(x) P(x)S(x)0 or 0Q(x)S(x) S(x)Q(x)
minus minus lt gt
Now solve as in Type 1
type iii For solving inequality of the form P(x) R(x) T(x)Q(x) S(x) M(x)
lt lt
step 1 Solve the inequalities P(x) R(x) 0Q(x) S(x)
minus lt and R(x) T(x) 0S(x) M(x)
minus lt
rArr P(x)S(x) R(x)Q(x) 0
Q(x)S(x)minus lt
helliphellip(i) and R(x)M(x) T(x)S(x) 0
S(x)M(x)minus lt
helliphellip(ii)
Intersection of solution set of equations (i) and (ii) gives the solution set of the given inequality
Remarks
(i) If we have inequality of form gtP( x )
0Q( x )
and Q(x) gt 0 forall x isin ℝ then P(x) gt 0Q(x) rArr P(x) gt 0
(ii) If P( x )0
Q( x )gt and Q(x) lt 0 forall x isin ℝ then P(x)Q(x) lt 0 is multiplying by +ve real number does not
change the sign of inequality where as multiplying by ndashve real number reverses the sign of inequality
(iii) For all positive a b x
a x a if a bb x ba x a if a bb x b
+ gt lt + + lt gt +
122 Mathematics at a Glance
121 POLyNOMIALS
An algebraic expression involving one or more variable that contains two mathematical operations multiplication and raising to a natural exponent (power) with respect to the variablevariables involved is called lsquomono-nomialrsquo
1211 Leading TermsLeading Coefficient The term containing highest power of variable x is called lsquoleading termrsquo and its coefficient is called leading coefficient Because it governs the value of f(x) where x rarr infin)
∵ n n 1 n 2 nn 2 n
a a af(x) x a x x xminus minus = + + + +
1212 Degree of PolynomialsHighest power of x in the polynomial expression is called lsquodegree of polynomialrsquo (ie power of x in leading term) Based on degree polynomials can be classfied as 0 (Constant) ax0 1 (linear) ax + b 2 (quadratic) ax2 + bx + c 3 (cubic) ax3 + ax2 + cx + d 4 (bi-quadratic) ax4 + bx3 + cx2 + dx + e
12121 Rational function and rational equation
An equation of the form f(x)g(x) where f(x) and g(x) are polynomials in x is known as rational function of x and when equated to zero it generates a rational equation
solving rational inequality While solving rational inequality the following facts must be always bear in mind
gt rArr gt lt
f (x) and g(x) have f (x) and g(x) havesame sign opposite sign
f (x) f(x)0 f(x)g(x) 0 0g(x) g(x)
rArr
f (x) and g(x) ofsamesign or f (x) 0
f (x)f(x)g(x) 0 0g(x)
=
lt ge
rArr
=
gt le = ne
f (x) g(x) ofandopposite sign or
f (x) 0
f (x)g(x) 0f(x)or 0g(x)
f(x) 0 andg(x) 0 rArr
lt = ne
f(x)g(x) 0orf(x) 0 and g(x) 0
1213 Wavy-curve Method
To find the set of solution for inequality f(x) gt 0 (f(x) is polynomial)Factorize the polynomial and find all the roots eg f(x) = (x ndash a)3 (x ndash b)2 (x ndash d) (x ndash g)5 say a gt b gt d gt gLocate the roots (with their multiplicity) on the real number line Keep the sign expression in the
right-most interval same as that of the leading coefficient
Foundation of Mathematics 123
Moving towards left change the sign of expression across the root with multiplicity odd and retain the same sign across the root with multiplicity even
there4 f(x) gt 0 rArr (a b) cup (b g) cup (d infin) Also f(x) ge 0 rArr (a b) cup (b g) cup (d infin) cup (a bgd)
rArr x isin [a g] cup [d infin) Similarly f(x) lt 0 rArr (ndashinfin a) cup (g d) and f(x) le 0 rArr (ndashinfin a) cup (g d) cup a b g d f(x) le 0 (ndashinfin a]
cup [g d] cup b
12131 Concept of continued sums and products
continued sum (sum) Sigma (Σ) stands for sum of indexed terms eg n
kk 1
a=sum = a1 + a2 + a3 ++ an
In the above symbol ak is called lsquogeneral termrsquo and k is known as index
Properties
1 n
k 1
a=sum = a + a + a ++ a (n terms) = na
2 Sigma distributes on addition and subtraction n
k kk 1
(a b )=
plusmnsum = (a1 plusmn b1) + (a2 plusmn b2) ++ (an plusmn bn)
3 Sigma does not distribute on product and ratio of terms ie n
k kk 1
(a b )=
timessum = (a1 times b1) + (a2 times b2)
++ (an times bn) ne n n
k kk 1 k 1
a b= =
sum sum and
n
k kk 1
(a b )=sum = (a1b1) + (a2b2) ++ (anbn) ne
n
kk 1n
kk 1
a
b
=
=
sum
sum
4 A constant factor can be taken out of sigma notation ien n
k kk 1 k 1
ma m a= =
=sum sum = m (a1 + a2 + a3
+ + an) cyclic and symmetric expressions An expression is called symmetric in variable x and y iff interchanging x and y does not changes the
expression x2 + y2 x2 + y2 ndash xy x3 + y3 + x2y + y2x x3 ndash y3 is not symmetric An expression is called cyclic in x y z iff cyclic replacement of variables does not change the
expression eg x + y + z xy + yz + zx etc Such expression can be abbreviated by cyclic sigma notation as follows Σx2 = x2 + y2 + z2 Σxy = xy + yz + zx
Σ(x ndash y) = 0 rArr x + y + z + x2 + y2 + z2 = Σx + Σx2
5 If sigma is defined for three variables say a b c occurring cyclically then it is evaluated as follows Σa = a + b + c = a + b + c Σ a b = ab + bc + ca Σa2 = a2 + b2 + c2
continued Products (π) Continued product of indexed termsn
kk 1
a=prod is defined as product
of n number of indexed terms as n
k 1 2 3 nk 1
a a a a a=
=prod
124 Mathematics at a Glance
Properties
1 =
=prodn
k 1
a aaaa (n times) = an
2 = =
λ = λ λ λ = λ = λprod prodn n
nn 1 2 n 1 2 n k
k 1 k 1
a ( a )( a )( a ) (a a a ) a
3 π distributes over product and ratio of indexed terms but not over sum and difference of terms
ie = = =
= =prod prod prodn n n
k k k k 1 2 n 1 2 nk 1 k 1 k 1
a b a b (a a a )(b b b )
=
=
=
= =
prodprod
prod
n
kn1 2 3 nk k 1
nk 1 k 1 2 3 n
kk 1
aa a a aa
b b b b bb
n n n
k k k kk 1 k 1 k 1
(a b ) a b= = =
plusmn ne plusmnprod prod prod
122 PARTIAL FRACTIONS
12231 Linear and non-repeating
Let D(x) = (x - a1) (x - a2) (x - a3) Then = + + + +minus minus minus minus
31 2 n
1 2 3 n
AA A AN(x) Q(x) D(x) x a x a x a x a
12232 Linear and repeated roots
Let D(x) = (x - a)K (x - a1) (x - a2)(x - an)
Then = + + + + + + + +minus minus minus minus minus minus
1 2 k 1 22 k
1 2 n
A A A B BN(x) BnQ(x) D(x) x a (x a) (x a) x a x a x a
12233 Quadratic and non-repeated roots
Let D(x) = (x2 + ax + b) (x ndash a1) (x ndash a2)(x ndash an) then+
= + + + + ++ + minus minus minus1 2 1 2 n
2 1 2 n
A x A B B BN(x) Q(x) D(x) (x ax b) x a x a x a
12234 Quadratic and repeated
Let D(x) = (x2 + a1 x + b1) (x2 + a2x + b2)(x2 + anx + bn)type V When both N(x) and D(x) contain only the even powers of x To solve these types of integrals follow the steps given belowstep 1 Put x2 = t in both N(x) and D(x) step 2 Make partial fractions of N(t)D(t)step 3 Put back t = x2 and solve the simplified integral now
123 THEOREMS RELATED TO TRIANGLES
Theorem 1 If two straight lines cut each other the vertically opposite angles are equalTheorem 2 If two triangles have two sides of the one equal to two sides of the other each to each and the angles included by those sides are equal then the triangles are equal in all respectsTheorem 3 If two angles of a triangle are equal to one another then the sides which are opposite to the equal angles are equal to one another
Foundation of Mathematics 125
Theorem 4 If two triangles have the three sides of which one side is equal to three sides of another then they are equal in all respectsTheorem 5 If one side of a triangle is greater than other then the angle opposite to the greater side is greater then the angle opposite to the smaller sideTheorem 6 If one angle of a triangle is a greater than another then the side opposite to greater angle is greater than the side opposite to lessTheorem 7 Any two sides of a triangle are together greater they third sideTheorem 8 If all straight lines drawn from a given point to a given point on a given straight line then the perpendicular is the leastTheorem 9 If a straight line cuts two straight lines to make (i) The alternate angles equal or (ii) Exterior angles equal to the interior opposite angles on the same side of the cutting line or (iii) The interior angles on the same is side equal to two right angles then in each case the two straight
lines are parallelTheorem 10 If a straight line cuts two parallel lines it makes (i) The alternate angles equal to one another (ii) The exterior angle equal to the interior opposite angle on the same side of the cutting line (iii) The two interior angles on the same side together equal to two right anglesTheorem 11 The three angles of a triangle are together equal to two right anglesTheorem 12 If two triangles have two angles of one equal to two angles of the other each to each and any side of the first equal to the corresponding side of the other the triangles are equal in all respects called lsquoconjugatersquoTheorem 13 Two right angled triangles which have their hypotenuses equal and one side of one equal to one side of the other are equal in all respectsTheorem 14 If two triangles have two sides of the one equal to two sides of the other each to each but the angles included by the two sides of one greater than the angle included by the corresponding sides of the other then the base of that which has the greater angle is greater than the base of the other
12331 Theorems related to parallelograms
Theorem 15 The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallelTheorem 16 The opposite sides and angles of a parallelogram are equal to one another and each diagonal bisects the parallelogramTheorem 17 If there are three or more parallel straight lines and the intersepts made by them on any transversal are equal then the corresponding intercept on any other transversal are also equalTheorem 18 Parallelograms on the same base and between the same parallels are equal in terms of area
12332 Theorems related to intersection of loci
The concurrence of straight lines in a triangle (i) The perpendiculars drawn to the sides of a triangle from their middle points are concurrent (ii) The bisectors of the angles of a triangles are concurrent (iii) The medians of a triangle are concurrentTheorem 19 Triangles on the same base and between the same parallel line are equal in area
126 Mathematics at a Glance
Theorem 20 If two triangles are equal in area and stand on the same base and on the same side of it they are between the same parallel lineTheorem 21 Pythagorasrsquos theorem In any right-angled triangle the area of the square on the hypotenuse equals to the sum of the area of the squares on the other two sides
1231 Theorems Related to the Circle Definitions and First Principles
12311 Chords
Theorem 22 If a straight line drawn from the centre of a circle bisects a chord which does not pass through the centre it cuts the chord at right angles Conversely if it cuts the chords at right angles the straight line bisects itTheorem 23 One circle and only one can pass through any three points not in the same straight lineTheorem 24 If from a point within a circle more than two equal straight lines can be drawn to the circumference that point is the centre of the circleTheorem 25 Equal chords of a circle are equidistant from the centre Conversely chords which are equidistant from the centre than the equalTheorem 26 Of any two chords of a circle which is nearer to the centre is greater than one more remote Conversely the greater of two chords is nearer to the centre than the lessTheorem 27 If from any external point straight lines are drawn to the circumference of a circle the great-est is that which passes through the centre and the least is that which when produced passes through the centre And of any other two such lines the greater is that which subtends the greater angle at the centre
12312 Angles in a circleTheorem 28 The angle at the centre of a circle is double of an angle at the circumference standing on the same arcTheorem 29 Angles in the same segment of a circle are equal Coverse of this theorem states ldquoequal angles standing on the same base and on the same side of it have their vertices on an arc of a circle of which the given base is the chordrdquoTheorem 30 The opposite angles of quadrilateral inscribed in a circle are together equal to two right angles coverse of this theorem is also trueTheorem 31 The angle in a semi-circle is a right angleTheorem 32 In equal circles arcs which subtend equal angles either at the centres or at the circumferences are equalTheorem 33 In equal circles arcs which are cut-off by equal chords are equal the major arc equal to the major arc and the minor to the minorTheorem 34 In equal circles chords which cut-off equal arcs are equal
1232 TangencyTheorem 35 The tangent at any point of a circle is perpendicular to the radius drawn to the point of contactTheorem 36 Two tangent can be drawn to a circle from an external pointTheorem 37 If two circles touch one another the centres and the point of contact are in one straight line
Foundation of Mathematics 127
Theorem 38 The angles made by a tangent to a circle with a chord drawn from the point of contact are respectively equal to the angles in the alternate segments of the circle Theorem 39 If two of straight lines one is divided into any number of parts the rectangle contained between the two lines is equal to the sum of the rectangles contained by the undivided line and the several parts of the divided lineTheorem 40 If a straight line is divided internally at any point the square on the given line is equal to the sum of the squares on the squares on the two segments together with twice the rectangle contained by the segmentsTheorem 41 If a straight line is divided externally at any point the square on the given line is equal to the sum of the squares on the two segments diminished by twice the rectangle contained by the segmentsTheorem 42 The difference of the squares on the two straight lines is equal to the rectangle contained by their sum and differenceTheorem 43 In an obtuse-angled triangle the square on the side subtending the obtuse angle is equal to the sum of the squares on the sides containing the obtuse angle together with twice the rectangle contained by one of those sides and the projection of the other side upon itTheorem 44 In every triangle the square on the side subtending an acute angle is equal to the sum of the squares on the sides containing that angle diminished by twice the rectangle contained by one of these sides and the projection of the other side upon itTheorem 45 stewardrsquos theorem If D is any point on the side BC of a then AB2DC + AC2 BD = AC (AD2 + BD DC)Theorem 46 In any triangle the sum of the squares on two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side (Appolonius theorem which is a special case of Stewardrsquos theorem)
1233 Rectangles in Connection with CirclesTheorem 47 If two chords of a circle cut a point within it the rectangle contained by their segments are equalTheorem 48 If two chords of a circle when produced cut at a point outside it the rectangles contained by their segments are equal And each rectangle is equal to the square on the tangent from the point of intersectionTheorem 49 If from a point outside a circle two straight lines are drawn one of which cuts the circle and the other meets it and if the rectangle contained by the whole line which cuts the circle and the part of it outside the circle is equal to the square on the line which meets the circle then the line which meets the circle is a tangent to it
1234 Proportional Division of Straight LinesTheorem 50 A straight-line drawn parallel to one side of a triangle cuts the other two sides or those sides produced proportionallyTheorem 51 If the vertical angle of a triangle is bisected internally into segments which have the same ratio as the other sides of the triangle Conversely if the base is divided internally or externally into segments proportional to the other sides of the triangle the line joining the point of section to the vertex bisects the vertical angle internally or externally AD and ADrsquo are internal and external angle bisectors of the triangle
1235 Equiangular TrianglesTheorem 52 I f two triangles are equiangular to each other their corresponding sides are proportionalTheorem 53 If two triangles have their sides proportional when taken in order the triangles are equiangular to one another and those angles are equal which are opposite to the corresponding sides
128 Mathematics at a Glance
Theorem 54 If two triangles have one angle of which one is equal to one angle of the other and the sides about the equal angles are proportionals then the triangles are similarTheorem 55 If two triangles have one angles of which one is equal to one angle of the other and the sides about another angle in one proportional to the corresponding sides of the other then the third angles are either equal or supplementary and in the former case the triangles are similarTheorem 56 In a right-angled triangle if a perpendicular is drawn from the right angle to the hypotenuse the triangles on each side of it are similar to the whole triangles and to each other
12351 Similar Figures
Theorem 57 Similar polygons can be divided into the same number of similar triangles and the lines joining the corresponding vertices in each figure are proportionalTheorem 58 Any two similar rectilinear figures may be placed in a way that the lines joining corre-sponding the vertices are concurrentTheorem 59 In equal circles angles whether at the centres or circumferences have the same ratio as the arcs on which they stand
12352 Proportion applied to area
Theorem 60 The areas of similar triangles are proportional to the squares on there corresponding sidesTheorem 61 The area of similar polygons are proportional to the squares on there corresponding sides
1236 Some Important Formulae 1 (a + b)2 = z2 + 2ab + b2 = (a ndash b)2 + 4ab 2 (a + b)2 = a2 ndash 2ab + b2 = (a ndash b)2 + 4ab 3 a2 ndash b2 = (a + b) (a ndash b) 4 (a + b)3 = a3 + b3 + 3ab (a + b) 5 (a ndash b)3 = a3 + b3 ndash 3ab(a ndash b) 6 a3 + b3 = (a + b)3 ndash 3ab(a + b) = (a + b) (a2 + b2 ndash ab) 7 a3 ndash b3 = (a ndash b)3 + 3ab (a ndash b) = (a ndash b) (a2 + b2 + ab)
8 2 2 2 2 2 2 2 1 1 1(a b c) a b c 1ab 2bc 2ca a b c 2abc
a b c + + = + + + + + = + + + + +
9 3 3 3 2 2 21a b c ab bc ca (a b) (b c) (c a)2 + + minus minus minus = minus + minus + minus
10 ( )( )3 3 2 2 2 2a b c 3abc a b c a b c ab bc ca+ + minus = + + + + minus minus minus = ( ) ( )2 2 21 a b c (a b) (b c) (c a)2
+ + minus + minus + minus 11 a4 ndash b4 = (a + b) (a ndash b) (a2 ndash b2)
12 a4 + a2 + 1 = (a2 + 1)2 ndash a2 = (1 + a + a2) (1 ndash a + a2)
13 2 2a b a b
ab2 2+ minus
= minus
14 a b (a b)(a b)minus = minus +
15 a2 + b2 + c2 ndash ab ndash bc ndash ca = (a ndash b2) + (b ndash c)2 + (c ndash a)2 16 (x + a) (x + b) = x2 + (a + b)x + ab 17 (a + b + c)3 = a3 +| b3 |+c3 + 3 (a + b) (b + c) (c + a) 18 a3 + b3 + c3 ndash3abc = (a + b + c) (a2 + b2 + c2 ndash ab ndash bc ndash ca) 19 (a + b)4 = (a +| b)2 |(a +| b)2 = a4 + b4 + 4a3b + 6a2b2 + 4ab3
20 (a ndash b)4 = (a ndash| b)2 |(a ndash| b)2 = a4 + b4 ndash 4a3b + 6a2b2 ndash 4ab3
21 (a + b)5 = (a +| b)3 |(a +| b)2 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
Chapter 2eXpONeNtIaL LOGarIthM
21 ExponEntial Function
If a is a positive real number then ax (a ne 1) is always positive and it is called lsquoexponential function of xrsquo Here a is called lsquobasersquo and x is called index
211 Properties of Exponential Functions
(i) As we know that = times times times forall isin
n
n times
a a a a n where a is called lsquobasersquo and n is index or exponent
Exponential function f(x) = ax is generalisation of this law to facilitate some useful applications with some imposed functional restrictions ie a gt 0 and a ne 1
(ii) Domain of f(x) is set of real number and range of f(x) is (0 infin) ie forall x isinℝ f(x) = ax associates x to some positive real number uniquely ie exponential function f(x) is defined such that it is invertible
(iii) For a lt 0 and a = 0 the function f(x) = ax loses its meaning for some values of x isin ℝ For instance for a = ndash1 ndash12 ndash3 etc
f(x) = ax becomes non-real forall =pxq
where p and q are co-prime and q is an even integer
eg (ndash3)32 (ndash1)14 etcSimilarly when base a = 0 then f(x) = 0x does not remain an one-to-one function which is required for invariability same restriction also holds for a = 1 Since then f(x) = 1x again becomes many one function as all inputs x get associated to single output 1Therefore we conclude that for f(x) = ax the base a gt 0 a ne 1 and x isin ℝ thus y isin (0 infin)
(iv) If the base a is Euler number lsquoersquo then the exponential function ex is known as natural exponential function
212 Laws of Indices
(i) ax is defined and ax gt 0 forall x isin ℝ (ii) a0 = 1 We can observe that rarrn a 1 as n assumes very large value (n rarr infin) and it is true for both
cases ie a gt 1 or a isin (0 1) therefore when n rarr infin = = =1n 0n a a a 1
230 Mathematics at a Glance
(iii) axtimesay = ax+y
(iv) minus=x
x yy
a aa
(v) (ax)y = axy = (ay)x
(vi) = qpq pa a where isinq and q ne 1 (vii) ax = ay rArr x = y or a = 1 (viii) ax = bx rArr either x = 0 or a = b
(ix) axbx = (ab)x and =
xx
x
a ab b
(x) ax ge ay rArr ge gt
le isin
x y if a 1x y if a (01)
213 Graphical Representation of an Exponential Function
1 ax where a gt 1 behaves as an increasing nature function For example when a = 2 the value of function 2x increases as the
input x increases It can be understood from the table given below
x ndash3 ndash2 ndash1 0 1 2 3 4 2ax 18 14 12 1 2 4 8 16 32
2 If 0 lt a lt 1 behaves like a decreasing nature function For example when a = 12 the value of function 2ndashx decreases as the
input x increases which can be observed in the following table
x ndash5 ndash4 ndash3 ndash2 ndash1 0 1 2 3ax 32 16 8 4 2 1 12 14 18
3 If the base a gt 1 then ax ge 1 for all x ge 0 and ax lt 1 when x lt 0 if 0 lt a lt 1 then 0 lt ax lt 1 for x gt 0 and ax gt 1 for x lt 0 The above fact as well as the relative position of graphs of exponenital functions with different bases can be understood with the help of following figure
If the base a gt 1 then ax ge 1 for all x ge 0
Exponential Logarithm 231
214 Composite Exponential FunctionsA composite function is a function in which both the base and the exponent are the functions of x Generally any function of this form is a composite exponential function This function is also called an exponential power function or a power exponential function ie y = [u(x)]v(x) = uv In calculus the domain consists of such values of x for which u(x) and v(x) are defined and u(x) gt 0
215 Methods of Solving Exponential Equation
To solve an exponential equation we make use of the following facts
(i) If the equation is of the form ax = ay(a gt 0) rArr x = y or a = 1 (ii) If the equation is of the form ax = bx (a b gt 0) rArr either x = 0 or a = b (iii) If the equation is of the form ax = k (a gt 0) then
Case I If b ge 0 rArr x isin Case II If b gt 0 k ne 1 rArr x = logak Case III If a = 1 k ne 1 rArr x isin Case IV If a = 1 k = 1 rArr x isin ℝ(Since 1x = 1 rArr 1 = 1 x isin ℝ)
(iii) If the equation is of the form af(x) = ag(x) where a gt 0 and a ne 1 then the equation will be equivalent to the equation f(x) = g(x)
Remarksax = 1 rArr x = 0 is an incomplete conclusion it is only true if the base a ne 0 plusmn 1if a = 0 so equality does not holds as 00 is meaningless
Where as when a = 1 then an = 1 rArr 1x = 1 Thus x isin ℝ
In case a = ndash1 then (ndash1)x = 1 is true for x = pq when p is even and GCD of p and q = 1
22 Solving ExponEntial inEquality
(i) The value of ax increases as the value of x increases when base a isin (1 infin) but the value of ax de-
creases as the value of x increases when base a isin (0 1) ge gtge rArr le isin
x y x y if a 1a a
x y if a (01)
(ii) The elementary exponential inequalities are inequalities of form ax gt k ax lt k where a and k are certain numbers (a gt 0 a ne 1) Depending on the values of the parameters a and k the set of solutions of the inequality ax gt k can be in the following forms1 x isin (logak infin) for a gt 1 k gt 02 x isin (ndashinfin logak) for 0 lt a lt 1 k gt 03 x isin ℝ for a gt 0 k lt 0Depending on the values of a and k the set of solutions of the inequality ax lt k can be in the fol-lowing forms1 x isin (ndashinfin logak) for a gt 1 k gt 02 x isin (logak infin ) for 0 lt a lt 1 k gt 03 x isin for a gt 0 k lt 0 (ie the inequality has no solutions)
(iii) + = forall isinminus =
f(x y) f(x) f(y)x y
f(x y) f(x) f(y)
232 Mathematics at a Glance
23 logaRitHMic Function
The logarithm of any number N to the given base a is the exponent or index or the power to which the base must be raised to obtain the number N Thus if ax = N x is called the logarithm of N to the base a It is denoted as logaN
loga N = x hArr ax = N a gt 0 a ne 1 and N gt 0
Notes
(a) The logarithm of a number is unique ie no number can have two different logarithms to a given base
(b) The base lsquoarsquo is a positive real number but excluding 1 ie a gt 0 a ne 1 As a consequence of the definition of exponential function we exclude a = 1
Since for a = 1 logax = y rArr x = ay = 1y which has no relevance to the cases of logax when x ne 1 ie for all values of exponent the value of x remains 1
(c) The number lsquoxrsquo represents result of exponentiation ie ay therefore it is also a positive real number ie x = ay gt 0
(d) The exponent lsquoyrsquo ie logarithm of lsquoxrsquo is a real number and neither a nor x equals to zero
(d) Domain of function y = logax is (0 infin) and the range (-infin infin)
when x rarr 0 then logax rarr-infin (for a gt 1) and logax rarr infin (for 0 lt a lt 1) because y = logax rArr x = ay which approaches to zero iff y rarr-infin as a-infin = 0 forall a gt 1 and when a isin (0 1) x = ay approaches to zero iff y rarr infin ∵ ainfin = 0 if 0 lt a lt 1
(e) Common Logarithms and Natural Logarithms The base of logarithm can be any positive number other than 1 but basically two bases are mostly used They are 10 and e (=2718 approximately) Logarithm of numbers to the base 10 are named as Common Logarithms whereas the logarithms of the numbers to the base e are called as Natural or Napierian logarithms
If a = 10 then we write log b instead of log10b
If a = e then we write ℓnb instead of logeb
We find logea = log10a loge10 or e10
e
log alog a 0434
log 10= = logea (this transformation is used to convert
natural logarithm to common logarithm)
231 Properties of Logarithm
P 1 loga 1 = 0 because 0 is the power to which a must be raised to obtain 1 P 2 logaa = 1 since 1 is the power to which a must be raised to obtain a P 3 alogaN = N and logaa
N = N as N is the power to which a must be raised to obtain aN P 4 logm(ab) = logm a + logmb (a gt 0 b gt 0) Logarithm of the product of two numbers to a certain base
is equal to the sum of the logarithms of the numbers to the same base
Exponential Logarithm 233
P 5 logm (ab) = logma-logmb logarithm of the quotient of two numbers is equal to the difference of their logarithms base remaining the same throughout
P 6 loga Nk = k logaN (k is any real number) Logarithm of the power of a number is equal to the product
of the power and logarithm of the number (base remaining the same) P 7 logak N = (1k)loga N
Note
(1) The property 4 5 6 7 are not applicable conditionally because logaM + logaN is defined only when M and N are both positive whereas logaMN is defined even if M and N are both negative Therefore logaMN cannot be always replaced by logaM + logaN Therefore such replacement can lead to loss of root while solving an equation
(2) Thus to avoid the loss of root we consider the following transformations
(a) logaMN = loga |M| + loga |N| (when MN gt 0)
(b) a a a
Mlog log |M | log |N |
N= minus (when MN gt 0 N ne 0)
(c) logaN2k = 2k loga |N| (when N ne 0 k an integer)
(d) k |a|a
1log N log N
2k2 = (when N gt 0 k is an integer ne 0 a ne 0 |a| ne 1)
(3) The transformation 2(a) 2(b) or also valid conditionally as LHS is defined when M and N have same sign whereas the RHS is defined for any arbitrary values of M and N other than zero So such replacement while solving an equation can generate extraneous roots but since extraneous roots can be counter checked (and those not satisfying the parent equation can always be discarded) on the other hand the loss of root is difficult be traced therefore it is suggested to use the results of 2(a) 2(b) 2(c) 2(d) in place of property number 4 5 6 7 While simplifying and solving equationinequations
P 8 logba = logca logbc
P 9 = =cb
c
log a log alog alog b log b
(base remaining the same in numerator and denominator)
P 10 logbalogab = 1
P 11 alogmb = blogma
24 logaRitHMic EquationS
If we have an equation of the form loga f (x) = b (a gt 0) a ne 1 is equivalent to the equation f(x) = ab (f (x) gt 0)
241 Some Standard Forms to Solve Logarithmic Equations
Type 1 An equation of the form logxa = b a gt 0 has (a) Only root x = a1b if a ne 1 and b ne 0 (b) Any positive root different from unity if a = 1 and b = 0 (c) No roots if a = 1 b ne 0 (d) No roots if a ne 1 b = 0
234 Mathematics at a Glance
Type 2 Equations of the form
(i) f (logax) = 0 a gt 0 a ne 1 (ii) g (logxa) = 0 a gt 0
Then equation (i) is equivalent to f(t) = 0 where t = loga xIf t1 t2 t3 tk are the roots of f(t) = 0 then logax = t1 logax = t2 logax = tk
rArr = 1 2 kt t tx a a a and equation (ii) is equivalent to g(y) = 0 where y = logxa If y1 y2 y3 yk are the roots of f(y) = 0 then logx a = y1 logxa = y2 logxa = yk = x
rArr = 1 2 k1y 1y 1yx a a a
Type 3 Equation of the form (i) loga f(x) = loga g(x) a gt 0 a ne 1 is equivalent to systems of equations and inequations as
given below
System 1 gt
=
g(x) 0f(x) g(x) System 2
gt =
f(x) 0f(x) g(x)
(Any one of the two systems can be used) (ii) logf(x)A = logg(x) A A gt 0 is equivalent to the systems of equations and inequations as given below
System 1 gt ne =
g(x) 0g(x) 1f(x) g(x)
System 2 gt ne =
f(x) 0f(x) 1f(x) g(x)
(Any one of the two systems can be used)
Type 4 Equation of the form
(i) logf(x) g(x) = logf(x) h(x) is equivalent to two systems of equations and inequations
System 1 gt gt =
g(x) 0f(x) 0g(x) h(x)
System 2 gt gt =
h(x) 0f(x) 0g(x) h(x)
(Any one of the two systems can be used) (ii) logg(x)f(x) = logh(x) f(x) is equivalent to two systems of equations and inequations
System 1
gt gt ne =
f(x) 0g(x) 0g(x) 1g(x) h(x)
System 2
gt gt ne =
f(x) 0h(x) 0h(x) 1g(x) h(x)
(Any one of the two systems can be used)
Type 5 An equation of the form logh(x) (logg(x) f(x)) = 0 is equivalent to the system
gt ne gt ne =
h(x) 0h(x) 1g(x) 0g(x) 1f(x) g(x)
Exponential Logarithm 235
Type 6 An equation of the form 2m loga f (x) = logag(x) a gt 0 a ne 1 m isin N is equivalent to the
system gt
=2m
f(x) 0[f(x)] g(x)
Type 7 An equation of the form (2m + 1) loga f (x) = logag (x) a gt 0 a ne 1 m isin N is equivalent to the
system +
gt =
2m 1
f(x) 0[f(x)] g(x)
Type 8 An equation of the form loga f(x) + logag(x) = logam(x) a gt 0 a ne 1 is equivalent to the
system gt gt =
f(x) 0g(x) 0f(x)g(x) m(x)
Type 9 An equation of the form loga f(x)ndashlogag(x) = logah(x)ndashlogat(x) a gt 0 a ne 1 is equivalent to the
equation loga f(x) + logat(x) = logag(x) + logah(x) which is equivalent to the system
gt gt gt gt =
f(x) 0t(x) 0g(x) 0h(x) 0f(x)t(x) g(x)h(x)
25 logaRitHMic inEqualitiES
When base a gt 1 then logax is an increasing function where as when 0 lt a lt 1 then logax is a decreasing function
We can observe this by simple taking log2x and log12x and evaluating their value for various positive inputs and thus plotting the approximate graph of both the functions
x 116 18 14 12 1 2 4 8log2x ndash4 ndash3 ndash2 ndash1 0 1 2 3
log12x 4 3 2 1 0 ndash1 ndash2 ndash3
To solve a logarithmic inequality following facts must be kept in mindGive any positive real number then
(a) For a gt 1 the inequality logax gt logay rArr x gt y (Since for a gt 1 logax is an increasing function)
236 Mathematics at a Glance
rArr If a gt 1 then logax lt a rArr 0 lt x lt aa
rArr If a gt 1 then logax gt a rArr x gt aa
(b) For 0 lt a lt 1 then inequality 0 lt x lt y rArr logax gt logay (∵ for 0 lt a lt 1 logax is a decreasing function) rArr If 0 lt a lt 1 then logax lt a rArr x gt aa
rArr If 0 lt a lt 1 then logax gt a rArr 0 lt x lt aa
251 Characteristic and Mantissa
Generally the logarithm of a number is partially integral and partially fractional The integral part to the logarithm of a number is called lsquocharacteristicrsquo and the decimal part is known as mantissa
252 Characteristic and Mantissa
(a) Given a number N Logarithms can be expressed as log10 N = Integer + fraction (+ ve) darr darr Characteristic Mantissa (i) The mantissa part of the log of a number is always kept positive the characteristic may be
positive or negative eg if loge x = ndash14 = ndash2 + 06 written as 26 (ii) If the characteristics of log10 N be n then the number of digits in N is (n + 1) (iii) If the characteristics of log10 N be (-n) then there exists (n ndash 1) number of zeros after the
decimal point of N (b) The number of positive integer having base lsquoarsquo and characteristic n = an+1-an (c) If the number and base are on the same side of the unity then the logarithm is positive If the number
and the base are on the opposite side of the unity then the logarithm is negative (d) Characteristic of the common logarithm of (i) any number greater than 1 is positive (ii) any positive number less than 1 is negative
Chapter 3SequenCe and progreSSion
31 Definition
ldquoSequence is a definite pattern of the numbers (defined by a function Tn ℕ rarr ℂ where ℕ is natural numbers ℂ is complex numbers) each of which is derived according to a definite law and whose general term (Tn) is expressible in terms of nrdquo It denoted by lt Tn gt where Tn is the general term eg ltngt 1 2 3 4ltn2gt 12 22 32
311 Types of Sequence rArr Increasing Sequence lttngt is called increasing sequence iff tn + 1 gt tn forall n isinℕ ie t1 lt t2 lt t3 lt t4hellip
eg lt 2n ndash 1gt 1 3 7 9 11 rArr Decreasing Sequence lt tngt is called decreasing iff tn+1 lt tn ie t1 gt t2 gt t3 gt t4hellip
eg n
1 1 1 1 1 1 2 2 4 8 16 32
lt gt
rArr Converging Sequence A sequence is called converging sequence iff its term at infin is a finite real number ie Tn = finite when nrarrinfin
eg 1 1 1 1 2 4 8 16
is converging as 1T = 02infin infin = Also 1 1 1 1
2 3 4 5 is converging as 1T 0infin = =
infin rArr Diverging Sequence A sequence is called diverging sequence iff Tn rarrinfin when n rarrinfin
eg lt 2n gt 2 4 8 16 32 Tinfin = 2infin = infin rArr Oscillating Sequence A sequence is called oscillating sequence iff its value oscillates between two
numbers eg lt (ndash1)n gt ndash1 1 ndash1 1 ndash1 1 rArr Periodic Sequence If the term of sequence repeats after a fixed interval then the sequence is
called a periodic sequence
eg Periodic repeating
terms
nsin 10 10 1 0 1 02π minus minus
312 Progression and SeriesProgression is a sequence in which each succeeding term bears a fixed relation with its proceeding one (ie a sequence following a certaindefinite pattern)
338 Mathematics at a Glance
Types of Progression Progressions are generally of the following types
(i) Arithmetic Progression (AP) (ii) Geometric Progression (GP) (iii) Harmonic Progression (HP) (iv) Arithmetico Geometric Progression (AGP) (v) Arithmetico Arithmetic Progression (AAP) (vi) Arithmetico Arithmetic Geometric Progression(AAAP) (vi) Arithmetico Arithmetico Geometric Progression (AAGP) etc
32 SerieS
The term of a sequence are separated by positive sign is called series Hence a series is the summation of
terms of sequence denoted as Sn n
n kk 1
S t=
=sum ie the sum of the first lsquonrsquo terms of a sequence
(i) Arithmetic Progression It is the progression in which the difference of successive terms remain constant and this constant is known as common difference (eg sequence of odd natural numbers 1 3 5 7 )
321 Properties of Arithmetic Progression P1 If a is the first term and d is the common difference of the AP then AP can be written as a a + d
a + 2d a + ( n -1)d P2 General Terms nth term from beginning Tn = a + (n - 1)d = l (last term) where d = Tn - Tnndash1 n
th term from last Tn = l + (n -1) (-d)
bull If d gt 0 rArrincreasing Arithmetic Progression (AP) bull If d lt 0 rArrdecreasing Arithmetic Progression (AP) bull If d = 0 rArrall the terms remain same P3 Hence the nth term can also be written as Tn = Sn - Sn-1 P4 Sum of first n terms Sn = n2 [2a + (n -1)d] = n2 [a + l )
bull Sum of the first n natural number is n(n 1)2+ bull Sum of the first n odd natural number is n2
bull Sum of the first n even natural number is n(n + 1) P5 Summation of equidistant terms from beginning and end of an AP is always constant and is equal
to sum of the first and last term rArr n 1 n 2 n 1n nS (T T ) (T T ) 2 2 minus= + = + +
P6 If the nth term tn = an + b then the series so formed is an AP P7 If the sum of first n terms of a series is Sn = an2 + bn + c then the series so formed is an AP (provided
c = 0) If c ne 0 then series formed will be AP from the 2nd term onward P8 If every term of an AP is increased or decreased by the same quantity the resulting terms will also
be in AP with no change in common difference P9 If every term of an AP (CD = d) is multiplied or divided by the same non-zero quantity K then the
resulting terms will be in AP with new common difference equal to dK or dK P10 If the corresponding terms of two APrsquos are added or subtracted the resulting is also an AP
lt tn gt is AP with CD = d1 lt an gt in AP with CD = d2 lt tn + an gt is AP with CD = d1 + d2
Caution lttn angt n
n
ta
n
1t
is not necessarily an AP
Sequence and Progression 1339
P11 If equal number of terms (say k terms of an AP) are dropped alternately the resulting terms lie in AP with CD = (k + 1)d
P12 If equal number of terms say lsquokrsquo terms of an AP are grouped together and sum of terms in each group is obtained then the sum is in AP with common difference k2d
P13 If terms a1 a2an an+1a2n+1 are in AP then the sum of these terms will be equal to (2n + 1)an+1 Here total number of terms in the series is (2n + 1) and the middle term is an+1
P14 If terms a1 a2hellip a2nndash1 a2n are in AP The sum of these terms will be equal to n n 1a a(2n)2
++
where = n n 1a a2
++
AM of the middle terms
P15 If the ratio of the sum of Ist n term of 2 different APrsquos is given as a f(n) n
n
Sie f (n)S
= prime
then the ratio
of their kth terms is given by ( )kk
t f 2k 1t
= minus
P16 If the ratio of nth terms of two APrsquos is given as f(n) n
n
Tie f (n)T
= prime
then the ratio of their sum
of k terms is given by k
k
S k 1fS 2
+ =
Points to Remember
bull Selection of terms in AP When sum of given number of terms in AP is known then terms must be selected as bellow
bull Odd Number of Terms in AP Let the middle term be lsquoarsquo and CD = d 3 terms in AP a ndash d a a + d 5 terms in AP a ndash 2d a ndash d a a + d a + 2d
bull Even number of terms in AP The two middle terms (a ndash d) and (a + d) Consider the cd as 2d 4 terms in AP a ndash 3d a ndash d a + d a + 3d 6 terms in AP a ndash 5d a ndash 3d a ndash d a + d a + 3d a + 5d
33 Arithmetic meAn
331 Arithmetic Means of Numbers
Arithmetic mean for any n positive numbers a1 a2 a3 an is + + + +
= 1 2 3 na a a aAM
n
332 Insertion of n AMrsquos between Two Numbers
Arithmetic Mean between Two Numbers n arithmetic means between x and y is defined as a set of n numbers A1 A2 A3 An such that x A1 A2 A3 An y in AP
rArr kb aA a kn 1minus = + +
340 Mathematics at a Glance
Remarks
1 Sum of n AMrsquos between a and b is equal to n times single AM between a and b =
= +sumn
ii 1
nA ( a b)
2
2 If Tk and Tp of any AP are given then formula for Tn is minusminus
= =minus minus
p Kn Kn
T TT TT
n k p k
3 If p Tp = q Tq of an AP then Tp + q = 0
4 If pth term of an AP is q and qth terms is p then Tp + q = 0 and Tn = p + q ndash n
5 If pth terms of an AP is 1q and qth term is 1p then its pqth term is 1
6 If number of terms in any series is odd then only one middle term exists which is +
thn 1
2term
34 Geometric ProGreSSion
Geometric progression is a progression in which the ratio of the successive term remains the constant Such ratio is known as common ratio eg 3 6 12 24 48hellip
Therefore a1 a2 a3 an is in GP iff 32 4 n
1 2 3 n 1
aa a aa a a a minus
= = = = = constant (r) is called as common ratio
341 Properties of Geometric Progression
P1 If a is the first term and r is the common ratio then GP can be written as a ar ar2 ar3 ar4 arn-1
P2 nth term from the beginning Tn = arnndash1 = l (last term) where n
n 1
TrT minus
=
P3 nth term from the last with last term n n 1T r minus=
P4 The product of equidistant term from both ends of GP is constant T1Tn = T2Tnndash1= T3Tnndash2 = = a2rnndash1
P5 Sum of first n term n n
na(r 1) a(1 r )S(r 1) (1 r)
minus minus= =
minus minus If arnndash1 = l then n
a rS1 rminus
=minus where l is the last term in
the series
P6 Sum of infinite GP n
n nn
a(1 r )S (S )1 rinfin rarrinfin
rarrinfin
minus= = minus
=
plusmn infinplusmn infin minus
minus ltminus
if |r| gt 1 if r = 1
a if r = 1 and n odd0 if r = 1 and n evena if |r | 1
1 r
Remark
Sum of infinite GP infin =minusa
S1 r
when | r | lt 1 ie ndash1 lt r lt 1 not finite when | r | gt 1
ie r gt 1 or r lt ndash1
Sequence and Progression 1341
P7 If every term of a GP is increased or decreased by the same non-zero quantity the resulting series may not be in GP
P8 If every term of a GP is multiplied or divided by the same non-zero quantity the resulting series is in GP with the same common ratio
P9 If a1 a2 a3 and b1 b2 b3 two GPrsquos of common ratio r1 r2 respectively then a1b1
a2 b2 31 2
1 2 3
aa a b b b
and will also form GP and the common ratio will be r1r2 and r1r2 respectively
P10 If each term of a GP be raised to the same power then the resulting series is also a GPie lt tn
kgt is also a the GP with CR = rk
P11 If a b c are in GP then logk a logk b logk c are in AP ie in general if a1 a2 a3 be a GP of positive terms then log a1 log a2 log a3 will be in AP and conversely
P12 If F a 1 a 2 a3 are in GP then 1 2 3
1 1 1 a a a
are in GP
P13 Value of a recurring decimal Let R denote the decimal representation of a number as given
x numbers y numbers
R 0XXXX YYYY=
where X0 denotes the figure consist of non-recurring digit appearing
x times whereas Y0 denote the recurring period consisting of y digits x0 010 R X Y=
where 0x times
X XXXX=
and 0y times
Y YYYY=
and x y0 0 010 R X Y Y+ times = Therefore by subtraction
0 0x y x
X YR
(10 10 )+
minus=
minus
P14 Selection of Terms in GP When product of given number of terms in GP is known then the terms must be selected as followsrArr Odd Number of Terms in GP Let the middle term be lsquoarsquo and CR = r 3 terms in AP a r a ar 5 terms in AP a r2 a r a ar ar2rArr Even number of terms in GP The two middle terms are ar ar and CR = r2
4 terms in GP 33
a a ararr r
6 terms in GP 3 5a a a ar ar arr r r
35 Geometric meAn
351 Geometric Means of Numbers
Geometric Mean If three or more than three terms are in GP then all the numbers lying between first and last term are called geometrical means between them Geometric mean (G) of lsquonrsquo numbers x1 x2
x3xn is defined as nth root of their product rArr 1n
1 2 3 nG (x x x x )=
352 Geometric Mean between Two Numbers
If a b c are three positive numbers in a GP then b is called the geometrical mean between a c and b2 = ac
If a and b are two positive real and G is the single GM between them then G2 = ab
342 Mathematics at a Glance
To insert lsquonrsquo GMrsquos between a and b Let a and b are two positive numbers and G1 G2hellip Gn are lsquonrsquo GMrsquos between them then a G1 G2 Gn b is a GP with lsquobrsquo as its (n + 2)th term
rArr b = arn+1
1n 1br
a+ =
rArr G1 = ar G2 = ar2 hellip Gn = arn
Notes 1 Product of n GMrsquos inserted between a and b is equal to the nth power of a single GM between them
2 If a is positive and r gt 1 then GP is increasing but if 0lt r lt 1 then it is a decreasing GP
3 If a is negative and r is positive (r gt 1) then it is a decreasing GP but if 0 lt r lt 1 it is an increasing GP
36 hArmonic ProGreSSion
A sequence is said to be a harmonic progression if and only if the reciprocal of its terms form an arithmetic progression (eg 12 14 16 form a HP because 2 4 6 are in AP)
361 Properties of Harmonic Progression
P1 General form of a harmonic progression + + ++ +
1 1 1 a a d a 2d
P2 General Term Tn of HP = reciprocal of Tn of its corresponding AP (eg in the above series
n1t
a (n 1)d=
+ minus)
P3 If a b are the first two terms of an HP then n1t
1 1 1(n 1)a b a
= + minus minus
P4 If all the terms of an HP are multiplie or divided by a constant non-zero quantity the resulting series remains in HP
P5 If the term of an HP is infin this means that the corresponding term of the AP is zero P6 There is no general formula for finding the sum to n terms of HP
P7 If a b c are in HP 1 1 1 a b c
are in AP 2 1 1b a c= + rArr 2acb
a c=
+or a a b
c b cminus
=minus
P8 If a b c are in GP then logak logbk logck in HP
Note
If terms are given in HP then the terms could be picked up in the following ways
(i) For three terms minus +1 1 1
a d a a d
(ii) For four terms minus minus + +1 1 1 1
a 3d a d a d a 3d
37 hArmonic meAn
If three or more than three terms are in HP then all the numbers lying between the first and last term are called harmonic means between them
Sequence and Progression 1343
371 Harmonic Mean of Numbers
(a) H of any two numbers a and b is given by 2abHa b
=+
where a b are two non-zero numbers
(b) Also the HM of n non-zero numbers a1 a2 a3 an n
j 11 2 n j
n nH1 1 1 1a a a a=
= =+ + + sum
(c) Insertion of n harmonic mean between two numbers Let a and b be two given numbers and H1 H2
H3 Hn are HMrsquos between them Then a H1 H2 H3 Hn b ie r
1 1 rdH a
= + where (a b)d(n 1)ab
minus=
+
NoteSum of the reciprocals of all the n HMrsquos between a and b is equal to n times the reciprocal of single HM (H) between a and b
38 inequAlity of meAnS
rArr If A and B are positive numbers then A ge G ge H rArr If A G H are respectively AM GM HM between a and b both being unequal and positive then rArr G2 = AH ie A G H are in GP rArr For any set of positive real numbers x1 x2 xn
1 2 n 1n1 2 n
1 2 n
x x x n(x x x )1 1 1n x x x
+ + +ge ge
+ + + ie AM ge GM ge HM
Condition of Application rArr Equality holds (ie A = G = H) iff x1 = x2 = hellip = xn rArr If sum of the variable x1 x2 xn be x1 + x2 + hellip + xn = S then product x1x2hellipxn = P can be
maximized A ge G rArr 1nS (P)
nge rArr
nSPn
le
rArr n
maxSPn
=
and it is obtained when x1= x2=hellip= xn= Sn
rArr Similarly if x1x2x3hellipxn = P is constant then minimum value of sum lsquoSrsquo can be obtained as
rArr 1
1 2 n n1 2 n
x x x(x x x )
n+ + +
ge rArr 1nS (P)
nge rArr S ge n(P)1n
rArr Smin ge n(P)1n and it is obtained when x1 = x2 =hellip = xn = (P)1n
Remarks
1 If A and G are two AM and GM between two positive questions a and b then the quadratic equation having a b as its roots is x2 ndash 2Ax + G2 = 0
2 If AG H are AM GH and HM respectively then the equation having three roots is
minus + minus =3 2 33Gx 3Ax x G 0
H
344 Mathematics at a Glance
39 Arithmetic-Geometric ProGreSSion
A series formed by multiplying the corresponding terms of an AP and GP is called an Arithmetic Geometric progression eg 1 + 3 + 5 + 7 + is an AP 1 + x + x2 + x5 + is a GP Multiplying together the terms of these series we get 1+ 3x + 5x2 +7x3 + which is an Arithmetic Geometric Series
391 Standard Form
ab + (a + d)br + (a + 2d) br2 + + [a + (n + 1)d]brnndash1 n 1
nn 2
ab dbr(1 r ) [a (n 1)d]S br1 r (1 r) 1 r
minusminus + minus= + minus
minus minus minus
392 Sum to Infinity Terms
When | r | lt 1 2
ab dbrS1 r (1 r)infin = +minus minus
310 (S) SiGmA notAtion
3101 Concept of Continued Sum [Sigma (S) Notation]
Continued Sum Sigma (S) stands for continued sum of indexed terms It is denoted as
Index
n
kk 1
general term
a=sum
where k is called lsquoindex of termrsquo and it varies from 1 to n (where maximum value of k is n and minimum value of k is 1) thus indicating n number of terms in the series
rArr n
k 1 2 3 nk 1
a a a a a=
= + + + +sum rArr n
k 1
a na=
=sum as the general term is independent of k
Q n
k 1
a a a a a n times na=
= + + + + =sum
rArr A constant factor from the general term can be factored out of sigma notation ie n n
k kk 1 k 1
a a= =
λ = λsum sum
LHS = la1 + la2 + la3 + hellip + lan = l(a1 + a2 + a3 + hellip + an) = n
kk 1
a=
λsum
rArr Sigma is distributive over addition and subtraction of terms ie n n n
k k k kk 1 k 1 k 1
(a b ) a b= = =
plusmn = plusmnsum sum sum
LHS = (a1 plusmn b1) + (a2 plusmn b2) + hellip + (an plusmn bn) = (a1 + a2 + hellip +an) plusmn (b1 + b2 +hellip +bn) = n n
k kk 1 k 1
a b s= =
plusmnsum sum rArr Sigma does not distributes on multiplication and division of terms
n n n
k k k kk 1 k 1 k 1
a b a b= = =
nesum sum sum Similarly
n
knk k 1
nk 1 k
kk 1
aab b
=
=
=
nesum
sumsum
Application of Sigma The concept of sigma is used to find sum of series whose general term is given or known For example let general term of a series be Tn = an2 + bn + c
Sequence and Progression 1345
Sn = T1 + T2 + T3 + +Tn = n
kk 1
T=sum = 2
1( )
n
kak bk c
=
+ +sum = n n n
2
k 1 k 1 k 1
a k b k c 1= = =
+ +sum sum sum
nn(n 1)(2n 1) n(n 1)S a b cn
6 2+ + + = + +
Usually sum of n terms of any series is represented by placing S before the nth term of the series But
if we have to find the sum of k terms of a series whose nth term is un then this will be represented by k
nn 1
u=sum
Note
Shortly S is written in place of Σn
1
311 ProPertieS
P1 n
r 1
r 1 2 3 n=
= + + + +sum = n(n 1)2+ P2
n2 2 2 2 2
r 1
r 1 2 3 n=
= + +sum = n(n 1)(2n 1)6
+ +
P3 2n
3
r 1
n(n 1)r2=
+ = sum P4
n4 2
r 1
nr (n 1)(2n 1)(3n 3n 1)30=
= + + + minussum
312 Double SiGmA notAtion
m n
iji 1 j 1
T= =sumsum stands for summation of elements of a two-dimensional array (arrangement) of terms
It can also be regarded as summation of summation of a series rArr General Element The general term is represented by Tij where i denotes the row index
(row position) and j denotes the column index (column position) of the term
row columnindex index
i jT is the element placed in the ith Row and jth column
3121 Representation
m n
iji 1 j 1
T= =
sumsum can be represented as a two dimensional array of
numbers on a rectangular matrix with m rows and n columns
For example T14 is element placed in the 1st row and the 4th column T41 is element placed in the 4th row and the 1st column rArr Now consider square matrix of size n times n Elements (terms)
on the principal diagonal are addressed with i = j bull Tij
i lt j is the term that lies above the principal diagonal
Tij i gt j is the term that lies below the principal diagonalTij i = j is the term that lies on the principal diagonal
346 Mathematics at a Glance
31211 Conclusion
rArr Total number of squares abovebelow the principal diagonal line
rArr Total number of squares on or below the diagonal = 2 2n n n nn2 2minus +
+ = = n 12
n(n 1) C 2
++=
3122 Properties of Double Sigma
P1 n n n n
j 1 i 1 j 1 i 1
a a= = = =
=
sumsum sum sum =
n n
i 1 i 1
na na 1= =
=sum sum = nan = n2a ie summation of a a in n2 places on matrix
P2 n n n n n
1 i j n j 2 j 3 j ni 1 i 2 i n 1
a a a ale lt le = = =
= = = minus
= + + +sumsum sum sum sum = (n ndash 1)a + (n ndash 2)a ++ a = a(1 + 2 + 3++ (n ndash 1)) =an(n 1)
2minus
(ie Number of terms above the Principle Diagonal)
P3 2n n
n 12
1 i j n
(n n)aa C a2
+
le le le
+= =sumsum
2n n n(n 1)a na a2 2
minus ++ =
(ie Look at the sum of all the terms on or above the principle diagonal = (number of terms)a = n+1C2a)
P4 n n n n
i j i ji 1 j 1 i 1 j 1
a a a a= = = =
=
sumsum sum sum let
n
k 1 2 nk 1
a a a a S=
= + + + =sum = 2n n n
2i i k
i 1 i 1 k 1
Sa S a SS S a= = =
= = = =
sum sum sum
| P5 Sum of Product taken two at a time of any set given n numbers a1 a2 a3 an
= i j 01 i j n
a a S (say)le lt le
=sumsum rArr 2n n
20 i k
i 1 k 1
2S a a= =
+ =
sum sum
2n n2
k kk 1 k 1
0
a aS
2= =
minus
=sum sum
P6
2n n2
k kn nk 1 k 1
i j1 j j n
a aa a
2= =
le le le
+
=sum sum
sumsum = Sum of terms on or above the diagonal
P7 n n n
i j ki 1 j 1 k 1
(a a ) 2n a= = =
+ =
sumsum sum Q Each term is written 2n times in the matrix = nS + nS = 2nS
Sequence and Progression 1347
P8 A constant factor can always be factored out of double sigma n n n n
i j i ji i j n i i j n
(a a ) (a a )le lt le le lt le
+ λ = λ +sumsum sumsum P9 Double sigma distributes on sum and difference of element provided the elements are
symmetric in the variable i and j i j i j i j i j0 i j n 0 i j n 0 i j n
(a a a a ) (a a ) (a a )le lt le le lt le le lt le
+ + = + +sumsum sumsum sumsum P10 i j
0 i j n
(a a )le lt lesumsum = Sum of product of n numbers a1 a2an taking two at a time
=
2n n2
k kk 1 k 1
i j1 i j n
a aa a
2= =
le lt le
minus
=sum sum
sumsum
313 methoDS of Difference
Given a series with nth term unknown eg
1 2 3 4 5 n 1
1 2 3 4 5 6 n 1 n
d d d d d d
t t t t t t t t minus
minus
If the differences of the successive terms (dkrsquos) of a series are in AP or GP then we can find nth term of the series by the following procedureStep 1 Denote the nth term and the sum of the series upto n terms of the series by Tn and Sn respectivelyStep 2 Rewrite the given series with each term shifted by one place to the rightStep 3 Then substract the second expression of Sn from the first expression to obtain Tn
Remarks
(a) Difference of successive terms is constant then nth term is given by Tn = a + bn (where a and b is constant)Sn = S Tn
(b) If difference of difference is constant then Tn = an2 + bn + c (where a b c are constant)
(c) If difference of difference is constant then Tn=an3 + bn2 + cn + d (where a b c d are constant)
314 Vn methoD
A method to find sum of an unknown series whose general term tn is known
eg to compute n1 1 1 1S
1middot2 2middot3 3middot4 4middot5= + + + +
Step 1 Write the general term n1t
n(n 1)=
+
Step 2 Express tn as difference of two consecutive terms of another series lt vn gt
eg n1 (n 1) n 1 1t
n(n 1) n(n 1) n n 1+ minus
= = = minus+ + +
rArr n n n 11 1t V Vn n 1 += minus = minus
+
Step 3 rArr 1 1 21 1t V V1 2
= minus = minus rArr 2 2 31 1t V V2 3
= minus = minus rArr 3 3 41 1t V V3 4
= minus = minus
rArr n 1 n 1 n1 1t V V
n 1 nminus minus= minus = minusminus
rArr n n n 11 1t V Vn n 1 += minus = minus
+rArr n 1 n 1
1 nS V V 1n 1 n 1+= minus = minus =+ +
Chapter 4InequalIty
41 InequalIty contaInIng modulus functIon
Type 1 The inequality of the type f(|x|) lt g(x) is equivalent to the collection (union) of
system lt ge
minus lt lt
f(x) g(x) if x 0f( x) g(x) if x 0
Type 2 The inequality of the form |f(x)| lt g(x) is equivalent to collection (union) of the
systems lt ge
minus lt lt
f(x) g(x) if f(x) 0f(x) g(x) if f(x) 0
Aliter ndashg(x) lt f(x) lt g(x) for g(x) gt 0 and no solution for g(x) le 0
In particular |f(x)| lt a has no solution for a le 0 and for a gt 0 it is equivalent to the
system lt ge
minus lt lt
f(x) a for f(x) 0f(x) a for f(x) 0 or ndasha lt f(x) lt a for a gt 0 and no solution for a le 0
Type 3 The inequation of the form |f(x)| gt g (x) is equivalent to the systems
Aliter f(x) lt ndashg(x) or f(x) gt g(x) for g(x) ge 0 and solution will be the domain set Df of f(x) for g(x) lt 0
In particular |f(x)| gt a has solution x isin domain of f(x) if a lt 0 and for a ge 0 equation is equivalent to
collection (union) of the system gt ge
lt minus lt
f(x) a for f(x) 0f(x) a for f(x) 0
Type 4 The inequation of the form | f (| x |)| gt g (x) or | f (| x |)| lt g(x) is equivalent to the collection
(union) of systems gt ge
minus gt lt
| f (x)| g(x) if x 0| f( x)| g(x) if x 0 or
lt ge minus lt lt
| f (x)| g(x) if x 0| f( x)| g(x) if x 0 respectively
Aliter minus lt lt gt le
g(x) f(| x |) g(x) for g(x) 0Nosolution for g(x) 0 or
minus lt lt gt geminus lt minus lt gt lt le
g(x) f(x) g(x) for g(x) 0 x 0g(x) f( x) g(x) for g(x) 0 x 0
Nosolution for g(x) 0
Inequality 449
Type 5 The inequation of the form | f (x) | ge | g (x) | is equivalent to the collection of system f 2 (x) ge g 2 (x)
Aliter f(x) le ndash|g(x)| or f(x) ge |g(x)| or
lt minus ge lt lt lt lt gt ge gt gt minus lt gt
f(x) g(x) for g(x) 0 f(x) 0f(x) g(x) for g(x) 0 f(x) 0f(x) g(x) for g(x) 0 f(x) 0f(x) g(x) for g(x) 0 f(x) 0
Type 6 The inequation of the form h (x | f (x) |) lt g (x) or h (x | f (x) |) gt g(x) is equivalent to the
collection of systems
lt gt ge ge minus lt minus gt lt lt
h(x f(x)) g(x) h(x f(x)) g(x)if f (x) 0 if f(x) 0
orh(x f(x)) g(x) h(x f(x)) g(x)if f (x) 0 if f(x) 0
42 IrratIonal InequalItIes
The inequalities which contain the unknown under the radical sign There are some standard forms to solve these irrational inequalities
Type 1 The equation of the type lt isin2n 2nf(x) g(x) n is equivalent to the system ge
gt
f (x) 0g(x) f(x) and
inequation of the type + +lt isin2n 1 2n 1f (x) g(x) n is equivalent to the f (x) lt g (x)
Type 2 An inequation of the type lt2n f(x) g(x) n isin ℕ is equivalent to the system
ge
gt lt
2n
f(x) 0g(x) 0f(x) g (x)
and
inequation of the type + lt2n 1 f (x) g(x) n isin ℕ is equivalent to the equation f(x) lt g2n + 1(x)
Type 3 An inequation of the form gt isin2n f(x) g(x)n is equivalent to the collection of two systems
of inequations ie ge
gt2n
g(x) 0f(x) g (x)
or lt
ge
g(x) 0f(x) 0 and inequation of the form + gt isin2n 1 f (x) g(x)n is
equivalent to the inequation f (x) gt g 2n + 1 (x)
421 Exponential Inequalities
Type 1 To solve exponential inequation af(x) gt b (a gt 0) we have
(i) x isin Df if b le 0
(ii) if b gt 0 then we have
lt lt lt gt gt = ge isin = lt lt
a
a
f
f (x) log b if 0 a 1f(x) log b if a 1no solution if a 1and b 1x D if a 1and 0 b 1
450 Mathematics at a Glance
Type 2 af(x) lt b (a gt 0)
(i) No solution for b le 0 (ii) x isin Df for a = 1 b gt 1 (iii) No solution for a = 1 (iv) f(x) lt logab for b gt 0 a gt 1 (v) f(x) gt logab for b gt 0 0 lt a lt 1
Type 3 An equation of the form f(ax) ge 0 or f(ax) le 0 is equivalent to the system of
collection gt = ge le
xt 0 where t af(t) 0 or f(t) 0
Type 4 An inequation of the form aaf(x) + bbf(x) + gcf(x) ge 0 or aaf(x) + bbf(x) + gc f(x) le 0 when a b g isin R a b g ne 0 and the bases satisfy the condition b2 = ac is equivalent to the inequation at2 + bt + g ge 0 or at2 + bt + g le 0 when t = (ab)f(x)
Type 5 An equation of the form aaf(x) + bbf(x) + g ge 0 or aaf(x) + bbf(x) + g le 0 where a b g isin R and a b g ne 0 and ab = 1
(a b are inverse + ve numbers) is equivalent to the inequation at 2 + gt + b ge 0 or at 2 + gt + b le 0 where t = af(x)
Type 4 If an inequation of the exponential form reduces to the solution of homogeneous algebraic inequation ie aof
n(x) + a1fnndash1(x) g(x) + a2f
nndash2(x) g2(x) + + anndash1 f(x) gnndash1(x) + angn (x) ge 0 when a0 a1 a2an
are constants (a0 ne 0) and f (x) and g (x) are functions of x
rArr minus minus
minus minus+ + + + gen n 1 n 2
0 1 2 nn n 1 n 2
f (x) f (x) f (x)a a a a 0g (x) g (x) g (x)
rArr a0tn + a1t
nndash1 + a2tnndash2 + hellip + an ge 0 where =
f(x)tg(x)
and hence gives n values of t = t1 t2 t3helliptn (say)
rArr = if (x) tg(x)
i = 1 2 3 helliphelliphellip n
rArr f(x) ndash tig(x) = 0 solve for x corresponding to each i
Type (iii) Logh(x) f(x) gt logh(x)g(x)
rArr lt lt lt gt
gt gt gt
0 h(x) 1 f(x) g(x) f(x) 0h(x) 1 f(x) g(x)g(x) 0
Type (iv) Logh(x) f(x) lt a
rArr α
α
lt lt lt lt
gt gt
0 h(x) 10 f(x) (h(x))h(x) 1 f(x) (h(x))
422 Canonical Forms of Logarithmic Inequality
(a) gt gt rArr gt gt
alog x 0 x 1a 1 a 1 (b)
gt lt lt rArr lt lt lt lt
alog x 0 0 x 10 a 1 0 a 1
(c) lt lt lt rArr gt gt
alog x 0 0 x 1a 1 a 1 (d)
lt gt rArr lt lt lt lt
alog x 0 x 10 a 1 0 a 1
Inequality 451
423 Some Standard Forms to Solve Logarithmic Inequality
Type 1 Equation of the type
Type Collection of Systems
(a) logg(x) f(x) gt 0 hArr gt lt lt
gt lt lt
f(x) 1 0 f(x) 1
g(x) 1 0 g(x) 1
(b) logg(x) f(x) ge 0 hArr ge lt le
gt lt lt
f(x) 1 0 f(x) 1
g(x) 1 0 g(x) 1
(c) logg(x) f(x) lt 0 hArr gt lt lt
lt lt gt
f(x) 1 0 f(x) 1
0 g(x) 1 g(x) 1
(d) logg(x) f(x) le 0 hArr ge lt le
lt lt gt
f(x) 1 0 f(x) 1
0 g(x) 1 g(x) 1
Type 2 Equation of the type
Type Collection of systems
(a) logf(x) f(x) gt logf(x) g(x) hArr
gt lt gt gt φ gt lt φ lt
f(x) g(x) f(x) g(x)g(x) 0 f(x) 0
(x) 1 0 (x) 1
(b) logf(x) f(x) ge logf(x) g(x) hArr
ge le gt gt φ gt lt φ lt
f(x) g(x) f(x) g(x)g(x) 0 f(x) 0
(x) 1 0 (x) 1
(c) logf(x) f(x) lt logf(x) g(x) hArr
lt gt gt gt φ gt lt φ lt
f(x) g(x) f(x) g(x)f(x) 0 g(x) 0
(x) 1 0 (x) 1
(d) logf(x) f(x) le logf(x) g(x) hArr
le ge gt gt φ gt lt φ lt
f(x) g(x) f(x) g(x)f(x) 0 g(x) 0
(x) 1 0 (x) 1
424 Inequalities of Mean of Two Positive Real Numbers
If a and b are two positive real numbers then AM ge GM ge HM ie +
ge ge+
a b 2abab2 a b
452 Mathematics at a Glance
Remarks
(i) AM gt GM gt HM if a ne b (ii) AM = GM = HM if a = b
425 Inequality of Means of n Positive Real Number
If = sum ixA
n = AM of x1 x2 x3 hellipxn
=
= prod
1nn
ii 1
G x = GM of x1 x2 x3 hellipxn
=
= sum
n
i 1 i
nH1x
= HM of x1 x2 x3 hellipxn then A ge G ge H
Remark
(i) A gt G gt H iff x1 x2 x3 hellipxn are not all equal
(ii) A = G = H iff x1 = x2 = x3 =hellip= xn
43 theorem of weIghted mean
Theorem of weighted mean implies + ++ + +ge
+ + +31 2 n 1 2 n
1mm m m m m m1 1 2 2 n n
1 2 3 n1 2 n
m a m a m a (a a a a )m m m
forall ai gt 0
where i = 1 2 3n and mi be +ve real numbers ( )sumgesum prodsum1
i i mi mii
i
m aa
m ge equality holds where airsquos are
equal Here a1 a2 a3 an are positive real numbers and m1m2mn are positive real numbers
431 Theorem
(a) (Inequality of the mean of mth power and mth power of mean) If a and b are two positive real numbers Then
(i) + + gt
mm ma b a b2 2
if m lt 0 or m gt 1 (ii) + + lt
mm ma b a b2 2
if 0 lt m lt 1
(iii) + + =
mm ma b a b2 2
if m = 0 or 1 or a = b
(b) If a1 a2 a3 hellipan are n positive real numbers then
(i)
gt
sum summm
i ia an n
if m isin (ndashinfin 0) cup (1 infin) (ii)
lt
sum summm
i ia an n
if m isin (0 1)
(iii)
=
sum summm
i ia an n
if m = 0 or 1 or all airsquos are equal
Inequality 453
432 Weighted Power Mean Inequality
If a1 a2an b1 b2bn are two sets of n rationals airsquos are not all equal m isin Q (rational)
Then + + + + + +
gt + + + + + +
mm m m1 1 2 2 n n 1 1 2 2 n n
1 2 n 1 2 n
b a b a b a b a b a b ab b b b b b
when m notin (01) and
+ + + + +lt
+ + + + + +
mm m1 1 n n 1 1 2 2 n n
1 2 n 1 2 n
b a b a b a b a b ab b b b b b
when m isin (0 1) Equality occurs when either a1 = a2 = = an or m isin 0 1
433 Cauchy-Schwarz Inequality
If a1 a2an and b1b2bn are two sets of n real numbers then (a1 b1 + a2b2 + + an bn)2 le
(a12 + a2
2 + + an2) (b2
1 + b22 ++ b2
n) with the equality holding if and only if = = =1 2 n
1 2 n
a a ab b b
434 Tchebysheffrsquos Inequality
If x1 x2 xn and y1 y2 yn are real numbers such that x 1 le x2 lele xn and y1 le y2 le leyn then n(x1y1 + + xn yn) ge (x1 + + xn) (y1 ++ yn) For i ne j xi ndash xj and yi ndash yi are both non-positive or non-negative
For the equality to hold at least one in every pair of xi - xj and yi - yj must be zero This certainly hap-pens if x1 = x2 = = xn or if y1 = y2 = = yn and these are the only possibilities
Corollary If x1xn and y1yn are any real numbers such that x1 le x2 lele xn and y1 le y2 lele yn
then + + + + + ge
1 1 n n 1 n 1 nx y x y x x y y
n n n
44 weIerstrass InequalIty
For all ai isin IR + and n gt 1 and ai lt 1 If Sn = a1 + a2 + a3 +an then (1 + Sn) lt (1 + a1) (1 + a2) (1 + a3)
(1 + an) lt minus n
11 S
if Sn lt 1 otherwise (1 - Sn) lt (1 - a1) (1 - a2) (1 - a3)(1 - an) lt + n
11 S
441 Application to Problems of Maxima and Minima
Suppose that a1 a2 a3 an are n positive variables and k is a constant then
(a) If a1 + a2 + a3 + + an = k (constant) the value of a1 a2 a3an is greatest when a1 = a2 = a3 = = an so that the greatest value of a1 a2 a3 an is (kn)n
(b) If a1 a2 a3an = k (constant) the value of a1 + a2 + a3 + + an is least when a1 = a2 = a3 = = an So the least value of a1 + a2 + a3 + + an is n (k)1n
(c) If a1 + a2 + a3 + + an = k (constant) then as m does not or does lie between 0 and 1 the least or the greatest value of + + + +
1 2 3 n
m m m ma a a a occurs when a1 = a2 = a3 = = an the value in question being n1ndashmkm
454 Mathematics at a Glance
(d) If + + + +1 2 3 n
m m m ma a a a = k then according as m does not or does lie between 0 and 1 the greatest or the least value of a1 + a2 + a3 + + an occurs when a1 = a2 = a3 = = an the value in question being n1ndash1mk1m
Theorem 4 If a b are two angles in the 1st quadrant with a given constant sum f then maximum value
of minus φ
α β =1 cossin sin
2 ie
φ2sin2
and that of φ
α + β =sin sin 2sin2
and it occurs when φ
α =β =2
and
similar result also holds good for cosine
Theorem 5 If a1a2a3 are n angles each lying between (0p2) whose sum is constant A To find maxi-
mum value of ==
α αsumprodn n
k kk 1K 1
sin and sin Suppose that any two of the angles (say) α1 and α2 are unequal then
if we replace two unequal factors sin α1 and sin α2 in the given product by two equal factors α +α1 2sin
2
α +α1 2sin2
the value of product is increasing but the sum of angles remains unaltered as long as any
two of the angles are unequal the product is not maxm this indicalies that the product is maxn when all the
angles are equal so each angle is An Therefore=
α =
prod
nn
kk 1 max
Asin sinn
=
α =
sum
n
kk 1 max
Asin nsinn
45 use of calculus In ProVIng InequalItIes
451 Monotonicity
A function f is defined on an interval [a b] said to be (a) Monotonically increasing function If x2 ge x1 rArr f(x2) ge f(x1) for all x1 x2 isin [a b] (b) Strictly increasing function If x2 gt x1 rArr f(x2) gt f(x1) for all x1 x2 isin [a b] (c) Monotonically decreasing function If x2 ge x1 rArr f(x2) le f(x1) for all x1 x2 isin [a b] (d) Strictly decreasing function If x2 gt x1 rArr f(x2) lt f(x1) for all x1 x2 isin [a b]
452 Test of Monotonicity
(a) The function f (x) is monotonically increasing in the interval [a b] if fprime(x) ge 0 in[a b] (b) The function f (x) is strictly increasing in the interval [a b] if fprime(x) gt 0 in [a b] (c) The function f (x) is monotonically decreasing in the interval [a b] if fprime(x) le 0 in [a b] (d) The function f (x) is strictly decreasing in the interval [a b] if fprime(x) lt 0 in [a b]
Chapter 5theory of equation
51 Polynomial ExPrEssion
An algebraic expression involving one or more variable that contains two mathematical operations multiplication and raising to a natural exponent (power) with respect to the variablevariables involved
is called lsquomono-nomialrsquo For example 2 22ax bx 3xy x yz3
etc
An expression that involves many such mono-nomials separated by positive sign is known as multinomial
For example 3 2 2 3 3 3ax bx yz cxy z dy z+ + + + etc A multinomial having single unknown variable is called lsquopolynomialrsquo An algebraic expression of
type f(x) = a0 + a1x + a2x2 + a3x
3 +hellip+ anxn is called lsquopolynomialrsquo in variable x provided that the powers of x are whole numbers The numerical constants a0 a1 a2hellip an are known as coefficients
511 Leading TermsLeading Coefficient
The term containing highest power of variable x is called leading term and its coefficient is called lsquoleading coefficientrsquo Because it governs the value of f(x) where x rarr infin
Q
n n 1 n 2 nn 2 n
a a af(x) x a x x xminus minus = + + + +
5111 Degree of polynomials
Highest power of x in the polynomial expression is called degree of polynomial (ie power of x in leading term)
5112 Root of polynomial
Roots are the value of the variable x for which the polynomial expression vanishesGeometrically roots are the x-coordinate of the points where the graph of the polynomial
meets axis of x
556 Mathematics at a Glance
52 ClassifiCation of Polynomials
521 Polynomial Equation
When a polynomials expression is equated to zero then it generates corresponding equation Roots of polynomial expression are the solution of its corresponding equation A Polynomial equation of nth degree has exactly n roots not necessarily all real (Because it can be
factorized into exactly n linear factors) Two polynomials are equal if they have same degree and same coefficients corresponding to same
power of x If sum of coefficients of a polynomial equation vanishes then x = 1 is one of its roots If sum of coefficients of odd power term of x is equal to the sum of coefficients of even power term
of x then x = ndash1 is one of its roots
522 Polynomials Identity
If an equation is true for all values of variable for which it is defined then it is called identity 2
Rational identity
x 3x 2 (x 1)(x 2)x 1 x 1minus + minus minus
=+ +
ax3 + bx2 + cx + d = 0 is identity rArr a = b = c = d = 0 If has more number of roots than its degree
Theory of Equation 557
5221 Conclusion
Therefore to prove a nth degree polynomial equation to be an identity there are two ways Either show that number of roots ge n + 1 Show that all the coefficients are zero
NotesIn an identity in x coefficients of similar powers of x on the two sides are equal
Thus if ax3 + bx3 + cx + d = 7x3 ndash 5x2 + 8x ndash 6 be an identity in x then a =7 b = ndash5 c = 8 d = ndash 6
53 Equation stanDarD Equation anD quaDratiC
ax2 + bx + c = 0 is known as quadratic equation if a is non-zero a b c isin R The roots of this equation can be obtained by ax2 + bx + c = 0 (i)
rArr 2b b 4acx
2a 2aminus minus
= plusmn (b2 ndash 4ac = D is known as Discriminant of quadratic)
531 Quadratic Equation
Consider the quadratic expression y = ax2 + bx + c (a ne 0) and a b c are real numbers Thus y = ax2 + bx + c
= 2 b ca x 2 x2a a
+ + =
2 22
2 2
b b c ba x 2 x2a 4a a 4a
+ + + minus
=
2 2
2
b 4ac ba x2a 4a
minus + +
rArr 2D by a x
4a 2a + = +
Where D = b2 ndash 4ac is the discriminant of the quadratic equation shifting the origin at
(- b2a - D4a) ie substituting bX x2a
= +
and DY y4a
= +
The parabola opens upwards or downwards as a gt 0 or a lt 0
54 naturE of roots
1 If a b c isin R and a ne 0 then (a) If D lt 0 then roots of equation (i) will be non-real complex conjugate
558 Mathematics at a Glance
(b) If D gt 0 then the roots of equation (i) are real and distinct namely b D2a
minus +α = b D
2aminus minus
β =
and then ax2 + bx + c = a(x - a) (x - b) (ii) (c) If D1 and D2 are discriminants of equation a1x2 + b1x + c1 = 0 (i) a2x2 + b2x + c2 = 0 (ii) Case I D1 + D2 ge 0 then (i) At least one of D1 or D2 ge 0 (must be greater than zero) (ii) If D1 lt 0 then D2 gt 0 and if D1 gt 0 then D2 lt 0 ie at least one of equation has both
roots real and distinct (d) If D1 + D2 lt 0 then (i) at least one of D1 and D2 lt 0 (ii) If D1 lt 0 then D2 gt 0 and if D1 gt 0 then D2 lt 0 (must be less than zero) ie at least one of equations has both roots imaginary (ie complex conjugates) (e) If D = 0 then equation (i) has real and equal rootsa + b = -b2a and then
ax2 + bx + c = a (x - a)2 (iii) 2 If a b c isin Q and D is a perfect square of a rational number then the roots are rational and in case
it is not a perfect square then the roots are irrational Conjugate Roots 3 If a b c isin R and p + iq is one root of equation (i) (q ne 0) then the other must be the conjugate p ndash iq
and vice versa (p q isin R and i = radicndash1) Irrational Roots
4 If a b c isin Q and p + q is one root of equation (i) then the other must be the conjugate p qminus
and vice versa (where p is a rational and q is irrational) 5 ax2 + bx + c = 0 equiv a(x - a) (x - b) (if a and b are roots of the equation) Q a ne 0 dividing both sides of the equation by a and comparing the coefficient a + b = - ba and ab = ca
rArr a - b (difference of roots) = radicDa 6 If the equation ax2 + bx + c = 0 has more than two roots then its degree then it will becomes an
identity and this implies a = b = c = 0 7 If a = 1 and b c are integers and the root of equation (i) are rational numbers ie D gt 0 and perfect
square then these roots must be integers Q a + b = - ba isin I and ab = ca isin I a and b must be integers 8 If a + b + c = 0 and a b c are rational then 1 is a root of the equation (i) and roots of the
equation (i) are rational
541 Formation of Quadratic EquationA quadratic equation whose summation of roots is S and product of roots is P can be written as x2 - Sx + P = 0 Hence a b be the roots of equation ax2 + bx + c = 0 then to obtain the equation whose roots are (i) 1a 1b (ii) -a -b (iii) ka kb (iv) - 1a -1b (v) pa + q pb + q
We proceed as below
Since a + b = -ba and ab = ca and the equation whose summation of roots is S and product of roots is P can be written as x2 ndash Sx + P = 0 Therefore
Theory of Equation 559
(i) S β+α=
αβ P = 1ab rArr 2x β+α
minusαβ
x + 1ab = 0
rArr abx2 - (b + a) x + 1 = 0 rArr cx2 + bx + a = 0 (The reciprocal equation of ax2 + bx + c = 0 can be obtained by replacing x with 1x in the
later equation ie i interchanging the coefficients of equidistant terms from beginning and end)
(ii) S = -(a + b) P = ab rArr x2 + (a + b)x +ab = 0 rArr ax2 - bx + c = 0 (The equation whose roots are negative of the roots of equation ax2 + bx + c = 0 can be obtained by
replacing x with ndashx is the ax2 + bx + c = 0) (iii) If a b g are roots the roots of cubic equation then the equation is x3 ndash (a + b + g) x2 +
(ab + bg + ag) x ndash abg = 0
542 Sum and Product of the Roots
Since a + b = - ba and ab = ca are the sum and product of the equation x2 ndash Sx + P = 0 where a and b are the roots of this equation
55 ConDition for Common roots
(i) One roots to be common Consider two quadratic equations ax2 + bx + c = 0 and aprimex2 + bprimex + cprime = 0 (where aaprime ne 0 and abprime ndash aprimeb ne 0) Let a be a common root then aa2 + ba + c = 0 (i) and aprimea2 + bprimea + cprime = 0 (ii)
Solving the above equations we get 2 1
bc b c ca c a ab a bα α
= =prime prime prime prime prime primeminus minus minus
From first two relations we get bc b cca c aprime primeminus
α =prime primeminus
and from last two relations we get ca c aab a bprime primeminus
α =prime primeminus
eliminating a we get bc b cca c aprime primeminusprime primeminus
= ca c aab a bprime primeminusprime primeminus
rArr 2(bc b c)(ab a b) (ca c a)prime prime prime prime prime primeminus minus = minus or
rArr 2a b b c c a
a b b c c atimes =
prime prime prime prime prime prime (Remember) this is the required condition for one root of two
quadratic equation to be common (ii) Both roots to be common
If a + b = ndashba = ndashbprimeaprime and ab = ca = cprimeaprime ie a b ca b c= =prime prime prime
this is the required condition for both
roots of two quadratic equations to be identical
NoteTo find the common root between the two equations make the same coefficient of x2 in both equations and then subtract the two equations
Detail Analysis of Quadratic Equation If b2 ndash 4ac gt 0 then
Coefficients Graphs Analysis of Nature of Roots
560 Mathematics at a Glance
a gt 0 b gt 0 c gt 0a gt 0 b gt 0 c = 0a gt 0 b gt 0 c lt 0
a + b lt 0 ab gt 0a + b lt 0 ab = 0a + b lt 0 ab lt 0
Both roots are negativeOne root is ndashve and the other is zeroRoots are opposite in sign and magnitude of negative root is more than the magnitude of positive root
a gt 0 b lt 0 c gt 0a gt 0 b lt 0 c =0a gt 0 b lt 0 c lt 0
a + b gt 0 ab gt 0a + b gt 0 ab = 0a + b gt 0 ab lt 0
Both roots are positiveOne root is +ve and the other is zeroRoots are opposite in sign and magnitude of positive root is more than the magnitude of negative root
56 symmEtriC funCtion of thE roots
A function of a and b is said to be a symmetric function if it remains unchanged when a and b are interchanged
In order to find the value of a symmetric function of a and b express the given function in terms of a + b and ab The following results might be useful 1 a2 + b2 = (a + b)2 ndash 2ab 2 a3 + b3 = (a + b)3 ndash 3 ab (a + b)
3 a4 + b4 = (a3 + b3) ndash (a2 + b2) -2a2b2 (a2 +b2) 4 2( ) 4αminusβ = α+β minus αβ
5 (a3 - b3) = (a + b) [(a - b)2 - ab] 6 (a4 ndash b4) = (a + b) (a - b) (a2 + b2) 7 a5 + b5 = (a2 + b2) (a2 + b2) ndasha2b2(a + b)
561 MaximumMinimum Value and Sign of Quadratic Equation
Extreme value of any quadratic expression y = ax2 + bx + c is given by y-coordinate of vertex of corresponding parabola and it occurs at x-coordinate of vertex
(i) For a gt 0 The curve y = ax2 + bx + c is a parabola opening upwards
such that minD by at x
4a 2aminus minus
= = and ymax rarr infin
(ii) For a lt 0 The curve y = ax2 + bx + c is a parabola opening downward such that
maxD by at x
4a 2aminus minus
= = and ymin rarr ndash infin
57 loCation of roots
Let f(x) = ax2 + bx + c where a b c isin R be a quadratic expression and k k1 k2 be real numbers such that k1 lt k2 and if a b be the roots of equation
f(x) = 0 Then b D2a
minus minusα = and b D
2aminus +
β = where D is the discriminant
of the equation
Theory of Equation 561
(a) Conditions for a number k to lie between the roots of a quadratic equation OR under what condition do the roots of akx2 + bx + c = 0 lie on either side of number k
If a number k lies between the roots of a quadratic equation f(x) = ax2 + bx + c = 0 then the equation must have real roots and the sign of f(k) must be opposite to the sign of lsquoarsquo as is evident from the
(i) D gt 0 and (ii) a f(k) lt 0 (b) Condition for both the roots of a quadratic equation to lie
between numbers k1 and k2 or in the interval k1lt x lt k2
If both the roots aand b of a quadratic equation lie between number k1 and k2
(i) D gt 0 (ii) a f(k1) gt 0 a f(k2) gt 0 and (iii) k1 lt ndashb2a lt k2 (c) Conditions for a number k to be less than the roots of a
quadratic equation or under what condition will both roots of ax2 + bx + c = 0 be greater than a certain specified number k
Thus a number k is smaller than the roots of a quadratic equation ax2 + bx + c = 0 iff (i) D gt 0 (ii) a f(k) gt 0 (iii) k lt ndashb2a
(d) Condition for exactly one root of a quadratic equation to lie in the interval (k1 k2) where k1 lt k2
If exactly one root of the equation ax2 + bx + c = 0 lies in the interval (k1 k2) then the equation ax2 + bx + c = 0 must have real roots and f(k1) and f(k2) must be of opposite signs Thus exactly one root of the equation ax2 + bx + c = 0 lies in the interval (k1 k2) if
(i) D gt 0 (ii) f(k1) f(k2) lt 0
562 Mathematics at a Glance
(e) Condition for a number lsquokrsquo to be more than the roots of a quadratic equation
If a number k is more than the roots of a quadratic equation ax2 + bx + c then (i) D gt 0 (ii) a f(k) gt 0 (iii) k gt ndashb2a
58 DEsCartEs rulE
Step 1 To check at most positive roots in f(x) = 0 Check change in sign = most positive roots eg f(x) = x9 + 5x8 ndash x3 + 7x + 2 = 0 There are 2 changes in sign at most 2 positive roots
Step 2 Check at most negative roots in f(x) = 0 The numbers of changes in sign = most negative roots eg f(x) = x9 + 5x8 - x3 + 7x + 2rArr f(- x) = - x9 + 5x8 + x3 - 7x + 2 There are 3 changes in sign at most 3 negative roots
581 Some Important Forms of Quadratic Equations
An equation f(x) = 0 cannot have more positive roots then there are changes of sign in f(x) and cannot have more negative roots than there are changes of sign in f(ndashx)
1 An equation of the form (x ndash a) (x ndash b) (x ndash c) (x ndash d) = A where a lt b lt c lt d a + b = c + d can be solved by a change of variable
ie (x a) (x b) (x c) (x d)y4
minus + minus + minus + minus= or (a b c d)y x
4+ + +
= minus
2 Equation of type (x ndash a) (x ndash b) (x ndash c)(x ndash d) = Ax2 where ab = cd can be reduced to a collection
of two quadratic equations by a change of variable aby x4
= +
3 An equation of the form (x - a)4 + (x - b)4 = A can also be solved by a change of variable
ie making a substitution (x a) (x b)y2
minus + minus=
4 A reciprocal equation of the standard form can be reduced to an equation of half of its dimensions
Theory of Equation 563
5 An equation of the form af(x) + bf(x) = c where a b c isin R and a b c satisfies the condition a2 + b2 = c then solution of the equation is f(x) = 2 and no other solution of the equation is possible
582 Position of Roots of a Polynomial Eqution
(a) If f(x) = 0 is an equation and a b are two real numbers such that f(a) f(b) lt 0 Then the equation f(x) = 0 has at least one real root or an odd number of real roots between a and b (b) If f(a) f(b) gt 0 then either no real root or an even number of real roots of f(x) = 0 lies
between a and b
59 Equation of highEr DEgrEE
The equation a0 + a1x + a2x2 + + an xn (an 0) when a0a1a2an are constant but an ne 0 is a polynomial of digree n a1a2an an be n roots then
a1 + a2 + a3 + + an = 1
0
aaminus a1a2 + a2a3 + a3a4 + + an- 1an = 2
0
aa
a1a2 a3 an = 1
0
aaminus
rArr Cubic and Biquadratic
Tips and TricksThe truth of the following statements will be readily admitted
1 If all the coefficients are real then the imaginary roots occurs in pairs (ie number of complex roots is always even)
2 If the degree of a polynomial equation is odd then the number of real roots will also be odd It follows than at least one of the roots will be real
3 Every equation of an odd degree has at least one real root whose sign is opposite to that of its last term
4 Every equation which is of even degree and has its last term negative has at least two real roots one positive and one negative
6 If a b c k are roots of the equation f(x) = 0 then = + + + +minus minus minus minus
f ( x ) f ( x ) f ( x ) f ( x )f ( x )
x a x b x c x k
8 If the coefficients are all positive the equation has no positive root Thus the equation x5 + x3 + 2x +1 = 0 cannot have a positive root
9 If the coefficients of the even powers of x are all of one sign and the coefficients of the odd powers are all of the contrary sign the equation has no negative root Thus thee quation x7 + x5
ndash 2x4 + x3 ndash 3x2 + 7x ndash 5 = 0 cannot have a negative root
10 If the equation contains only even powers of x and the coefficients are all of the same sign the equation has no real root Thus the equation 2x6 + 3x4 + x2 + 7 = 0 cannot have a real root
11 If the equation contains only odd powers of x and the coefficients are all of the same sign the equation has no real root except x = 0 Thus the equation x9 + 2x5 + 3x3 + x = 0 has no real root except x = 0
12 If there is no change in sign then all the roots are imaginary
13 If in the polynomial of degree n the maximum number of possible positive real roots is k1 and maximum number of possible negative real roots is k2 and zero is not the root of polynomial then the minimum number of complex roots will be equal to n ndash (k1 + k2)
Chapter 6permutation and
Combination
61 introduction
Permutations and combinations is the art of counting without counting ie we study various principles and techniques of counting to obtain the total number of ways an event can occur without counting each and every way individually
62 Fundamental PrinciPles oF counting
621 Addition Rule
If an event (operation) E1 can occur in n1 ways E2 can occur in n2 ways hellip and Ek can occur in nk ways (where k ge 1) And these ways for the above events to occur are pair-wise disjoint then the number of
ways for at least one of the events (E1 E2 E3 hellip or Ek) to occurs is (n1 + n2 + n3 + hellip + nk) = i k
ii 1
n=
=sum
bull An equivalent form of above rule using set-theoretic terminology is given belowLet A1 A2 hellip Ak be any k finite sets where k ge 1 If the given sets are pairs wise disjoint
ie Ai cap Aj = f for i j = 1 2 hellip k i ne j then k k
i 1 2 k ii 1i 1
A | A A A | | A |==
= cup cup cup =sum
where |Ai|
denotes the number of elements in the set Ai
622 Multiplication RuleIf an event E can be decomposed into n ordered event E1 E2 hellip Er and that there are n1 ways for the event E1 to occurs n2 ways for the event E2 to occur hellip nr ways for the event Er to occur Then the total number
of ways for the event E to occur is given by n(E1 and E2 and hellip and Er) = r
1 2 r ii 1
n n n n=
times times times =prod
bull An equivalent form of (MP) using set-theoretic terminology is stated belowr
i 1 2 ri 1
A A A A=
= times times timesprod = (a1 a2 hellip an) | ai isin Ai i = 1 2 hellip r denote the cartesian product of the
finite sets A1 A2 Ar Then r r
i 1 2 r ii 1 i 1
A | A | | A | | A | A= =
= times times times =prod prod
Permutation and Combination 1665
Notes
bull And stands for intersection (cap) or multiplication
bull Or stands for union (cup) or addition
bull Both addition and multiplication rules can be extended to any finite number of mutually exclusive operations
623 Complementation Rule
If A and A are two complementary sets and S be universal set thenQ ( )n(A) + n A = n(S) rArr = minusn(A) n(S) n(A)
So we count n(A) or n(A) whichever is easier to count then subtract from n(S) to get the other
624 Principles of Inclusion-Exclusion
Let X be a finite set of m elements and x1 x2 x3 xr be some properties which the elements of X may or may not have if the subset of X having the property xi (where i = 1 2 3 r) is Xi and those having both
properties xi and xj is denoted by i jX Xcap and so on
Then the number of elements of X which have at least one of the properties x1 x2 x3 xr is given
by r
ii 1
n X=
= S1 - S2 + S3 - S4 + + (-1)rndash1Sr and the number of elements of U which have none of the
properties x1 x2 x3 xr is given byr
ci
i 1
n X m=
=
- S1 - S2 + S3 - S4 + + (-1)rndash1Sr where
r
1 ii 1
S n(X )=
=sum r r
2 i j1 i 1 r
S n(X X )le lt le
= capsumsumeg For r = 2 n(X1 cup X2) = n(X1) + n(X2) ndash n(X1 cap X2)For r = 3 n(X1 cup X2 cup X3) = n(X1) + n(X2) + n(X3) ndash n(X1 cap X2) ndash n(X1 cap X3) ndash n(X2 cap X3) +
n(X1 cap X2 cap X3)
625 Injection and Bijection Principles
Suppose that a group of n students attend a lecture in a lecture theater which has 100 seats assuming that no student occupies more than one seat and no two students share a seat if it is known that every student has a seat then we must have n le 100 If it known furthermore that no seat is vacant then we are sure that n = 100 without actually counting the number of students
6251 Injection principle (IP)
Let A and B be two finite sets if there is an injection from A to B then |A| le |B|
6252 Bijection principle (BP)
Let A and B be two finite sets if there is a bijection from A to B then |A| = |B|
666 Mathematics at a Glance
63 combinations and Permutations
Each of the groups or selections which can be made by taking some or all of a number of things without considering the order in which the objects are taken is called a combination Whereas a selection of objects where the order in which the objects are taken is also taken into account is called as an arrangementpermutation
To understand the concept of combination and permutation let us consider the combinations which can be made by taking the letters from a b c d two at a time namely
Combinations (total no 6)
Permutations (totalnumber 12)
ab ac ad da ca babc bd bd cb
cd dc
=
=
Number of combinations of lsquonrsquo distinct objects taken r at a time denoted as nr
nCr(n r)
=minus
NoteFrom the above illustration it is simply clear that in combinations we are only concerned with the number of things each selection contains without taking into account the order in which the objects are being selected (ie ab and ba are regarded as same selection) Whereas in permutation the order of objects is taken into account
64 Permutation oF diFFerent objects
Case I When repetition of objects is not allowedNumber of permutation of n distinct things taken r at a time (0 le r le n) is denoted by nPr and it is equivalent to filling up of r vacancies by n different person clearly first place can be filled in n ways and after which 2nd place can be filled in (n -1) ways and 3rd place can be filled in (n - 2) ways and similarly rth place can be filled in (n - r + 1) ways
rArr nPr = n(n - 1) (n - 2) (n - r + 1) = n(n r)minus
= nr
nr C rr(n r)
= timesminus
Case II When repetition of objects is allowedNumber of permutation =
times
r
n n n ntimes times times times = nr because now each of the vacancies can be filled
up in n ways
Notes
bull The word indicating permutation are arrangement standing in a line seated in a row problems on digits word formation rank of word number of vectors joining given points and number of greetings sent among a group etc
bull The number of permutations of n distinct objects taken all at a time = n
bull The number of all permutations of n different object taken r at a time when a particular set of k objects is to be always included in each arrangement is r nndashkCrndashk
bull Number of permutations of n different things taken all at a time when r specified things always remain together is r(n ndash r +1)
Permutation and Combination 1667
bull Number of permutations of n different things taken all at a time when r specified things never occur together is n ndash r(n ndash r +1)
bull The number of permutations of n different things taken all at a time when no two of the r particular things come together is nndashr+1Cr (n ndash r) r
65 Permutation oF identical objects (taking all oF them at a time)
Number of permutations (N) of lsquonrsquo things taken all at a time when lsquoprsquo are of one kind lsquoqrsquo of a second kind
lsquorrsquo of a third kind and so on is given by nNpqr
=
Explanation let N be the required number of permutations From any of these if the p like things were different we could make p new permutations Thus if the p like things were all different we would have got N(p) new permutations Similarly if the q like things were different we would get N (q) new permutations from each of the second set of permutations
Thus if the p like things and the q like things were all different we would have got Npq permutations in all The process is continued untill all the sets of like things are different and we then get the number of permutations of n things taken all at a time when they are all different (which is n)
Npqr n= rArr nNpqr
=
66 rank oF words
When all the letters of a word are arranged in all possible ways to form different words and the words formed are further arranged as per the order of ordinary dictionary then the position occupied by that word is called as its rank eg rank of the word MAT is 3 because it occupied third position in the alpha-betical list (AMT ATM MAT MTA TAM TMA) of words formed using letters A M T
Shortcut to Find Rank of a Word Example Banana Example Large
1 Write the letters of the word in alphabetical order
AAABNN A E G L R
2 Pick the letters one-by-one in the order in which they are heard while speaking
B A N A N A L A R G E
3 For each of the letters in this order using representation in Step (1) find
n
number of letter in left on nx
pq=
p q are number of identical letters
Cross the letters as done with it
13x AAABNN
32= rarr
x2 = 0 rarr AANN
32x AAN
22= rarr
x4 = 0x5 = 1 rarr A
1x 3 AEGER= rarr
2x 0 AEGER= rarr
3x 2 EGR= rarr
x4 = 1
Rank = (x1)5 + (x2)4 + (x3)3 + (x4)2 + x5 + 0 5 3 3 2 1 1 0 3432 22times times
+ + times + =Rank 3 times 4 + 0 times 3 + 22 + 11 + 0 = 78
668 Mathematics at a Glance
67 circular Permutation
The arrangement of objects around a circle is called lsquocircular permutationrsquo Two circular permutations are called identical iff one of them can be super imposed on the other by a suitable rotation without overturn-ing and without changing the relative position of object eg following 5 circular permutations are identical
671 Circular Permutation of n ObjectsWhen lsquonrsquo distinct objects (A1 A2 A3 An) are to be arranged around a circle then each circular arrangement generates lsquonrsquo number of distinct linear arrangements by rotating the objects around the
circle by 0360
n
at a time (keeping their relative position fixed)
rArr Each circular array generates lsquonrsquo linear permutation Let the total number of circular array be x
rArr Number of linear arrays = nx rArr nx = n rArr nx = n 1n
= minus
Remark bull As in circular permutation (unlike linear permutations) there is no initial and terminal position
therefore fixing the position of one object around the circle its position acts as a terminal consequently the remaining (n ndash 1) positions become as distinct as in linear permutations Therefore rest of (n ndash 1) object can be arranged in these position in (n ndash 1) ways
Permutation and Combination 1669
ExplanationIn a circular permutation the relative position among the things is important whereas the place of a thing has no significance Thus in a circular permutation the first thing can be placed anywhere This operation can be done only in one way then relative order begins Thus the ways for performing remaining parts of the operation can be calculated just like the calculation of linear permutation for an example to place 8 different things round a circle first we place any one thing at any place there will be only one numbers of ways = 7 Thus required number of circular permutations if 7
bull Since each circular arrangement has its unique counter-clockwise arrangement therefore the number
of clockwise array = number of counter-clockwise arrays = minus( n 1)2
bull In a garland of flowers or a necklace of beads (since the overturning of permutations is possible) It is difficult to distinguish clockwise and anti-clockwise orders of things so a circular permutation under both these orders (the clockwise and anti-clockwise) is considered to be the same
Therefore the number of ways of arranging n beads along a circular wire is minus( n 1)2
bull The total number of circular arrangements of n distinct objects taken r at a time is
(i) minus =n
n rr
PC ( r 1)
r when clockwise and anticlockwise orders are treated as different
(ii) minus =n
n rr
P1C ( r 1)
2 2r when the above two orders are treated as same
68 number oF numbers and their sum
Case I Number of r digit numbers formed using n digits D1 D2 Dn when repetition allowed bull Number of numbers = nr
bull Sum of all numbers = r r
r 1k
k 1
10 1 D n9
minus
=
minus
sum
Proof When all the numbers formed are arranged vertically for summation Any digit gets repeated nrndash1 times in each column keeping a particular digit say Dk
fixed at some place out of r then remaining (r ndash 1)
places can be arranged using n digits in nrndash1 ways
Summation of digits in any column = sum of all digits times repetition of digit ( )n
r 1k
k 1
D n minus
=
sum
rArr Sum of all numbers ( )n
r 1k
k 1
D n minus
=
sum (1 + 10 + 102 + hellip + 10rndash1)
Case II Number of r digit numbers formed using n digits D1 D2 Dn when repetition not allowed
bull Number of numbers = n
rP if r n0 if r n
le gt
bull Sum of all numbers = r r
n 1k r 1
k 1
10 1 D P9
minusminus
=
minus
sum
Proof When all the numbers formed are arranged vertically for summation Any digit gets repeated nndash1Prndash1
times in each column keeping a particular digit say Dk
fixed at some place out of r
then remaining (r ndash 1) places can be arranged using n ndash 1 digits in nndash1Prndash1 ways
670 Mathematics at a Glance
Summation of digits in any column = sum of all digits times repetition of digit = minusminus
=
sum
nn 1
k r 1k 1
D P
rArr Sum of all numbers = ( )n
n 1k r 1
k 1
D Pminusminus
=
sum (1 + 10 + 102 + hellip + 10rndash1) =
r rn 1
k r 1k 1
10 1 D P9
minusminus
=
minus
sum
681 Divisor of Composite NumberA natural number x = pa qb rg is called divisor of N = pa qb rc iff N is completely divisible by x For Example when all the prime factors of x are present in N which is possible only if 0 le a le a 0 le b le b and 0 le g le c where a b g isin ℕ cup 0
bull Set of all divisors of N is given as x x = pa qb rg where 0 le a le a 0 le b le b 0 le g le c
bull Number of divisor number of divisors = n(a b g) 0 le a le a 0 le b le b 0 le g le c= na 0 le a le a times nb 0 le b le b times ng 0 le g le c = (a + 1) (b + 1) (c + 1)
bull Number of divisors are given by number of distinct terms in the product= (1+ p + p2 + + pa) (1+ q + q2 + + qb) (1+ r + r2 + + rc)= (a + 1) (b + 1) (c + 1) (which includes 1 and the N it self)
682 Sum of Divisor
Since each individual divisor is given as terms of the expansion (p0 + p1 + p2 + + pa) (1+ q + q2 + + qb) (1+ r + r2 + + rc) therefore the sum of all divisors is = 1 + p + q + r + p2 + q2 + r2 + pq + pr ++ pa qb rc
= a 1 b 1 c 1p 1 q 1 r 1p 1 q 1 r 1
+ + + minus minus minus minus minus minus
(i)
Notes bull ImproperProper divisors of N = pa qb rc When a = b = g = 0
rArr x = 1 which is divisor of every integer and a = a b = b and g = c then x becomes number N itself These two are called lsquoimproper divisorrsquo
rArr The number of proper divisors of N = (a + 1)(b + 1)(c + 1) ndash 2
bull If p = 2 then number of even divisors = a(b + 1)(c + 1) number of odd divisors = (b + 1)(c + 1)
683 NumberSum of Divisors Divisible by a Given NumberIf x = pa qb rg is divisor of N = pa qb rc and completely divisible by 1 1 1y = p q r α β γ
bull Set of all divisors of N is given as x x = pa qb rg where a1 le a le a b1 le b le b g1 le g le c rArr Number of divisors = n(a b g) a1 le a le a b1 le b le b g1 le g le c = (a ndash a1 + 1) (b ndash b1 + 1)
(c ndash g1 + 1)
684 Factorizing a Number into Two Integer FactorsIf x and y be two factors of the Natural Number N = pa qb rc N = xy
rArr x and y are divisors of N
Permutation and Combination 1671
Case I If number N is not a perfect square
Number of two factor products (number of total divisors)= 2
Case II If number N is a perfect square
Number of two factor products (number of total divisors) + 1= 2
Case III Number of integer solution of equation xy = pa qb rc sd = 2 times total number of divisor
Since number of natural number solution of the equation
xy = pa qb rc sd = Number of divisors = (a + 1) (b + 1) (c + 1) (d + 1)
rArr Number of integer solution of the equation = 2(a + 1) (b + 1) (c + 1) (d + 1)
69 combination
Combination of n objects taken r at a time is denoted as nCr and defined as nr
nCr(n r)
=minus
691 Properties of Combinations
1 The number of combination of n different things taken r at a time is denoted by nCr or C(n r)
or nr
and it is empirically calculated as =minus
nr
nC
r(n r) (0 le r le n) where n isin N and r isin W
whole numbers = 0 (if r gt n) 2 nCr is always an integer The following important conclusions can be made out of the above statement (a) Product of r consecutive integers is always divisible by r
∵ nr
n(n 1)(n 2)(n 3)(n r 1)C Ir
minus minus minus minus += isin
Clearly the numerator is completely divisible by r
(b) 0 = 1 n n0 n
nC C 1
n 0 = = =
and nC1 = n
(c) k = infin if k lt 0 (Think why) 3 nCr = nCnndashr this is simply selection of r things means rejection of n ndash r at the same time 4 nCx + nCy rArr x = y or x + y = n 5 nCr + nCrndash1 = n+1Cr (1 lt r lt n) this is also known as Pascal Rule
672 Mathematics at a Glance
6 rnCr = nnndash1Crndash1 rArr n n 1 n 2
r r 1 r 2n n n 1C ( C ) C r r r 1
minus minusminus minus
minus = = = minus Thus we can work out as
Choosing r MPrsquos from n citizens (nCr ways)Choosing 1 PM from r Choosen MPrsquos (r ways)r times rCr waysMP Member of Parliament
equivChoosing 1 PM from n citizens (n ways) and Choosing remaining (rndash1) MPrsquos fromremaining (n ndash 1) citizens (nndash1Crndash1) waysMP Prime Minister
7 n n n 2r r 1 r 2
r 1 (r 1(r 2)C C Cn 1 (n 1)(n 2)
++ +
+ + + = = + + +
8 nCr rCs = nCs
nndashsCrndashs (n ge r ge s) This we can work out as
Choosing r MPrsquos (nCr ways) andChoosing s ministers out of rMPrsquos (rCs ways) nCr times rCs
equivChoosing s ministers (rCs ways) andChoosing remaining (r ndash s) MPrsquos out ofremaining (n ndash s) citizens nCs times nndashsCrndashs
9 n
rn
r 1
C n r 1C rminus
minus +=
10 nC0 + nC1 + nC2 + + nCn = 2n this is selection of any number of objects out of given n objects For each object we have only two possibilities selection or rejection which is 2n
11 nC0 + nC2 + nC4 + = nC1 + nC3 + nC5 + hellip = 2nndash1 12 nCm + nndash1Cm + nndash2Cm + + mCm = n+1Cm+1
692 Restricted Combinations
The number of combinations of n different things taking r at a time (a) When p particular things are always to be excluded = nndashpCr (b) When p particular things are always to be included = nndashpCrndashp (c) When p particular things are always included and q particular things are always excluded = nndashpndashqCrndashz
693 Combination of Objects Taking any Number of Them at a Time
bull Number of selections of objects when any number of them can be selected is given as nC0 + nC1 + + nCn = 2n
Where nCr corresponds to the case when r objects are selected out of n different objects In above case r varies from 0 to n The right hand side value 2n can be explained as number of ways of dealing with all n objects each in exactly two ways either selected or rejected
bull Number of selection of objects (at least one) out of n different objects n
n n n n nr 1 2 n
r 0
C C C C 2 1=
= + + + = minussum
bull Number of selection of atleast two object out of n = 2nndashnC0 ndash nC1
Permutation and Combination 1673
694 Combination when Some Objects are Identical (Taking any Number of Them at a Time)
1 Combination when some objects are identical The total number of ways in which it is possible to make a selection taking some or all out of (p + q + r) things where p are alike of the first kind q are alike of the second kind and r alike of the third kind and s are different = (p + 1) (q + 1) (r + 1) 2s ways
Explanation Out of p alike things we may select none or one or two or three or all p Hence they may be disposed off in (p + 1) ways Similarly q alike things may be disposed of in (q + 1) ways similarly for r And s different things may be disposed of in 2s ways (This includes the case in which all of them are rejected)
bull Number of ways (if at least one object to be selected) = (p + 1) (q + 1) (r + 1) 2s ndash1 bull Number of ways (if at least one from s different object to be selected) = (p + 1) (q + 1)(r + 1) (2s ndash1) bull Number of ways (if at least one object of each identical type lot is to be selected) = (p q r)2s
695 Combination when Some Objects are Identical (Taking specific number of them at a time)
Case 1 If a group has n things in which p are identical then the number of ways of selecting r things
from a group is r
n pk
k 0
Cminus
=sum or
rn p
kk r p
Cminus
= minussum according as r le p or r gt p
Explanation It can be obtained by assuming the selection of k distinct object and rest r ndash k objects identical and taking the values of variable k from 0 to r (or p) whichever is less
For an instance when no object is selected from identical objects (k = 0) then the number of selection = nndashpCr
And when one object is selected from identical objects (k = 2) then the number of selection = nndashpCrndash1Similarly for k = 3 the number of selection = nndashpCrndash2 and so on
Notes
(i) The number of ways of selecting r objects out of n identical objects is 1
(ii) The number of ways of selecting any number of objects out of n identical objects is n + 1
Case 2 If there are p1 objects of one kind p2 objects of second kind pn objects of nth kind then the number of ways of choosing r objects out of these (p1 + p2 + + pn) objects
= coefficients of xr in 1 2 np p p2(1 x x )(1 x x )(1 x x )+ + + + + + + + +If one object of each kind is to be included in such a collection then the number of ways of choosing r objects
= coefficients of xr in the product 1 2 np p p2(x x )(x x )(x x )+ + + + + +This problem can also be stated as Let there be n distinct objects x1 xn x1 can be used at the most p1 times x2 at the
most p2 times xn at the most pn times then the number of ways to have r things
Renarks bull Given n distinct points in a plane no three of which are collinear then the number of line segments
they determine is nC2
674 Mathematics at a Glance
bull The number of diagonals in n-polygon (n sides closed polygon) is nC2 ndash n
If in which m points are collinear (m ge 3) then the number of line segments is (nC2 ndash mC2) + 1
bull Given n distinct points in a plane no three of which are collinear then the number of triangles formed = nC3 If in which m points are collinear (m ge 3) then the number of triangles is nC3 ndash mC3
bull Given n distinct points of which no three points are collinear
(i) Number of straight lines = nC2
(ii) Number of triangles = nC3
(iii) Number of quadrilaterals = nC4
(iv) Number of pentagon = nC5
bull Given n points in a plane out of which r of them are collinear Except these r points no other three points are collinear Then number of different geometric figures constructed by joining these points are expressed as below
(i) number of line segments (LS) = nC2
(ii) number of directed line segments vectors (DLS) = nP2
(iii) number of lines formed = nC2 ndash rC2 + 1 or nndashrC2 + (n ndash r)r + 1
(iv) number of triangles formed = nC3 ndash rC3 or nndashrC3 + (n ndash rC2)r + (nndashr) rC2
(v) number of quadrilateral = nC4 ndash (rC4 + (n ndash r) rC3 )) or nndashrC4 + nndashrC3
rC1 + nndashrC2rC2
(vi) number of rectanglessquares formed put of m horizontal lines and n vertical lines such that distance between conjugative line both set of parallel lines is unity
bull Given A1 A2 A3 An are horizontal lines B1 B2 B3 Bm are vertical lines as shown in figure
(i) Number of rectangles = number of ways of
choosing two lines from each set = ( )i j k ln A A and B B
= nC2 times mC2 (ii) Number of square of size k times k = number of
ways of choosing two lines i j jA A + horizontal
line = ( ) ( )i j k j j k
1 i n k 1 j m k
n A A n B B+ +
le le minus le le minus
times
= (n ndash k) (m ndash k)
(iii) Total number of squares = ( ) ( )k r
k 1
n ndash k m ndash k=
=sum where r = min m ndash 1 n ndash 1
610 distribution
6101 Distribution Among Unequal Groups
To find the number of ways in which m + n things can be divide into two groups containing m and n things respectively This is clearly equivalent to finding the number of combinations of m + n things taking m at a time for every time we select a group of m things we leave a group of n things behind
Thus the required number = (m n)mn+
Permutation and Combination 1675
6102 To Find the Number of Ways in Which the m + n + p Things Can be Divided into Three Groups Containing m n p Things Separately
First divide the m + n + p things into two groups containing m and n + p things respectively the number
of ways in which this can be done is m+n+pCm = (m n p)m(n p)+ +
+ And the number of ways in which the group of
n + p things can be divided into two groups containing n and p things respectively is n pn
(n p)C np
+ += Hence
the number of ways in which the subdivision into three groups containing m n p things can be made follows
6103 Distribution Among Equal Groups
When name of groups is not specified If 2m objects are to be distributed among two equal groups
containing m objects each Then it can be done in (m m)m m2
+ = 2
2m(m) 2
because each division it is possible
to arrange the groups into 2 ways without obtaining new distributionExplanation Then we divide the total number of arrangements obtained normally by k where k is |
number of groups among which the objects are distributed If we put n = p = m we obtain3m
m m m
but since this include 3 times the actual number of divisions because of the arrangement of groups among them selves therefore the number of different ways in which subdivision into three equal groups can
be made is =3m
mm m 3
6104 When Name of Groups Specified If the name of groups among which the objects are distributed are specified (eg distributing books to students dividing soldiers into regiment distributing students into sections etc) If we put n = p = m
we obtain 3mm m m
bull The number of ways of dividing pq objects among p groups of same size each group containing q
objects = p
(pq)(q) p
bull The number of ways of distributing pq objects among n people each person getting q
objects = p
(pq)(q)
611 multinomial theorem
The expansion of [x1 + x2 + x3 + + xn]r where n and r are integers (0 lt r le n) is a homogenous
expression in x1 x2 x3 xn and given as [x1+ x2 + x3 + + xn]r = 31 2 n1 2 3 n
1 2 3 n
r x x x x
λλ λ λ λ λ λ λ
sum
676 Mathematics at a Glance
(where n and r are integers 0 lt r le n and l1 l2 ln are non-negative integers) Such that l1 + l2 + + ln = r (valid only if x1 x2 x3 xn are independent of each other) coefficient of 31 2
1 2 3x x x λλ λ = total number of arrangements of r objects out of which l1 number of x1rsquos are identical l2 number of x2rsquos are identical and
so on = 1 2 3 n
1 2 3 n 1 2 3 n
( ) (r)
λ +λ +λ + λ=
λ λ λ λ λ λ λ λ
6111 Number of Distinct TermsSince (x1 + x2 + x3 + + xn)r is multiplication of (x1 + x2 + x3 + + xn)r times and will be a homogeneous expansion of rth degree in x1 x2 xn So in each term sum of powers of variables must be r
So number of distinct terms will be total number of non-negative integral solution of equation is l1 + l2 + l3 + + ln = r = Number of ways of distributing r identi-cal objects among n persons = number of ways of distributing r balls among n people
= number of arrangements of r balls and n ndash 1 identical separators = (n 1 r)(n 1)rminus +minus
= n+rndash1Cr = n+rndash1Cnndash1
612 dearrangements and distribution in Parcels
Any change in the order of the things in a group is called a derangement If n things are arranged in a row the number of ways in which they can be dearranged so that none of them occupies its original position
is n1 1 1 1 1n 1 ( 1)1 2 3 4 n
minus + minus + minus + minus
bull If r objects go to wrong places out of n thing then (n ndash r) objects go to their original place If Dn rarr number of group and if all objects go to the wrong places and Dr rarr number of ways if r objects go to wrong places out of n then (n ndash r) objects go to correct places
Then Dn = nCnndashr Dt where Dr = r1 1 1 1 1r 1 ( 1)1 2 3 4 r
minus + minus + minus + minus
bull Derangement of a given n-permutations minus
minus
1 2 3 n 1 n
n permutation
P P P P P is an arrangement in which at least one
object does not occupy its assigned position rArr Total number of dearrangements = n ndash 1 bull Let Ai denotes set of arrays when ith objects occupies ith place n(Ai) = (n ndash 1)
rArr n(A1 cup Aj) = (n ndash 2)rArr Number of arrays in which atleast one object occupies its correct place = n(A1 cap A2 cap A3 hellip cap
An) = Σn(Ai) ndash Σn(Ai cup Aj) + Σn(Ai cup Aj cup Ak) ndash hellip + (ndash1)nndash1 n (A1 cup A2 cup A3 hellip cup An)= nC1 (n ndash 1) ndash nC2(n ndash 2) + nC3 (n ndash 3) ndash hellip + (ndash1)nndash1 nCnO
= minusminus + minus +
nn n n ( 1) n1 2 3 n
= minus minus
minus + minus +
n 11 1 1 ( 1)n 1 2 3 n
the total number of dearrangement in which no object occupies its correct place = n ndash n (A1 cap A2 cap A3 hellip cap An)
= minus minus
minus minus + +
n 11 1 1 ( 1)n n 1 2 3 n
= minusminus + minus + +
n1 1 1 1 ( 1)n 1 1 2 3 4 n
= minus
minus + +
n1 1 1 ( 1)n 2 3 4 n
(n 1)separators
| | | | |minus
Permutation and Combination 1677
bull Number of dearrangement in which exactly r objects occupy their assigned places
=
minus
minus
minustimes minus minus + minus + minus
n rn
r
choo singr objectsand placing them Arrangingg n r objects so that noneat their correct places oft hem occupies their assigned positions
1 1 1 ( 1)C (n r) 2 3 4 (n r)
613 distribution in Parcels
6131 Distribution in Parcels When Empty Parcels are Allowed The number of ways in which n different objects can be distributed in r different groups (here distributed means order of objects inside a group is not important) under the condition that empty groups are allowed = rn Take any one of the objects which can be put in any one of the groups in r ways Similarly all the objects can be put in any one of groups in r number of ways So number of ways = r r rn times = rn
= coefficient of xn in n (ex)r = r 1
k r nk
k 0
( 1) C (r k)minus
=
minus minussum
6132 When at Least One Parcel is EmptyNumber of distribution when at least one parcel is empty
= n (A1 cup A2 cup A3 hellip cup Ar) Ai is the set of distribution when ith parcel is emptyn(Ai) = (r ndash n)n and n (Ai cap Aj) = (r ndash 2)n = Sn (Ai) ndash S n (Ai cap Aj) + Sn (Ai cap Aj cap Ak) + hellip + (ndash1)rndash1 n (A1 cap A2 cap cap Ar)
= nC1 (r ndash1)n ndash rC2(r ndash2)n + rC3 (r ndash 3)n + helliphellip +(ndash1)rndash1 rCrndash1 = r 1
k 1 r nk
k 1
( 1) C (r k)minus
minus
=
minus minussumThe number of ways in which n different objects can be arranged in r different groups= n r 1
r 1n C+ minusminustimes if empty groups are allowed = n 1
r 1n Cminusminustimes if empty groups are not allowed
The number of ways in which n different things can be distributed into r different places blank roots being admissible is rn
RemarksGiven two sets A = a1 a2 an and B = b1 b2 b3 br then following holds good
(i) n(A times B) = n(A) n(B) = n times r (ii) Number of relation R A rarr B = number of subsets of A times B = 2nr (iii) Number of functions f A rarr B = number of ways of distributing n elements
(objects) of A in to elements (boxes) of B = rn (iv) Number of injective functions f A rarr B = number of permutations of n elements
of A (objects) over r elements of B (places) =r
nP if r n
0 if r n
ge
lt
(v) Number of into (non surjective) functions f A rarr B = number of ways of distributing n elements
(objects) of A into elements (boxes) of B such that atleast one box is empty = r 1
k 1 r nk
k 1
( 1) C ( r k )minus
minus
=
minus minussum
(vi) Number of on-to (surjective) functions f A rarr B = number of ways of distributing n elements (objects)
of A in to elements (boxes) of B such that no box is empty= r 1
k r nk
k 0
( 1) C ( r k )minus
=
minus minussum
678 Mathematics at a Glance
614 exPonent oF a Prime in n
Exponent of prime p in n is denoted by Ep (n) where n is natural number so the last integer amongest 1 2(n - 1)n which is divisible by p is [np] p when [n] le x
rArr s
p 2
n n nE (n) p p p
= + + +
where s is the largest number such that ps le n lt ps+1
6141 Exponent of Prime lsquoPrsquo in n
Exponent of prime number lsquoprsquo in n is defined as power of p when n is factorized into prime factor using unique factorization theorem and it is denoted as Ep (n)
Theorem The largest natural number divisible by p is less than or equal to lsquonrsquo is given as n pp
Proof Division algorithm as n le p thus there existTwo natural number q and r such that n = pq + r
where 0 le r le p rArr n rqp p= + where r0 1
ple lt
q is called integer part of number np denoted as n randp p
is known as fractional part of
number np denoted as n p
Observe the situation on ℝ number lies
Conclusion ie np
is the quotient in the division of n by p
Theorem The number of natural numbers divisible by p less than or equal to lsquonrsquo is equal to np
rArr The number of natural numbers divisible by p2 less than or equal to lsquonrsquo is equal to 2
np
rArr The number of natural numbers divisible by p3 less than or equal to lsquonrsquo is equal to 3
np
Exponent of prime p in n p 2 3
n n nE (n) p p p
= + + +
Chapter 7binomial theorem
71 IntroductIon
We have dealt with expansions of (x + a)2 while dealing with quadratic equations Herein we will study expansions of the form (x + a)n Any power of binomial expression (a + x)2 can be expanded in the form of a series which is obtained the by process of continuous multiplication as shown here (a + x)2 = (a + x) (a + x) = a2 + ax + ax + x2 = a2 + 2ax + x2 which can be explained as the terms of expansion are obtained when any one of two terms a or x are selected from each factor and finally they are multiplied together
72 BInomIal
Any algebraic expression containing two terms is called lsquobinomial expressionrsquo [Bi (two ) + Nomial (terms)] is an expression containing sum of two different terms
721 Binomial Expansion (Natural Index)
Binomial expansion is a polynomial equivalent of powers of a given binomial expression The expressions for (a + x)n has been obtained as (a + x)n = nC0 a
n x0 + nC1 anndash1 x1 + nC2 a
n ndash2 x2 + + nCr an - r xr + +nCn a
0 xn
bull Where n is a positive integer which is given by n
n n n r rr
r 0
(a x) C a xminus
=
+ =sum and
nn r n n r r
rr 0
(a x) ( 1) C a xminus
=
minus = minussum
bull n
n r nr
r 0
(1 x) x C=
+ =sum n
n r n rr
r 0
(1 x) ( 1) C x=
minus = minussum where n isin I+ is known as index of binomial
and nCr is binomial coefficient) bull nCr are known as binomial coefficients bull n is called index of binomial bull The binomial expansion is homogenous in a and x ie sum of powers of a and x in each term
remains constant and this constant is equal to index of binomial bull Number of distinct terms in the expansion is equal to (n + 1)
780 Mathematics at a Glance
bull The equidistant binomial coefficients from beginning and end are equal
bull The number of terms in the expansion (a + x)n + (a ndash x)n will be n2+1 when n is even n 12+ and
when n is odd bull The number of terms in the above expansion (a + x)n ndash (a ndash x)n will be n2 when n is even
and n 12+ when n is odd
73 General term
A general term is known as representative term of binomial and it is (r + 1)th term of the expansion and is given by Tr +1 = nCr a
n - r xr in expansion of (a + x )n
731 rth Term from Beginning
The term nCr xnndashr y r is the ( r + 1)th term from beginning in the expansion of (x + y)n It is usually called
the general term and it is denoted by Tr+1 ie Tr +1 = nCr xnndashr yr
732 kth Term from End
kth term from end in the expansion of (x + y)n = (n - k + 2)th term from beginning
74 mIddle term
The middle term depends upon the value of n
Case I If n is even Then total number of terms in the expansion of (x + y)n is n+1 (odd) So there is only one middle term ie (n2 +1)th term is the middle term ie Tn2 +1= nCn2 x
n2 yn2
Case II If n is odd Then total number of terms in the expansion of (x + y)n is n+1 (even) So there are
two middle terms ie n 12+
th and n 32+
th are two middle terms They are given by n 1 n 1
2 2n 1C x y
minus +
and n 1 n 1
n 2 2n 1
2
C x y+ minus
+
75 numBer of terms In expansIons
bull (a + x)n = nC0an + nC1a
nndash1 x + nC2anndash2x2 + hellip + nCnndash1a
1xnndash1 + nCna0xn = n
n n r rr
r 0
C a xminus
=sum
bull (a ndash x)n = nC0anx0 ndash nC1a
nndash1x + nC2anndash2 x2 + hellip + nCnndash1a(ndashx)nndash1 + nCna0 (ndashx)n =
nr n n r r
rr 0
( 1) C a xminus
=
minussum
bull (a + x)n + (a ndash x)n = m
n n 2r 2r2r
r 0
2 C a xminus
=sum where
n 2m if n is evenn 1 2m if n is odd
= minus =
Binomial Theorem 1781
rArr Number of terms
n 2 if n is even2m 1
n 1 if n is odd2
++ = +
bull (a + x)n ndash (a ndash x)n = m
n n 2r 1 2r 12r 1
r 0
2 C a xminus minus ++
=sum where
n 2m 1 if n is oddn 1 2m 1 if n is even
= + minus = +
rArr Number of terms
n if n is even2m 1
n 1 if n is odd2
+ = +
76 Greatest term
If Tr and Tr+1 be the rth and (r + 1)th terms in the expansion of (1 + x)n then n r
r 1 rn r 1
r r 1
T C x n r 1 xT C x r+
minusminus
minus += =
Let numerically Tr+1 be the greatest term in the above expansion Then Tr+1 ge Tr or r 1
r
T 1T+ ge
n r 1 | x| 1
rminus +
ge to find the value of r ie (n 1)r | x|(1 | x |)
+le
+
Now substituting values of n and x in (i) we get r le m + f or r le m where m is a positive integer and f is a fraction such that 0 lt f lt 1 In the first case Tm+1 is the greatest term while in the second case Tm and Tm+1 are the greatest terms and both are equal
761 To Find the Greatest Term in the Expansion of (1 + x)n
bull Calculate m = bull If m is integer then Tm and Tm+1 are equal and both are greatest term bull If m is not integer then T[m]+1 is the greatest term where [] denotes the greatest integral part
NoteTo find the greatest term in the expansion of (x +y)n since (x +y)n = xn(1+ yx)n and then find the greatest term in (1+yx)n
77 Greatest coeffIcIent
To determine the greatest coefficient in the binomial expansion of (1 + x)n consider the following
r 1 r
r r 1
T C n r 1 n 1 1T C r r+
minus
minus + += = = minus
Now the (r + 1)th binomial coefficient will be greater than the rth binomial
coefficient when n 1 1 1r+
minus gt
782 Mathematics at a Glance
rArr n 1 r2+
gt (i)
But r must be an integer and therefore when n is even the greatest binomial coefficient is given by the greatest value of r consistent with (i) ie r = n2 and hence the greatest binomial coefficient is nCn2
bull If n is even then greatest coefficient = nCn2 bull If n is odd then greatest coefficients are nC(n ndash 1)2 and nC(n + 1)2
78 propertIes of BInomIal coeffIcIent
The binomial coefficient for general term of the expansion (a + x)n is given as nCr which states the number of ways the term an - r xr occurs in the expansion
781 Properties of nCr
It is defined as number of selections of r objects out of n different objects and is given by
nr
nC
r(n r)=
minus when n gt r (= 0 if n lt r)
bull nCr is always an integer Product of r consecutive integers is always divisible by r
nr
n(n 1)(n 2)(n 3)(n r 1)C Ir
minus minus minus minus += isin (Clearly the numerator is completely divisible by r)
bull nCr = nCnndashr
bull nCx = nCy rArr x = y or x + y = n bull nCr + nCr-1 = n+1Cr
bull ( )n n 1 n 2r r 1 r 2
n n n 1C C Cr r r 1
minus minusminus minus
minus = = minus = helliphelliphelliphellip
bull n n 1 n 2r r 1 r 2
(r 1)(r 2)r 1C C Cn 1 (n 1)(n 2)
+ ++ +
+ ++ = = + + +
bull r nCr = n nndash1Crndash1 and n n 1
r r 1C Cr 1 n 1
++
= + +
79 propertIes of coeffIcIents
Properties of binomial expression are derived from
bull n
n n rr
r 0
(1 x) C x=
+ =sum = nC0 + nC1x + nC2x2 + +nCr x
r + + nCnxn (i)
bull n
n r n rr
r 0
(1 x) ( 1) C x=
minus = minussum = nC0 - nC1x + nC2x2 -+ (ndash1)n nCn x
n (ii)
bull n
n r n n n n nr 0 1 2 n
r 0
C x C C C C 2=
= + + + + =sum
Binomial Theorem 1783
bull n
r n n n n n nr 0 1 2 n
r 0
( 1) C C C C ( 1) C 0=
minus = minus + minus + minus =sum bull The sum of the binomial coefficients of the odd terms in the expansion of(1 + x)n is equal to the sum
of the coefficients of the even terms and each is equal to 2nndash1 bull C0 + C2 + C4 + hellip = C1 + C3 + C5 + hellip = 2nndash1
bull n
nr
r 0
r C=sum =1C1 + 2C2 + 3C3 + + nCn = n 2n ndash 1
bull n
2 n 2 2 2 2r 1 2 3 n
r 0
r C 1 C 2 C 3 C n C=
= + + + +sum
bull 0r 1 2 nCC C C Cr 1 1 2 3 n 1
= + + + ++ +sum
bull nn n
rn
r 1 k 0r 1
C n(n 1)r kC 2= =minus
+= =sum sum
710 multInomIal theorem
bull The general term in the multinomial expansion is 1 2 kr r r1 2 k
1 2 k
n x x xr r r
bull The total number of terms in the multinomial expansion = number of non -negative integral solutions of the equation r1 + r2 + + rk = n = n + k ndash 1Cn or n + k ndash 1Ckndash 1
bull Coefficient of x1r1 x2
r2 x5r5 in the expansion of a1x1 + a2x2 + + akxk =
bull Greatest coefficient in the expansion of (x1 + x2 + + xk)n where q is the quotient and r the remainder when n is divided by k
bull The number of terms in the expansion of (x + y + z)n where n is a positive integer is 12 (n + 1) (n + 2) bull Sum of all the coefficients is obtained by putting all the variables xi equal to 1 and it is equal to nm
711 tIps and trIcks
1 (x + y)n = sum of odd terms + sum of even terms
2 In the expansion of (x + y)n r 1
r
T n r 1 yn NT r x+ minus + isin =
3 The coefficient of xn ndash 1 in the expansion of (x + 1) (x + 2) (x + n) = n(n 1)2+
4 The coefficient of xn ndash 1 in the expansion of (x + 1) (x ndash 2) (x ndash n) = n(n 1)2
minus +
5 Greatest term in (x +y)n = xn Greatest terms in ny1
x +
6 The number of terms in the expansion of (x1 + x2 + + xn) n = n+rndash1Crndash1
7 If the coefficients of the rth (r + 1) and (r + 2) th terms in the expansion of (1 + x)n are in HP then n + (n ndash 2r)2 = 0
8 If the coefficients of the rth (r + 1) th and (r + 2) th terms in the expansion of (1 + x)n are in AP then n2 ndash n(4r + 1) + 4r2 ndash 2 = 0
Chapter 8InfInIte SerIeS
81 Binomial theorem for any index (n)
|x| lt 1 ie ndash1 lt x lt 1minus minus minus minus minus minus +
+ = + + + + + + infin2 3 r
n n(n 1)x n(n 1)(n 2)x n(n 1)(n 2)(n r 1)x(1 x) 1 nx to2 3 r
General term of (1 + x)nrn(n 1)(n 2)(n r 1)x
rminus minus minus +
Expansion of (x + a)n for any index
Case I When x gt a ie ax lt 1
In this case (x + a)n = x (1 + ax)n = x n (1 + ax)n = xn
2
3
a n(n 1)1 n (ax)x 2
n(n 1)(n 2) (ax) 3
minus + + + minus minus +
Case II When x lt a ie xa lt 1
In this case (x + a)n = a (1 + xa)n = a n (1 + xa)n = an
2
3
x n(n 1) x1 na 2 a
n(n 1)(n 2) x 3 a
minus + + +
minus minus +
Remarks
q nCr cannot be used because it is defined only for natural number
q If x be so small then its square and higher powers may be neglected then the approximate value of (1 + x)n = 1 + nx
82 Greatest term
To find the greatest term numerically in the expansion of (1 + x)n |x| lt 1 If Tr + 1 is the required term
then | Tr + 1| ge |Tr| or r 1
r
T 1T+ ge gives
| x |(x 1)r m| x | 1
+le =
+(say)
Infinite Series 885
(a) Calculate | x |(n 1)m
| x | 1+
=+
(b) If m is integer then Tm and Tm+1 are equal and both are greatest terms (c) If m is not integer then T[m]+1 is the greatest term where [] denotes the greatest integer
Remarks
1 (1 ndash x)ndash1 = 1 + x + x2 + x3 + + xr +
2 (1 ndash x)ndash2 = 1 + 2x + 3x2 + + (r + 1)xr + and
3 (1 ndash x)ndash3 = 1 + 3x + 6x2 + + + +( r 1)( r 2)
2 xr + helliphellip
83 taylor expansion
For any function f(x) we have
(i) 2 3hf (a) h f (a) h f (a)f(a h) f(a)
1 2 3+ = + + + +
(ii) replacing (a + h) by x 2 3(x a)f (a) (x a) f (a) (x a) f (a)f(x) f(a)
1 2 3minus minus minus
= + + + +
That is function f(x) expressed as a polynomial of infinite degree in (x ndash a)
831 Maclaurins Expansions
In taylors expansions replace a by 0 and h hy x we have 2 3xf (0) x f (0) x f (0)f(x) f(0)
1 2 3= + + + +
That is 2 3x cos0 x ( sin(0)) x ( cos0)sin x sin0
1 2 3minus minus
= + + + +
(i) 3 5x xsin x x
3 5= minus + +
(ii) 2 4 6x x xcos x 1
2 4 6= minus + minus +
(iii) 3 5 7x 2x 17xtan x x
3 15 315= + + + +
(iv) 3 5 7
1 x x xtan x x 3 5 7
minus = minus + minus +
(v) + = minus + minus + minus lt le
2 3 4x x xn(1 x) x ( 1 x 1)2 3 4
832 Eulerrsquos Number
The summation of the infinite series + + + + + infin1 1 1 11 1 2 3 4
is denoted by e which is equal to the limiting
value of (1 + 1n)n as n tends to infinity
886 Mathematics at a Glance
833 Properties of e
(a) e lies between 2 and 3 ie 2 lt e lt 3 n 1
1 1since for n 2n 2 minus
le ge
(b) The value of e correct to 10 places of decimals is 27182818284 (c) e is an irrational (incommensurable) number (d) e is the base of natural logarithm (Napier logarithm) ie ln x = loge x
834 Expansion of ex
For x isin R 2 3 r
x x x x xe 1 1 2 3 r
= + + + + + + infin or n
x
n 0
xen
infin
=
= sum
The above series is known as exponential series and ex is called exponential function Exponential function is also denoted by exp ie exp A = eA exp x = ex
835 Important Deduction from Exponential Series
(i) 2 3 r
x
r 0
x x x xe 1 1 2 3 r
infin
=
= + + + + +infin =sum (ii) 2 3 r r
x
r 0
x x x ( 1) xe 1 1 2 3 r
infinminus
=
minus= minus + minus + +infin =sum
(iii) x x 2 4 6 2r
r 0
e e x x x x1 2 2 4 6 (2r)
minus infin
=
+= + + + + =sum (iv)
x x 3 5 2r 1
r 0
e e x x x x2 1 3 5 (2r 1)
minus +infin
=
minus= + + + =
+sum
(v) r 0
1 1 1e 1 1 2 r
infin
=
= + + + +infin =sum (vi) r
1
r 0
1 1 1 ( 1)e 1 1 2 3 r
infinminus
=
minus= minus + minus + +infin =sum
(vii) 1
r 0
e e 1 1 1 11 2 2 4 6 (2r)
minus infin
=
+= + + + + +infin =sum (viii)
1
r 0
e e 1 1 1 12 1 3 5 (2r 1)
minus infin
=
minus= + + + +infin =
+sum
(ix) n 2 3n(n 1) n(n 1)(n 2)(1 x) 1 nx x x 2 3minus minus minus
+ = + + + + If given that x is so smalll as compared to 1
that x2 and higher powers of x can be neglected then it is called as binomial approximation of Binomial expression
84 loGarithmic series
For ndash1 lt x le 1 loge (1 + x) = ln (1 + x) = 2 3 4 r 1 r
r 1
x x x ( 1) xx 2 3 4 r
minusinfin
=
minusminus + minus + infin =sum
841 Important Deduction from Logarithmic Series
(i) 2 3 4
ex x xlog (1 x) x ( 1 x 1)2 3 4
minus = minus minus minus minus minus le lt
(ii) 2 4 6
ex x xlog (1 x)(1 x) 2 ( 1 x 1)2 4 6
+ minus = minus + + + minus lt lt
(iii) 3 5
e(1 x) x xlog 2 x ( 1 x 1)(1 x) 3 5
+ = + + + minus lt lt minus
Chapter 9trigonometriC ratios and identities
91 INTRODUCTION
The word lsquotrigonometryrsquo is derived from two Greek words (i) trigon (means a triangle) and (ii) metron (means a measure) Therefore trigonometry means science of measuring the sides of angles and study of the relations between side and angles of triangle
92 ANgle
Angle is defined as the measure of rotation undergone by a given revolving ray OX in a plane about its initial point O The original ray OX is called the initial side and the final position (OP) of the ray after rotation is called the terminal side of the angle angXOP The point of rotation (O) is called the vertex
921 Rules for Signs of Angles
bull If initial ray OA rotates to terminal ray OA then angle = q (rotation anti clockwise)
bull If initial ray OA rotates to terminal ray OB then angle = ndashq (rotation clockwise) where q is the measure of rotation
922 Measurement of AngleThe measurement of angle is done under the following three systems of measurement of angles
9221 Sexagesimal or english system
1 right angle = 900 (degrees) 10 = 60 (minutes)1 = 60 (seconds)
9222 Centesimal or french system (Grade)
1 right angle = 100g (grades) 1g = 100 (minutes)1 = 100 (seconds)
988 Mathematics at a Glance
RemarkThe minutes and seconds in the sexagesimal system are different them the respective minutes and seconds in the centesimal system Symbols in both there systems are also different
9223 Radian measure or circular measurement
One radian corresponds to the angle subtended by arc of length r (radius) at the centre of the circle with radius r Since the ratio is independent of the size of a circle it follows that a radian is a dimensionless quantity
Length of an arc of a circle θ = =arc lengthlAngle (in radians)
r radiusRelation between radian and degree πc = 180deg
In hand working tips
bull The unit radian is denoted by c (circular measure) and it is customary to omit this symbol c Thus
when an angle is denoted as 2π
it means that the angle is 2π
radians where p is the number with approximate value 314159
bull D G R
180 200= =
deg π where D G and R denotes degree grades and radians respectively
bull The angle between two consecutive digits in a clock is 30deg (p6 rad) The hour hand rotates through an angle of 30deg in one hour
bull The minute hand rotate through an angle of 6deg in one minute
93 POlygON AND ITs PROPeRTIes
A closed figure surrounded by n straight lines is called a polygon It is classified in two ways bull A closed figure surrounded by n straight lines bull If all sides of a polygon are equal then it is regular polygon bull Convex Polygon A polygon in which all the internal angles are smaller than 180deg bull Concave Polygon A polygon in which at least one internal angles is larger
than 180deg
Properties bull An angle is called reflexive angle if it is greater than or equal to 180deg or p radians bull Sum of all internal angles of a convex polygon = (n ndash 2) pc = (n ndash 2) 180deg
bull Each internal angle of regular polygon of n sides = (n 2)
nminus π
Nomenclature of Polygons
Name of Polygon
Number of Sides
Name of Polygon
Number of Sides
Name of Polygon
Number of Sides
1 Triangle 3 7 Nonagon 9 13 Penta-decagon 152 Quadrilateral 4 8 Decagon 10 14 Hexa-decagon 163 Pentagon 5 9 Hendecagon 11 15 Hepta-decagon 17
Trigonometric Ratios and Identities 1989
Name of Polygon
Number of Sides
Name of Polygon
Number of Sides
Name of Polygon
Number of Sides
4 Hexagon 6 10 Duodecagon 12 16 Octa-decagon 185 Heptagon 7 11 Tri-decagon 13 17 Nona-decagon 196 Octagon 8 12 Tetra-decagon 14 18 Ico-sagon 20
Circular Sector bull Perimeter of a circular sector of sectoral angle qc = r(2 + q)
bull Area of a circular sector of sectoral angle c 21q r q2
=
94 TRIgONOmeTRIC RATIOs
Consider an angle q = angXOA as shown in figure P be any point other than O on its terminal side OA and let PM be perpendicular from P on x-axis Let length OP = r OM = x and MP = y We take the length OP = r always positive while x and y can be positive or negative depending upon the position of the terminal side OA of angXOA
In the right-angled triangle OMP we have Base = OM = x perpendicular = PM = y and Hypotenuse = OP = r
We define the following trigonometric ratios which are also known as trigonometric functions
Perpendicular ysinHypotenuse r
θ = = Base xcos
Hypotenuse rθ = =
Perpendicular ytanBase x
θ = = Hypotaneuse rcosecPerpendicular y
θ = =
Hypotaneuse rsecBase x
θ = = Base xcot
Perpendicular yθ = =
990 Mathematics at a Glance
941 Signs of Trigonometric Ratios
Consider a unit circle with centre at origin of the coordinate axes Let P(a b) be any point on the circle with angle AOP = x radian ie length of arc AP = x as shown in the following figure
We defined cos x = a and sin x = b Since DOMP is a right triangle we have OM2 + MP2 = OP2 or
a2 + b2 = 1Thus for every point on the unit circle we have a2 + b2 = 1 or cos2x + sin2x = 1 Accordingly we can
judge the sign of a trigonometric function by comparing it with the sign of respective coordinates in that particular quadrant
Remark
The sign conventions can be kept in mind by the sentence ldquoAfter School To Collegerdquo where A stands for All S stands for Sine T stands for Tangent C stands for Cosine
942 Range of Trigonometric Ratios
Trigonometric Ratios and Identities 1991
943 Trigonometric Ratios of Allied Angles
9431 Trigonometric ratios of ndashq
Sin(ndashq) = ndashsinq
cos(ndashq) = cosq
tan(ndashq) = ndashtanq
cot(ndashq) = ndashcotq
sec(ndashq) = secq
cosec(ndashq) = ndashcosecq
9432 Trigonometric ratios of p ndash q
Sin(p ndash q) = sinq
cos(p ndashq) = ndashcosq
tan(p ndash q) = ndashtanq
cot(p ndash q) = ndashcotq
sec(pndash q) = ndashsecq
cosec(pndashq) =cosecq
Similarly
Sin(p + q)= ndashsinq cos(p+ q) = ndashcosq tan(p+ q) = tanq cot(p+ q) = cotq
sin cos2π minusθ = θ
cos sin
2π minusθ = θ
tan cot
2π minusθ = θ
cot tan
2π minusθ = θ
sin cos2π + θ = θ
cos sin
2π + θ = minus θ
tan cot
2π +θ =minus θ
cot tan
2π +θ =minus θ
sec(p+ q) = ndashsecq cosec(p+ q)= ndashcosec q
sec cosec2π minusθ = θ
cosec sec
2π minusθ = θ
cosec sec
2π + θ = θ
992 Mathematics at a Glance
Think yourself Try to evaluate the conversions for f(270 ndash q) f(270 + q) f(360 ndash q) f(360 + q) where f is a trigonometric function
Generalized Results The values of trigonometric functions of any angle can be represented in terms of
an angle in the first quadrant as follows Let A n2π
= plusmnθ where n isin Z 02π
le θ lt Then
(i) Sin p = 0 cosn p= (ndash1)n (ii)
(n 1)2
n2
( 1) cos if n is oddsin n2 ( 1) sin if n is even
minusπ minus θ+ θ = minus θ
(iii) ( )(n 1)
2
n2
1 sin if n is oddcos n2 ( 1) cos if n is even
+π minus θ+ θ = minus θ
(iv) tan if n is even
tan ncot if n is odd2
plusmn θπ plusmn θ = plusmn θ
(v) cot if n is even
cot ntan if n is odd2
plusmn θπ plusmn θ = plusmn θ (vi)
sec if n is evensec n
cosec if n is odd2plusmn θπ plusmn θ = plusmn θ
Think and fill up the blank blocks in the following table
Angles Functions
0
π6 4
π3π
2π 2
3π 5
6π
p 76π 4
3π
53π 11
6π
2p
sinq 0 12 1radic2 radic32 1cosq 1 radic32 1radic2 12 0tanq 0 1radic3 1 radic3 ND (infin)cotq infin radic3 1 1radic3 0secq 1 2radic3 radic2 2 ND (infin)tanq infin 2 radic2 2radic3 1
95 gRAPhs Of DIffeReNT TRIgONOmeTRIC RATIOs
951 y = sin x
x 0 p6 p4 p3 p2 2p3 3p4 5p6 psin x 0 12 1radic2 radic32 1 radic32 1radic2 12 0
Properties
P1 Domain of sinx is R and range is [ndash1 1] P2 sinx is periodic function with period 2p
P3 Principle domain 2 2π π minus
P4 It is an odd function P5 It is a continuous function and increases in first and fourth quadrants while decreases in second and
third quadrants
Trigonometric Ratios and Identities 1993
952 y = cos x
X 0 p6 p4 p3 p2 2p3 3p4 5p6 pcos x 1 radic32 1radic2 12 0 ndash12 ndash1radic2 ndashradic32 ndash1
Properties P1 The domain of cosx is R and the range is [ndash1 1] P2 Principle domain is [0 p] P3 cosx is periodic with period 2p P4 It is an even function so symmetric about the
y-axis
9521 y = tan x
X 0 p6 p4 p3 p2 2p3 3p4 5p6 ptan x 0 1radic3 1 radic3 infin ndashradic3 ndash1 ndash1radic3 0
Properties P1 The domain of tanx is R ndash (2n + 1) p2 and range
R or (ndashinfin infin) Principal domain is (ndashp2 p2) P2 It is periodic with period p P3 It is discontinuous x = R ndash (2n + 1) p2 and it is
strictly increasing function in its domain
953 y = cot x
X 0 p6 p4 p3 p2 2p3 3p4 5p6 pcot x infin radic3 1 1radic3 0 ndash1radic3 ndash1 ndashradic3 ndashinfin
Properties
P1 The domain of f(x) = cotx is domain isin R ~ np Range isin ℝ P2 It is periodic with period p and has x = np n isin z as its
asymptotes P3 Principal domain is (0 p) P4 It is discontinuous at x = np P5 It is strictly decreasing function in its domain
954 y = cosec x
x 0 p6 p4 p3 p2 2p3 3p4 5p6 pcosec x infin 2 radic2 2radic3 1 2radic3 radic2 2 infin
994 Mathematics at a Glance
Properties P1 The domain is R ~ np | n isin z P2 Range of cosecx is R ndash (ndash1 1)
P3 Principal domain is 02 2π π minus minus
P4 The cosecx is periodic with period 2p
955 y = sec x
X 0 p6 p4 p3 p2 2p3 3p4 5p6 psec x 1 2radic3 radic2 2 infin ndash2 ndashradic2 ndash2radic3 ndash1
Properties
P1 The domain of sec x is R (2n 1) n z2π minus + isin
and
range is R ndash (ndash1 1) P2 The sec x is periodic with period 2p P3 Principal domain is [0 p] ndash p2 P4 It is discontinuous at x = (2n + 1) p2
956 Trigonometric Identities
9561 Pythagorean identities
The following three trigonometric identities are directly derived from the pythagoras theorem
1 sin2x + cos2 x = 1 x isin ℝ rArr cos2 A = 1 ndash sin2 x or sin2 x = 1 ndash cos2 x or cos x 1 sin x
1 sin x cos x+
=minus
2 1+ tan2 x = sec2 x x ~ (2n 1) n2π isin + isin
rArr sec2x ndash tan2x = 1 or 1sec x tan x
sec x tan xminus =
+
3 cot2 x + 1 = cosec2 x x isin ℝ ~ np n isin ℤ rArr cosec2 x ndash cot2 x = 1 or1cosec x cot x
cosec x cot xminus =
+
NoteIt is possible to express trigonometrical ratios in terms of any one of them as
θ =+ θ2
1sin
1 cot
2
cotcos
1 cot
θθ =
+ θ 1
tancot
θ =θ
2cosec 1 cotθ = + θ
Remember sign of the dependent function will depend upon the location of angle in one or the other quadrant
957 Trigonometric Ratios of Compound AnglesAn angle made up of the sum of the algebraic sum of the two or more angles is called a lsquocompound anglersquo Some of the formulae on various trigonometric functions are given below
Trigonometric Ratios and Identities 1995
1 sin (A + B) = sin A cos B + cos A sin B 2 sin (A ndash B) = sin A cos B - cos A sin B 3 cos (A + B) = cos A cos B ndash sin A sin B 4 cos (A ndash B) = cos A cos B + sin A sin B
5 tan A tanBtan(A B)
1 tan A tanB+
+ =minus
6 tan A tanBtan(A B)
1 tan A tanBminus
minus =+
7 cot A cot B 1cot(A B)cot B cot A
minus+ =
+ 8
cot A cot B 1cot(A B)cot B cot A
+minus =
minus 9 sin(A + B) sin (A ndash B) = sin2 A ndash sin2 B = cos2 B ndash cos2 A
10 cos (A + B) cos (A ndash B) = cos2 A ndash sin2 B = cos2 B ndash sin2 A
958 Trigonometric Ratios of Multiples of Angles
1 22
2 tan Asin A 2sin A cos A1 tan A
= =+
2 2
2 tan Atan2A1 tan A
=minus
where A (2n 1)4π
ne +
3 1 cos A Atan
sin A 2minus =
where A ne (2n + 1)p 4
1 cos A Acotsin A 2+ =
where A ne (2np)
5 21 cos A Atan1 cos A 2minus = +
where A ne (2n + 1)p 6 21 cos A Acot1 cos A 2+ = minus
where A ne2np
7 A Asin cos 1 sin A2 2+ = plusmn + 8
A Asin cos 1 sin A2 2minus = plusmn minus
9 cos2A = cos2 A ndash sin2 A = 1 ndash 2 sin2 A = 2
22
1 tan A2cos A 11 tan Aminus
minus =+
10 1 + cos 2A = 2 cos2 A 1 ndash cos2A = sin2A or 21 cos2A cos A2
+= 21 cos2A sin A
2minus
=
11 sin 3A = 3 sin A ndash 4 sin3 A = 4 sin (60deg ndash A)sin Asin( 60deg + A) 12 cos 3A = 4cos3 A ndash 3cosA = 4 cos (60deg ndash A) cos Acos (60deg + A)
13 3
2
3tan A tan Atan3A1 3tan A
minus=
minus= tan (60deg ndash A)tan Atan (60deg + A)
14 sin A cos A 2 sin A 2 cos A2 4π π plusmn = plusmn =
959 Transformation Formulae
9591 Expressing the product of trigonometric ratio sum or difference
(i) 2 sin A cos B = sin (A + B) + sin (A ndash B) (ii) 2 cos A sin B = sin (A + B) ndash sin (A ndash B) (iii) 2 cos A cos B = cos (A + B) + cos (A ndash B) (iii) 2 sin A sin B = cos (A ndash B) ndash cos (A + B)
9592 Expressing the sum or difference of trigonometric ratios into product
1 C D C DsinC sinD 2sin cos
2 2+ + + =
2
C D C DsinC sinD 2cos sin2 2+ minus minus =
3 C D C DcosC cosD 2cos cos
2 2+ minus + =
4
C D C DcosC cosD 2sin sin2 2+ minus minus =
996 Mathematics at a Glance
5 sin(A B)tan A tanBcos AcosB
++ = where AB n
2π
ne π+
6 sin(A B)tan A tanBcos AcosB
minusminus = where A B ne np+ AB n
2π
ne π+
7 sin(A B)cot A cot Bsin AsinB
++ = where A B nen n isinz
8 sin(A B)cot A cot Bsin AsinB
minusminus = where A B ne npn isinz
9510 Conditional Identities
If A + B + C = p then
(i) sin2A + sin2b + sin2C = 4sinA sinB sinC (ii) A B Csin A sinB sinC 4cos cos cos22 2 2
+ + =
(iii) cos2A + cos2B + cos2C = ndash1 ndash 4 cosA cosB cosC (iv) A B Ccos A cosB cosC 1 4sin sin sin2 2 2
+ + = +
(v) tanA + tanB + tanC = tanA tanB tanC (vi) A B B C C Atan tan tan tan tan 12 2 2 2 2 2
+ + =
(vii) A B C A B Ccot cot cot cot cot cot2 2 2 2 2 2+ + = (viii) cotA cotB + cotB cotC + cotC cotA = 1
(ix) A B C2π
+ + = then tanA tanB + tanB tanC + tanC tanA = 1
96 sOme OTheR UsefUl ResUlTs
(i) sin a + sin (a + b) + sin (a + 2b) + hellip + hellip to n terms =
( )n 1 nsin sin2 2
sin2
minus β β α +
β
(ii) cos a + cos (a + b) + cos (a + 2b) + hellip + hellip to n term =
( )n 1 ncos sin2 2
sin2
minus β β α +
β
(iii) cos A cos 2A cos23 A hellip n
n 1n
sin2 Acos2 A2 sin A
minus = when n rarr infin minus
θ θ θ θinfin =
θ2 n 1
sincos cos cos 2 2 2
(iv) If A B C = π then bull cosA + cosB + cosC le 32 bull sinA2 sinB2 sinC2 le 18 equality holds good if A = B = C = 60deg bull tan2A2 + tan2B2 + tan2C2 ge 1
Trigonometric Ratios and Identities 1997
97 sOme OTheR ImPORTANT VAlUes
SNo Angle Value SNo Angle Value
1 sin 15deg 3 12 2minus 2 cos 15deg 3 1
2 2+
3 tan 15deg 2 3minus = cot 75deg 4 cot 15deg 2 3+ = tan 75deg
5 sin 2212
deg ( )1 2 22
minus 6 cos 2212
deg ( )1 2 22
+
7 tan 2212
deg 2 1minus 7 cot 2212
deg 2 1+
9 sin 18deg5 14minus
= cos 72deg 10 cos 18deg 10 2 54+
= sin 72deg
11 sin 36deg10 2 5
4minus = cos 54deg 12 cos 36deg 5 1
4+ = sin 54deg
13 sin 9deg 3 5 5 54
+ minus minus
or cos 81deg
14 cos 9deg3 5 5 5
4+ + minus
or sin 81deg 15 cos 36deg ndash cos 72deg 12 16 cos 36degcos 72deg 14
98 mAxImUm AND mINImUm VAlUes Of A COs q + b sIN q
Consider a point (a b) on the cartesian plane Let its distance from origin be r and the line joining the point and the origin make an angle a with the positive direction of x axis Then a = r cos a and b = r sin a
Squaring and adding 2 2r a b= + So a cos q + b sin q = r [cos a cos q + sin a sin q] = r cos (a ndash q)
but ndash 1 le cos (a ndash q)le 1rArr ndash r le a cos q + b sin q le r
So maximum value is 2 2a b+ and minimum value is 2 2a bminus +
99 TIPs AND TRICs
1 if x = secq + tanq Then 1x = secq ndash tanq 2 if x = cosecq + cotq Then 1x = cosecq ndash cotq
3 cos A cos2Acos22A n
n 1n
sin2 Acos2 A2 sin A
minus = if A ne n p = 1 if A = 2n p = (ndash1)n if A = (2n + 1) p
4 sinA2 plusmn cosA2 = radic2 = sin[p 4 plusmn A] = radic2cos [A p4]
5 cos a+ cos b + cos g + cos (a+b+g) = ( ) ( ) ( )4cos cos cos
2 2 2α+β β+ γ γ +α
6 sin a+ sin b + sin g ndash sin (a+b+g) = ( ) ( ) ( )4sin sin sin
2 2 2α+β β+ γ γ +α
Chapter 10trigonometriC
equation
101 IntroductIon
The equations involving trigonometric functions of one or more unknown variables are known as lsquotrigonometric equationsrsquo For example cosq = 0 cos2q - 4 cosq = 1 sin2q + sinq = 2 cos2q - 4sinq = 1 etc
102 SoLutIon oF trIGonoMEtrIc EQuAtIon
A solution of a trigonometric equation is the value of the unknown variable (angle) that satisfies the
equation For example 1sin2
θ = rArr 4π
θ = or 3 9 11
4 4 4 4π π π π
θ =
Thus the trigonometric equation may have infinite number of solutions
1021 Classification of Solutions of Trigonometric Equations
(i) Particular solution (ii) Principal solution (iii) General solution
103 PArtIcuLAr SoLutIon
Any specific solution that satisfies a given trigonometric equation is called a particular solution
For example sin x = has a particular solution π
=x3
104 PrIncIPAL SoLutIon
The solutions of a trigonometric equation having least magnitude that is belonging to principal domain of
trigonometric function are called principal solution For example sin 1x2
= has principal solution 6π
Parallely cos 1x2
= minus has principal solutions 23π
The following figures represent principal domains of trigonometric functions
Trigonometric Equation 1099
Principal Domain 2 2π π minus
Principal Domain [0 π]
Principal Domain 2 2π π minus
Principal Domain (0 π)
Principal Domain [0 ]~2π π
10100 Mathematics at a Glance
Principal Domain ~ 02 2π π minus
105 GEnErAL SoLutIon
Since trigonometric functions are periodic a solution can be generalized by means of periodicity of the trigonometric functions An expression which is a function of integer n and a particular solution a representing all possible particular solutions of a trigonometric equation is called its lsquogeneral solutionrsquo We use the following results for solving the trigonometric equations
Result 1 sinq = 0 hArr q = n p n isin ℤ
General Solutions for Some Standard Equations
Sin q = 0 rArr q = n π sin 1 (4n 1)2π
θ = rArrθ = + π
θ = minus rArrθ = minussin 1 (4n 1)2
Result 2 cos q = 0 hArr (2n 1) n2π
θ = + isin
General Solutions for Some Standard Equations
cos 0 (2n 1)2π
θ = rArrθ = + cos q = 1 rArr q = 2nπ cos q = ndash1 rArr q =(2n + 1)π
Result 3 tan q = 0 hArr q = n p n isin ℤ
Trigonometric Equation 10101
General Solutions for Some Standard Equations
tan q = 0
rArr q = nπ tan 1 (4n 1)4π
θ = rArrθ = + tan 1 (4n 1)4π
θ = minus rArrθ = minus
Result 4 sin q = sin a hArr q = n p +(-1)n a where n isin ℤ and a is a particular solution preferably taken least non-negative or
that having least magnitude
Result 5 cos q = cos a hArr q = 2n p plusmn a n isin ℤ
ndash +ndash
Result 6 tan q = tan a hArr q = n p + a n isin ℤ
Result 7 sin2 q = sin2 a cos2 q = cos2 a tan2 q = tan2 a hArr q = n p plusmn a n isinℤ
10102 Mathematics at a Glance
RemarkIn formulae if we take any of a the set of all possible solutions represented by general solution remains unique
Theorem 1 sin q = k where k isin [ndash1 1] has general solution q = nπ +(ndash1)na
Where 2 2π π αisin minus
st sin a = k
Theorem 2 cos q = k where k isin [ndash1 1] has general solution q = 2nπ plusmn α where α isin [0 π] st cos α = k
+ndash
Theorem 3 tan q = k where k isin ℝ has general solution q = nπ + α where 2 2π π αisin minus
st tan α = k
Theorem 4 sin2q = k where k isin [0 1] has general solution q = nπ plusmn α where 02π αisin
st sin2α = k
Theorem 5 cos2 q = k where k isin [0 1] has general solution q = nπ plusmn α where 02π αisin
st
cos2 α = k
ndash +
Trigonometric Equation 10103
Theorem 6 tan2 q = k where k isin [0 infin) has general solution q = nπ plusmn α where 02π αisin
st
tan2 α = k
ndash+ndash
106 SuMMAry oF thE AbovE rESuLtS
1 sin q = 0 hArr q = np n isin ℤ
2 cos q = 0 hArr (2n 1) n2π
θ = + isin
3 tan q = 0 hArr q = np n isin ℤ 4 sin q = sin a hArr q = n p +(-1)n a n isin ℤ 5 cos q = cos a hArr q = 2n p plusmn a n isin ℤ 6 tan q = tan a hArr q = n p + a n isin ℤ 7 sin2 q = sin2 a cos2 q = cos2 a tan2q = tan2a n isin ℤ
8 sin q = 1 hArr (4n 1) n2π
θ = + isin
9 sin q = ndash1 hArr q =(4n + 3)2π
n isin ℤ
10 cos q = 1 hArr q = 2n p 11 cos q = -1 hArr q =(2n + 1) p n isin ℤ 12 sin q = sin a and cos q = cos a hArr q = 2n p + a n isin ℤ
Notes
1 The general solution should be given unless the solution is required in a specified interval or range
2 a is a particular solution preferably taken least positive or that having least magnitude
10104 Mathematics at a Glance
107 tyPE oF trIGonoMEtrIc EQuAtIonS
Type 1 Trigonometric equations which can be solved by use of factorization eg (2 cos x ndash sin x)(1 ndash sin x) = cos2 x rArr (2 cos x ndash sin x)(1 + sin x) = 1 ndash sin2 x
rArr (1 + sin x)(2 cos x ndash 1) = 0 rArr sin x = ndash1 or 1cos x2
=
rArr x (4n 3)2π
= + or 2n n3π
πplusmn isin are the general solutions
Type 2 Trigonometric equations which can be solved by reducing them to quadratic equations eg 2 sin2 x + 2 sin x = 5 cos2 x rArr 2 sin2 x + 2 sin x = 5(1 ndash sin2 x)
rArr 7 sin2 x + 2 sin x ndash 5 = 0 rArr sin x = ndash1 or 5sin x7
=
rArr x (4n 3) n2π
= + isin or x = nπ +(ndash1)n α n isin ℤ
And 5sin7
α = are the required general solutions
Type 3 Trigonometric equation which can be solved by transforming a sum or difference of trigonomet-ric ratios into their product
eg cosx ndash sin3x = cos2x rArr cosx ndash cos2x = sin3x
rArr 3x x 3x 3x2sin sin sin3x 2sin cos2 2 2 2
= =
rArr 3x x x2sin sin cos3 02 2 2
minus = rArr
3xsin 02=
rArr 3x n n2= π isin rArr
2nx n3π
= isin helliphellip(i)
or x 3xsin cos 02 2minus = rArr
x 3xcos cos 02 2 2π minus minus =
rArr x2sin sin x 0
4 2 4π π + minus =
rArr x m4π
= π+ hellip(ii)
Combining equation (i) and (ii) general solutions are given by 2nx 2n n n
3 2 4π π π
= πminus π+ isin
Type 4 Trigonometric equations which can be solved by transforming a product of trigonometric ratios into their sum or difference For example sin x cos 5x = sin4x cos2 x
rArr sin6x + sin(ndash4x) = sin6x + sin2x rArr sin2x + sin4x = 0 rArr 2sin(3x) cos x = 0
rArr nx3π
= or x (2n 1) n Z2π
= + isin
Type 5 Trigonometric equations of the form a sinx + b cosx = c where a b c isin ℝ can be solved by
dividing both sides of the equation by 2 2a b+
To solve the equation a cosq + b sinq = c put a = r cos f b = r sin f such that 2 2r a b= + 1 btana
minusφ =
ie take 2 2π π φisin minus
such that
btana
φ =
Trigonometric Equation 10105
Substituting these values in the equation we have r cos f cos q + r sin f sin q = c
rArr ccos( )r
θminusφ = rArr 2 2
ccos( )a b
θminusφ =+
Notes
1 If gt +2 2c a b then the equation a cos q + b sin q = c has no solution
2 If 2 2c a ble + then put 2 2
|c |
a b+ = cos a so that cos(q ndash f) = cos a
rArr (q - f) = 2n p plusmn a rArr q = 2n p plusmn a + f where n isin ℤ eg sin x cos x 2+ =
rArr a = b = 1 Let a = r cosq b = r sinq
rArr 2 2r a b 2= + = 1 2 cos 1 2 sinθ θ= =
rArr tanq = 1 rArr q = tanndash11 rArr 4πθ =
2 cos x 24π minus =
rArr cos x 1
4π minus =
rArr x 2n n
4ππ= + isin
3 Trigonometric equation of the form a sinx + cosx = c can also be solved by changing sinx and cosx into their corresponding tangent of half the angle and solving for tan x2 ie we substitute
2
2
x1 tan
2cos xx
1 tan2
minus=
+ and
2
x2tan
2sin xx
1 tan2
=+
Type 6 Equation of the form R(sin x plusmn cos x sin x cos x) = 0 Where R is a rational function of the arguments in the brackets Put sin x + cos x = t (i) and use the following identity (sin x + cos x)2 = sin2 x + cos2 x + 2 sin x cos x = 1 + 2 sin x cos x
rArr 2t 1sin x cos x2minus
= (ii)
Taking equation (i) and (ii) into account we can reduce given equation into R(t(t2 ndash 1)2) = 0 Similarly by the substitution(sin x - cos x) = t we can reduce the equation of the form R(sin x - cos x sin x cos x) = 0 to an equation R(t(1 ndash t2)2) = 0
Type 7 Trigonometric equations which can be solved by the use boundedness of the trigonometric
ratios sinx and cosx eg 5xsin cos x 24+ = Now the above equation is true if
5xsin 14= and cos x = 1
rArr 5x 2n n4 2
π= π+ isin and x = 2mp m isin z
rArr (8n 2)x n
5+ π
= isin helliphelliphelliphellip(iii)
and x= 2mp m isin ℤ helliphelliphellip(iv)
Now to find general solution of equation (i) (8n 2) 2m
5+ π
= π
10106 Mathematics at a Glance
rArr 8n + 2 = 10 m rArr 5m 1n
4minus
=
If m = 1 then n = 1 m = 5 then n = 6 hellip helliphellip helliphellip hellip helliphellip helliphellip If m = 4p ndash 3 p isin ℤ then n = 5p ndash 4 p isin ℤ General solution of a given equation can be obtained by
(8n 2)x 2m m n ~ 2m m 4p 3p
5+ = π isin cup π isin π = minus isin
or (8n 2) (8n 2)x 2m m n ~ n 5p 4p
5 5+ + = π isin cup π isin π = minus isin
Type 8 A trigonometric equation of the form R(sin kx cos nx tan mx cot l(x) = 0 l m n then use the following formulae
=+ 2
2 tan x 2sin x
1 tan x 2 2
2
1 tan x 2cos x
1 tan x 2minus
=+
2
2 tan x 2tan x
1 tan x 2=
minus
21 tan x 2cot x
2tan x 2minus
=
108 hoMoGEnEouS EQuAtIon In SInx And coSx
The equation of the form a0 sinn x + a1 sinn-1 x cos x + a2 sinn-2 x cos2 x + + an cosn x = 0 where a0 a1 an are real numbers and the sum of the exponents in sin x and cos x in each term is equal to n are said to be homogeneous with respect to sin x and cos x For cos x ne 0 the above equation can be written as a0 tann x + a1 tann-1 x + + an = 0
109 SoLvInG SIMuLtAnEouS EQuAtIonS
Here we discuss problems related to the solution of two equations satisfied simultaneously We may divide the problems into two categories as shown by the following diagram
When number of equations is more than or equal to number of variables
Trigonometric Equation 10107
∎ Single variable problems with intermediate values
Step 1 Find the values of variable x satisfying both equations
Step 2 Find common period of function used in both the equation say T and obtain x = α isin(0 T] sat-isfying both the equations
Step 3 Generalizing the value of α we get x = nT + α
∎ Single variable problem with extreme values
Step 1 When LHS and RHS of a equation have their ranges say R1 and R2 in common domain and R1 cap R2 = f then the equations have no solution
Step 2 If R1 cap R2 have finitely many elements and the number of elements are few then individual cases can be analyzed and solved
Step 3 Generalizing the value of α we get x = nT + α
1091 More Than One Variable Problems
bull Substitute one variable (say y) in terms of other variable x ie eliminate y and solve as the trigonometric equations in one variables
bull Extract the linearalgebraic simultaneous equations from the given trigonometric equations and solve as simultaneous algebric equations
bull Many times you may need to make appropriate substitutions bull When number of variables is more than number of equations To solve an equation involving more than one variable definite solutions can be obtained if extreme
values (range) of the functions are used
10911 Some important results
1 While solving a trigonometric equation squaring the equations at any step should be avoided as far as possible If squaring is necessary check the solution for extraneous values
2 Never cancel terms containing unknown terms on the two sides which are in product It may cause loss of the genuine solution
3 The answer should not contain any such values of angles which make any of the terms undefined or infinite
4 Domain should not change If it changes necessary corrections must be made 5 Check that denominator is not zero at any stage while solving equations
1010 trAnScEdEntAL EQuAtIonS
To solve the equation when the terms on the two side (LHS and RHS) of the equation are of different nature eg trigonometric and algebraic we use inequality method Which is used to verify whether the given equation has any real solution or not In this method we follow the steps given below
10108 Mathematics at a Glance
Step I If given equation is f(x) = g(x) then let y = f(x) and y = g(x) ie break the equation in two parts
Step II Find the extreme values of both sides of equation giving range of values of y for both side If there is any value of y satisfying both the inequalities then there will be a real solution otherwise there will be no real solution
1011 GrAPhIcAL SoLutIonS oF EQuAtIonS
For solution of equation f(x) ndash g(x) = 0
Let a is root rArr α = α =f( ) g( ) k(say)
rArr y f(x) and y g(x)= =
have same output for input x = α
rArr ( k) satisfying both the curves y f(x) and y g(x)α = =
Solutions of equation f(x) ndash g(x) = 0 are abscissa (x-co-ordinate) of the point of intersection of the graph y = f(x) and y = g(x)
Algorithm To solve the equation f(x) ndash g(x) = 0 eg 10sinx ndash x = 0
Step 1 Write the equation f(x) = g(x) ie sinx = x10
Step 2 Draw the graph of y = f(x) and y = g(x) on same x ndash y plane
Let f(x) = sinx and g(x) = x
10
also we know that -1 le sinx le 1
-1 le x
10 le 1
rArr -10 le x le 10
Thus sketching both the curves when x isin [minus10 10]
Step 3 Count the number of the points of intersection If graphs of y = f(x) and y = g(x) cuts at one two three no points then number of solutions are one two three zero respectively
From the given graph we can conclude that f(x) = sinx and g(x) = x
10 intersect at 7 points So number
of solutions are 7
1012 SoLvInG InEQuALItIES
To solve trigonometric inequalities including trigonometric functions it is good to practice periodicity and monotonicity of functions Thus first solve the inequality for one period and then get the set of all solutions by adding numbers of the form 2np n isin ℤ to each of the solutions obtained on that interval
-1
-3π -2π2π
3π-π
y
f(x) = sinxg(x) = x10
O π
(-frac1234)
frac12 10
(101)(3 3 10)ππ(2 2 10)ππ
( 10)ππ
Trigonometric Equation 10109
For example Find the solution set of inequality sinx gt 12
Solution When sinx = 12 the two values of x between 0 and 2p are p6 and 5p6 from the grpah of y = sinx it is obvious that between 0 and 2p
sinx gt 12
for p6 lt x lt 5p6
Hence sinx gt 12 rArr 52n x 2n
6 6π π
π+ lt lt π+
The required solution set is n Z
52n 2n6 6isin
π π π+ π+
10121 Review of Some Important Trigonometric Values
1 3 1sin15
2 2minus
deg = 2 3 1cos 15
2 2+
deg =
3 tan 15deg = 2 - radic3 = cot 75deg 4 cot 15deg = 2 + radic3 = tan 75deg
5 ( )1 1sin 22 2 22 2
= minus
6 ( )1 1cos22 22 2 22 2
deg = = +
7 1tan 22 2 12
= minus
8 1cot 22 2 12
= +
9 5 1sin18 cos724minus
deg = = deg 10 10 2 5cos18 sin72
4+
deg = = deg
11 10 2 5sin36 cos54
4minus
deg = = deg 12 5 1cos36 sin544+
deg = = deg
13 3 5 5 5sin9 cos81
4+ minus minus
deg = = deg 14 3 5 5 5cos9 sin81
4+ + minus
deg = = deg
15 cos 36deg - cos 72deg = 12 16 cos 36deg cos 72deg = 14
Chapter 11properties of
triangles
111 IntroductIon
Here we shall discuss the various properties of tringels
1111 Sine Formula
In any triangle ABC the sides are proportional to the sines of the opposite angles
ie a b c 2Rsin A sinB sinC
= = = R = circumradius of DABC
1112 Cosine Formula
In any triangle ABC to find the cosine of an angle in terms of the sides
2 2 2b c acos A
2bc+ minus
= 2 2 2a c bcosB
2ac+ minus
= 2 2 2a b ccosC
2ab+ minus
=
1113 Projection FormulaIn any triangle ABC a = c cos B + b cos C b = a cos C + c cos A c = a cos B + b cos A the sine cosine and Tangent of the half-anlges in terms of the sides
(i) (s b)(s c)Asin2 bc
minus minus= (s a)(s c)Bsin
2 acminus minus
= (s a)(s b)sin
2 abminus minus
(ii) s(s a)Acos
2 bcminus
= s(s b)Bcos
2 acminus
= s(s c)Ccos
2 abminus
=
(iii) (s b)(s c) s(s a)A sin A 2tan
2 cos A 2 bc bcminus minus minus
= = divide (s b)(s c)Atan
2 s(s a)minus minus
=minus
(s a)(s c)Btan
2 s(s b)minus minus
=minus
and (s a)(s b)Ctan
2 s(s c)minus minus
=minus
Properties of Triangles 11111
11131 sin A in terms of the sides of the triangle
(s b)(s c) s(s a)A Asin A 2sin cos 22 2 bc bc
minus minus minus= = times
rArr 2 2sin A s(s a)(s b)(s c)bc bc
∆= minus minus minus = Similarly 2 2sinB s(s a)(s b)(s c)
ca ca∆
= minus minus minus = minus
2 2sinC s(s a)(s b)(s c)ab ab
∆= minus minus minus = D = area of D ABC
112 nAPIErrsquoS AnALoGY
In any triangle ABC (A B) a b Ctan cot2 a b 2minus minus
=+
(B C) b c Atan cot2 b c 2minus minus
=+
(C A) c a Btan cot2 c a 2minus minus
=+
1121 Solution of Triangle
Case 1 When three sides of a triangle are givenIn this case the following formulae are generally used
(i) minus minus
=(s b)(s c)Asin
2 bc (ii)
s(s a)Acos2 bc
minus=
(iii) (s b) (s c)Atan
2 s(s a)minus minus
=minus
(iv) 2 2 2b c acos A
2bc+ minus
= etc
Case 2 When two sides and the included angle of the triangle are given Let b c and A be given then lsquoarsquo can be found from the formula a2 = b2 + c2 ndash 2bc cos A
Now angle B can be found from the formulae 2 2 2c a bcosB
2ac+ minus
= or bsin AsinBa
= and C from
C = 180deg ndash A ndash B
Another way to solve such triangle is first to find B C2minus by using the formulae
B C b c Atan cot2 b c 2minus minus = +
and therefore by addition and subtraction B and C and the third side lsquoarsquo by
cosine formula a2 = b2 + c2 ndash 2bc cos A or bsin A
asinB
= or a = b cos C + c cos B
Case 3 When two angles and the included side of a triangle are givenLet angle B C and side a be given The angle A can be found fromA = 180deg - B - C and the sides b and c from sine rule
a b csin A sinB sinC
= = ie a sinBbsin A
= and a sinCcsin A
=
Case 4 Ambiguous CaseWhen two sides (say) a and b and the angle (say) A opposite to one side a are given There are following three possibilities
11112 Mathematics at a Glance
(i) Either there is no such triangle (ii) One triangle (iii) Two triangles which have the same given elements
We have b asinB sin A
= rArr bsin AsinB
a= hellip (1)
Also c2 ndash 2 (b cos A) c + b2 ndash a2 = 0 (2)
gives 2 2 2c bcos A a b sin A= plusmn minus (3)Now the following cases may raise
(a) When a lt b sin A rArr sin B gt 1 form equation (1) or from equation (3) c is imaginary which is impossible Hence no triangle is possible
(b) When b sin A = a rArr from equation (1) sin B = 1 rArr B = 90deg and from equation (3) c = b cos A This value of c is admissible only when b cos A is positive ie when the angle A is acute In such a
case a lt b (b sin A = a) or A lt B Hence only one definite triangle is possible
Note
In this case a = b is not possible since A = B = 90deg which is not possible Since no triangle can have two right angles
(c) When b sin A lt a and sin B lt 1 from (4) In this case there are three possibilities (i) If a = b then A = B and from equation (3) we get c = 2b cos A or 0 Hence in this case we get
only one triangle (since in this case it is must that A and B are acute angles) (ii) If a lt b then A lt B Therefore A must be an acute angle b cos A gt 0 Further a2 lt b2 rArr a2 lt b2 (cos2 A + sin2 A)
rArr 2 2 2a b sin A bcos Aminus lt From equation (3) it is clear that both values of c are positive so we get two triangles such that
and 2 22c bcos A a b sin A= minus minus
It is also clear from equation (1) that there are two values of B which are supplementary
(iii) If a gt b then A gt B also a2 - b2 sin2 A gt b2 cos2 A or 2 2 2a b sin A bcos Aminus gt
Hence one value of c is positive and other is negative for any value of angle A Therefore we get only one solution Since for given values of a b and A there is a doubt or ambiguity in the determination of the triangle Hence this case is called ambiguous case of the solution of triangles
113 GEomEtrIc dIScuSSIon
Let a b and the angle A be given Draw a line AX At A construct angle angXAY = A Cut a segment AC = b from AY Now describe a circular arc with its centre C and radius a Also draw CD perpendicular to AX
CD = b sin A The following cases may arise
Properties of Triangles 11113
(a) If a lt b sin A ie a lt CD then the circle will not meet AX and hence there is no triangle satisfying the given condition
(b) If a = b sin A the circle will touch AX at D (or B) and only one right angled triangle is possible In this case B = 90deg and A lt 90deg
(c) If a = b (angA ne 90deg) then the circle will cut AX at B and passes through A Hence here we get only one solution of given data (as shown in the figure)
(d) If a gt b sin A then the circle will cut AX at two distinct points (other than A) Let the point be B1 and B2Sub-case 1 If b sin A lt a lt b then both B1 and B2 are on the same side of A as shown in the figure and we get two distinct triangles ACB1 and ACB2
Sub-case 2 If a gt b then the two points B1 and B2 are on the opposite sides of A and only one of the triangle ACB1 or ACB2 will satisfy the given data If A is an acute angle then DCAB2 is the required triangle and if A is obtuse angle then DAB1C is the required triangle
114 ArEA of trIAnGLE ABc
If D represents the area of a triangle ABC then D = 12 (BCAD) 1 AD 1a(csinB) as sinB acsinB2 c 2
= = =
Also ADsinCb
= rArr AD = b sin C
1 a bsinC2
∆ = Similarly 1 bcsin A2
∆ =
1 1 1absinC bcsin A ca sinB2 2 2
∆ = = =
(i) Area of a triangle in terms of sides (Herorsquos formula)
1 1 A Abcsin A bc2sin cos2 2 2 2
∆ = = = (s b)(s c) s(s a)
bcbc bc
minus minus minus
rArr s(s a)(s b)(s c)∆ = minus minus minus
11114 Mathematics at a Glance
(ii) Area of triangle in terms of one side and sine of three angles
1 1bcsin A (k sinB)(k sinC)sin A2 2
∆ = = = 21 k sin AsinBsinC2
= 21 a sin AsinBsinC
2 sin A
= 2a sinBsinC
2 sin A
Thus 2a sinBsinC
2 sin A∆ = =
2 2b sin AsinC c sin AsinB2 sinB 2 sinC
=
115 mndashn thEorEm
In any triangle ABC if D is any point on the base BC such that BD DC m n angBAD = α angCAD = b angCDA = q then (m + n) cot q = m cot α ndash n cot b = n cot B ndash m cot C
1151 Some Definitions
11511 Circumcircle
The circle which passes through the angular points of a triangle is called its circumscribing circle or more briefly circumcircle The centre of this circle is called circumcentre Generally it is denoted by O and its radius always denoted by R Another property of circum centre is that it is the point of concurrency of perpendicular bisectors of sides of a triangle
11512 Radius of circum circle lsquoRrsquo of any triangle
In DABC we have 2sin sin sin
a b c RA B C= = =
The circumradius may be expressed in terms of sides of the trianglea abc abcR
2sin A 2bcsin A 4= = =
∆
1 sin2
bc A ∆ = Thus abcR
4=
∆
11513 Incircle
The circle which can be inscribed within the triangle so as to touch each of the sides is called its in-scribed circle or more briefly its incircle The centre of this circle is called incentre It is denoted by I and its radius always denoted by r In-centre is the point of concurrency of internal angles bisectors of the triangle
Properties of Triangles 11115
Radius r of the incircle of triangle ABCSince D = Area DIBC + ar(D ICA) + ar(D IAB)rArr D = (12) ar + (12) br + (12) cr = 12 (a + b + c)r
rArr D = sr rArr r = Ds a b cs2
+ += = semi-perimeter
The radius of incircle in terms of sides and tangent of the half angleAr (s a)tan2
= minus = B(s b)tan2
minus = C(s c)tan2
minus
The radius of incircle in terms of one side and the functions of the half anglesa sin(B 2)sin(A 2)r
cos(A 2)= = bsin(B 2)sin(C 2)
cos(B 2) = Csin(A 2)sin(B 2)
cos(C 2)since a = 2R sinA = 4R sinA2 cosA2 r = 4R sinA2 sin B2 sin C2
11514 Escribed circles
The circle which touches the sides BC and two sides AB and AC (produced) of triangle ABC is called escribed circle opposite the angle A The centre of escribed circle is called ex-centre and is denoted by I1 or IA and radius by r1 or rA
Radii of escribed circles of a triangle 1r s a∆
=minus
2r s b∆
=minus
3r s c∆
=minus
Radii of the Escribed circles in terms of sides and the tangents of half angler1
= s tan A2 r2 = s tan B2 r3 = s tan C2
Radii of the escribed circles in terms of one side and function of half angles
1a cos(B 2)cos(C 2)r
cos(A 2)= 2
bcos(C 2)cos(A 2)rcos(B 2)
= 3ccos(A 2)cos(B 2)r
cos(C 2)=
Now Since a = 2R sin A = 4R sin A2 cosA2rArr r1 = 4R sin A2 cosB2 cosC2 r2 = 4R cosA2 sinB2 cos C2 and r3 = 4RcosA2 cosB2 sinC2
116 orthocEntrE And PEdAL trIAnGLE
Let ABC be any triangle and let D E F be the feet of the perpendiculars from the angular points on the opposite sides of the triangle ABC DEF is known as Pedal Triangle of ABC
The three perpendiculars AD BE and CF always meet in a single point H which is called the ortho-centre of triangle
11116 Mathematics at a Glance
1161 Sides and Angles of the Pedal Triangle
angFDE = 180deg ndash 2A angDEF = 180deg ndash 2B angDFE = 180deg ndash 2CFD = b cos B DE = c cos C FE = a cos Aor FD = R sin 2B DE = R sin 2C FE = R sin 2A
11611 Perimeter of pedal triangle
R(sin 2A + sin 2B + sin 2C) = 4R sinA sinB sinC
NoteIf the angle ACB of the given triangle is obtuse the expressions 180deg ndash 2C and c cosC are both negative and the values we have obtained require some modification In this case the angles are 2A 2B 2C ndash 180deg and the sides are a cosA b cos B ndash c cos C
Distance of the orthocentre from the angular points of the triangleAH = 2R cos ABH = 2R cos B CH = 2R cosC
11612 Distances of the orthocentre from the sides of the triangle
HD = 2R cosB cosC HE = 2R cosA cosC HF = 2R cosA cosB
Cor
sin AAH 2R cos A cosBcosCHD 2R cosBcosC sin A cos A
= = =
sin(B C)tanB tanCcosBcosC
tan A tan A
++
=
Area and Circum-radius of the Pedal Triangle (a) Area of triangle = 12 (product of two sides)times (sin of included angle) = 12 (Rsin 2B) (R Sin 2C)
sin(180deg - 2A) 21 R sin2Asin2Bsin2C2=
(b) Circumradius = EF R sin2A R2sinFDE 2sin(180 2A) 2
= =degminus
(c) The in-radius of the Pedal Triangle Area of ( DEF)
DEFSemi Perimeter of DEF
∆=
∆
= 21 R sin2Asin2Bsin2C
2 2R cos A cosB cosC2R sin AsinBsinC
=
117 In-cEntrE of PEdAL trIAnGLE
Since HD HE and HF bisect the angles FDE DEF and EFD respectively So that H is the in-centre of the triangle DEF Thus the orthocentre of a triangle is the in-centre of the pedal triangle
Properties of Triangles 11117
118 cIrcumcIrcLE of PEdAL trIAnGLE (nInE-PoInt cIrcLE)
The circumcircle of pedal triangle for any DABC is called a nine-point circle
1181 Properties of Nine-point Circle
1 If passes through nine points of triangle L M N (feet of altitudes) D E F (mid points of sides) and midpoints of HA HB HC where H is orthocentre of triangle ABC
2 Its centre is called nine-points centre (N) It is the circumcentre of a pedal triangle
3 Its radius is 91 2
R R=
4 O (orthocentre) N G C (circumcentre) are collinear bull N divides OC in ratio 11 bull G divides OC in ratio 21 5 If circumcentre of triangle be origin and centroid has coordinate (x y) then coordinate of
orthocentre = (3x 3y) coordinate of nine point centre 3 32 2x y =
119 thE Ex-cEntrAL trIAnGLE
Let ABC be a triangle and I be the centre of incircle Let IA IB IC be the centres of the escribed circles which are opposite to A B and C respectively then IA IB IC is called the ex-central triangle of D ABC By geometry IC bisects the angle ACB and IBC bisects the angle ACM
ang ICIB = angACI + angACIB = 12
ang ACB + 12
ang ACM = 12
ang (180deg) = 90deg
Similarly ang ICIA = 90deg
11118 Mathematics at a Glance
Hence IA IB is a straight line perpendicular to IC Similarly AI is perpendicular to the straight line IBIC and BI is perpendicular to the straight line IA IC
Also since IA and IAA both bisect the angle BAC hence A I and IA are collinear Similarly BIIB and CIIC are straight lines
Hence IA IB IC is a triangle thus the triangle ABC is the pedal triangle of its ex-central triangle IA IB IC The angles IBIA and ICIA are right angles hence the points B I C IA are concyclic Similarly C I A IB and the points A I B IC are concyclic
The lines AIA BIB CIC meet at the incentre I which is therefore the orthocentre of the ex-central triangle IA IB IC
Remarks
1 Each of the four points I IA IB IC is the orthocente of the triangle formed by joining the other three points
2 The circumcentre the centroid the centre of the nine point circle and the orthocentre all lie on a straight line
1110 cEntroId And mEdIAnS of AnY trIAnGLE
In triangle ABC the midpoint of sides BC CA and AB are D E and F respectively The lines AD BE and CF are called medians of the triangle ABC the point of concurrency of three medians is called centroid Generally it is represented by G
By geometry 2 2AG AD BG BE3 3
= = and 2CG CF3
=
1111 LEnGth of mEdIAnS
= + minus2 2 21AD 2b 2c a2
2 2 21BE 2c 2a b2
= + minus and 2 2 21CF 2a 2b c2
= + minus
The angles that the median makes with sides
Let angBAD = b and angCAD = g we have sin DC asinC AD 2x
γ= = (Let AD = x)
2 2 2
a sinC a sinCsin2x 2b 2c a
γ = =+ minus
2 2 2
a sinBsin2b 2c a
β =+ minus
Again sin AC bsinC AD x
θ= =
2 2 2
bsinC 2bsinCsinx 2b 2c a
θ = =+ minus
11111 The Centroid Lies on the Line Joining the Circumcentre to the Orthocentre
Let O and H represent the circum-centre and orthocenter respectively OM is perpendicular to BC Let AM meets HO at G The two triangles AHG and GMO are equiangular
AH = 2R cosA and in DOMC OM = RcosA
rArr AH 2R cos A 2OM R cos A
= =
Properties of Triangles 11119
Hence by similar triangles AG HG AH 2GM GO OM
= = =
rArr G divides AM in the ratio 2 1 Clearly G is the centroid of DABC and G divides HA in the ratio 2 1 Thus centroid lies on the
line joining the orthocentre to the circum-centre and divides it in the ratio 2 1
The distance of the orthocentre from the circum-centre
OH R 1 8cos A cosBcosC= minus
The distance between the incentre and circumcentre
OI R 1 8sinB 2sinC 2sin A 2= minus
The distance of an ex-centre from the circum-centre
OI1 = A B CR 1 8sin cos cos2 2 2
+ OI2 = A B CR 1 8cos sin cos2 2 2
+
OI2 = R 1 8cos(A 2)cos(B 2)sin(C 2)+
111111 The length of angle bisector and the angle that the bisector makes with the sides
Let AD be the bisector of angle A and x and y be the portions of base BC From geometry BD ABDC AC
=
or x y x y ac b b c b c
+= = =
+ +
acxb c
=+
and abyb c
=+
(i)
Further DABC = DABD + DADC
rArr 1 1 A 1 Abcsin A czsin bzsin2 2 2 2 2
= +
bc sin A 2bcz cos A2b c sin A 2 b c
= = = + + (ii)
Also q = angBAD + B = A2 + B
The Perimeter and Area of a Regular Polygon of n-sides Inscribed in a circle of radius r
Perimeter of polygon = nAB = 2nR sin pn
Area of polygon = n(Area of triangle AOB) = 2nR 2sin
2 nπ
The Perimeter amp Area of Regular Polygon of n-sides Circumscribed about a given circle of radius lsquorrsquo
Perimeter of Polygon = n AB = 2n AL = 2nOL tannπ = 2n tan
nπ
π
11120 Mathematics at a Glance
Area of Polygon = n(Area of triangle AOB) = 2(OLAB)n nr tan2 n
π=
The Radii of the inscribed and circumscribing circles of a regular polygon having n sides each of length lsquoarsquo
a aR cosec2sin n 2 n
π= =
π a ar cot
2tan n 2 nπ
= =π
1112 rESuLt rELAtEd to cYcLIc quAdrILAtrAL
(a) Ptolemyrsquos Theorem In a cyclic quadrilateral ABCD ACBD = ABCD + BCDA ie the product of diagonals is equal to the sum of product of opposite sides
(b) D = area of cyclic quadrilateral
= 1 (ab cd)sinB2
+ = (s a)(s b)(s c)(s d)minus minus minus minus where a b c d2
+ + +
(c) (ac bd)(ad bc)AC(ab cd)+ +
=+
(d) Circum-radius (R) of cyclic-quadrilateral ACABCD
2sinB= = AC (ab cd)AC
2 4A2ab cd
+=
∆ +
1 (ac bd)(ad bc)(ab cd)R4 (s a)(s b)(s c)(s d)
+ + +=
minus minus minus minus
(e) 2 2 2 2a b c dcosB
2(ab cd)+ minus minus
=+
Chapter 12Inverse trIgonometrIC FunCtIon
121 INVerse FuNctIoN
If a function is one-to-one and onto from A to B then function g which associates each element y isin B to one and only one element x isin A such that y = f(x) hArr x = g(y) then g is called the inverse function of f denoted by g = fndash1 [Read as f inverse] Thus if f A rarr B then g B rarr A
1211 Inverse Trigonometric Functions
The equation sin x = y and x = sinndash1 y are not identical because the former associates many values of x of a single value of y while the latter associates a single x to a particular value of y To assign a unique angle to a particular value of trigonometric ratio we introduce a term called principle range
We list below the domain (values of x) and principle ranges (values of y) of all the inverse trigonometric functions and their graph
Remarks
1 sin 5π6 = 12 But 5π6 ne sinminus1(12) there4 sinndash1x cosndash1x tanndash1x denotes angles or real number lsquowhose sine is xrsquo lsquowhose cosine is xrsquo and lsquowhose tangent is xrsquo provided that the answers given are numerically smallest available
2 If there are two angles one positive and the other negative having same numerical value Then
we shall take the positive value For example cos 1
4 2
π= and cos 1
4 2
π minus =
But we write cosndash1
142
π =
and cosndash1 1
2
ne minus4π
3 I quadrant is common to all the inverse functions
4 III quadrant is not used in inverse function
5 IV quadrant is used in the clockwise direction ie minusπ2 le y le 0
12122 Mathematics at a Glance
122 DomaIN aND raNge oF INVerse FuNctIoNs
Function Domain Range Principal Value Branch
y = sinndash1x [ndash1 1] [ndashπ2π2] ndashπ2 le y le π2
y = cosndash1x [ndash1 1] [0π] 0 le y le π
y = tanndash1x ℝ (ndashπ2π2) minusπ2 lt y lt π2
y = cotndash1x ℝ (0 π) 0 lt y lt π
y = secndash1x (ndashinfinndash1]cup[1infin) [0 π] ndash π2 0 le y le π y ne π2
y = cosecndash1x (ndashinfinndash1] cup[1infin) [ndashπ2 π2]ndash0 ndashπ2 le y le π2 y ne 0
RemarkIf no branch of an inverse trigonometric function is mentioned then it means the principal value branch of the function
123 graphs oF INVerse cIrcular FuNctIoNs aND theIr DomaIN aND raNge
1 Graph of function y = sin x y = sinndash1x
Y
X
1
ndash1
ndashπ2 π2
y=sinx
ndashinfin infinO
π2 (1π2)
y = sinndash1 x(ndash1ndashπ2) ndashπ2
ndash1 O 1x
y
y = sinx and y = sinndash1 x(shown in single graph)
Y
ndash110
(1 π2)
(ndashπ2ndash1)
(ndash1ndashπ2)
(π21)
π2ndashπ2 X
2 Graph of function y = cos x y = cosndash1x
Y
X
1
ndash10
π2
y = cos x
π
Y(ndash1π) π
π2
(1 0)X0
y = cosndash1x
1ndash1
y=cosx and y=cosndash1x(shown in single graph)
y
(ndash1π)
(01) 1
y=x
ndash1ndash1
1
(0π2) π2
π20
(π20) (πndash1)(1 0)
π
πx
Inverse Trigonometric Function 112123
3 Graph of function y = tan x y = tanndash1x
Y infin
Xndashπ2
ndashinfin
π20
1
ndash1
ndashπ2
π2
ndashinfin infinO x
y
π2
π2
ndashπ2
ndashπ2ndashinfin
ndashinfin
+infin
infin xx
y
0
y = tanx y = tanndash1x y = tanx and y = tanndash1x
4 Graph of function y = cot x y = cotndash1x
xndashπ2 π
ndashinfin
+infin
π2
y
0
πndashinfin
infin
π2
0 x
y
ndashπ2π2
0
π
π
ndashinfin
ndashinfin
+infin
+infin
X
Y
y = cot x y = cotndash1 x y = cotx and y = cotndash1x
4 Graph of function y = sec x y = secndash1x
Y
1
0ndash1
π2 xπ
+infin
ndashinfin
Y
X
π2
O 1ndash1
π
infinndashinfin
X
Y+infin
+infin
ndashinfin
ndashinfin
π
π
π2
π2
1
0 1ndash1
ndash1
y = sec x y = secndash1x y = secx and y =secndash1x
5 Graph of function y = cosec x y = cosecndash1x
ndashπ2ndash10 π2
1
π
Y
X
+infin
ndashinfin
π2(1π2)
(π21)
(ndashπ2ndash1)
(ndash1 ndashπ2)
π2ndashπ2
ndashπ2
ndash1
ndash1
1
0 1
Y
X
+infin
ndashinfin y = cosec x y = cosecndash1x y = cosecx and y = cosecndash1x
12124 Mathematics at a Glance
124 composItIoNs oF trIgoNometrIc FuNctIoNs aND theIr INVerse FuNctIoNs
1241 Trigonometric Functions of Their Corresponding Circular Functions
(i) sin (sinminus1 x) = x for all x isin [minus1 1]
(ii) cos (cosminus1 x) = x for all x isin [minus1 1]
(iii) tan (tanminus1 x) = x for all x isin ℝ
(iv) cot (cotminus1 x) = x for all x isin ℝ
(v) cosec (cosecminus1 x) = x for all x isin (minusinfin minus1] cup [1 infin)
(vi) sec (secminus1 x) = x for all x isin (minusinfin minus1] cup [1 infin)
125 INVerse cIrcular FuNctIoNs oF theIr correspoNDINg trIgoNometrIc FuNctIoNs oN prINcIpal DomaIN
(i) sinndash1 (sin x) = x for all x isin [minusπ2 π2] (ii) cosndash1 (cos x) = x for all x isin [0 π] (iii) tanndash1 (tanx) = x for all x isin (minusπ2 π2) (iv) cotndash1 (cot x) = x for all x isin (0 π) (v) secndash1 (sec x) = x for all x isin [0 π] ~ π2 (vi) cosecndash1 (cosec x) = x for all x isin [minusπ2 π2] ~ 0
126 INVerse cIrcular FuNctIoNs oF theIr correspoNDINg trIgoNometrIc FuNctIoNs oN DomaIN
1 sinminus1 (sin x) =
minusπminus isin minus π minusπ isin minusπ ππminus isin π πminus π+ isin π π
x if x [ 3 2 2]x if x [ 2 2]
x if x [ 2 3 2]2 x if x [3 2 5 2]
and so on as shown below
x
y
1
1
y=x
Ondash1
ndash1
y=sin(sinndash1x)=cos(cosndash1x)=x
x
y
y=x
O45deg
ndash1
ndash11
1
y=tan(tanndash1x)=cot(cotndash1x)=x
Inverse Trigonometric Function 112125
X
y
minusπ 2 y=x
y=x
πminus
y=-(x)
π+
y=3x
πminusy=x-2
π
minus3π
y=ndash(3π+x)
2
y=2π
+x
minus5π2π2
3π25π2πminusπ
minus2πminus3π2π 3π0
y=sinndash1(sinx)
Domain ℝ Range 2 2π π minus
Period 2π
Remarky = sinndash1 (sinx) can be formed by tangents of y = sinx at x = nπ as shown below
ndash3π
ndash3π2
ndash5π2 ndashπ2π2 y=πndashx y=
xndash2π y=3πndashx
π23π2
5π22π 3πx
ndashπ2ndashπ πO
y=π
y y=sinndash1(sinx)
y=2π+
x
y=(3π+x
y=ndash(π+x)ndash2π
2 cosminus1 (cos x) =
x if x [ 0]x if x [0 ]2 x if x [ 2 ]
2 x if x [2 3 ]
minus isin minusπ isin π πminus isin π πminus π+ isin π π
and so on as shown
π
π 2πndashπndash2πndash3π 3πx
π2
y=2πndashx
y=xndash
2π
y=x
y=x+
2πy=ndash(x+2π)
y=ndashx
y
O
y = cosndash1(cosx)
Domain ℝ Range [0 π] Period 2π
3 tanminus1 (tan x) =
( )( )( )( )
x if x 3 2 2
x if x 2 2
x if x 2 3 2
x 2 if x 3 2 5 2
π+ isin minus π minusπ
isin minusπ π
minusπ isin π π minus π isin π π
and so on as shown
x
y
minusπ 2minusπ 2minus3π2
π 2
π2 3π2π
minusπminus2π2πO
y=tanndash1(tanx)
Domain ~ (2n 1)2π +
Range 2 2π π minus
Period π
12126 Mathematics at a Glance
4 1
x 2 for x ( 2 )x for x ( 0)
y cot (cot x) x for x (0 )x for x ( 2 )x 2 for x (2 3 )
minus
+ π isin minus π minusπ + π isin minusπ= = isin π minusπ isin π π minus π isin π π
The graph of cotndash1 (cotx) is as shown Domain x isin R minus n π n isin ℤ Range y isin (0 π) Period periodic with period π and and cotminus1 (cot x ) = x forall x isin (0π)
5 y = secndash1 (sec x) =
x for x [ 0]
x for x [0 ]~232 x for x [ 2 ]~2
minus isin minusπ π isin π
π πminus isin π π
x
y
π2
π2
minusπ 2minus3π 2 3π2
y=2πndashx
y=x
y=x+
2π y=ndashx
π
π
ndashπminus2π 2πO
y=secndash1(secx)
The graph of y = secndash1 (secx) is as shown
Domain x isin ℝ minus (2n 1) n2π + isin
Range y isin [0 π2) cup (π2 π]
Period Periodic with period 2π and secminus1(sec x) = x forall x isin [0 π2) cup (π2 π]
6 y = cosecndash1 (cosec x)
3( x) for x ~ 2 2
x for x ~ 02 2
3x for x ~ 2 2
minus π minusπ minus π+ isin minusπ minusπ π = isin π π πminus isin π
y
x
minusπ 2
y=x
y=x
πminus
y= ndash (x)
π+y=2x
π+
y=x ndash2ππ 2
πminusπminus2π 2π0
y=cosecndash1(cosecx)
Domain x isin ℝ sim nπn isin ℤ Range y isin [ndash π2 π2] sim 0 Period Periodic with period 2π and cosecminus1(cosec x) = x for x isin [ndash π2 π2] sim 0
127 INVerse trIgoNometrIc FuNctIoNs oF NegatIVe INputs
(i) sinndash1 (ndashx) = ndashsinndash1 (x) for all x isin [ndash1 1] (ii) cosndash1 (ndashx) = π ndash cosndash1 (x) for all x isin [ndash1 1] (iii) tanndash1(ndashx) = ndashtanndash1 x for all x isin R (iv) cosecndash1(ndashx) = ndashcosecndash1 x for all x isin (minusinfin minus1] cup[1 infin) (v) secndash1 (ndashx) = π ndash secndash1x for all x isin (minusinfin minus1] cup [1 infin) (vi) cotndash1(ndashx) = π minus cotndash1 x for all x isin R
x
y
y=x
πminusπminus2π 2π0
y=xndash
π
y=x+
π
y=x+
2π
y=cotndash1(cotx)
Inverse Trigonometric Function 112127
128 INVerse trIgoNometrIc FuNctIoNs oF recIprocal INputs
(i) sinminus1 (1x) = cosecminus1 x for all x isin (minusinfin ndash1] cup [1 infin) (ii) cosminus1 (1x) = secndash1 x for all x isin (minusinfin ndash1] cup [1 infin)
(iii) tanminus1(1x) = 1
1
cot x for x 0cot x for x 0
minus
minus
gtminusπ+ lt
129 INter coNVersIoN oF INVerse trIgoNometrIc FuNctIoNs
(a) sinndash1x = 1 2
1 2
cos 1 x if 0 x 1
cos 1 x if 1 x 0
minus
minus
minus le le minus minus minus le le
= 1
2
xtan1 x
minus
minus if forall x isin (ndash1 1)
=
21
21
1 xcot if 0 x 1x
1 xcot if 1 x 0x
minus
minus
minuslt le
minus minusπ+ minus le lt
= )
(
1
2
1
2
1sec if x 0 11 x
1sec if x 1 01 x
minus
minus
isin
minus
minus isin minus minus
= 1 1cosecx
minus
if x isin [ndash1 1] ~ 0
(b) minus
minus
minus
minus isin = πminus minus isin minus
1 21
1 2
sin 1 x for x 0 1cos x
sin 1 x for x 1 0 =
21
21
1 xtan for x (0 1]x
1 xtan for x [ 1 0)x
minus
minus
minus isin
minusπ+ isin minus
= 1
2
xcot for x ( 1 1)1 x
minus isin minus
minus = 1 1sec for x 1 1 ~ 0
xminus isin minus
=
1
2
1
2
1cosec for x [0 1)1 x
1cosec for x ( 1 0]1 x
minus
minus
isin
minus
πminus isin minus minus
(c) 1 1
2
xtan x sin for x1 x
minus minus = isin
+
=
1
2
1
2
1cos for x [0 1]1 x
1cos for x [ 1 0]1 x
minus
minus
isin
+
minus isin minus +
=
1
1
1cot for x 0x
1cot for x 0x
minus
minus
gt
minusπ+ lt
= ( )( )
1 2
1 2
sec 1 x for x 0
sec 1 x for x 0
minus
minus
+ gtminus + lt
= 2
1 1 xcosec for x ~ 0x
minus + isin
12128 Mathematics at a Glance
(d) 1
21
1
2
1sin for x 01 xcot x
1sin for x 01 x
minus
minus
minus
ge
+ = πminus le +
= 1
2
xcos x1 x
minus forall isin
+
=
1
1
1tan for x 0x
1tan for x 0x
minus
minus
gt
π+ lt
= 2
1 1 xsec x ~ 0x
minus + forall isin
=
( )( )
1 2
1 2
cosec 1 x for x 0
cosec 1 x for x 0
minus
minus
+ gtπminus + lt
(e)
21
1
21
x 1sin for x 0x
sec xx 1sin for x 0
x
minus
minus
minus
minus gt =
minus π+ lt
= 1 1cos x ~ 0x
minus forall isin
= ( )1 2
1 2
tan x 1 for x 0
tan x 1 for x 0
minus
minus
minus gtπminus minus lt
=
1
2
1
2
1cot for x 0 x 1x 1
1cot for x 0 x 1x 1
minus
minus
gt ne
minus
πminus lt ne minus minus
=
1
2
1
2
xcosec for x 0x 1
xcosec for x 0x 1
minus
minus
gt
minus
π+ lt minus
(f) 1 1 1cosec x sin for x ~ 0x
minus minus= isin =
21
21
x 1cos for x 0x
x 1cos for x 0x
minus
minus
minus gt
minus minusπ+ lt
=
1
2
1
2
1tan for x 0 1x 1
1tan for x 0 1x 1
minus
minus
gt ne
minus
minus lt ne minus minus
= ( )( )
1 2
1 2
cot x 1 for x 0
cot x 1 for x 0
minus
minus
minus gtminus minus lt
=
1
2
1
2
xsec for x 0 1x 1
xsec for x 0 1x 1
minus
minus
gt ne
minus
minusπ+ lt ne minus minus
1210 three ImportaNt IDeNtItIes oF INVerse trIgoNometrIc FuNctIoNs
(i) sinndash1x + cosndash1 x = π2 for all x isin[ndash1 1] (ii) tanndash1x + cotndash1 x = π2 for all x isin R (iii) secndash1x + cosecndash1 x = π2 for all x isin(ndashinfin ndash1] cup [1 infin)
Inverse Trigonometric Function 112129
1211 multIples oF INVerse trIgoNometrIc FuNctIoNs
Property (1)
1 2
1 1 2
1 2
1 1sin (2x 1 x ) if x2 2
12sin x sin (2x 1 x ) if x 12
1sin (2x 1 x ) if 1 x2
minus
minus minus
minus
minus minus le le
= πminus minus le leminusπminus minus minus le le minus
Property (2)
1 3
1 1 3
1 3
1 1sin (3x 4x ) if x2 2
13sin x sin (3x 4x ) if x 12
1sin (3x 4x ) if 1 x2
minus
minus minus
minus
minus minus le le= πminus minus le leminusπminus minus minus le le minus
Property (3) 2cosndash1 x = 1 2
1 2
cos (2x 1) if 0 x 12 cos (2x 1) if 1 x 0
minus
minus
minus le leπminus minus minus le le
Property (4) 3 cosndash1 x =
1 3
1 3
1 3
1cos (4x 3x) if x 121 12 cos (4x 3x) if x2 2
12 cos (4x 3x) if 1 x2
minus
minus
minus
minus le le πminus minus minus le le π+ minus minus le le minus
Property (5)
12
12
1
12
2xtan if 1 x 11 x
2xtan if x 11 x2tan x
2xtan if x 11 x
for x 12
minus
minus
minus
minus
minus lt lt minus π+ gt minus =
minusπ+ lt minus minus π =
Property (6) 3 tanndash1 x =
31
2
31
2
31
2
3x x 1 1tan if x1 3x 3 3
3x x 1tan if x1 3x 33x x 1tan if x1 3x 3
1for x2 3
minus
minus
minus
minusminus lt lt minus
minus π+ gt minus minusminusπ+ lt minus minus
π=
12130 Mathematics at a Glance
Property (7) 2 tanndash1 x =
12
12
12
2xsin if 1 x 11 x
2xsin if x 11 x
2xsin if x 11 x
minus
minus
minus
minus le le + πminus gt + minusπminus lt minus +
Property (8) 2 tanndash1 x =
21
2
21
2
1 xcos if 0 x1 x
1 xcos if x 01 x
minus
minus
minusle ltinfin +
minusminus minusinfin lt le +
1212 sum aND DIFFereNce oF INVerse trIgoNometrIc FuNctIoNs
Property (1)
2 21 2 2
2 2
1 1 1 2 2 2 2
1 2 2 2 2
if x y 1sin x 1 y y 1 x
or if xy 0 and x y 1 where x y 11
sin x sin y sin x 1 y y 1 x if 0 x y 1 and x y 1
sin x 1 y y 1 x if 1 x y 0 and x y 1
minus
minus minus minus
minus
+ leminus + minus
lt + gt isin minus + = πminus minus + minus lt le + gtminusπminus minus + minus minus le lt + gt
Property (2)
2 21 2 2
2 2
1 1 1 2 2 2 2
1 2 2 2 2
if x y 1sin x 1 y y 1 x
or xy 0 and x y 1 where x y 1 1
sin x sin y sin x 1 y y 1 x if 0 x 1 1 y 0 and x y 1
sin x 1 y y 1 x if 1 x 0 0 y 1 and x y 1
minus
minus minus minus
minus
+ leminus minus minus
gt + gt isin minus minus = πminus minus minus minus lt le minus le le + gtminusπminus minus minus minus minus le lt lt le + gt
Property (3)
cosndash1 x + cosndash1y =
1 2 2
1 2 2
cos xy 1 x 1 y if 1 x y 1 and x y 0
2 cos xy 1 x 1 y if 1 x y 1 and x y 0
minus
minus
minus minus minus
minus le le + geπminus minus minus minus
minus le le + le
Inverse Trigonometric Function 112131
Property (4)
cosndash1x ndash cosndash1y =
1 2 2
1 2 2
cos xy 1 x 1 y if 1 x y 1 and x y
cos xy 1 x 1 y if 1 x y 1 and x y
minus
minus
+ minus minus
minus le le leminus + minus minus minus le le ge
Property (5)
tanndash1x + tanndash1 y =
1
1
1
x ytan if xy 11 xy
x ytan if x 0 y 0 and xy 11 xy
x ytan if x 0 y 0 and xy 11 xy
for x 0 y 0 and xy 12
for x 0 y 0 and xy 12
minus
minus
minus
+lt minus
+π+ gt gt gt minus +minusπ+ lt lt gt minus π gt gt =
πminus lt lt =
Property (6)
tanndash1x ndash tanndash1 y =
1
1
1
x ytan if xy 11 xy
x ytan if x 0 y 0 and xy 11 xy
x ytan if x 0 y 0 and xy 11 xy
for x 0 y 0 and xy 12
for x 0 y 0 and xy 12
minus
minus
minus
minusgt minus +
minusπ+ gt lt gt minus + minusminusπ+ lt gt gt minus + π gt gt = minus
πminus lt lt = minus
Chapter 13point and
Cartesian system
131 IntroductIon
The study of co-ordinate geometry begins with the study of ldquoconcept of pointrdquo which is defined as a geometrical construction having no dimensions Several methods have been developed by mathematicians to uniquely locate the position of a point in the space
132 FrAME oF rEFrEncE
It is a set of fixed pointslinesurfaces with respect to which the following observations are made ∎ Rectangular co-ordinate System ∎ Oblique co-ordinate System ∎ Polar co-ordinate System
1321 Rectangular Co-ordinate SystemAny point P in (x y) plane can be represented by unique ordered pair of two real numbers (x y) Here x is abscissa of point (OM or PN) Y is ordinate of point (ON or PM)
Sign ConventionTherefore the x-y plane (Cartesian plane) is algebraically represented as Cartesian product of two set of real numbers
So called as ℝ times ℝ (ℝ2) plane ℝ times ℝ = (x y) x isin ℝ and y isin ℝ
ℝ+ times ℝ+ = 1 quadrant ℝndash times ℝ+ = 2nd quadrant ℝndash times ℝndash = 3rd quadrant ℝ+ times ℝndash = 4th quadrant
1322 Polar Co-ordinate SystemIt consist of a fixed point O which is known as pole and semi-infinite ray OX which is called initial line ∎ The polar coordinate of any point P is given as (r q) where r is the distance
of point P from pole O is lsquorrsquo and the angle angXOP = q
Point and Cartesian System 13133
133 dIstAncE ForMulA
The distance between any two points P and Q when coordinate of two points is given in Cartesian form Let P(x1 y1) and Q(x2 y2) be two given points then
2 22 1 2 1PQ d (x x ) (y y )= = minus + minus
1331 Applications of Distance Formula
Position of three points Let A B C are points lying in a plane then two condition arises either they are collinear or they form a triangle
13311 Collinearity of three given points
The three given points A B C are collinear ie lie on the same straight line if ∎ any of the three points (say B) lie on the straight line joining the other two points
∎ area of DABC is zero It means 1 1
2 2
3 3
x y 1x y 1 0x y 1
∆ = =
rArr [x1(y2 ndash y3) + x2(y3 ndash y1) + x3(y1 ndash y2)] = 0 ∎ slope of line AB(mAB) = slope of line BC(mBC) = slope of line AC(mAC) ∎ coordinates of any of the points x1 and y1 can be written as linear combination of other two x2 x3 and
y2 y3 as x1=lx2+mx3 and y1 =ly2+my3 such that l + m = 1
134 sEctIon ForMulA IntErnAl dIvIsIon
Co-ordinates of a point which divides the line seg-ment joining two points P(x1 y1) and Q(x2 y2) in the
ratio m n internally are 2 1 2 1mx nx my nym n m n+ +
+ +
Notes
∎ If P is the mid-point of AB then it divides AB in the ratio 11 so its coordinates
are 1 2 1 2x x y y
2 2+ +
∎ The given diagram helps in remembering the section formula
Coordinates of a point which divides the line segment joining two points P(x1 y1) and Q(x2 y2) in the
ratio m n externally are 2 1 2 1mx nx my ny
m n m nminus minus
minus minus
∎ To get the point of the external division only replace the n of internal division by -n
∎ Co-ordinates of any point on the line segment joining two points P(x1 y1) and Q(x2 y2) and dividing it
in the ratio l1 is given by 1 2 1 2x x y y ( 1)
1 1λ λ λ
λ λ+ + ne minus + +
13134 Mathematics at a Glance
∎ Lines formed by joining (x1 y1) and (x2 y2) is divided by
(a) x-axis in the ratio hArr -y1y2 (b) y-axis in the ratio hArr -x1x2
If the ratio is positive the axis divide it internally and if negative then divides externally
∎ Line ax + by + c = 0 divides the line joining the points P(x1 y1) and Q(x2 y2) in the ratio l 1
then 1 1
2 2
ax by c
ax by cλ
+ += minus + +
If l is positive it divides internally if l is negative then externally
135 slopE oF lInE sEgMEnt
Slope of a line segment is a physical quantity that measures the amount of inclination of the line with respect to the x axis and defined as rate of change of ordinate with respect to the abscissa
Denoted as ym x
∆ = ∆ bull Slope can be obtained as tangent of angle that line
segment makes with positive direction of x axis in anticlockwise sense
rArr 2 1
2 1
y ym tanx xminus
= θ =minus
bull If Line is horizontal rArr q = 0 rArr m = 0 bull If line vertical rArr q = 90deg rArr m rarr infin bull If the points A and B coincide rArr Slope is indeterminate
1351 Area of Triangle
Area of triangle when the coordinates of vertices A B C of triangle are A(x1 y1) B(x2 y2) and C(x3 y3) is given as
1 2 3 2 3 1 3 1 21[ x (y y ) x (y y ) x (y y )]2
∆ = minus + minus + minus
This expression for the area can also be written in the
determinant form 1 1
2 2
3 3
x y 11 x y 12
x y 1∆ =
Notes
∎ If area of D is zero then the point are collinear Hence for three points to be collinear the essential
condition is area of D = 0 rArr 1 1
2 2
3 3
x y 1
x y 1 0
x y 1
=
∎ If the coordinate of vertices of D are given in polar form (r1 q1) (r2 q2) (r3 q3) then the area of D will
be given by [ ]2 3 1 1 3 1 1 3 1 2 2 1
1r r sin( ) r r sin( ) r r sin( )
2θ θ θ θ θ θ∆ = minus + minus + minus
Point and Cartesian System 13135
1352 Area of General Quadrilateral
If A(x1 y1) B(x2 y3) C(x3 y3) and D(x4 y4) are vertices of a quadrilateral then its area will be given
by
1 1
2 2
3 3
4 4
x y 1x y 11x y 12x y 1
= 1 2 2 1 2 3 3 2 3 4 4 3 4 1 1 41[(x y x y ) (x y x y ) (x y x y ) (x y x y )]2
minus + minus + minus + minus =
1 1
2 2
3 3
4 4
x y 1x y 11x y 12x y 1
NoteIf area of a quadrilateral joining four points is zero then four points are collinear
1353 Area of Polygon
The area of polygon whose vertices are (x1 y1) (x2 y2) (x3 y3)(xn yn) is |(x1y2 ndash x2y1) + (xny3 ndash x3y2) ++(xny1 ndash x1yn)|
Stair method Repeat first co-ordinate one time in last for down arrow use +ve sign and for up arrow use -ve sign
Area of polygon =
1 1
2 2
3 3
n n
1 1
x yx yx y
1 2
x yx y
= 1 2 2 3 n 1 1 2 2 3 n 11 |(x y x y x y ) (y x y x y x )|2
+ + + minus + + +
bull Area of a triangle can also be found by easy method ie stair method
1 1
2 2
3 3
1 1
x yx y1x y2x y
∆ = = 1 2 2 3 3 1 1 2 2 3 3 11 |(x y x y x y ) (y x y x y x )|2
+ + minus + +
bull If one vertex (x3 y3) is at (0 0) then D = 1 2 2 11 |(x y x y )|2
∆ = minus
13136 Mathematics at a Glance
136 locus oF poInt And EquAtIon oF locus
The path traced by a moving point P(x y) is called locus of P The equation of locus is a relation in the variable x and y which is satisfied by the coordinates of the moving point P(which moves under given geometrical restriction) at any position on its path
That is if f(x y) = 0 is satisfied by (a b) forall (a b) lying on the path then its called equation of locus
Method to Find Equation of Locus
Step I Let coordinate of point P be (h k) and apply the condition given to express h and k as a function of some parameter (q a b t l etc )
Step II Eliminate the parameters to relate h and k
Step III In the equation between h and k therefore obtained replace h by x and k by y to get equation of locus
1361 Union of LociLocus is a set of points that follow a given relation in x and y
Given two loci S1 and S2 defined as belowS1 (x y) S = f (x y) = 0 and S2 (x y) S = g(x y) =0Union of loci S = 0 and S = 0 is set of those points which lie ether
on S = 0 or S = 0 rArr S1 cup S2 = (x y) f(x y) = 0 or g(x y) = 0 And its equation is given
as S S = 0 ie f(x y) g(x y) = 0
1362 Intersection of LociIntersection of loci S = 0 and S = 0 is defined as set of those points which lie on both the curves S = 0 and S = 0 That is set of common points
rArr S1 cap S2 = (x y) f(x y) = 0 or g(x y) = 0 and its equation is given as
rArr |f(x y)| + |g(x y)| = 0 or |S| + |S| = 0 or radicS + radicS = 0 or S2 + S2 = 0
1363 Locus Passing Through Intersection of Two Locus
Given two loci S = 0 and S = 0 defined as ( )( )x y S f(x y) 0x y S g(x y) 0
= =
= = The equation S + lS = 0 represents a family of curve
passing through A and B that is intersection of S = 0 and S = 0 where l is a real parameterDiscussion S + lS = 0 rArr f(x y) + lg(x y) = 0Represent infinitely many curve due to parameter l and since both point A and B satisfy the above equation because f(a b) = g(a b) = f(g d) = g(g d) = 0
rArr f(a b) + lg(a b) = 0 + l0 = 0
Point and Cartesian System 13137
Ellipse Ellipse is a locus of a point which moves so that the summation of its distances from two fixed points A and B remains constant l
Hyperbola Hyperbola is locus of a point which moves so that the difference of its distances from two fixed points A and B remains constant l
Parabola It is the locus of all points such that the distance from a fixed point and perpendicular distance from a fixed line is always equal
Circle Locus of all points which are equidistant from a given point in a plane
137 cHoIcE oF orIgIn And sElEctIon oF coordInAtE AXEs
In order to solve any general geometric problem conveniently a suitable choice of origin and proper selection of coordinate axes can be considered but care must be taken that during such selection the generality of the problem is not lost So any assumption is regarded as perfectly general iff by shifting the origin to a suitable point and rotating the coordinate axes by some angle the most general case can be transformed to assumed case
138 gEoMEtrIcAl trAnsForMAtIons
Any geometric operation undergoing through which the coordinate of the point changes It is of two types (i) Linear Transformation A transformation in which the origin of reference frame does not
change and the new coordinate obtained are linear function of old coordinate ie xrsquo = ax + by and yrsquo = cx + dy is called linear transformation
(ii) Non-linear Transformation In such a transformation the straight line remains straight The remaining transformations are called non-linear transformation
1381 Transformations in Cartesian Plane
T1 Reflection of point in x-axis
1T(x y) (x y)rarr minus
T2 Reflection of point in y-axis
2T(x y) ( x y)rarr minus
T3 Reflection of point in origin
3T(x y) ( x y)rarr minus minus
T4 Reflection of point in the line y = x
4T(x y) (x y)rarr
T5 Rotation of point about origin
5T(x y) (x y )rarr
T6 Reflection of point in the line y = xtan q
6T(x y) (x y )rarr
13138 Mathematics at a Glance
1382 Transformation of Coordinates Axis
Shifting of origin without rotating axes If origin of coordinate frame is shifted to O to O (h k) keeping the coordinate axis respectively parallel regional axes
Conclusion ∎ New coordinate of point P in terms of old x = x ndash h and
y = y ndash k ∎ Old coordinate of point P in terms of new x = x + h and
y = y + k ∎ The transformation equation of a locus f(x y) = 0 is
obtained by replacing x by x + h and y by y + k x x h
y y kf(x y) 0 f(x h y k) 0rarr +rarr += rarr + + =
1383 Rotation of the Axes (Without Changing Origin)
To change the direction of the axis of coordinates without changing the origin let OX and OY be the old axes and OX and OY be the new axes obtained by rotating the old axes through an angle q in anti-clock wise sense about origin
The old coordinate of P(x y) with respect to new coordi-nate axes will be given by
x = ON ndash NL y = PQ + QL
x x cos y siny y cos x sin
prime prime= θminus θprime prime= θ+ θ
helliphellip (i)
139 gEoMEtrIcAl tIps And trIcks
Method to Find Circum Centre
Step I Consider (OA)2 = (OB)2 = (OC)2 rArr (x ndash x1)2 + (y ndash y1)
2 = (x ndash x2)
2 + (y ndash y2)2 = (x ndash x3)
2 + (y ndash y3)2
Step II Solving two linear equations obtained we can get coordinates of circum-centre
Step III The obtained value of x and y always satisfy third equation that indicates the concurrency of ^ bisectors
1391 The Coordinates of Centroid
In a DABC the coordinates of centroid are given by a b c a b cx x x y y y
3 3+ + + +
bull If mid-points of the sides of a triangle ABC are D E F respectively of BC CA AB as shown in the figure then A(xE + xF ndash xD yE + yF ndash yD) B(xD + xF ndash xE yD + yF ndash yE) and C(xD + xE ndash xE yD + yE ndash yF)
Point and Cartesian System 13139
Area of DABC = 4 times Area of DDEF ie area of a D is four times the area of the D formed by joining the midpoints of its sides
bull If two vertices of a D are (x1 y1) and (x2 y2) and the coordinates of Centroid are (a b) then co-ordinates of the third vertices are (3a ndash x1 ndash x2 3b ndash y1 ndash y2)
1392 Coordinates of Incentre
If A(x1 y1) B(x2 y2) and C(x3 y3) are the vertices of the DABC with sides BC CA AB of lengths a b c
respectively then the coordinates of the incentre 1 2 3 1 2 3ax bx cx ay by cyI
a b c a b c+ + + + = + + + +
1393 Coordinates of Ex-centre
The coordinates of ex-centres of the triangle are given by
rArr A
D CI
(b a)y cyy
b a cminus +
=minus +
rArr A
B A CI
bx ax cxx
b a cminus +
=minus +
and minus +=
minus +A
B A CI
by ay cyy
b a c
Chapter 14Straight line and
pair of Straight line
141 Definition
A straight line is a curve such that every point on the line segment joining any two points lie on it or in other words straight line is the locus of a point which moves such that the slope of line segment joining any two of its position remains constant
1411 Equation of Straight Line
A relation between x and y which is satisfied by coordinates of every point lying on a line is called the equation of straight line Every first degree equation in x y ie ax + by + c = 0 represents a line Thus a line which is also defined as the locus of a point satisfying the condition ax + by + c = 0 where a b c are constant
∎ Equation of straight line parallel to axes (i) Equation of a straight line which is parallel to x-axis and at a distance b units from it is given by
y = b b gt or lt 0 according as it is in positive or negative side
equation of x-axis is y = 0 (ii) Similarly for any line parallel to the y axis and at a distance a unit from it is given by x = a
a gt or a lt 0 according as the line lies on positive or negative sides of the x-axis ∎ The combined equation of the coordinate axis is xy = 0
Straight Line and Pair of Straight Line 14141
1412 Different Forms of the Equation of Straight Line
Two Point From Straight line passing through A(x1 y1) and B(x2
y2)
2 11 1
2 1
y yy y (x x )x xminus
minus = minusminus
or 1 1
2 2
x y 1x y 1 0x y 1
= in
determinant form
Slope Point From Straight line passing through A(x1 y1) and having slope m
y ndash y1 = m(x ndash x1)
Slope Intercept Form Equation of line having slope lsquomrsquo and making an intercept c on y-axis
y = mx + c where q is the angle made by line with +ve direction of x-axis in counter-clockwise sense
Two Intercept From Equation of line making intercepts a and b respectively on x and y axis
x y 1a b+ = or
x y 1a 0 1 00 b 1
=
in determinant form
14142 Mathematics at a Glance
PerpendicularNormal Form Equation of line upon which the length of perpendicular form origin is p and perpendicular makes a angle with +ve direction of x-axis
x cos a + y sin a = prArr If equation of line be x cos a + y sin a = ndashp (p gt 0) the equation will not be in normal form to convert it to normal form multiply both sides by ndash1rArr x(ndashcosa) + y(ndashsina)
= prArr x cos(p + a) +
ysin(p+a) = p
Symmetric (Parametric) From Straight line passing through A(x1 y1) and making angle q with positive x-axis
1 1x x y y rcos sinminus minus
= =θ θ
where r is distance of the point P(x y) from the fixed point A(x1 y1)
rArr Using symmetric form bull To find coordinate of any point P(x y) from the fixed point A(x1y1) on the line if AP is given as r rArr x = x1+ r cosq and y = y1 + r sinq bull To find distance of a point from a fixed point on the line along the line bull To find distance r if qis known q if r is given
1413 Angle Between Two Lines
Given two lines
11 1 1 1 1
1
aL a x b y c 0 m tanb
+ + = = minus = α
22 2 2 2 2
2
aL a x b y c 0 m tanb
+ + = = minus = β
The angle between L1 = 0 and L2 = 0 q = b ndash a
rArr 2 1
1 2
m mtan1 m m
minusθ =
+
rArr 1 2 2 1
1 2 1 2
a b a btana a b b
minusθ =
+
Straight Line and Pair of Straight Line 14143
14131 Conclusion
∎ There are two angles formed between any pair of line q and p ndash q (say) then tangent of acute angle q
2 1
1 2
m mtan1 m m
minusθ =
+ and 2 1
1 2
m mtan( )1 m m
minusπminusθ = minus
+
∎ Lines are parallel if rArr tanq = 0 rArr m1 = m2
∎ Lines are perpendicular rArr tanq rarr infin rArr m1 m2 = ndash1
∎ Lines are coincident if they have same slope and intercept
rArr 1 1 1
2 2 2
a b ca b c
= =
∎ Lines L1 = 0 and L2 = 0 are perpendicular when q = 90 ∎ If m1m2 = 1 then angle of L1 with x-axis is same as angle of L2 with y-axis Hence both lines make same angle with y = x + k and y = ndashx + k ∎ If m1 + m2 = 0 Lines L1 and L2 make supplementary angles with x and y-axis when extended to
intersect they form an isosceles triangle with the coordinates axis (x or y)
1414 Equation of a Line Perpendicular and Parallel to Given Line
rArr Let m be the slope of the line ax + by + c = 0 Then m = -ab Since the required line is parallel to the given line The slope of the required line is also m Let C1 be the intercept by the line on y-axis Then its equation is y = mx + c1
rArr 1ay x c
bminus
= +
rArr ax + by - bc1 = 0 rArr ax + by + l = 0 where l = - bc1 = constant The equation of line parallel to a given line is ax + by + l = 0
Note
To find the equation of a line parallel to a given line keep the expression containing x and y same and simply replace the given constant by a new arbitrary constant l The value of a l can be determined by same given condition
rArr The equation of line perpendicular to given line ax+ by + cz = 0 is bx ndash ay + l = 0
ie interchange the coefficient of x and y by reversing the sign of exactly of them one and replace the constant term by parameter l
1415 Straight Line Through (x1 y1) Making an Angle α with y = mx + c
Equation of line passing through a point A(x1 y1) and making a given angle q with the line y = mx + c
14144 Mathematics at a Glance
Let slope of the line be mrsquo
m mtan1 mm
minusθ =
+ rArr m m tan
1 mmminus
= plusmn θ+
rArr m ndash m = plusmn tan q plusmn m m tan q rArr m tan m(1 m tan )θ = plusmn θ
rArr m tanm1 m tan
θ=
plusmn θ
So equation of lines are 1 1m tany y (x x )
1 m tanθ
minus = minusplusmn θ
1416 Position of Two Points wrt a Straight Line
Two points P(x1 y1) and Q(x2 y2) lie on the same side or on the opposite side of the line ax + by + c = 0 according as ax1 + by1 + c and ax2 + by2 + c are of the same sign or opposite signs respectively The coordinates
of the point R which divides the line joining P and Q sides in the ratio mn are 2 1 2 1mx nx my nym n m n+ +
+ +
If this point lie on (i) then 2 1 2 1mx nx my nya b c 0m n m n+ + + + = + +
rArr m(ax2 + by2 +c) + n(ax1 + b1 + c) = 0
rArr 1 1
2 2
ax by cmn ax by c
+ += minus
+ +
If the point R is between the points P and Q
Then the ratio m n is positive So from the above equation we get 1 1
2 2
ax by c 0ax by c
+ +lt
+ + rArr ax1 + by1 + c and ax2 + by2 + c are of opposite sign
If point R is not between P and Q then the ratio m n is negative
rArr 1 1
2 2
ax by c 0ax by c
+ +gt
+ +
rArr ax1 + by1 + c and ax2 + by2 + c are of same sign
Straight Line and Pair of Straight Line 14145
NoteIf the location of a single point is to be defined then the other point is taken as the origin and wrt the origin The location of the point wrt the line is defined
Two points P(x1 y1) and Q(x2 y2) will be located at the same side of the line If they give the same sign of the expression when they are used in the line otherwise they will lie on the opposite side of the line
1417 Distance of a Point From a LineLet the given line be ax + by + c = 0 then the distance of any point P(x1 y1) from the given line be
rArr 1 12 2
|ax by c |PNa b
+ +=
+
Note
The length of the perpendicular from the origin to the line ax + by + c = 0 is 2 2
| c |
a b+
1418 Distance Between Two Parallel Straight LinesLet ax + by + c = 0 and ax + by + c = 0 be the parallel straight lines then the distance between them is
given by 2 2
c - c
a b+ rArr Oblique distance of a point from a line Distance of a point
P(x1 y1) from a line L1 = ax + by + c = 0 along L2 = y = mx + cMethod I Let line parallel to y = mx + c through P cuts ax + by + c = 0 at Q(x0 y0) rArr equation of PQ y ndash y1 = m(x ndash x1) hellip(i)
Solving (i) and (ii) get coordinates of Q and applying distance
formula ( ) ( ) ( )2 21 0 1 0pqd x x y y= minus + minus
Method II Let m = tan qEquation of PQ is 1 1x x y y rcos sinminus minus
= =θ θ
For Q (x1 + r cosq y1 + rsin q) Must satisfy L1 ax + by + c = 0rArr a(x1 + r cosq) + b(y1 + r sinq) + c = 0
rArr 1 1(ax by c)ra cos bsin
+ + = minus θ+ θ The sign of r indicates the position of point wrt Line and |r| is
required distance
1419 Intersection of Two LinesThe point of intersection of two lines a1x + b1y + c1 = 0 and a2x + b2y +c2 = 0 is 1 2 2 1 1 2 2 1
1 2 2 1 1 2 2 1
b c b c c a c aa b a b a b a b
minus minus
minus minus ∎ Condition for concurrency of Lines Three lines are said to be concurrent if they pass through a
common point Thus if three lines are concurrent the point of intersection of two lines lies on the third line Let a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0
rArr 1 1 1
2 2 2
3 3 3
a b ca b c 0a b c
= which is the required condition for concurrency of lines
14146 Mathematics at a Glance
NoteAnother condition of concurrency of three lines L1 a1x + b1y + c1 = 0 L2 a2x + b2y + c2 = 0 and
L3 a3x + b3y + c3 = 0 are concurrent iff there exists constants l1 l2 l3 not all zero such that
l1L1 + l2 L2 + l3 L3 = 0 l1(a1x + b1y + c1) + l2 (a2x + b2y + c2) + l3(a3x + b3y + c3) = 0
14110 Equation of the Bisectors of the Angles Between LinesMethod 1 Let L1 equiv a1x + b1y + c1 = 0 and L2 equiv a2x + b2y + c2 = 0 be two intersecting lines then the
equations of the lines bisecting the angles between L1 and L2 are given by 1 1 1 2 2 22 2 2 21 1 2 2
a x + b y c a x + b y c
a + b a + b
+ += plusmn
If a1a2 + b1b2 = 0 then the given lines are perpendicular to each other else they will contain acute and obtuse angle
ie a1a2 + b1b2 ne 0 Let q be the angle between L1 and L2 which is bisected by one of the bisectors say L3 Then angle between L1 and L3 is q2 Now find tan q2
Two Cases Arise
(i) If tan 1 then 2 2θ πlt θ lt Thus L3 will be bisecting the acute angles between L1 and L2
(ii) If tan 12θgt then
2π
θ gt Thus L3 will be bisecting the obtuse angle between L1 and L2
Method 2 If c1 ne 0 c2 ne 0 then origin must lie in one of the angles between L1 and L2 Let us assume
c1 c2 gt 0 Then 1 1 1 2 2 22 2 2 21 1 2 2
a x b y c a x b y c
a b a b
+ + + += +
+ + is one of the bisectors of L1 and L2 If a1a2 + b1b2 gt 0 the given
equation represents obtuse angle bisector otherwise it represents acute angle bisector (if a1 a2 + b1b2 lt 0)
141101 Bisector of angle containing the origin
Let the equations of the two lines be a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 To find the bisectors of the angle containing the origin the following steps are taken
Step 1 See whether the constant terms c1 and c2 in the equations of two lines are positive or not If not then multiply both the sides of the equations by ndash1 to make the constant term positive
Step 2 Now obtain the bisector corresponding to the positive sign 1 1 1 2 2 22 2 2 21 1 2 2
a x b y c a x b y c a b a b
+ + + += +
+ +
This is the required bisector of the angle containing the origin and negative sign bisector of that angle which does not contain origin
141102 Bisector of acute and obtuse angle
Let the equations of the two lines be a1x + b1y +c1 = 0 and a1x + b2y + c2 = 0 To separate the bisectors of the obtuse and acute angles between the lines we proceed as follows
Step 1 See whether the constant terms c1 and c2 in the equations of two lines are positive or not If not then multiply both the sides of the equations by -1 to make the constant term positive
Straight Line and Pair of Straight Line 14147
Step 2 Determine the sign of the expression a1a2 + b1b2
Step 3 If a1a2 + b1b2 gt 0 then the bisector corresponding to + sign gives the obtuse angle bisector and the bisector corresponding to ndash sign is the bisector of acute angle between the lines
ie 1 1 1 2 2 22 2 2 21 1 2 2
a x b y c a x b y c
a b a b
+ + + +=
+ + and 1 1 1 2 2 2
2 2 2 21 1 2 2
a x b y c a x b y c
a b a b
+ + + += minus
+ + are the bisectors of
obtuse and acute angles respectively
Step 4 If a1a2 + b1b2 lt 0 then the bisector corresponding to + sign gives the acute and obtuse angle
bisectors respectively 1 1 1 2 2 22 2 2 21 1 2 2
a x b y c a x b y c
a b a b
+ + + +=+
+ +
and 1 1 1 2 2 22 2 2 21 1 2 2
a x b y c a x b y c
a b a b
+ + + += minus
+ +
are
the bisectors of acute and obtuse angles respectively
141103 Whether the origin lies in the obtuse angle or acute angle
Let the equations of the two lines be a1x + b1y +c1 = 0 and a2x + b2y + c2 = 0 To determine whether the origin lies in the acute angle or obtuse angle between the lines we proceed as follows
Step 1 See whether the constant terms c1 and c2 in the equations of two lines are positive if not then multiply both the sides of the equations by ndash1 to make the constant term positive
Step 2 Determine the sign of the expression a1a2 + b1b2
Step 3 If a1a2 + b1b2 gt 0 then the origin lies in the obtuse angle and the lsquo+rsquo sign gives the bisector of obtuse angle If a1a2 + b1b2 lt 0 then the origin lies in the acute angle and the lsquo+rsquo sign gives the bisector of acute angle
Tips and Tricks
Equation of a Reflected Ray in a Mirror Given a line mirror LM = ax + by + c = 0 and a ray is incident along the line L1 = a1x + b1y + c1 = 0
The equation of the reflected ray is LR = (y ndash b) ndash m0 (x ndash α) = 0
In general if a point (x2 y2) lies at a distance k times the distance of P(x1 y1) from M (xm ym) then
2 1 2 1 1 12 2
y y x x (ax by c)(k 1)b a a bminus minus + +
= = minus ++
Foot of perpendicular and image of a point in a line If point P is reflected with respect to line Lm then the coordinates of its reflection are given by Q (xQ = 2xm ndash xp yQ = 2ym ndash yp) bull Equation of a Reflected Ray in a Mirror Choose a point P(p q) on the incident ray (preferably
any one of p or q taken zero) and get the image in line mirror Q(r s) In the line mirror
rArr ( )
2 2
2 ap bq cr p s qa b a b
minus + +minus minus= =
+
14148 Mathematics at a Glance
Equation of reflected ray is sy (x )rminusβ
minusβ = minusαminusα
rArr Yet another way the equation of the reflected ray is given as LI + lLM = 0 ie (a1x + b1y + c1) + l (ax + by + c) = 0
rArr minus +λ = λ = +1 1
2 2
2(aa bb )0 (incidentray) or (reflectedray)a b
Equation of reflected ray is minus ++ =
+1 1
I M2 2
2(aa bb )L L 0a b
14111 Family of Straight Lines
The general equation of line has two effective parameters Therefore two conditions are needed to repre-sent a line uniquely But if only one condition is given then the resulting equation consist of a parameter and termed as lsquofamily of straight linesrsquo ∎ If L1 equiv a1x + b1y + c1 = 0 and L2 equiv a2x + b2y + c2 = 0 are two straight lines (not parallel) then
L1 + lL2 equiv a1x + b1y + c1 + l (a2x + b2y + c2) = 0 represents family of lines passing through the point of intersection of L1 = 0 and L2 = 0 (Here l is a parameter)
∎ Family of straight lines parallel to the line ax + by + c = 0 is given by ax + by + k = 0 where k is a parameter
∎ Family of straight lines perpendicular to the line ax + by + c = 0 is given by bx ndash y + k = 0 where k is a parameter
∎ If a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 a3x + b3y + c3 = 0 are concurrent then p (a1x + b1y + c1)
+ q(a2x + b2y + c2) + r(a3x + b3y + c3) = 0 rArr p + q + r = 0 ie 1 1 1
2 2 3
3 3 3
a b ca b b 0a b c
=
142 General equation of SeconD DeGree anD Pair of StraiGht lineS
The general equation of pair of a straight lines is represented by the most general equation of second degree in x and y but any equation in x and y in degree two does not always represent pair of straight lines
Considering the following equation as a quadratic equation in y
rArr by2 + 2(hx + f) y + ax2 + 2gx + c = 0
rArr = minus + plusmn + minus + +2 2by (hx f ) (hx f ) b(ax 2gx c)
+ + = plusmn minus minusα minusβ2hx by f (h ab) (x )(x ) (1)
where a and b are roots of quadratic (h2 ndash ab)x2 + 2(hf ndash bg)x + f2 ndash bc
This equation (1) represents pair of straight lines if a = b ie D = 0
rArr D = 4 (hf ndash bg)2 ndash 4(h2 ndash ab) (f2 ndash bc) = 0
Straight Line and Pair of Straight Line 14149
rArr b2g2 ndash 2hfgb + h2bc + abf2 ndash ab2c = 0
D= abc + 2fgh ndash af2 ndash bg2 ndash ch2 = 0rArr
a h gh b f 0g f c
∆ = =
rArr The lines represented are given as + + = plusmn minus minusα2hx by f h ab(x )
ConclusionsIf h2 ndash ab gt 0 rArr two real and distinct linesIf h2 ndash ab lt 0 rArrtwo imaginary linesIf h2 ndash ab = 0 rArrtwo parallel lines if atleast one of bg ndash hf ne 0 af ndash gh ne 0If h2 ndash ab = 0 and bg ndash hf = 0 af ndash gh = 0 rArrtwo coincident linesa + b = 0 rArrboth lines are perpendicular
1421 Pair of Straight Lines Through the Origin
The homogenous equation of second degree ax2 + 2hxy + by2 = 0 always represent a pair of straight lines through the origin ax2 + 2hxy + by2 = 0
rArr b(yx) 2 + 2h(yx) + a = 0 rArr 2y 2h 4h 4ab
x 2bminus plusmn minus
=
rArr y = m1x or y = m2x where2
1h h abm
bminus + minus
= and 2
2h h abm
bminus minus minus
=
Since h2 le ab therefore values of m1 and m2 are real Clearly y = m1x and y = m2x are straight lines passing through the origin Hence ax2 + 2hxy + by2 = 0 represents a pair of straight lines passing through the origin
rArr According to the value of m1 and m2 then line are Real and distinct if h2 gt ab = 0 and h2 gt ab If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents two straight lines they can be found by considering
the lines as (lx + my + n) (lprimex + mprimey + nprime) = 0 After multiplying and comparing the coefficients of like power we can find l lprime m mprime n nprime to find the required equations
1422 Angle Between the Pair of Straight Lines ax2 + 2hxy + by2 = 0 (i)
2
1 2 (h ab)tan|a b |
minus minus
θ = +
(i) Condition for the lines to be parallel If the two lines are parallel then q = 0 ie tanq = 0 Hence the two lines are parallel if h2 = ab (ii) Condition for the lines to be perpendicular If the two lines are perpendicular then q = 90deg ie tanq=infina + b = 0 ie coefficient of x2 + coefficient of y2 = 0
NoteThe above conditions are also valid for general equation of second degree
14150 Mathematics at a Glance
1 Equation of angle bisector of the pair of straight lines ax2 + 2hxy + by2 = 0 is given by minus=
minus
2 2x y xya b h
rArr Condition for coincidence of lines The lines will be coincident if 1 1 1
2 2 2
l m nl m n= = Taking the
above ratios in pairs the conditions are h2 - ab = 0 g2 - ac = 0 and f 2 - bc = 0 rArr Point of intersection of the lines The point of intersection of ax2 + 2hxy + by2 + 2gx + 2fy
+ c = 0 is 2 2
bg hf af ghh ab h ab
minus minus minus minus
or 2 2
2 2
f bc g cah ab h ab
minus minus minus minus
2 Bisectors of the angles between the lines given by ax2+ 2hxy + by2 + 2gx + 2fy + c = 0 If (xprime yprime) be the point of intersection of the lines then we shift the origin to the point (xprime yprime) The
transformed equation will be ax2 + 2hxy + by2 = 0 of the bisectors which are given by 2 2x y xya b hminus
=minus
The above bisectors are referred to (xprime yprime) as origin Now we have to write x - xprime from x and y -yprime for y Hence the equation of the bisectors of the angle between the lines is
2 2(x x ) (y y ) (x x )(y y )a b hprime prime prime primeminus minus minus minus minus
=minus
2 2 2 2
2 2 2 2shifting origin to
( )
ax 2hxy by 2gx 2fy c 0 ax 2hxy by 0(x ) (y ) (x )(y ) x y xy
a b h a b hα β
+ + + + + = + + = rarr minusα minus minusβ minusα minusβ minus
= = minus minus
Tips and Tricks
rArr Point of Intersection Given a pair of straight lines S = ax2 + 2hxy + by2
+ 2gx + 2fy + c = 0
Let (ab) be the point of intersection of both lines represented by S = 0
Shifting origin to (a b) the equation S = 0 must transform to homogenous form
ie a(x + a)2 + b (y + b)2 + 2h (x + a)(y + b) + 2g(x + a) +2 f(y + b) + c = 0
coefficient of x 0 a h g 0coefficient of y 0 h b f 0
= rArr α+ β+ = = rArr α+ β+ =
rArr ( )
S 0x α β
part = part and
( )
S 0y
α β
part= part
The point of intersection of POSL if D = 0
rArr 2
hf bgab hminus
α =minus
and 2
af gh h abminus
β =minus
Homogeneous equation of degree 2 in x and yax2 + 2hxy + by2
= 0 always represents POSL
(real or imaginary) passing through origin
Straight Line and Pair of Straight Line 14151
A homogeneous equation of degree n represents n straight lines through origin If two POSL have same homogeneous part of degree two in their equation then they always
construct a parallelogram If two POSL S = 0 (L1L2 = 0) and Srsquo = 0 (L1L2 = 0) have common angle
bisectors (B1B2 = 0) then their lines are iso-inclined to each other respectively ie angle between L1 and L1 is equal to angle between L2 and L2 also angle between L1 and L2 is equal to angle between L2 and L1 angle between L1 = 0 and L1 = 0 = angle between L2 = 0 and L2 = 0 = f ndash q also angle between L1 = 0 and L2 = 0 = angle between L2 = 0 and L1 = 0 = f + q
Equation of POSL joining origin to the point of intersection of a curve and a straight line
( ) + +
22 2
Homogeneous LinearHomogeneous
Homogeneous
lx my lx myS = ax + 2hxy + by +2 gx + fy +c = 0n n
Equation of POSL joining origin to the point of intersection of a curve and a straight line Given a straight line lx + my = n hellip (i) and a conic S = ax2 + 2hxy + by2 +2gx + 2fy +c = 0 hellip (ii)
Required a homogeneous equation of degree two that satisfies the coordinates of A(a b) and B(g d)
Since la + mb = n and S(a b) = aa2 + 2hab + bb2 + 2ga + 2fb +c = 0 ∎ If A (x1 y1) B( x2 y2) and C(x3 y3) are the vertices of a DABC
rArr equation of median through A is given by 1 1 1 1
2 2 3 3
x y 1 x y 1x y 1 x y 1 0x y 1 x y 1
+ =
rArr equation of the internal bisector of angle A is (where b = AC and c = AB)
1 1 1 1
2 2 3 3
x y 1 x y 1b x y 1 c x y 1 0
x y 1 x y 1+ =
Chapter 15CirCle and
Family oF CirCle
151 introduction
A circle is the most regular object we know Each point on a circlersquos circumference is equidistant from its centre The shape and symmetry of circle has been fascinating mathematicians since ages
152 definiton of circle
A circle is the locus of a point moving in a plane so that its distance from a fixed point remains constant The fixed point is called centre of the circle and the constant distance is called the radius of the circle
1521 Equation of a Circle in Various Forms
Centrendashradius form Equation of a circle with Centre at (h k) and radius lsquorrsquo is (x ndash h)2 + (y ndash k)2 = r2 Standard Form When centre is (0 0) and radius is lsquoarsquo then the standard
form becomes x2 + y2 = a2
1522 General EquationThe equation x2 + y2 + 2gx + 2fy + c = 0 is called general equation of circle in canonical form Comparing with equation x2 + y2 ndash 2αx ndash 2βy + α2+ β2 ndash r2 = 0 The equation x2 + y2 + 2gx + 2fy + c = 0
can also be written as ( ) ( )22 2 2 2x g (y f ) g f c+ + + = + minus
Hence centre equiv (ndashg ndashf) ie 1 1 coefficient of x coefficient of y2 2
minus minus
and radius equiv + minus2 2g f c
g2 + f2 ndash c gt 0 rArrreal circle with positive radiusg2 + f2 ndash c = 0 rArrrepresent a point circleg2 + f2 ndash c lt 0 rArrrepresent an imaginary
Y
X
r
(h k)
O
Circle and Family of Circle 15153
NoteA general equation of second degree non-homogenous is ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 in x y represents a circle if
Coefficients of x2 = coefficients of y2 ie a = b ne 0
Coefficient of xy is zero ie h = 0
g2 + f2 ndash c le 0
The general equation may be of the form Ax2 + Ay2 + 2Gx + 2Fy + c = 0 represent a equation of circle
Centre = G F
A A
minus minus
and radius = 2 21G F AC
A+ minus
1523 Diametric FormIf (x1 y1) and (x2 y2 ) are the extremities of one of the diameter of a circle then its equation is (x ndash x1) (x ndash x2) + (y ndashy1) (y ndashy2 )= 0
1524 Equation of Circle Thorugh Three Points The equation of circle through three non-collinear points
A(x1 y1) B(x2 y2) C(x3 y3) is
2 2
2 21 1 1 12 22 2 2 22 33 3 3 3
x y x y 1x y x y 1
0x y x y 1x y x y 1
+
+=
+
+
1525 The Carametric Coordinates of any Point on the CircleParametric Equation of Circle When both x and y coordinates of the point on the circle are expressed as a function of single parameter eg t or θ etc then the equation is called parametric equation of circle
Case 1 Standard Equation x2+ y2 = r2 parametric equation x = r cosθ and y = r sinθBy restricting the values of parameter we can express the part of curve (the arc of circle
line segment etc) very conveniently which is not as easy in case of Cartesian equation of curveqisin[0 2p) full circle qisin(0p) upper semicircleqisin(p2p) lower semicircle qisin(a b) circular arc
Case 2 General equation (x ndash a)2 + (y ndash b)2 = r2 parametric equation x = a + r cosq and y = b + r sinq
x y rcos sinminusα minusβ
= =θ θ
where q is parameter and constant represents circle
x y rcos sinminusα minusβ
= =θ θ
where r is parameter and q is constant represents
straight line
Parametric coordinates of any point on the circle x2 + y2 + 2gx + 2fy + c = 0 are 2 2x g g f c cos= minus + + minus θ2 2y f g f c sin= minus + + minus θ (ndashg ndash f) is the centre and 2 2g f c+ minus is the radius of the circle
(x1y1) (x3y3)
(x2y2)
15154 Mathematics at a Glance
1526 Position of a Point with Respect to a CirclePoint P( x1 y1) lies inside on or outside the circle
S = x2 + y2 + 2gx + 2fy +c = 0 accordingly as S1 = x12 + y1
2 + 2gx1 + 2fy1 + c is lt 0 = 0 or gt 0 respectively
rArr 2 2 2 21 1(x g) (y f ) g f c+ + + hArr + minus
rArr (x1 + g)2 + (y1 + f)2 hArr g2 + f 2 ndash crArr x1
2 + y12 + 2gx1 + 2fy1 + c hArr 0 or S1 hArr 0 where
S1 = x12 + y1
2 + 2gx1 + 2fy1 + c So S1 gt 0 rArr (x1 y1) is outside the circle S1 = 0 rArr (x1 y1) is on the circle S1 lt 0 rArr (x1 y1) is inside the circle
Length of tangent from point P to the circleS = x2 + y2 + 2gx + 2fy + c = 0
2 2 2 2 2 2T 1 1L PT PC r (x g) (y f ) (g f c)= = minus = + + + minus + minus
= 2 21 1 1 1 1(x y 2gx 2fy c S+ + + + =
If S1 is called power of point P wrt circle S = 0 radic S1= length of tangent drawn from P to circlebull If P lies outside S1 then is + ve rArrtwo tangents drawnbull If P lies on circle S1 = 0 rArr only one tangent bull If P lies inside circle S1 lt 0 rArrno (imaginary) tangent
1527 Position of a Line with Respect to a CircleLet L = 0 be a line and S = 0 be a circle if lsquorrsquo be the radius of a circle and p be the length of perpendicular from the centre of circle on the line then if
p gt r rArrLine is outside the circle p = r rArrLine touches circlep lt r rArrLine is the chord of circle p = 0 rArrLine is diameter of circle
Notes
(i) Length of the intercept made by the circle on the line is 2 22 r pminus
(ii) The length of the intercept made by the line y = mx +c with the circle x2 + y2 = a2 is 2 2 2
2
a (1 m ) c21 m+ minus+
15271 Condition for Tangency
(i) The line y = mx + c is tangent to the circle x2 + y2 = a2 if and only if c2 = a2(1 + m2) If it is tangent
then the point of contact is given by 2 2ma a
c c minus
(ii) The line lx + my + n = 0 is tangent to the circle x2 + y2 = a2 if and only if n2 = a2 (l2 + m2) If it is
tangent then point of contact is given by 2 2la ma
n n minus minus
Circle and Family of Circle 15155
Note2y mx a 1 m m= plusmn + forall isin is called family of tangents or tangent in term of slope In case the slope of
tangent is given or tangents passing from a given point are to be obtained this formula can be applied
153 equation of tangent and normal
1531 TangentsTangent line to a circle at a point P(x1 y1) is defined as a limiting case of a chord PQ where Q is (x2 y2) such that Q rarr P As Q rarr P ie x2 rarr x1 and y2 rarr y1
Then chord PQ rarrtangent at P rArr Slope of chord PQ rarrslope of tangent at P
rArr 2 12 1
2 1t x x
2 1y y
y ym limx xrarr
rarr
minus=
minus =
2 12 1
1 2 1
x x1 2 1y y
x x xlimy y yrarr
rarr
+minus = minus
+
( )11 1
1
xy y x xy
minus =minus minus rArr T = xx1 + yy1 ndash a2 = 0
Q 2 2 21 1x y a+ = (1)
2 2 22 2x y a+ = (2)
rArr 2 2 2 22 1 2 1(x x ) (y y )minus = minus minus rArr 2 1 1 2
2 1 1 2
y y x xx x y y
minus += minus
minus + If the equations of the circle are given in general form then the equation of tangent to S = x2 + y2 + 2gx + 2fy + c = 0 at a point (x1 y1) is T = xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
1532 Parametric FormEquation of tangent to circle x2 + y2 = a2 at (a cos a a sin a) is x cos a + y sin a = a
Point of intersection of the tangent drawn to the circle x2 + y2= a2 at the point P(a) and Q(b) is
a cos2x
cos2
α+β
=αminusβ
a sin2y
cos2
α+β
=αminusβ
1533 Pair of Tangents
Combined equation of the pair of tangents drawn from an external point lsquoPrsquo to a given circle is SS1 = T2 2 2 2
1 1 1S x y a= + minus and T equiv xx1 + yy1 ndash a2 = 0
Q(h k)
C(00)
PR
T
(x1 y1)
15156 Mathematics at a Glance
1534 Normals
Normal is defined as a line perpendicular to the tangent line to the circle at the point of tangency P(x1 y1)
If the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0
rArr slope of the normal 1
1
y fmx g+
=+
rArr Equation of normal +minus = minus
+1
1 11
y f(y y ) (x x )x g
Equation of normal in determinant form is given by 1 1
x y 1x y 1 0g f 1
=minus minus
Director Circle The locus of point of intersection of two perpendicular tangents is called the director circle The director circle of the circle x2 + y2 = a2 is x2 + y2 = 2a2
Diameter of a circle The locus of middle points of a system of parallel chords of a circle is called the diameter of a circle The diameter of the circle x2 + y2 = r2 corresponding to the system of parallel chords y = mx + c is x + my = 0
Notes
(i) Every diameter passes through the centre of the circle
(ii) A diameter is perpendicular to the system of parallel chords
1535 Equation of Chord with Mid-point as (x1y1)
Slope of chord = 1
1
xy
minus rArr equation of chord minus = minus minus11 1
1
x(y y ) (x x )y
rArr 2 21 1 1 1yy y xx xminus = minus + rArr 2 2
1 1 1 1yy xx x y+ = +
rArr 2 2 2 21 1 1 1xx yy a x y a+ minus = + minus ie T = S1
For any conic section the equation of chord whose mid point is (x1 y1) is given by T = S1
154 chord of contact
From a point P(x1 y1) exterior to a circle two tangents can be drawn to the circle Let these tangents be PA and PB Then the line segment AB is a chord of the circle and is called chord of contact of P(x1 y1) with respect to the circle
If S = 0 is the circle then equation of the chord of contact of P(x1 y1) wrt the circle S = 0 is T = 0
Equation of locus through intersection of S = 0 and Sprime = 0 is S + lSprime = 0 ie (x2 + y2 ndash a2) + l(x2 + y2 ndash xx1 ndash yy1) = 0
Circle and Family of Circle 15157
For l = ndash1 the curve becomes x x1 + y y1 = a2
1541 Relative Position of Two CirclesS1 = x2 + y2 + 2g1x + 2f1y +c1 = 0 and S2 = x2 + y2 + 2g2x + 2f2y + c2 = 0
Case 1 Two circle lies outside each other Distance between centres d gt r1 + r2 Four common tangent (two direct two transverse) PQ divides C1C2 in ratio r1 r2 externallyinternally
Equation of direct common tangent
= =
minusβ = minusα α β
1 2
Two values of m can be obtained from condition thatthis line touches both the circles S 0 and S 0
y m(x ) where P is( )
Similarly we get equation of TCT
bullDirect Common Tangent Length of direct common tangent is defined as distance between
point of contacts ie ( )22D 1 2L MN d r r = = minus minus
Angle between DCT = 2q = 1 1 2| r r |2sind
minus minus
bullTransverse Common Tangent Length of transverse common tangent is defined as distance between point of
contacts ie S and T ( )22T 1 2L ST d r r = = minus + Angle
between TCT = 2a = 1 1 2r r2sind
minus +
Case 2 Two circles touch each other externally C1 C2 = d = r1 + r2
Three common tangent (two DCT and one TCT)
Equation of DCT (obtained as in case I) Equation of TCT is S1 ndash S2 = 0
DR Q
S direct commontangentstransverse
common tangents
Tr1
P
C1
S1
R1 Q1
C2
r2
P1
15158 Mathematics at a Glance
Direct Common Tangent ( )22D 1 2 1 2 1 2L (r r ) r r 2 r r = + minus minus = Angle between DCT = 2q
= 1 1 2
1 2
r r2sinr r
minus minus
+
Transverse Common Tangent ( )22T 1 2 1 2L (r r ) r r 0= + minus + = Angle between TCT = 2a
= 1 1 2
1 2
r r2sinr r
minus minus= π
+
Case 3 Two Circles intersect each other |r1 ndash r2 | lt C1C2 lt r1 + r2 Two common tangent (two DCT and no TCT) Equation of common chord is S1 ndash S2 = 0
2 1 1 2 2 1 1 2
1 2 1 2
r g r g r f r fP r r r r
minus minus
minus minus Equation of DCT
= =
minusβ = minusα α β
1 2Two values of m can be obtained fromcondition that this
line touches both the circles S 0 and S 0
y m(x ) where P is ( )
Direct Common Tangent ( )22D 1 2L MN d r r= = minus minus
Angle between DCT 1 1 2| r r |2 2sind
minus minus θ =
Case IV Two Circles touch each other internally C1 C2 = |r1 ndash r2| Two direct common tangents Equation of DCT S1 ndash S2 = 0
1542 Direct Common Tangent
( ) ( )2 22 2D 1 2 1 2 1 2L d r r (r r ) r r 0= minus minus = minus minus minus =
Angle between DCT = 2q = 1 1 2
1 2
r r2sinr r
minus minus= π
minus
Case V If 0 lt C1C2 = d lt |r1 ndash r2| then the circle lies completely inside other bullAngle of Intersection Angle of intersection (q) between two curve is defined as angle between
their tangents at their point of intersection which is same as angle between their normals at the point of intersection
2 2 21 2
1 2
r r dcos2r r+ minus
θ =
2 2 21 1 2
1 2
r r dcos2r r
minus + minus
rArr θ =
bullOrthogonal Intersection If the angle of intersection is p2 then it is called as orthogonal intersection Condition of orthogonality of the above two circles is
2 2 21 2 r r d+ = 2 2 2 2 2 2
1 2 1 2 2 2 2 1 2 1g f c g f c (g g ) (f f )rArr + minus + + minus = minus + minus
1 2 1 2 1 22(g g f f ) c crArr + = +
Circle and Family of Circle 15159
155 intercept made on coordinate axes by the circle
The intercept made by the circle x2 + y2 + 2gx + 2fy + c = 0Let circle intersect x-axis at two points (x1 0) and ( x2 0) then x1 x2 are roots
x2 + 2gx + c = 0
Length of x-intercept = |x1 ndash x2| = 2 2g cminus
Similarly length of y-intercept = | y1 ndash y2| = 2 2f cminus Conditions that given circle touches
(i) x-axis is g2 = c(ii) y-axis is f2 = c
NotesCircle x2 + y2 + 2gx + 2fy + c = 0 cuts
(i) x-axis in two real coincident or imaginary points according as g2 gt = lt c
(ii) y-axis in two real coincident or imaginary points according as f2 gt = lt c
156 family of circles
General Equation of Circle x2 + y2 + 2gx + 2fy + c = 0 contains three unknown parameters (effective) Therefore three conditions are necessary in order to determine a circle uniquely and if only two conditions are given then the obtained equation contains a parameter and it is described as family of circle
Following are the ways of expressing some known family of circles
1 Equation of circle through intersection of a circle S = 0 and a line L = 0 S + lL = 0
2 Equation of family of circle passing through intersection of two circles S1 = 0 and S2 = 0 is given as 1 1 2S (S S ) 0+λ minus =
3 Family of concentric circles The family of circles with the same centre and different radii is called a family of concentric circles (xndasha)2 + (y ndash b)2 = r2 where (a b) is the fixed point and r is a parameter
15160 Mathematics at a Glance
4 Equation of any circle passes through two points (x1y1) and
(x2y2) 1 2 1 2 1 1
2 2
x y 1(x x )(x x ) (y y )(y y ) x y 1 0
x y 1minus minus + minus minus +λ =
5 Equation of family of circle touching the line with slope m at the point (x1y1) is
2 21 1 1 1(x x ) (y y ) (y y ) m(x x ) 0minus + minus +λ minus minus minus = and if m is
infinite the family of circle is 2 21 1 1(x x ) (y y ) (x x ) 0minus + minus +λ minus =
where lis a parameter
6 Equation of circle circumscribing a triangle with sides L1= 0 L2 = 0 and L3 = 0 is 1 2 2 3 3 1L L L L L L 0+λ +micro = where l m is obtained by applying the condition that coefficient x2 = coefficient y2 and coefficient of xy = 0
7 Family of conic circumscribing a quadrilateral with sides L1 = 0 L2 = 0 L3 = 0 and L4 = 0 taken in order is 1 3 2 4L L L L 0+λ = and condition of concyclic ness and equation of possible circumcircle can be obtained by applying the condition that coefficient of x2 = coefficient of y2 and coefficient xy = 0 and analyzing the outcome mathematically
Circle and Family of Circle 15161
157 radical axes and radical centre
Radical axis of S = 0 and Sprime = 0 is the locus of the point from which the tangents drawn to the two circles are of equal lengths Its equation is given by S ndash Sprime = 0 ( only if coefficients of x2 y2 in both circles are same)
Remarks
(i) If the circles S = 0 and Sprime= 0 intersect each other then their common chord and their radical axis coincide Thus they have the same eqn S ndash Sprime = 0
(ii) If two circles touch each other then their radical axis coincides with their common tangent at their point of contact The equation is again S ndashSprime= 0
bullRadical Centre The common point of intersection of the radical axes of three circles taken two at a time is called the radical centre of the three circles
Tips and Tricks
1 If two circles do not intersect (c1c2 gt r1 + r2) then they have two transverse and two direct common tangents
2 If two circles intersect (c1c2 lt r1 + r2) then they have two direct tangents only
3 If two circles touch externally (c1c2 = r1 + r2) then they have one transverse and two direct common tangents
4 If two circles touch internally (c1c2 = r1 ndash r2) then they have only one common tangent
5 If the point P lies outside the circle then the polar and the chord of contact of this point P are same straight line
6 If the point P lies on the circle then the polar and the tangent to the circle at P are same straight line
7 The coordinates of the pole of the line lx + my + n = 0 with respect to the circel x2 + y2 = a2 are
8 If (x1 y1) is the pole of the line lx + my + n = 0 wrt the circle x2 + y2 + 2gx + 2fy +c = then where r is the radius of the circle
Chapter 16parabola
161 IntroductIon to conIc SectIonS
A conic section or conic is the locus of a point which moves in a plane so that its distance from a fixed point is in a constant ratio to its perpendicular distance from a fixed straight line Conic sections are section obtained when a pair of two vertical cones with same vertex are intersected by a plane in various orientation The point V is called vertex and the line L1 is Axis
The rotating line L2 is called as generator of the cone the vertex separates the cone into two parts known as nappes
Nature of conic sections depends on the position of the intersecting plane with respect to the cone and the angle f made by it with the vertical axis of the cone
Circle When f = 90deg the section is a circle
Ellipse When q ltflt 90deg the section is an ellipse
Parabola If plane is parallel to a generator of the cone (ie when f = q) then section is a parabola
Parabola 16163
Hyperbola When 0 le f lt q the plane cuts through both the nappes and the curves of intersection is hyperbola
Degenerated Conics
When the plane cuts at the vertex of the cone we have the different cases
When q lt f le 90deg then the section is a point
When 0 le f lt q then the section is a pair of intersecting straight lines It is the degenerated case of a hyperbola
Whenf= q then the plane contains a generator of the cone and the section is a coincident straight line
1611 Definition of Various Terms Related to Conics
Focus The fixed point is called the focus of the conic section
Eccentricity The constant ratio (e) is called the eccentricity of the conic section
Directrix The fixed straight line is called the directrix
Axis The straight line passing through the focus and perpendicular to the directrix is called the axis of the conic section
Vertex The point of intersection of conic and the axis are called vertices of conic section
Centre The point which bisects every chord of the conic passing through it is called centre of the conic
Double Ordinate A chord perpendicular to the axis is called double ordinate (normal chord) of the conic section The double ordinate passing through the focus is called the latus rectum
16164 Mathematics at a Glance
1612 General Equation of a ConicIf the focus is (a b) and the directrix is ax + by + c = 0 then the equation of the conic section whose eccentricity is e is given by
According to the definition of conic SP costant ePM
= = or
SP = e PM 2 2
2 2
|ax by c |(x ) (y ) e(a b )
+ +minusα + minusβ =
+ where P(x y) is a
point lying on the conic or 2
2 2 22 2
(ax by c)(x ) (y ) e(a b )+ +
minusα + minusβ =+
The equation of conics is represented by the general equation of second degree ax2 +2hxy + by2 + 2gx + 2fy + c = 0
We know that the discriminant of the above equation is represented by D where
2 2 2abc 2fgh af bg ch∆ = + minus minus minus or a h gh b fg f c
∆ =
Case I When D = 0 then the equation represents degenerate conic
Condition Conic
D = 0 and h2 ndash ab = 0 A pair of coincident lines or parallel linesD = 0 and h2 ndash ab gt 0 A pair of intersecting straight linesD = 0 and h2 ndash ab lt 0 Imaginary pair of straight lines with real point of intersection also
known as point locus
Case II When D ne 0 the equation represents non-degenerate conic
Condition ConicD ne 0 and h = 0 a = b A circleD ne 0 and h2 ndash ab = 0 A parabolaD ne 0 and h2 ndash ab lt 0 An ellipse or empty setD ne 0 and h2 ndash ab gt 0 A hyperbola D ne 0 and h2 ndash ab gt 0 and a + b = 0 A rectangular hyperbola
162 PArABoLA
A parabola is the locus of a point which moves in a plane so that its distance from a fixed point (called focus) is equal to its distance from a fixed straight line (called directrix) It is the conic with e = 1
1621 Standard EquationGiven S(a 0) as focus and the line x + a = 0 as directrix Standard Equation Given S(a 0) as focus and the line x + a = 0
as directrix
Parabola 16165
Focal distance SP PM a h= = + ( )2 2h a k a hrArr minus + = +
2 2 2 2 2a h 2ah k a h 2ahrArr + minus + = + + 2k 4ahrArr = 2y 4axrArr =
Equation of parabola y2 = 4ax a gt0 Opening rightwards passing through origin Parametric equation x = at2y = 2 at where t isin ℝ Focus S(a o) vertex (0 0) Axis y = 0 Directrix x + a = 0 TV x = 0 Focal distance=a + h Latus rectum Equation x ndash a = 0 and length 4a extremities (a plusmn2a)
Equation (a gt 0) Axis Focus Directrix Latus rectum Graph
y2 = 4axx = at2
y = 2at
y = 0 (a 0) x + a = 0 x = a 4a (a plusmn2a)
y2 = ndash4axx = ndashat2
y = 2at
y = 0 (ndasha 0) x ndash a = 0 x = ndasha 4a(ndasha plusmn2a)
x2 = 4ayy = at2
x = 2at
x = 0 (0 a) y + a = 0 y = a 4a(plusmn2a a)
x2 = ndash4ayy = ndashat2
x = 2at
x = 0 (0 ndasha) y ndash a = 0 y = ndasha 4a(plusmn2a ndasha)
Equation of parabola with length of LR (latus rectum) = 4a vertex at (a b) and axis is given as (y ndash β)2 = plusmn 4a(x ndash α)
16166 Mathematics at a Glance
Focus (α plusmn a β) Axis y ndash β = 0 TV (transverse axis) x ndash α = 0 Parametric equation 2 2( at 2at)( at 2at)α+ β+ αminus β+ Directrix x a= α Extremetric ( a 2a)αminus βplusmn ( a 2a)αminus βplusmn
Focus lies at 14th of the latus rectum away from vertex along axis towards parabola
Equation of parabola with length of LR = 4a vertex at (a b) and axis parallel to y-axis is given as (x ndash α)2 = plusmn4a(y ndash β) Focus (α β plusmn a) Axis x ndash α = 0 TV y ndash β = 0 Directrix y a=β
Ends of LR ( 2a a)( 2a a)αplusmn β+ αplusmn βminus
Parametric equation 2 2( 2at y at )( 2at y at )α+ =β+ α+ =βminus
NoteEquation of general parabola with axis lx + my + n = 0 and TV is mx ndash ly + k = 0 and LR is of length 4a
is given as 2 2 2( lx my n) LR l m ( mx ly k )+ + = plusmn + minus +
1622 Position of Point wrt ParabolaThe region towards focus is defined as inside region of parabola and towards directrix is outside region of parabola
Given a parabola y2 = 4ax and a point P(x1 y1) Point P lies inside hArr S1 lt 0 Point P lies on parabola hArr S1 = 0 Point P lies outside parabola hArr S1 gt 0
1623 Position of Line wrt ParabolaWhether the straight line y = mx + c cutstoucheshas no contact with the parabola y2 = 4ax can be determined by solving the parabola and straight line together
2 y cy 4a 0mminus minus =
(mx + c)2 ndash 4ax = 0 which is m2x2 + (2cm ndash 4a) x + c2 = 0
2 4a 4acy y 0m m
rArr minus + = (i) D gt 0 rArr line cuts at two distinct point
1 2 1 24a 4acy y and y ym m
rArr + = =
(ii) D = 0 rArr line touches the parabolaCondition of tangency D =0
( )2
16a a cm 0m
minus =
acm
rArr = (iii) D lt 0 rArr line has no contact
Parabola 16167
rArr ay mx m ~ 0m
= + forall isin known as family of tangent with slope m is tangent to the parabola
y2 = 4ax
rArr Point of contact 2
a 2am m
22
a 2a 1 (at 2at) mm m t
hArr rArr =
rArr Parametric equation of tangent at point lsquotrsquo is given as 2xy at yt x att
= + rArr = +
163 chordS of PArABoLA And ItS ProPertIeS
Given a parabola y2 = 4ax let AB be the chord joining A(x1 y1) and B(x2 y2)
Q 2 21 1 2 2y 4ax and y 4ax= = rArr y2
2 ndash y12 = 4a(x2 ndash x1) rArr 2 1
2 1 1 2
y y 4ax x y yminus
=minus +
rArr Slope of chord 2 1
2 1
y yABx xminus
=minus
= 1 21 2
4a 2ay yy y
2
=++
Equation of chord ( )1 11 2
4ay y x xy y
minus = minus+
Condition to be a focal chord rArr y1y2 = ndash4a2 and x1x2 = a2 ie t1t2 = ndash1
1631 Chord of Parabola in Parametric Form
1 2
2Slope of chordt t
=+
Equation of chord ( )21 1
1 2
2y 2at x att t
minus = minus+
For focal chord Put y = 0 x = a rArr 0 = 2a(1 + t1t2) rArr t1t2 = ndash1
1632 Properties of Focal Chord
A focal chord is basically a chord passing through the focus of the parabola
Extremeties of focal chord P(at2 2at) and minus 2
a 2aQ t t
Segments of focal chord SP = l1 = a + at2 2 2
aSQ l at
= = +
HM of segments of focal chord is semi latus-rectum 2a
Length of focal chord 21L a t
t rArr = +
Slope of focal chord 2tt 1
Equation of focal chord ( )2
2ty x at 1
= minusminus
16168 Mathematics at a Glance
Notes
(i) Equation of chord with mid-point M (x1 y1)
rArr 21 1 1 1yy 2a( x x ) y 4axminus + = minus ie T = S1
(ii) Equation of a chord of contact formed by joining the points of contacts of the tangents drawn form point A to the parabola
Chord of contact is yy1 ndash 2a(x + x1) = 0 ie T = 0
164 tAnGent of PArABoLA And ItS ProPertIeS
Tangent to a parabola at P(x1 y1) rArr T yy1 ndash 2a(x + x1) = 0
Tangent to the parabola at P(at2 2at)
rArr yt = x + at2
1641 Properties of Tangents of a Parabola
If the point of intersection of tangents at t1 and t2 on the parabola be T then T (at1 t2 a (t1 + t2))
If T be the point of intersection of tangent at P and Q then SP ST SQ are in GP
ie ST = SPSQ
Consider the parabola shown in the diagram below
Coordinate of T (ndashat2 0) coordinate of Y (0 at) SP = ST = PM = SG = a + at2
angMPT = angSTP = angSPT = q
Parabola 16169
Reflection Property of Parabola Light rays emerging from focus after reflection become parallel to the axis of parabolic mirror and all light rays coming parallel to axis of parabola converge at focus
Foot of perpendicular from focus upon any tangent lies at Y(0 at) on the tangent at vertex (TV)
SY is median and DSPT is isosceles SY is altitude ie SY is perpendicular to PT angTSY = angYSP = p2 ndash q and SY = MY rArr SPMT is rhombus
Points A B and C lie on the parabola y2 = 4ax The tangents
to the parabola at A B and C taken in pairs intersect at points P Q and R respectively then the ratio of the areas of the DABC and DPQR is 2 1
Tangent at any point on parabola bisects the internal angle between focal distances SP and PM
Normal at P bisects the external angle between SP and PM The portion of the tangent intercepted between axis and point
of contact is bisected by tangent at vertex Y is the mid-point of PT SY is median and DSPT is isosceles
SY is altitude angTSY = ang YSP = p2 ndash q and SY = MY rArr SPMT is rhombus
Equation of a pair of tangents to the parabola form P(x1y1) SS1 =T2
2 2 21 1 1 1(y 4ax)(y 4ax ) [yy 2a(x x )]minus minus = minus +
165 norMALS And theIr ProPertIeS
Given a parabola y2 = 4ax at point lsquotrsquoSlope of normal m = ndashtEquation of normal y ndash 2at = ndasht(x ndash at2)rArr y + xt = 2at + at3
16170 Mathematics at a Glance
1651 Properties
Coordinate of G = (2a + at2 0)
If the normal at P(t) meets the parabola at Q(t1) then = minus minust t
If the normal to the parabola y2 = 4ax at point P(t1) and Q(t2) cuts the parabola at some point R (t3) then
(i) t1t2 = 2 (ii) t3 = ndash(t1 + t2)
1652 Normals in Terms of SlopeSince Equation of normal y + xt = 2at + at3 at (at2 2at)Put t = ndashm rArr y = mx ndash 2am ndash am3 where foot of normal is (am2 ndash2am) From any point P(h k) in the plane of the parabola three normals can be drawn to the parabola The
foot of these normals are called co-normal points of the parabola rArr Sum of ordinate of foot of conformal points yP + yQ + yR = ndash2a (m1 + m2 + m3) = 0 where m1 m2 m3
are the slopes of the three normals Sum of the slopes of the concurrent normals to a parabola is zero Centroid of the triangle joining the
co-normal point P Q R lies on the axis of the parabola Necessary condition for existence of three real normal through the point (h k) is h gt 2a if a gt 0
and h lt 2a if a lt 0But the converse of statement is not true ie if h gt 2a if a gt 0 and h lt 2a if a lt 0 does not necessarily implies that the three normals are real
Sufficient condition for 3 real normals from (h k) f(m) = am3 + (2a ndash h)m + k it has 3 real and distinct rootsIf f rsquo(m) = 3am2 + 2a ndash h = 0 has 2 real and distinct roots ie
h 2am say 3aminus
= plusmn α βsufficient
condition for 3 real slopes is f (a) f(b) lt 0 rArr f(a) f(b) lt 0 rArr f(a)(ndasha) lt 0 rArr 27ak2 lt 4(h ndash 2a)3
Atmost there are four concylic point on the parabola and sum of ordinates of these points vanishes
rArr Sum of ordinates of four concyclic points on parabola Since 2a(t1 + t2 + t3 + t4) = 0
Parabola 16171
Pair of chord obtained by joining any four concyclic points are equally inclined to the axis of the parabola
Circle passing through any three co-normal points on the parabola also passes through the vertex of the parabola
Table representing the equations of tangents in different forms and related terms
Equation y2 = 4ax y2 = ndash4ax x2 = 4ay x2 = ndash4ay
Tangent in point form yy1 = 2a(x + x1) yy1 = ndash2a(x + x1) xx1 = 2a(y + y1) xx1 = ndash2a(y + y1)
Parametric co-ordinate (at2 2at) (ndashat2 2at) (2at at2) (2at ndashat2)
Tangent in parametric form
ty = x + at2 ty = ndashx + at2 tx = y + at2 tx = ndashx + at2
Point of contact in terms of slope (m) 2
a 2am m
2
a 2am m
minus minus
(2am am2) (ndash2am ndasham2)
Condition of tangency acm
=acm
= minusc = ndasham2 c = am2
Tangent in slope form ay mxm
= +ay mxm
= minusy = mx ndash am2 y = mx + am2
Table representing the equations of tangents in different forms and related terms to parabolas having vertex at (h k) and axes parallel to co-ordinate axes
Equation (y ndash k)2 = 4a(x ndash h) (y ndash k)2 = ndash4a(x ndash h) (x ndash h)2 = 4a(y ndash k) (x ndash h)2 = ndash4a(y ndash k)
Tangent in point form
(y ndash y1)(y ndash k) = 2a(x ndash x1)
(y ndash y1)(y ndash k) = ndash2a (x ndash x1)
(x ndash x1)(x1 ndash h) = 2a (y ndash y1)
(x ndash x1)(x1 ndash h) = ndash2a(y ndash y1)
Parametric co-ordinate
(h + at2 k + 2at) (h ndash at2 k + 2at) (h + 2at k + at2) (h + 2at k ndash at2)
Tangent in parametric form
t(y ndash k) = (x ndash h) + at2
t(y ndash k) = ndash(x ndash h) + at2
t(x ndash h) = (y ndash k) + at2
t(x ndash h) = ndash(y ndash k) + at2
Point of con-tact in terms of slope (m)
2
a 2ah km m
+ + 2
a 2ah km m
minus minus
(h + 2am k + am2) (h ndash 2am k ndash am2)
Condition of tangency
ac mh km
+ = +ac mh km
+ = minusc + mh = k ndash am2 c + mh = k + am2
Tangent in slope form
ay mx mh km
= minus + +ay mx mh km
= minus + minusy = mx ndash mh + k ndash am2
y = mx ndash mh + k + am2
16172 Mathematics at a Glance
Tabl
e re
pres
entin
g th
e eq
uatio
ns o
f nor
mal
and
rela
ted
term
s to
stan
dard
par
abol
as in
diffe
rent
form
s
Equa
tion
of P
arab
ola
y2 = 4
axy2 =
ndash4a
xx2 =
4ay
x2 = ndash
4ay
Equa
tion
of n
orm
al in
po
int f
orm
11
1y
yy
(xx
)2aminus
minus=
minus1
11
yy
y(x
x)
2aminus
=minus
11
12ay
y(x
x)
xminusminus
=minus
11
12ay
y(x
x)
xminus
=minus
Para
met
ric co
-ord
inat
e(a
t2 2at
)(ndash
at2 2
at)
(2at
at2 )
(2at
ndashat
2 )N
orm
al in
par
amet
ric fo
rmy
+ tx
= 2
at +
at3
y ndash
tx =
2at
+ at
3x
+ ty
= 2
at +
at3
x ndash
ty =
2at
+ at
3 Po
int o
f con
tact
in te
rms o
f slo
pe (m
)(a
m2 ndash
2am
)(ndash
am2 2
am)
2
2aa
m
m
minus
2
2aa
m
m
Con
ditio
n of
nor
mal
ityc =
ndash2a
m ndash
am
3c =
2am
+ a
m3
2ac
2am
=+
2a
c2a
m=minus
minus
Nor
mal
in sl
ope
form
y =
mx
ndash 2a
m ndash
am
3y
= m
x +
2am
+ a
m3
2ay
mx
2am
=+
+2a
ym
x2a
m=
minusminus
Equa
tion
of P
arab
ola
(y ndash
k)2 =
4a(
x ndash
h)(y
ndash k
)2 = ndash
4a(x
ndash h
)(x
ndash h
)2 = 4
a(y
ndash k)
(x ndash
h)2 =
ndash4a
(y ndash
k)
Equa
tion
of n
orm
al in
po
int f
orm
11
1(y
k)y
y(x
x)
2aminus
minusminus
=minus
11
1(y
k)y
y(x
x)
2aminusminus
=minus
11
1
2ay
y(x
x)
xh
minusminus
=minus
minus1
112a
yy
(xx
)x
hminus
=minus
minus
Nor
mal
in p
aram
etric
fo
rm(y
ndash k
) + t(
x ndash
h) =
2a
t + at
3(y
ndash k
) ndash t(
x ndash
h) =
2a
t + at
3(x
ndash h
) + t(
y ndash
k) =
2a
t + at
3(x
ndash h
) ndash t(
y ndash
k) =
2a
t + at
3
Poin
t of c
onta
ct in
term
s of
slop
e (m
)(h
+ a
m2 k
ndash 2
am)
(h ndash
am
2 k +
2am
)2
2aa
hk
)m
m
minus
+
2
2aa
hk
)m
m
+
minus
Con
ditio
n of
no
rmal
ityc =
k ndash
mh
ndash 2a
m ndash
am
3c =
k ndash
mh
+ 2a
m +
am
3
2ac
km
h2a
m=
minus+
+2a
ck
mh
2am
=minus
minusminus
Nor
mal
in sl
ope
form
(y ndash
k) =
m(x
ndash h
)ndash 2
am
ndash am
3(y
ndash k
) = m
(x ndash
h)+
2am
+
am3
2
(yk)
m(x
h)a
2am
minus=
minus+
+2
(yk)
m(x
h)a
2am
minus=
minusminus
minus
Chapter 17ellipse
171 Definition
Ellipse is the locus of a point which moves in a plane such that the ratio of its distance from a fixed point (Focus) to its distance from the fixed line (Directrix) is always constant and equal to a quantity which is less than 1
172 StAnDARD eQUAtion of eLLiPSe
Given focus S(ae 0) and the x ndash (ae) = 0 as directrix
1721 Focal DistanceFocal distance (SP) of a point P is given as
Q
aSP ePM e h a ehe
= = minus = minus
rArr 2 2(h ae) k a ehminus + = minus
rArr 2 2 2 2 2 2 2a e h 2aeh k a e h 2aeh+ minus + = + minus
17174 Mathematics at a Glance
rArr 2 2 2 2 2 2 2a e h k a e h+ + = + rArr 2 2 2 2 2 2 2h e h k a a eminus + = minus
rArr 2 2 2 2 2h (1 e ) k a (1 e )minus + = minus rArr 2 2
2 2 2
h k 1a a (1 e )
+ =minus
Let 2 2 2a (1 e ) bminus = rArr 2 2
2 2
x y 1a b
+ =
173 tRAcing of eLLiPSe
Equation of Ellipse 2 2
2 2
x y 1a b
+ =
Eccentricity 2
2
be 1a
= minus
Symmetry Since curve is even wrt variable x and y the graph is symmetric about both the co-ordinate axes There are two foci and two directrices
Foci S1 (ae 0) S2 (ndashae 0) Directrices D1 x = ae D2 x = ndashae
Focal distances S1P = ePM = a ndash eh 2aS P = ePM = e he
+
= a + eh
AAprimeis called major axis length = 2a equation y = 0 BBprimeis called minor axis length= 2b equation x = 0 The point of intersection of major and minor is called
centre All the chords passing through the centre get bisected at the centre
Normal chord Chord normal to the major axis is called normal chord or double ordinate If it passes through the focus it is called latus rectum
Length of 22bLR =
a equation of LR x = ae
Ellipse 17175
Ellipse is a locus of the point that moves in such a manner so that the summation of its distances from two fixed points S1 and S2 (foci) remains constant (2a)
S1P + S2P = 2a where 2a is length of major axis Case I If 2a gt S1S2 = 2ae locus ellipse Case II S1P + S2P = S1S2 locus segment SSprime Case III S1P + S2P lt S1S1 no locus
If equation of ellipse is 2 2
2 2
x y 1a b
+ = where b gt a
Eccentricity e = radic1ndash(a2b2) Major axis x = 0
Length of Major axis 2b Minor axis y = 0Length of Minor axis 2a foci (0 plusmn be)LR y = plusmn be length of LR = 2a2b Extremities (plusmna2b be)
Equation of ellipse where centre lies at (a b) and major axis is parallel to the x-axis of length 2a and
minor axis of 2b (a gt b) 2 2
2 2
(x ) (y ) 1a bminusα minusβ
+ =
Major axis y = b Length of Major axis 2aMinor axis x = a Length of Minor axis 2bFoci S1 = (a + ae b)
S2 = (a ndash ae b)Directrix x = a + ae
x = a ndashae
Auxiliary Circle of an Ellipse A circle drawn on major axis of the ellipse as diameter is called
Auxiliary circle of ellipse Given the equation of ellipse 2 2
2 2
x yS 1a b
+ =
The equation of auxiliary circle 2 2 2x y a + =
Eccentric Angle Of any point P on the ellipse is angle (q) made by CPprimewith positive direction of major axis in anti-clockwise sense (where C is centre and Pprimeis corresponding point of P on Auxiliary circle)
17176 Mathematics at a Glance
Q P Px x a cos= = θ rArr 2 2 2
2 2
a cos y 1a b
θ+ = rArr y2 = b2 sin2q
Parametric equation xp = a cosq and yp = b sinq isin [0 2p) (a cosq b sinq) is called point q an the ellipse
174 PRoPeRtieS ReLAteD to eLLiPSe AnD AUxiLiARy ciRcLe
The ratio of ordinate of point P on ellipse and its corresponding point on AC is constant PM bsin bPM a cos a
θ= =
θ
The ratio of area of triangle inscribed in ellipse (DPQR) to the area of triangle (DPprimeQprimeRprime) formed by its corresponding point an AC is constant = ba
The above property is valid even for an n-sided polygon inscribed in the ellipse As n rarrinfin is the polygon that coincides with the ellipse and its corresponding polygon coincides with auxiliary circle
Ellipse 17177
1741 Position of a Point with Respect to Ellipse + minusyxS 1 = 0a b
22
2 2
A point P(x1y1) lies insideonoutside of ellipse as S1 lt 0S1 = 0S1 gt 0
1742 Position of a Line with Respect to EllipseThe Straight line y = mx + c cutstoucheshas no contact with ellipse
2 2
2 2
x yS 1 0a b
+ minus = as the equation b2x2 + a2 (mx + c) 2 ndash a2b2 = 0 has D gt 0D = 0D lt 0
Condition of tangency 2 2 2a m bplusmn + Thus all lines of the form 2 2 2y mx a m b= plusmn + will always be tangent to the ellipse where m is real
Equation of tangent in terms of slope also known as ever tangent 2 2 2y mx a m b= plusmn + and point
of contact is 2 2a m b c c
minus
Chord of ellipse joining point q and f
Slope of chord of joining point q and f b cota 2
θ+ φ = minus
Equation of chord x ycos sin cosa 2 b 2 2
θ+ φ θ+ φ θminusφ+ =
Condition of focal chord If Passes through (ae 0)
or (ndashae 0) rArr e 1 e 1tan tan or2 2 e 1 e 1θ φ minus +
=+ minus
Equation of tangent at q (a cosq b sinq) x ycos sin 1a b
θ+ θ =
Equation of Normal at q Slope am tanb
= θ rArr Equation a siny bsin (x a cos )b cos
θminus θ = minus θ
θ rArr ax sec q ndash by cosec q = a2 ndash b2 = a2e2
Equation of tangent 1 12 2
xx yyT 1 0a b
+ minus = and equation of Normal 2 2
2 2 2 2
1 1
a x b y a b a e x y
minus = minus =
175 PRoPeRtieS of tAngentS AnD noRmALS
The slopes and equations of various tangents and normals are given by
Construction Slope Equation
Tangent at (x1y1)2
12
1
b xa y
minus 1 12 2
xx yy 1 0a b
+ minus =
Tangent at qb cota
minus θx ycos sin 1a b
θ+ θ =
17178 Mathematics at a Glance
Construction Slope Equation
Normal at (x1y1)2
12
1
a yb x
2 22 2
1 1 2 2a e
a x b y a bx y
minus = minus
Normal at qa tanb
θ2 2
2 2a eax sec bycosec a bθminus θ = minus
Point of Intersection of Tangent Point of intersection of tangent at
lsquofrsquoand lsquoqrsquoon the ellipse 2 2
2 2
x y 1a b
+ = is
a cos bsin2 2
cos cos2 2
θ+ φ θ+ φ
θminusφ θminusφ
Locus of foot of perpendicular from either foci upon any tangent is auxiliary circle of ellipse
Locus of point of intersection of a perpendicular tangents is the director circle of ellipse in fact the locus of point of intersection of perpendicular tangents (in case of conic sections other than parabola) is called lsquodirector circle of conic sectionrsquo
Product of length of perpendiculars from both foci upon any tangent is constant (b2) where b is length of semi-major axis of ellipseproduct of the lengthrsquos of the perpendiculars from either foci on a variable tangent to an EllipseHyperbola = (semi minor axis)2(semi conjugate axis)2 = b2
Tangent at any point (P) bisects the external angle and nor-mal at same point bisects the internal angle between fo-cal distances of P This refers to the well-known reflection property of the ellipse which states that rays from one are reflected through other focus and vice-versa
Ellipse 17179
In general four normals can be drawn to an ellipse from any point and if a b d g are the eccentric angles of these four co-normal points then a + b + d + g is an odd multiple of p
In general there are four concyclic points on an ellipse and if a b d g are the eccentric angles of these four points then a + b + d + g is an even multiple of p
The circle on any focal distance as diameter touches the auxiliary circle The straight lines joining each focus to the foot of the perpendicular from the other focus upon the
tangent at any point P meet on the normal PG and bisects it where G is the point where normal at P meets the major axis
Chord of contact 1 12 2
xx yyT 1 0a b
= + minus =
Pair of tangents SS1 = T2 2 2
2 2
x y 1a b
+ minus
21 1
2 2
xx yy 1a b
+ minus
Chord with a given middle point T = S1 2 2
1 1 1 12 2 2 2
xx yy x y1 1a b a b
+ minus = + minus
rArr 2 2
1 1 1 12 2 2 2
xx yy x ya b a b
+ = +
Diameter The locus of the mid points of a system of parallel chords of an ellipse is called the diameter and the point where the diameter intersects the ellipse is called the vertex of the diameter
If y = mx + c is the system of parallel chords to 2 2
2 2
x y 1a b
+ =
then the locus of the midpoint is given
by 2
2
b xya m
= minus
Conjugate diameter Two diameters are said to be conjugate if each bisects all chords parallel to the other
If conjugate diameters are perpendicular to each other then ellipse becomes a circle The eccentric angles of the ends of a pair of conjugate diameters of an ellipse differ by a right angle The sum of squares of any two conjugate semi-diameters of an ellipse is constant and is equal to sum
of the squares of the semi-axes of the ellipse The product of the focal distances of a point on an ellipse is equal to the square of the semi-diameter
which is conjugate to the diameter through the point The tangents at the extremities of a pair of conjugate diameters form a parallelogram whose area is
constant and is equal to the area of rectangle formed by major and minor axis lengths
Chapter 18hyperbola
181 Definition
It is the locus of a point P whose ratio of distance from a fixed point (S) to a fixed line (Directrix) remains constant (e) is known as the eccentricity of hyperbola (e gt 1)
1811 Standard EquationGiven S(ae 0) as focus and the line x ndash (ae) = 0 as directrix
Focal Distance Focal distance of a point P is given as Q SP = ePM = eh ndash a rArr a2e2 + h2 ndash 2aeh + k2 = a2 + e2h2 ndash 2aeh
rArr h2(1 ndash e2) + k2 = a2(1 ndash e2) rArr minus =minus
2 2
2 2 2
h k 1a a (e 1)
rArr minus =2 2
2 2
x y 1a b
where a2(e2 ndash 1) = b2
1812 Tracing of Hyperbola
Equation of hyperbola minus =2 2
2 2
x y 1a b
Eccentricity = +2
2
be 1a
Symmetry Since equation is even wrt variable x and y so graph is symmetric about both co-ordinate axes Hence there should be two foci and two directrix
Hyperbola 18181
Foci S1(ae 0) S2 (ndashae 0) Directrices D1 x = ae D2 x = ndashae Intersection with x-axis y = 0 rArr x = plusmn a rArr A(a 0) Aprime(ndasha 0)
AAprime is called transverse axis of hyperbola length = 2a equation y = 0 Intersection with y-axis x = 0 rArr y = plusmnbi rArr B(0 b) Bprime(0 ndashb)
BBprime is called conjugate axis length = 2b equation x = 0The point of intersection of transverse and conjugate is called centre
Normal chord Chord normal to transverse axis is called normal chord or double ordinate If it passes through focus it is called latus rectum
Extremities of Latus rectum
=
2
1bL aea
and
= minus
21
bL aea
Length of =22bRR
aequation x = +ae ndashae
Focal distances S1P = ePM = eh ndash a S2P = ePMprime = eh + a|S2P ndash S1P| = 2a where 2a is length of transverse axisCase I If 2a lt S1S2 = 2ae rArr hyperbolaCase II If S1P + S2P = S1S2 rArr union of two raysCase III If S1P + S2P gt S1S2
rArr No locus
Conjugate hyperbola of a hyperbola H = 0 is a hyperbola C = 0 whose transverse axis is conjugate axis of H = 0 and conjugate axis is transverse axis of H = 0 both in the sense of length and equation
Equation hyperbola minus =2 2
2 2
x y 1a b
Conjugate hyperbola minus = minus2 2
2 2
x y 1a b
Eccentricity = + 2 22e 1 (a b ) Foci (0 plusmnbe2)
18182 Mathematics at a Glance
Transverse axis x = 0 Length = 2b Conjugate axis y = 0 Length = 2a
Latus Rectum y = plusmn be2 LprimeL plusmn
2
2a beb
and length = 22a
b
+ = + =+ +
2 2
2 2 2 2 2 22 1
1 1 b a 1e e a b a b
The foci of a hyperbola and its conjugate are con-cylic and form the vertices of a square
If a = b hyperbola is said to be equilateral or rectangular and has the equation x2 ndash y2 = a2 Eccentricity for such a hyperbola is radic2Equation of hyperbola whose centre lies at (a b) and trans-verse axis is parallel to x-axis of length 2a and conjugate axis of
length 2b equation minusα minusβminus =
2 2
2 2
(x ) (y ) 1a b
Transverse axis y = b Length = 2a Conjugate axis x = a Length = 2b Foci S1 = (a + ae b) S2 = (a ndash ae b) Directrix D1 x = a + ae x = a ndashae
Equation of Hyperbola Referred to two perpendicular straight lines as their axes but not parallel to coordinate axes
( ) + +minus + + + minus =
2 2
1 1 11 1 22 2 2 21 1 1 1
2 2
l x m y nm x l y nm l l m
1a b
Centre C is the point of intersection of line l1x + m1y + n1 = 0 and m1x ndash l1y + n2 = 0
Equations of Directrices If (x y) is any point on a directrix then its ^r distance from conjugate axis ie m1x ndash l1y + n2 = 0 is ae
Equation of directrices are given by minus +
= plusmn+
1 1 22 21 1
m x l y n aem l
Hyperbola 18183
Foci Foci can be obtained by solving the equation l1x + m1y + n1 = 0 and the pair of normal chords
(Latera Recta) minus +
= plusmn+
1 1 22 21 1
m x l y n aem l
Length of each Latera Recta =22b
a Equations of Latera Recta are given by
minus += plusmn
+1 1 2
2 21 1
m x l y n aem l
1813 Auxiliary Circle of HyperbolaA circle drawn on transverse axis of the hyperbola as diameter is called auxiliary circle of hyperbola for
hyperbola minus =2 2
2 2
x y 1a b
auxiliary circle is given by x2 + y2 = a2
Eccentric Angle Of any point P on the hyperbola is angle (q) made by CPprime with positive direction of transverse axis in anticlockwise (where C is centre and Pprime is point of contact of tangent drawn from foot of ordinate of P to the Auxiliary circle)
Parametric Equation x = a secq and y = btanq π π θisin π
3[02 ) 2 2
and (a secq b tanq) is called
point q an the hyperbola The ratio of ordinate of point P on hyperbola and length of tangent from the foot of ordinate (M) to
the Auxiliary circle is constant (ba) θ= =
θPM bsin bPM a cos a
182 Director circle
The locus of the point of intersection of the tangents to the
hyperbola minus =2 2
2 2
x y 1a b
which are perpendicular to each other
is called director circleThe equation of director circle is P(h k) is x2 + y2 = a2 ndash b2
(a gt b)
1821 Position of a Point with Respect to Hyperbola
Given hyperbola minus minus =2 2
2 2
x y 1 0a b
rArr = minus2
2 2 22
by (x a )a
A point P(x1 y1) lies inside (towards centre)on
outside (towards focus) of hyperbola as S1 lt 0S1 = 0S1 gt 0
1822 Position of a Line with Respect to Hyperbola minus minus22
2 2
yxS 1 = 0a b
The straight line y = mx + c cutstoucheshas no contact with hyperbola minus =2 2
2 2
x y 1a b
as the equation
b2x2 ndash a2 (mx + c)2 ndash a2b2 = 0 has D gt 0D = 0D lt 0
18184 Mathematics at a Glance
Condition of tangency = plusmn minus2 2 2c a m b
Equation of tangent in terms of slope = plusmn minus2 2 2y mx a m b and
point of contact is minus minus
2 2a m b c c
Chord of Hyperbola Joining Point q and fEquation of chord of hyperbola Joining Point q and f
is θ θ =φ φ
x y 1a sec btan 1 0a sec btan 1
which can also be written as
θminusφ θ+ φ θ+ φminus =
x ycos sin cos a 2 b 2 2
Condition for Focal Chord Chord becomes focal chord if it passes through (ae0) or (ndashae 0) Sup-
pose it passes through (ae0) then rArr θ φ minus = +
1 etan tan 2 2 1 e
or +minus
1 e1 e
if it passes through (ndashae 0)
1823 Properties of Tangents and Normals
Construction Slope Equation
Tangent at (x1y1)2
12
1
b xa y
minus minus =1 12 2
xx yy 1 0a b
Tangent at q θb coseca
θminus θ =x ysec tan 1a b
Normal at (x1y1) minus2
12
1
a yb x
+ = +
2 22 2
1 1 2 2a e
a x b y a bx y
Normal at q minus θa sinb
θ+ θ = +
2 2
2 2a eax cos by cot a b
Point of intersection of tangent at q and f on
the hyperbola minus =2 2
2 2
x y 1a b
θminusφ
=θ+φ1
cos2x a
cos2
θ+ φ
=θ+φ1
sin2y b
cos2
Tangent at any point (P) bisects the internal angle and normal at same point bisects the external angle between focal distances of P This refers to reflection property of the hyperbola which states that rays from one Focus are reflected such that they appear to be coming from other focus
Hyperbola 18185
An ellipse and hyperbola if con-focal always intersect orthogonally
Chord of contact = minus minus =1 12 2
xx yyT 1 0a b
Pair of tangents = minus minus minus minus = minus minus
22 22 21 1 1 1
1 2 2 2 2 2 2 2
x y xx yyx ySS T 1 1 1a b a b a b
Chord with a given middle point = minus minus = minus minus
2 21 1 1 1
1 2 2 2 2
xx yy x yT S 1 1a b a b
rArr minus = minus2 2
1 1 1 12 2 2 2
xx yy x ya b a b
1824 Asymptote Hyperbola
Asymptote to any curve is straight line at finite distance that touches the curve at infinity (micro)
Let y = mx + c be asymptote to hyperbola then both roots of the equation (b2 ndash a2m2)x2 ndash 2a2cmx ndash a2 (c2 + b2) = 0 approach to micro
rArr minus =
2 2 2
sum of root infinity
b a m 0 and 2 2 2 2
condition of tangency
c a m b= minus
rArr = plusmnbma
and c = 0
rArr
= minus =
by xa
by xa
are pair of asymptotes
18241 Properties of asymptote hyperbola
Both the asymptotes are pair of tangents to a hyperbola from its centre Axis of Hyperbola bisects the angle between asymptotes
If lines be drawn through A Aprime parallel to C axis and through B Bprime parallel to T axis then asymptotes lie along the diagonal of rectangle thus formed
Combined equation of asymptotes (A = 0) differs from equation of hyperbola (H = 0) and conjugate hyperbola (C = 0) by same constant ie A = H + l and A = C ndash l
As minus =2 2
2 2
x yH 1a b
and minus =2 2
2 2
x yA 0 2a b
minus = minus2 2
2 2
x yC 1a b
Relation between A C H+
=C HA
2 Angle between Asymptote Included angle between two asymptotes is
minus minus = minus 1 1
2 2
2ab btan 2tana b a
or 2 Secndash1(e)
If the angle asymptotes is 90deg then b = a and hyperbola is called rectangular hyperbola The product of the perpendicular drawn from any point on a hyperbola to its asymptotes is constant The foot of the perpendicular from a focus to an asymptote is a point of intersection of the auxiliary
circle and the corresponding directix
18186 Mathematics at a Glance
The portion of any tangent to hyperbola intercepted between asymptote is bisected at the point of contact
Any tangent to the hyperbola makes with asymptote a triangle of constant area
183 rectangular hyperbola
A hyperbola whose asymptotes include a right angle is called rectangular hyperbola or if the lengths of transverse and conjugate axes of a hyperbola be equal it is called rectangular or equilateral hyperbola Equation x2 ndash y2 = a2 TA y = 0 Length 2a CA x = 0 Length 2a Eccentricity (e) = radic2 Foci (plusmnaradic2 0) Directrix x = plusmnaradic2 Asymptote y = x and y = x
1831 Rectangular Hyperbola where Asymptote are Coordinate Axis
Given rectangular hyperbola x2 ndash y2 = a2 If axes rotating by p4 about the origin+
rarrx yx
2 and minus +
rarrx yy
2 the equation transforms to + minus
minus =2 2
2(x y) (x y) a2 2
Eccentricity = radic2 Transverse axis Equation
y = x Length 2radic2c
Conjugate axis Equation y + x = 0 Length 2radic2c
Foci S(cradic2 cradic2) and Sprime(-cradic2 ndash cradic2)
Directrix x + y = plusmncradic2
Parametric equation x = ct y = ct t isin Rndash0 Centre (0 0)
Vertex (c c) and (ndash
c ndashc)
Conjugate Hyperbola of Rectangular Hyperbola xy = c2
It is given by xy = ndashc2 Centre (0 0)
Hyperbola 18187
Vertex (ndashc c) and (c ndashc) Eccentricity = radic2
TA Equation y + x = 0 Length 2radic2c CA Equation y = x Length 2radic2c
Foci S(ndashcradic2 cradic2) and Sprime(cradic2 ndashcradic2) Directrix x ndash y = plusmncradic2 Parametric equation x = ct y = ndashct t isin R ndash 0
1832 Parametric Equations of Chord Tangents and Normal
Slope of chord joining the points P(t1) and Q(t2) = minus1 2
1m t t
Equation of chord x + t1t2
y = c
(t1 + t2)
Condition for focal chord += plusmn
+1 2
1 2
t t 21 t t
Equation of the tangent at P(x1 y1) + =
1 1
x y 2x y
Equation of tangent at P(t) x + yt2 = 2ct
Equation of normal at P (t) minus = minus2cy t (x ct)
t
rArr xt3 ndash yt = c(t4 ndash 1) If normal of hyperbola xy = c2 at the point P(T) meet the hyperbola again at Tprime the T3Tprime = ndash1
Chord with a given middlepoint as (h k) is kx + hy = 2hk
1833 Co-normal Points
In general four normals can be drawn on a hyperbola each passing through a common point The feets of perpendicular of these four normals lying on the hyperbola are called co-normal points
18188 Mathematics at a Glance
18331 Properties of co-normal points
1 In general four normals can be drawn to a hyperbola from any point and if a b g d be the eccentric angles of these four co-nomal points then a + b + g + d is an odd multiple of p
2 If a b g are the eccentric angles of three points on the hyperbola minus =2 2
2 2
x y 1a b
the normals at which
are concurrent then sin (a + b) + sin (b + g) + sin (a + g) = 0
18332 Diameter of a hyperbola
The locus of the middle points of a system of parallel chords of a hyperbola is called a diameterThe equation of a diameter bisecting a system of parallel chords of slope m of the hyperbola
minus =2 2
2 2
x y 1a b
18333 Conjugate diameters
Two diameters are said to be conjugate when each bisects all chords parallel to the othersTwo diameters y = mx and y = kx are said to be conjugate if their gradients are related as
km = b2a2
1834 Properties of Conjugate Diameters
If a pair of diameters are conjugate with respect to a hyperbola then they are also conjugate with respect to its conjugate hyperbola
If a pair of diameters be conjugate with respect to a hyperbola then one of those diametsrs will meet the hyperbola in real points while the other diameter will meet the conjugate hyperbola in real points
If a pair of conjugate diameters meet the hyperbola
minus =
2 2
2 2
x y 1a b
and its conjugate hyperbola
minus + =
2 2
2 2
x y 1 0a b
in PPprime and D Dprime respective then
(i) CP2 ndash CD2 = a2 ndash b2
(ii) The parallelogram formed by the tangents at the extremities of conjugate diameters has its vertices lying on the asymptotes and its of constant area
(iii) Show that the asymptotes to the hyperbola bisect PD PDprime PprimeD and PprimeDprime
Chapter 19Complex Number
191 IntroductIon
While working with real numbers (ℝ) we would not find relations to equations such as x2 + 9 = 0 () So to look forward we have to difine another set of number systems
1911 Imaginary Numbers (Non-real Numbers)
A number whose square is non-positive is termed as an imaginary number eg 2 or (1 2)minus + minus
Iota Euler introduced the symbol i for the number 1minus It is known as iota (a Greek word for
lsquoimaginaryrsquo) Thus 2 2iminus = and + minus = +1 2 1 2i are imaginary numbers
Remark (i) Imaginary numbers do not follow the property of order ie for z1 and z2 imaginary numbers we
cannot say which one is greater Since i is neither positive nor negative nor zero
(ii) Here non-possible does not imply negative eg 1 2+ minus is also non-positive
1912 Purely Imaginary Numbers (I)
The number z whose square is non positive real number (negative or zero) is termed as purely imaginary
number For example 5minus ie I = z z = ai where a isin ℝ and i =
19121 Geometrical representation of purely imaginary numbers
Single multiplication by i is equivalent to geometrical rotation of number by p2 radians anti-clockwise
Therefore purely imaginary numbers are represented as points lying on y axis of argand plane For example z = ai is represented by point (0 a) on y axis as shown here
19190 Mathematics at a Glance
Remarks 1 The plane formed by real and imaginary axes is called ArgandGaussianComplex Plane
2 It should be kept in mind that any equation not having real roots does not necessarily posses imaginary roots For example the equation x + 5 = x + 7 is neither satisfied by real numbers nor is satisfied by imaginary numbers
1913 Properties of Iota
1 i0 = 1 i2 = ndash1 i3 = ndashi i4 = 1 2 Periodic properties of i i4n = 1 i4n + 1 = i i4n + 2 = ndash 1 i4n + 3 = ndashi forall n isin ℤ 3 i ndash 1 = ndash i 4 Sum of four consecutive power terms of i is zero that is in + in + 1 + in + 2 + in + 3 = 0 forall n isin ℤ 5 For any two real numbers a and b times =a b ab is true only when atleast one of a and b is
non-negative real number ie both a and b are non-negative
192 complex number
A number z resulting as a sum of a purely real number x and a purely imaginary number iy is called a
complex number ie a number of the form z = x + iy where x y isin ℝ and = minusi 1 is called a complex number Here x is called real part and y is called imaginary part of the complex number and they are expressed as Re(z) = x Im (z) = y A complex z = x + iy number may also be defined as an ordered pair of real numbers and may be denoted by the symbol (x y)
The set of complex numbers is denoted by ℂ and is given by = z z = x + iy where x y isin ℝ and = minusi 1
193 ArgAnd plAne
Any complex number z = a + ib can be written as an ordered pair (a b) which can be represented on a plane by the point P(a b) (known as affix of point P) as shown in the figure This plane is called Argand plane complex plane or the Gaussian plane
1931 Representation of Complex Numbers
Complex numbers can be represented by following forms 1 Cartesian form (rectangular form) A complex number z = x + iy can be represented by the
point P having coordinate (x y) 2 Vector form (Algebraic form) Every complex number z is regarded as a position vec-
tor
(OP) which is sum of two position vectors Purely real vector x
(OA) and purely imaginary
vector iy
(OB)
= + = +
OP OA AP OA OB rArr z = x + iy
Modulus of z Distance of point P from the origin is called modulus of complex number z and is denoted by |z| It is length of vector
(OP) It is distance of P(z) from origin
Complex Number 19191
( )( ) ( )( )there4 = = + = + 2 22 2z OP x y Re z Im z
Argument of z Argument of z is the angle made by
OP with the positive direction of real axis Also known as amplitude z and is denoted by amp (z)
Arg(z) = q where θ =ytanx
q lies in the quadrant in
which complex number z lies
NoteThe principal arguments q isin (ndashp p]
3 Polar form (amplitude modulus form) In DOAP OP = |z| = r rArr OA = x = r cosq and AP = y = r sinq rArr z = x + iy = r (cosq + i sinq) = r cisq
Remarkcis q is unimodular complex number and acts as unit vector in the direction of q where q is arg z
4 Euler form (Exponential form) Euler represented complex number z as an exponential function of its argument q (radians) and described here As we know that using Taylorrsquos series expansion cos q and sinq can be expanded in terms of polynomial in q as given below
θ θ θ
θ = minus + minus +2 4 6
cos 1 2 4 6
and θ θ θθ = θminus + minus +
3 5 7
sin 3 5 7
(cosq + isinq) = ( ) ( ) ( ) θθ θ θ
+ θ+ + + + infin =2 3 4
ii i i1 i to e
2 3 4 rArr z = x + iy = r (cosq + i sinq) = reiq
Advantages of using Euler form Convenient for division and multiplication of complex numbers Suitable for exponential logarithmic and irrational functions involving complex numbers
19311 Inter-conversion from polartrigonometric to algebraic form
(i) Algebraic form to polar form Given z = x + iy then
= +2 2r x y θ = θ =x ycos sinr r
gives q = f (say)
In polar form = + φ+ φ2 2z x y (cos isin )
(ii) Polar form to algebraic form Given z = r(cosq + isinq) = rcosq + i(rsinq)
rArr z = x + iy where x = rcosq and y = rsinq
1932 Properties of Complex Numbers
(i) Equality Two complex numbers z1 and z2 are equal only when their real and imaginary parts are respectively equal ie Re(z1) = Re(z2) and I(z1) = I(z2) or |z1| = |z2| and arg (z1) = arg (z2)
19192 Mathematics at a Glance
RemarksStudents must note that x y isin ℝ and x y ne 0 If x + y = 0 rArr x = ndash y is correct but x + iy = 0 rArr x = ndash iy is incorrect (unless both x and y are zero)
Hence a real number cannot be equal to the imaginary number unless both are zero
(ii) Inequality Inequality in complex number is not defined because lsquoirsquo is neither positive zero nor negative So 4 + 3i gt 1 + 2i or i lt 0 or i gt 0 is meaningless
(iii) If Re(z) = 0 then z is purely imaginary and if Im (z) = 0 then z is purely real (iv) z = 0 rArr Re(z) = Im (z) = 0 therefore the complex number 0 is purely real and purely imaginary or both (v) If z = x + iy then iz = ndashy + ix rArr Re(iz) = ndash Im(z) and Im(iz) = Re(z) (vi) Conjugate of complex number z = x + i y is denoted
as z = (x ndash iy) ie a complex number with same real part as of z and negative imaginary part as that of z
(vii) If z is purely real positive rArr Arg(z) = 0 (viii) If z is purely real negative rArr Arg(z) = p (ix) If z is purely imaginary with positive imaginary part
rArr Arg(z) = p2 (x) If z is purely imaginary with negative imaginary part
rArr Arg(z) = ndashp2 (xi) Arg(0) is not defined
19321 Binary operations defined on set of complex numbers
Binary operation on set of complex number is a function from set of complex numbers to itself That is if z1 z2 isin C and is a binary operation on the set of complex numbers then z1 z2 isin C Following binary operations are defined on set of complex numbers
Addition of two complex numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 rArr z1 + z2 = (x1 + iy1) + (x2 + iy2)= (x1 + x2) + i (y1 + y2) ie z1 + z2 = [R(z1) + R(z2)] + i[I(z1) + I(z2)] isin C
19322 Geometric representation
Consider two complex numbers z1 = (x1 + iy1) and z2 = (x2 + iy2) represented by
vector =
1z OA
z OB as shown in figure
Then by parallelogram law of vector addition + = + =
1 2z z OA OB OC Hence C represents the affix of z1 + z2
NotesIn DOAC [Since sum of two sides of a D is always greater than the third side] OA + AC ge OC
rArr |OA| |OB| |OC|+ ge
rArr | z1 | + | z2 | ge | z1 + z2| This is called triangle inequality Also considering OAB OA + OB ge AB
rArr + ge rArr + ge minus
1 2 1 2|OA| |OB| |BA| |z | |z | |z z |
Subtraction of two complex numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 then z1ndash z2 =(x1 + iy1) ndash (x2 + iy2) = (x1 ndash x2) + i (y1 ndash y2) ie z1 ndash z2 = [R(z1) ndash R(z2)] + i[I(z1) ndash I(z2)] isin C
Complex Number 19193
19323 Geometric representation
Using vector representation again we have = minus = minus =
1 2BA OA OB z z OC Hence the other diagonal of the parallelogram represents the difference
vector of z1 and z2
Notes
1 While BA
represents the free vector corresponding to z1 ndash z2 OC
represents the position vector of z1 ndash z2
rArr C is affix of complex number z1 ndash z2
2 In a triangle the difference of two sides is always less than the third side
rArr OB OA ABminus le
rArr ||z2| ndash |z1|| le |z2 + z1|
3 Triangle Inequality ||z1| ndash |z2|| le |z1 plusmn z2| le |z1| + |z2 |
Multiplication of two complex numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 then z1z2 = (x1 + iy1) (x2 + iy2) = [R(z1)R(z2) ndash I(z1)I(z2)] + i[R(z2)I(z1) + R(z1) I(z2)] isin C
Geometric representation Let A and B are two points in the complex plane respectively affixes of z1 and z2 where z1 = r1(cos q1 + i sinq1) and z2 = r2(cos q2 + i sinq2) z1z2 = r1r2(cosq1 + isin q1) (cosq2 + i sinq2)
1933 ResultThe product rule can be generalized to n complex numbers Let zn = rn(cosqn + i sinqn) where n = 1 2
Now |z1 z2zn| = r1r2rn = |z1| | z2 ||zn| and arg (z1 z2zn) = q1 + q2 + + qn = arg z1 + arg z2 + + arg zn
As special case arg zn = n arg z
Division of two complex numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 rArr z1z2 = (x1 + iy1)(x2 + iy2)
= + minus+ isin
+ +1 2 1 2 2 1 1 2
2 2 2 22 2 2 2
(x x y y ) i(x y x y ) C(x y ) (x y )
Geometric representation Let A and B are two points in the complex plane which are affixes of z1 and z2 respectively where z1 = r1(cos q1 + i sinq1) and z2 = r2(cos q2 + i sinq2)
Then we get =2 2
1 1
z rz r
[cos(q2 ndash q1) + i sin(q2 ndash q1)]
Notes
1 If q1 and q2 are principal values of argument of z1 and z2 then q1 + q2 may not necessarily be the principal value of argument of z1 z2 and q1 ndash q2 may not necessarily be principal value of argument of z1z2 To make this argument as principal value add or subtract 2np where n is such an integer which makes the argument as principal value
2 Note that angle a between two vectors OA
and OB
is a = q2 ndash q1 a = arg z2 ndash arg z1
19194 Mathematics at a Glance
194 AlgebrAIc Structure of Set of complex numberS
(i) Complex numbers obey closure law (for addition subtraction and multiplication) commutative law (for addition and multiplication) associative law (for addition and multiplication) existence of additive and multiplicative identitiy and inverse
(ii) Existence of conjugate element Every complex number z = x + iy has unique conjugate denoted as x ndash iy
1941 Conjugate of a Complex NumberConjugate of a complex number z = x + iy is defined as = minusz x iy It is mirror image of z in real axis as mirror shown in the figure given here
Let z = r (cosq + isinq) rArr = θminus θz r(cos isin ) = r [cos(ndashq) + isin(ndashq)]rArr z has its affix point having magnitude r and argument (ndashq)
1942 Properties of Conjugate of a Complex Number 1 = = minusR(z) R(z) I(z) I(z)
2 = = = +2 2 2 2zz | z | | z | (R(z)) (I(z))
3 = =(z) z (z) z and so on
4 = minus =|z| |z|and Agr z Arg z
5 If =z z ie arg z = arg z rArr z is purely real 6 If z = ndash z ie arg (ndashz) = arg( z ) rArr z is purely imaginary
7 += = =
z zR(z) x R(z)2
minus
= = = minusz zIm(z) y Im(z)
2i
8 θ minus θ +
θ =
i ie ecos2
θ minus θ minus
θ =
i ie esin2i
9 plusmn plusmn plusmn plusmn = plusmn plusmn plusmn plusmn1 2 3 n 1 2 3 n(z z z z ) z z z z
10 =1 2 3 n 1 2 3 n(z z z z ) (z )(z )(z )(z )
11 = 11 2
2
(z )(z z )(z )
12 =n n(z ) (z) 13 If w = f(z) then ω= f(z) where f(z) is algebraic polynomial
14 + =1 2 2 1 2 1z z z z 2R(z z )
15 + = + +2 21 2 1 2 1 2| z z | | z | | z | 2Re(z z )
16 |z1 + z2|2 + |z1 ndash z2|
2 = 2(|z1|2 + |z2|
2)
1943 Modulus of a Complex NumberModulus of a complex number z = x + iy is denoted by |z| If point p(x y) represents the complex number
z on Argandrsquos plane then = = +2 2z OP x y = distance between origin and point = +2 2P [R(z)] [I(z)]
Complex Number 19195
19431 Properties of modulus of complex numbers
1 Modulus of a complex numbers is distance of complex number from the origin and hence is non-negative and |z| ge 0 rArr | z | = 0 iff z = 0 and | z | gt 0 iff z ne 0
2 ndash| z | le Re(z) le | z |and ndash | z | le Im(z) le | z | 3 = = minus = minus| z | | z | | z | | z |
4 = 2zz | z | 5 | z1z2 | = | z1 || z2 | In general |z1z2z3 zn| = |z1||z2|| z3 ||zn| 6 (z2 ne 0) 7 Triangle inequality | z1 plusmn z2 | le | z1 | + | z2 | In general | z1 plusmn z2 plusmn z3plusmn zn| le | z1 |plusmn| z2 |
plusmn | z3 | plusmnplusmn | zn | 8 Similarly | z1 plusmn z2 | ge | z1 | ndash | z2 | 9 | zn | = | z |n
10 || z1 | ndash | z2 || le | z1 plusmn z2| le | z1 | + | z2 | Thus | z1 | + | z2 | is the greatest possible value of | z1 plusmn z2 | and || z1 | ndash | z2 || is the least possible value of | z1 plusmn z2|
11 plusmn = + plusmn +2 2 21 2 1 2 1 2 1 2| z z | | z | | z | (z z z z ) or + plusmn2 2
1 2 1 2| z | | z | 2Re(z z ) or | z1 |2 + | z2 |
2 plusmn 2 | z1 || z2 |
cos (q1 ndash q2)
12 + = θ minusθ21 2 1 2 1 2 1 2| z z z z | 2 |z | | z | cos( ) where q1 = arg (z1) and q 2 = arg (z2)
13 | z1 + z2 |2 = | z1 |
2 + | z2 |2 hArr 1
2
zz
is purely imaginary
14 | z1 + z2 |2 + | z1 ndash z2 |
2 = 2| z1 |2 + | z2 |
2 15 | az1 + bz2 |
2 + | bz1 ndash az2 |2 = (a2 + b2)(| z1 |
2 + | z2 |2) where a b isin R
16 Unimodular If z is unimodular then | z | = 1 Now if f (z) is a unimodular then it can always be expressed as f (z) = cosq + isinq qisinℝ
19432 Argument and principal argument of complex number
Argument of z (arg z) is also known as amp(z) is angle which the radius vector
OP makes with positive direction of real axis
Principle Argument In general argument of a complex number is not unique if q is the argument then 2nπ + q is also the argument of the complex number where n = 0 plusmn 1 plusmn 2 Hence we define principle value of argument q which satisfies the condition ndashp lt q le p Hence Principle value of arg(z) is taken as an angle lying in (ndashp p] It is denoted by Arg(z) Thus arg(z) = Arg(z) plusmn 2kp k isin ℤ
A complex number z given as (x + iy) lies in different quadrant depending upon the sign of x and y Based on the quadrantal location of the complex number its principle argument are given as follows
19196 Mathematics at a Glance
Sign of x and y Location of z Principal Argument
x gt 0 y gt 0 Ist quadrant minusθ = α = 1 ytanx
x lt 0 y gt 0 IInd quadrant minusθ = πminusα = πminus 1 y( ) tanx
x lt 0 y lt 0 IIIrd quadrant 1 ytanx
minusθ = minusπ+
x gt 0 y lt 0 IVth quadrant minusθ = minusα = minus 1 ytanx
19433 Caution
An a usual mistake is to take the argument of z = x + iy as tanndash1 (yx) is irrespective of the value of x and y
Remember that tanndash1 (yx) lies in the interval π π minus
2 2
Whereas the principal value of argument of z (Arg(z)) lies in the interval (ndashπ π]
Thus if z = x + iy then
1
1
1
tan (yx) if x 0 y 0tan (yx) if x 0 y 0tan (yx) if x 0 y 0Arg(z)
2 if x 0 y 02 if x 0 y 0
Not defined for x 0 y 0
minus
minus
minus
gt ge
+ π lt ge minusπ lt lt= π = gtminusπ = lt
= =
19434 Properties of argument of complex number
1 arg (z1z2) = arg z1 + arg z2
2 arg(zn) = n (argz)
3
= minus
11 2
2
zarg arg z arg zz
4 arg(z) = 0 hArr complex number z is purely real and positive 5 arg(z) = p hArr complex number z is purely real and negative 6 arg(z) = p2 hArr complex number z is purely imaginary with positive Im(z) 7 arg(z) = ndash p2 hArr complex number z is purely imaginary with negative Im(z) 8 arg(z) = not defined hArr z = 0 9 arg(z) = p4 hArr z = (1 + i) or (x + xi) etc for (x gt 0)
Properties of Principal Arguments (Principal argument of complex number is denoted by arg (z))
1 If θ= θ + θ = kik k k k kz r (cos isin ) r e are number of complex numbers then
==
= plusmn π
sumprod
n n
k kk 1k 1
Arg z Arg z 2k
where k isin ℤ choose k suitably such that principal Arg of the resultant number lies in principal range
Complex Number 19197
2 =
zArg 2Arg (z)z
3 Arg (zn) = n Arg z plusmn 2kp 4 Arg (ndashz) = ndashp + Arg z or p + Arg z respectively when Arg z gt 0 or lt 0 5 Arg (1z) = ndashArg z
Method of Solving Complecs EquationsLet the given equation be f(z) = g(z) To solve this equation we have the following four methods
Method 1 Put z = x + iy in the given equation and equate the real and imaginary parts of both sides and solve to find x and y hence z = x + iy
Method 2 Put z = r(cosq + isinq) and equate the real and imaginary parts of both sides solve to get r and q hence z
Method 3 Take conjugate of both sides of given equations Thus we get two equations f(z) = g(z) (1) and =f(z) g(z) (2)
Adding and Subtracting the above two equations we get two new equations solving then we get z
Method 4 Geometrical Solution From the given equation we follow the geometry of complex number z and find its locus
1944 Square Roots of a Complex Number
Square roots of z = a + ib are given by + minus
plusmn +
| z | a | z | ai
2 2 b gt 0 and
+ minusplusmn minus
| z | a | z | ai
2 2 b lt 0
19441 Shortcut method
Step 1 Consider =0Im(z ) b2 2
Step 2 Factorize b2 into factors x yx2 ndash y2 = Re(z0) = a
Step 3 Therefore a + ib = (x + iy)2
rArr + = plusmn +a ib (x iy) eg minus8 15i a = 8 b = ndash15 lt 0
rArr = minusb 152 2
= xy such that x2 ndash y2 = 8 rArr = = minus5 3x y2 2
rArr
minus = plusmn minus = plusmn minus
5 3i 18 15i (5 3i)2 2 2
19442 Cube root of unity
Let 3 1 = cube root of unity
rArr x3 = 1 where minus +ω=
1 3i2
and minus minusω =2 1 3i
2 Cube roots of unity are 1 w w2 and w w2 are called the
imaginary cube roots of unity
19198 Mathematics at a Glance
19443 Properties of cube root of unity
P(1) |w| = |w2| = 1 P(2) ω=ω2
P(3) w3 = 1 P(4) w3n = 1 w3n + 1 = w and w3n + 2 = w2 forallnisinℤP(5) Sum of cube roots of unity is 0 That is 1 + w + w2 = 0
Remarks
∵ 2ω ω= ∵ 1 0ω ω+ + =
∵ 2ω ω= and 3 4 2 2 21 ( ) ( )ω ω ωω ω ω ω= = = = =
∵ 2 21 1 ( )ω ω ω ω+ + = + + ∵ 21 ( ) 0ω ω+ + =
P(6) +ω +ω =
n 2n 3 when n is multiple of 31
0 when n is not a multiple of 3
P(7) 1 w w2 are the vertices of an equilateral D having each side = radic3
P(8) Circumcentre of D ABC with vertices as cube roots of unity lies at origin and the radius of circumcircle is 1 unit Clearly OA = OB = OC = 1
RemarkFrom the above properties clearly cube roots of unity are the vertices of an equilateral D having each side = radic3 and circumscribed by circle of unit radius and having its centre at origin
P(9) π
ω = minus + =
1 3iarg( ) arg2 2 3
π
ω = minus minus =
2 1 3 4arg( ) i2 2 3
P(10) Any complex number a + ib for which =1(a b)3
or 3 1 can always be expressed in
terms of i w w2
eg + = minus ω21 i 3 2 + minus + ω+ = + = = =
i 1 i 3 2 1 i 3 23 i (1 i 3) 2i2 2i i 2 i
19444 Important relation involving complex cubic roots of unity
(a) x2 + x + 1 = (x ndash w) (x ndash w2) (b) x2 ndash x + 1 = (x + w) (x + w2) (c) x2 + xy + y2 = (x ndash yw) (x ndash yw2) (d) x2 ndash xy + y2 = (x + yw) (x + yw2) (e) x2 + y2 = (x + iy) (x ndash iy) (f) x3 + y3 = (x + y) (x + yw) (x + yw2) (g) x3 ndash y3 = (x ndash y) (x ndash yw) (x ndash yw2) (h) x2 + y2 + z2 ndash xy ndash yz ndash zx = (x + yw + zw2) (x + yw2 + zw) (i) x3 + y3 + z3 ndash 3xyz = (x + y + z)(x + yw + zw2)(x + w2y + wz)
Complex Number 19199
195 de moIVerrsquoS tHeorem
This theorem states that (i) (cosq + isinq)n = cosnq + isinnq if n is an rational number (ii) (cosq + isinq)1n = [cos(q + 2kp) + isin (q + 2kp)]1n
(∵ period of sinq and cosq is 2p) = π θ π θ+
(2k + ) (2k + )cos i sin n n
k = 0 1 2 n ndash 1
1951 nth Root of Unity
Let x be an nth root of unity then ( ) ( )= = +1 1n nx 1 cos0 isin0 = π+ π+ + =
2r 0 2r 0cos isin r 0n n
1 2 n ndash 1
= ππ+ π+ + = =
i2rn2r 0 2r 0cos isin r 0 e r 0
n n 12n ndash 1=
π π minus π2 4 2(n 1)i i in n n1 e e e = 1 a a2
an ndash 1 where π
α =2ine
1952 Properties of nth Root of Unity
P(1) nth roots of unity form a GP
P(2) 1 + a + a2 + + an ndash 1 = 0
P(3) 1 aa2an ndash 1 = (ndash1)n ndash 1
P(4) nth roots of unity are vertices of n-sided regular polygon circumscribed by a unit circle having its centre at the origin
Case (i) When n is oddLet n = 2m + 1 m is some positive integers then only one root is real that is 1 and remaining 2m roots are non real complex conjugates
The 2m non-real roots are (a a2m) (a2 a2mndash1) (a3 a2mndash2) (am am+1) where the ordered pairs are (z z) ie non-real roots and their
conjugates and π
α =2ine
NoteThe nth roots given as ordered pairs are conjugate and reciprocal of each other
m2m 1 2m 11 2m 2m 1 m m 1
m m
1 1 11
α αα α α α αα α α α α
+ +minus + +
= = = = = = = = =
Case (ii) When n is even
Let n = 2m π π α = =
2cis cisn m
except 1 and ndash1 other roots are non-real
complex conjugate pairs
19200 Mathematics at a Glance
NoteThe nth roots arranged vertically below are conjugate and reciprocal of each other and diagonally (passing through origin) are negative of each other
19521 nth root of a complex number n z
Let z = r cis q z1n = (r1n) (cis(2kπ + q))1n = (r1n) π θ +
2kcisn n
where r1n is positive nth root of r
= π θ
1n 2k(r ) cis cisn n
where π2kcisn
is the nth root of unity k = 0 1 2 n ndash 1
19522 To find logarithm of a complex number
Consider z = x + iy converting lsquox + iyrsquo into Eulerrsquos form such that q = principal value of argument of z then we get loge (x + iy) = loge (|z|eiq)
rArr loge(x + iy) = loge |z| + logeeiq rArr loge (x + iy) = loge |z| + iq
In general loge(x + iy) = loge|z| + i(q + 2np) nisinℤ To find (x + iy)(a+ib) ie 2z
1(z )
Let u + iv = (x + iy)(a+ib)
rArr ln (u + iv) = (a + ib) ln (x + iy) rArr (u + iv) = e (a + ib) ln (x + iy) now solve for u and v by expressing (x + iy) in polar form
For example x = (i)i rArr lnx = ilni = ππ π π + = =
i 2 2i n cos isin i n(e ) i ne2 2 2
rArr π= minusnx
2 rArr
πminus
= 2x e Thus (i)i = endashp2
Alternatively ππ π π π+ minus
= = = = = =
ii 2
i n cos isin iii n(i) i ni i n(e )2 2 2 2(i) e e e e e e
196 geometry of complex number
1961 Line Segment in Argandrsquos Plane
Any line segment joining the complex numbers z1 and z2 (directed towards z2) represents a complex number given by z2 ndash z1 Since every complex number has magnitude and direction therefore z2 ndash z1 also
|z2 ndash z1| represents the length of line segment BA ie the distance between z1 and z2 and arg(z2 ndash z1) represents the angle which line segment AB (on producing) makes with positive direction of real axis
19611 Angle between to lines segments (Roation theorm or conirsquos theorem)
Consider three complex numbers z1 z2 and z3 such that the angle between line segments joining z1 to z2 and z3 to z1 is equal to q
Complex Number 19201
Then q = a ndash b = Arg(z3 ndash z1) ndash Arg(z2 ndash z1) = 3 1
2 1
z z Post-rotation vectorArg Argz z Pre-rotation vector
minus = minus
rArr i3 1
2 1
z zArg Arg( e )
z zθ minus
= θ = ρ minus
rArr (z3 ndash z1) = (z2 ndash z1) r eiq where minus
ρ =minus
3 1
2 1
z zz z
If z1 = 0
rArr z3 = rz2 eiq arg(z3z2) is an angle through which z2 is to be rotated to
coincide it with z3If a complex number (z2 ndash z1) is multiplied by another complex number reiq then the complex
number (z2 ndash z1) gets rotated by the argument (q) of multiplying complex number in anti-clockwise direc-tion (It is called Rotation Theorem or Conirsquos Theorem)
1962 Application of the Rotation Theorem (i) Section Formula Let us rotate the line BC about the point C so that it becomes parallel to
the line CA The corresponding equation of rotation will be ( )πminus minus= = minus
minus minusi1 1
2 2
z z | z z | m e 1z z |z z | n
rArr nz1 ndash nz = ndash mz2 + mz rArr +=
+1 2nz mzzm n
Similarly if C(z) divides the segment AB externally in the ratio of m n
then minus=
minus1 2nz mzzm n
In the specific case if C(z) is the mid point of AB then += 1 2z zz
2
(ii) Condition for Collinearity If there are three real numbers (other than 0) l m and n such that lz1 + mz2 + nz3 = 0 and l + m + n = 0 then complex numbers z1 z2 and z3 will be collinear
(iii) To find the conditions for perpendicularity of two straight lines Condition that angA of DABC where A(z1) B(z2) C(z3) is right angle and can be obtained by applying Rotation Theorem at A
minus π π= minus
minus 3 1
2 1
z zArg
z z 2 2 (i)
rArr π
plusmn minus minus= ρ = plusmnρ ρ = minus minus
i3 1 3 12
2 1 2 1
z z z ze iz z z z
rArr minus
= minus
3 1
2 1
z zR 0
z z
rArr minus minus
+ =minus minus
3 1 3 1
2 1 2 1
z z z z0
z z z z rArr |z2 ndash z3|
2 = |z3 ndash z1|2 + |z2 ndash z1|
2
If ABC is right-angled isosceles triangle with AB = AC
rArr r = 1 rArr minus
= plusmnminus
3 1
2 1
z zi
z z
19202 Mathematics at a Glance
(iv) Conditions for ∆ABC to be an equilateral triangle Let the DABC where A(z1) B(z2) C(z3) be an equilateral triangle
The following conditions hold
(i) |z1 ndash z2| = |z2 ndash z3| = |z3 ndash z1|
(ii) minus π= plusmn minus = minus
minus 3 1
3 1 2 12 1
z zArg and |z z | |z z |
z z 3
(Applying the rotation theorem at A and knowing CA = BA)
(iii) minus minus π
= = minus minus
3 1 1 2
2 1 3 2
z z z zArg Argz z z z 3 (Applying rotation theorem at A and B)
(iv) + + = + +2 2 21 2 3 1 2 2 3 3 1z z z z z z z z z
(v) πminus
= = +minus
i1 2 3
3 2
z z 1 3e iz z 2 2
(vi) + + =minus minus +2 3 3 1 1 2
1 1 1 0z z z z z z
(vii) Conditions for four points to be concyclic or condition for points z1 z2 z3 z4 to represent a cyclic quadrilateralIf points A(z1) B(z2) C(z3) D(z4) are con-cyclic then the following two cases may occur
Case I If z3 and z4 lies on same segment with respect to the chord joining z1 and z2
minusminusminus =
minus minus 2 32 4
1 4 1 3
z zz zArg Arg 0z z z z
rArr minusminus
= minus minus
1 32 4
1 4 2 3
w
z zz zArg 0z z z z
rArr w is real and positive or Im(w) = 0 and Re(w) gt 0
Case II If z3 and z4 lie on opposite segment of circle with respect to chord joining z1 and z2
minus minus
+ = π minus minus
2 3 1 4
1 3 2 4
z z z zArg Argz z z z
rArr Arg (1w) = π rArr Arg (w) = ndashπ So the principal argument of w = πrArr w is negative real number or Im(w) = 0 and Re(w) lt 0
Conclusion Four complex numbers z1 z2 z3 z4 to be concyclic
minus minus
= π minus minus
1 3 2 4
2 3 1 4
w
(z z )(z z )Arg 0 or
(z z )(z z ) rArr w is purely real I(w) = 0 rArr =w w
1963 Loci in Argand Plane
A(1) Straight line in Argand plane A line through z0 making angle a with the positive real axisArg(z ndash z0) = α or ndash π + α
Complex Number 19203
The given equation excludes the point z0 Arg (z ndash z0) = a represents the right-ward ray Arg (z ndash z0) = ndashp + a represents the left-ward ray
A(2) Line through points A(z1) and B(z2) Consider a straight line passing through A(z1) and B(z2) taking a variable point P(z) on it
∵ for each position of P
AP is collinear with
AB rArr = λ
AP AB rArr = λ minus
2 1AP (z z )
∵ = +
OP OA AP z = z1 + l(z2 ndash z1) z = z1(1 ndash l) + lz2
19631 Conclusion
1 if z = xz1 + yz2 x + y = 1 and x and yisinℝ then z z1 z2 are collinear
2 Equation represents line segment AB if l isin [0 1] 3 Right-ward ray through B if lisin (1 infin) 4 Left-ward ray through A if lisin (ndashinfin 0)
(i) Equation of straight line with the help of coordinate geometry
Writing + minus= =
z z z zx y2 2i
etc in minus minus
=minus minus
1 1
2 1 2 1
y y x xy y x x
and re-arranging the terms we find that the
equation of the line through z1 and z2 is given by minus minus
=minus minus
1 1
2 1 2 1
z z z zz z z z
or =1 1
2 2
z z 1z z 1 0z z 1
(ii)Equation of a straight line with the help of rotation formulaLet A(z1) and B(z2) be any two points lying on any line and we have to obtain the equation of this line For this purpose let us take any point C(z) lying on
this line Since Arg minus
= minus
1
2 1
z z 0z z
or p
minus minus=
minus minus1 1
2 1 2 1
z z z zz z z z
hellip (i)
This is the equation of the line that passes through A(z1) and B(z2) After rearranging the terms
it can also be put in the following form =1 1
2 2
z z 1z z 1 0z z 1
(iii) Line segment AB The equation of the line segment AB is given as minus
= π minus
1
2
z zArgz z
(iv) Equation of two rays excluding the line segment AB minus
= π minus
1
2
z zArgz z
19204 Mathematics at a Glance
(v) Complete line except z1 and z2 (general equation of line)
The equation is given as 1 1
2 2
z z z zArg 0 ie I 0z z z z
minus minus= π =
minus minus
rArr minus minus
=minus minus
1 1
2 2
z z z zz z z z
rArr minus minus +2 1 1 2zz z z z z z z = minus minus +1 2 2 1zz z z z z z z
rArr minus + minus + minus =1 2 2 1 1 2 2 1(z z )z (z z )z z z z z 0 rArr minus minus
+ + =1 2 2 11 2
(z z ) (z z )z z I(z z ) 02i 2i
rArr + + =az az b 0 where rArr where minus= 2 1z za
2i and minus minus
= =minus2 1 1 2z z z za
2i 2i
RemarkTwo points P(z1) and Q(z2) lie on the same side or opposite side of the line + +az az b accordingly as
+ +1 1az az b and + +2 2az az b have the same sign or opposite sign
197 tHeorem
Perpendicular distance of P(c) (where c = c1 + ic2) from the straight line is
given by+ +
=|ac ac b |p
2 |a | Slope of a given line Let the given line be + + =za za b 0
Replacing z by x + iy we get ( ) ( )+ + minus + =x iy a x iy a b 0
rArr ( ) ( )+ + minus + =a a x iy a a b 0
Itrsquos slope is = ( ) ( )+
= = minusminus 2
a a 2Re(a) Re(a)i a a 2i lm a lm(a)
Equation of a line parallel to a given line Equation of a line parallel to the line + + =za za b 0 is + +λ =za za 0 (where l is a real number)
Equation of a line perpendicular to a given line Equation of a line perpendicular to the line + + =za za b 0 is minus + λ =za za i 0 (where l is a real number)
Equation of perpendicular bisectorConsider a line segment joining A(z1) and B(z2) Let the line lsquoLrsquo be itrsquos perpendicular bisectorIf P(z) be any point on the lsquoLrsquo then we havePA = PB rArr | z ndash z1 | = | z ndash z2 |rArr ( ) ( )minus + minus + minus =2 1 2 1 1 1 2 2z z z z z z z z z z 0
Complex Number 19205
198 complex Slope of tHe lIne
If z1 and z2 are two unequal complex numbers represented by points P and Q then minusminus
1 2
1 2
z zz z
is called the
complex slope of the line joining z1 and z2 (ie line PQ) It is denoted by w Thus minus
=minus
1 2
1 2
z zwz z
Notes
1 The equation of line PQ is 1 1z z w( z z )minus = minus Clearly 1 2 1 2
1 2 1 2
z z z zw 1
z z z z
minus minus= = =
minus minus
2 The two lines having complex slopes w1 and w2 are parallel if and only if w1 = w2
3 Two lines with complex slopes w1 and w2 are perpendicular if w1 + w2 = 0
1981 Circle in Argand PlaneA(1) Centre radius form
The equation of circule with z0 as centre and a positive real number k as radius as given as |z ndash z0| = k
rArr |z ndash z0|2 = k2
rArr minus minus = 20 0(z z )(z z ) k rArr minus minus + minus =2 2
0 0 0zz z z z z |z | k 0 (1)
If z0 = 0 then |z| = K
A(2) General Equation of CircleReferring to equation (1) thus we can say
+ + + =zz az az b 0 (2)where a is a complex constant and bisinℝ represents a general circle
Comparing (2) with (1) we note that centre = ndasha and radius = minus2a b
A(3) Diametric Form of CircleAs we know that diameter of any circle subtends right angle at any point on the circumference Equation of circle with A(z1) and B(z2) as end points of diameter
π minus = πminus minus
2
1
Case Iz z 2Argz z Case II
2
rArr minus minus
= plusmn =minus minus
2 2
1 1
z z z zki where kz z z z rArr
minus minus= minus
minus minus2 2
1 1
z z z zz z z z
rArr minus minus + minus minus =1 2 2 1(z z )(z z ) (z z )(z z ) 0 further minus minus
+ =minus minus
2 2
1 1
z z z z 0z z z z
is diametric form
rArr |z ndash z1|2 + |z ndash z2|
2 = |z1 ndash z2|2
19206 Mathematics at a Glance
199 AppoloneouS cIrcle
If minus=
minus1
2
z z kz z
ie |z ndash z1| = k |z ndash z2| Then equation represents apploloneous
circle of A (z1) B(z2) with respect to ratio k when k = 1 this gives |z ndash z1| = |z ndash z2| which is straight line ie perpendicular bisector of line segment joining z1 to z2
1910 eQuAtIon of cIrculAr Arc
As per the figure equation of circular arc at which chord AB (where A(z1) and B(z2)) subtends angle a is
given as minus
= α minus
2
1
z zArgz z
Case I If 0 lt a lt p2 or ndash p2 lt a lt 0 (Major arc of circle)
Case II πα = plusmn
2 (Semicircular arc)
Case III π π αisin minusπ cup π
2 2
(Minor arc of circle)
Case IV a = 0 (Major arc of infin radius)
Case V a = p (Minor arc of infin radius)
19101 Equation of Tangent to a Given Circle
Let | z ndash z0 | = r be the given circle and we have to obtain the tangent at A(z1) Let us take any point P(z) on the tangent line at A(z1)
Clearly angPAB = p2 arg minus π
= plusmn minus
1
0 1
z zz z 2
rArr minusminus
1
0 1
z zz z
is purely imaginary
rArr ( ) ( )minus + minus + minus minus =20 1 0 1 1 1 0 1 0z z z z z z 2 |z | z z z z 0
In particular if given circle is | z | = r equation of the tangent at z = z1 would be + = =2 21 1 1zz zz 2 |z | 2r
If minus= λ
minus1
2
z zz z
(l isin R+ l ne 1) where z1 and z2 are given complex numbers and z is a arbitrary
complex number then z would lie on a circle
19102 ExplanationLet A(z1) and B(z2) be two given complex numbers and P(z) be any arbitrary complex number Let PA1 and PA2 be internal and external bisectors of angle angAPB respectively Clearly angA2PA1 = p2
Now minus minus= = = λ
minus minus1 1
2 2
| z z | z zAPBP |z z | z z
(say)
Complex Number 19207
Thus points A1 and A2 would divide AB in the ratio of l 1 internally and externally respectively Hence P(z) would be lying on a circle with A1A2 being itrsquos diameter Note If we take lsquoCrsquo to be the mid-point of A2A1 it can be easily prove that CA CB = (CA1)
2 ie | z1 ndash z0 || z2 ndash z0 | = r2 where the point C is denoted by z0 and r is the radius of the circle
Notes (i) If we take lsquoCrsquo to be the mid-point of A2A1 it can be easily proved that CA CB = (CA1)
2 ie | z1 ndash z0 || z2 ndash z0 | = r2 where the point C is denoted by z0 and r is the radius of the circle
(ii) If l = 1 rArr | z ndash z1 | = | z ndash z2 | hence P(z) would lie on the right bisector of the line A(z1) and B(z2) Note that in this case z1 and z2 are the mirror images of each other with respect to the right bisector
19103 Equation of Parabola
Equation of parabola with directrix + + =az az b 0 and focus z0 is given as SP = PM
+ +minus =0
|az az b || z z |2 |a |
rArr minus = + +2 2 204 | z z | |a | |az az b | rArr minus minus = + + 2
0 04aa(z z )(z z ) (az az b)
rArr minus minus + = + + 20 0 0 04aa(zz zz z z z z ) (az az b)
19104 Equation of EllipseEllipse is locus of point P(z) such that sum of its distances from two fixed points A(z1) and B(z2) (ie foci of ellipse) remains constant (2a)
rArr PA + PB = 2a rArr |z ndash z1| + |z ndash z2| = 2a where 2a is length of major axis
Case I If 2a gt |z1 ndash z2| = AB (Locus is ellipse)
Case II 2a = |z1 ndash z2| (Locus is segment AB)
Case III 2a lt |z1 ndash z2| (No locus)
Case IV If |z ndash z1| + |z ndash z2| gt 2a 2a gt |z1 ndash z2| (Exterior of ellipse)
Case V If |z ndash z1| + |z ndash z2| lt 2a 2a gt |z1 ndash z2| (Interior of ellipse)
1911 eQuAtIon of HyperbolA
Hyperbola is locus of point P(z) such that difference of its distances from two fixed point A(z1) and B(z2) (foci of hyperbola) remains constant (2a)
rArr PA ndash PB = 2a rArr ||z ndash z1| ndash |z ndash z2|| = 2a where 2a is length of major axis
Case I If 2a lt |z1 ndash z2| = AB (locus is branch of hyperbola)
19208 Mathematics at a Glance
Case II 2a = |z1 ndash z2| (Locus is union of two rays)
Case III 2a gt |z1 ndash z2| (No locus)
Case IV If ||z ndash z1| ndash |z ndash z2|| gt 2a 2a lt |z1 ndash z2| (Exterior of hyperbola)
Case V If |z ndash z1| ndash |z ndash z2| lt 2a 2a lt |z1 ndash z2| (Interior of hyperbola)
1912 Some ImpotAnt fActS
A (1) If A B C are the vertices of a triangle represented by complex numbers z1 z2
z3 respectively in anti-clockwise sense and DBAC = a then αminus minus
=minus minus
i3 1 2 1
3 1 2 1
z z z z ez z z z
A(2) If z1 and z2 are two complex numbers representing the points A and B then
the point on AB which divides line segment AB in ratio m n is given by ++
1 2nz mzm n
A(3) If a b c are three real numbers not all simultaneously zero such that az1 + bz2 + cz3 = 0 and a + b + c = 0 then z1 z2 z3 will be collinear
A(4) If z1 z2 z3 represent the vertices ABC of DABC then
(i) Centroid of + +∆ = 1 2 3z z z
ABC3
(ii) In centre of + +∆ =
+ +1 2 3az bz cz
ABCa b c
(iii) Orthocentre of + +
∆ =+ +
1 2 3(a secA)z (bsecB)z (csecC)zABC
(a secA) (bsecB) (csecC)= + +
+ +1 2 3(z tan A z tanB z tanC)
tan A tanB tanC
(iv) Circumcentre of + +
∆ =+ +
1 2 3z sin2A z sin2B z sin2CABC
sin2A sin2B sinC (v) If z1z2z3 are the vertices of an equilateral triangle then the circumcentre z0 may be given
as z21 + z2
2 + z23 = 3z0
2 (vi) If z1z2z3 are the vertices of an isosceles triangle right angled at z2 then z2
1 + z22 + z2
3 = 2z2( z1 + z3) (vii) If z1z2z3 are the vertices of right-angled isosceles triangle then (z1 ndash z2)
2 = 2 (z1 ndash z3)(z3 ndash z2)
(viii) Area of triangle formed by the points z1 z2 and z3 is 1 1
2 2
3 3
z z 11 z z 14i
z z 1
19121 Dot and Cross ProductLet z1 = x1+ iy1 and z2 = x2 + iy2 be two complex numbers ie (vectors) The dot product (also called the
scalar product) of z1 and z2 is defined by z1 z2 = |z1| |z2| cosq = x1x2 + y1y2 = Re = +1 2 1 2 1 21z z z z z z 2
Complex Number 19209
Where q is the angle between z1 and z2 which lies between 0 and p
If vectors z1 z2 are perpendicular then z1z2 = 0 rArr + =1 2
1 2
z z 0z z
ie Sum of complex slopes = 0
The cross product of z1 and z2 is defined by z1z2 = |z1| |z2| sinq = x1y2ndashy1x2 = = minus1 2 1 2 1 2Imz z z z z z 2i
If vectors z1 z2 are parallel then z1 z2 = 0 rArr =1 2
1 2
z zz z
ie complex slopes are equal
A(5) amp(z) = q represents a ray emanating from the origin and inclined at an angle q with the positive direction of x-axis
Also arg(z ndash z1) = q represents the ray originating from A(z1) inclined at an angle q with positive direction of x-axis as shown in the above diagram
A(6) |z ndash z1| = |z ndash z2| represents perpendicular bisector of line segment joining the points A(z1) and B(z2) as shown here
A(7) The equation of a line passing through the points A(z1) and B(z2) can be expressed in determinant
form as =1 1
2 2
z z 1z z 1 0z z 1
it is also the condition for three points z1 z2 z3 (when z is replaced by z3) to be
collinear
A(8) Reflection Points for a Straight LinesTwo given points P and Q are the reflection points of a given straight line if the given line is the right bisector of the segment PQ Note that the two points denoted by the complex number z1 and z2 will be the reflection points for the straight line α +α + =z z r 0 if and only if α +α + =1 2z z r 0 where r is real and a is non-zero constant
19122 Inverse Points wrt a CircleTwo points P and Q are said to be inverse wrt a circle with centre O and radius r if
(i) The point O P Q are collinear and P Q are on the same side of O (ii) OP OQ = r2
NoteThat the two points z1 and z2 will be the inverse point wrt the circle zz z z r 0α α+ + + = if and only if
1 2 1 2z z z z r 0α α+ + + =
19123 Ptolemys Theoremrsquos It states that the product of the length of the diagonal of a convex quadrilateral in scribed in a circle is equal to the sum of the products of lengths of the two pairs of its opposite sides ie |z1ndashz3||z2ndashz4| = |z1ndashz2| |z3ndashz4| + |z1ndashz4| |z2ndashz3|
19210 Mathematics at a Glance
A(8) |z ndash z1| = a represents circle of radius a and having centre at z1 |z ndash z1| lt a represents interior of the given circle |z ndash z1 |gt a represents exterior of the given circle
A(9) The equation |z ndash z1|2 + |z ndash z2|2 = k will represent a circle if k ge 12 |z1 ndash z2|
2
A(10) a lt |z| lt b represents points lying inside the circular annulus bounded by circles having radii a and b and having their centres at origin as shown below
A(11) |z + z1| = |z| + |z1| represents the ray originating from origin and passing through the point A(z1) as shown below |z + z1| = PPprime = PO + OPprime = |z| + OA = |z| + |z1| (∵ OPprime = OA)
A(12) |z ndash z1| = |z| ndash |z1| represents a ray originating from A(z1) but not passing through the origin as shown below |z ndash z1| = OP ndash OA = |z| ndash |z1|
A(13) Re(z) ge a represents the half-plane to the right of straight line x = a including the line itself as shown below
Re(z) le a represents the half-plane to the left of straight line x = a including the line itself as shown here
Complex Number 19211
Im(z) le a represents the half-plane below the straight line y = a including the line itself as shown below
Im(z) ge a represents the half-plane above the straight line y = a including the line itself as shown below
A(13) Inverse points wrt a circleTwo points A and B are said to be inverse wrt a circle with its centre lsquoOrsquo and radius a if
(i) The points O A B are collinear and on the same side of O and (ii) OAOB = a2
RemarkTwo points z1 and z2 will be the inverse points wrt the circle zz z z r 0β β+ + + = if and only
if 1 2 1 2z z z z r 0β β+ + + =
A(14) If l is a positive real constant and z satisfies minus= λ
minus1
2
z zz z
then the point z describes a circle of
which A B are inverse points unless l = 1 in which case z describes the perpendicular bisector of AB
A(15) To convert an equation in cartesian to complex form put +=
z zx2
and minus=
z zy2i
and to convert
an equation complex form to Cartesian form put z = x + iy and = minusz x iy
Chapter 20SetS and
relationS
201 SetS
lsquolsquoA set is any collection of distinct and distinguishable objects of our intuition or thoughtrsquorsquo By the term lsquodistinctrsquo we mean that no object is repeated By the term lsquodistinguishablersquo we mean that given an object we can decide whether that object is in our collection or not
202 RePReSeNtAtION OF SetS
A set is represented by listing all its elements between braces and by separating them from each other by commas (if there are more than one element)
203 NOtAtION OF SetS
Sets are usually denoted by capital letters of the English alphabet while the elements are denoted in gen-eral by small letters eg set of vowels = A = a e i o u
204 NOtAtION FOR SOMe SPeCIAL SetS
W Whole Number ℤ Integer ℚ Rational Numbers ℝ Real Numbers
ℕ Nutural Numbers I Integer Number ℚc Irrational Number C Complex Numbers
205 NOtAtION FOR SOMe SPeCIAL SetS
If x is an element of a set A we write x isin A (read as lsquox belongs to Arsquo) If x is not an element of A we write x notin A (read as lsquox does not belong to Arsquo) The symbol lsquoisinrsquo is called the membership relation a isin A but d notin A
206 MetHOD RePReSeNtAtION OF SetS
(i) Tabular Form or Roster Form Under this method elements are enclosed in curly brackets after separating them by commas For example if A is a set of naturals number which is less than 5 then A = 1 2 3 4
Sets and Relation 20213
(ii) Set Builder Method Under this method set may be represented with the help of certain property or properties possessed by all the elements of that set
A = x | P(x) or A = x P(x) This signifies A is the set of element x such that x has the property P For example the set
A = 1 2 3 4 5 can be written as A = x | x isin N and x le 5
207 CARDINAL NuMbeR OF A SetS
Cardinal number of a set X is the number of distinct elements in a set and it is denoted by n(X) For example for X = x1 x2 x3 n (X) = 3
208 tyPeS OF SetS
Finite Set A set lsquoXrsquo is called lsquofinitersquo if it haslimited number of elements in it That is ifits all elements are labeled with the helpof natural numbers the processterminates at certain finite naturalnumber eg set of living people on earth
Null Set A set lsquoXrsquo iscalled nullvoidemptyif it has no element init It is denoted By φ or For example A = x x isin amp x2 + 2 = 0B = xx isin amp x2 lt0
Singletion Set A set Xis called singleton set if ithas only one element init For example A = xx isin and x2 + 4 = 0B = xx isin and x2 le0
Infinite Set A set lsquoXrsquo is calledinfnite if it has unlimited numberof elements in it For exampleset of rational numbers or set
of points in a plane
Classification of Set
Countably infiniteSet A set lsquoXrsquo is called countableif its elements can belabeled with the helpof natural numbersThat is its elementsare function ofnatural numbers Forexample a set of oddnatural numbers
Uncountable A set lsquoXrsquo is calleduncountable if itselements cannot belabeled with the helpof Natural numbersie Its elements cannot be written asfunction of naturalnumbers eg set ofreal numbers set ofirrational numbers
Some Important Remarks
Equivalent Sets Two finite sets A and B are equivalent if their cardinal numbers are same That is n (A) = n (B)
Equal Sets Two sets A and B are said to be equal if every element of A is a member of B and every element of B is a member of A That is A = B if A and B are equal and A ne B if they are not equal
Every finite set is countable but every countable set is not necessarily finite
Infinite sets may or may not be countable
Uncountable sets are always infinite
Every subset of a countable set is countable
Every superset of an uncountable set is also uncountable
Intersection of countable sets is always countable
Countable union of countable sets is always countable
20214 Mathematics at a Glance
209 SubSetS
A set A is said to be a subset of B if all the elements of A are present in B and is denoted by A sube B (read as A is subset of B) and symbolically written as x isin A rArr x isin B hArr A sube B
2010 NuMbeR OF SubSetS
Consider a set X containing n elements as x1 x2 xn then the total number of subsets of X = 2n
Proof Number of subsets of the above set is equal to the number of selections of elements taking any number of them at a time out of the total n elements and it is equal to 2n Q
nC0 + nC1 + nC2++ nCn = 2n
2011 tyPeS OF SubSetS
(i) Proper Subset A non-empty set A is said to be a proper subset of a set B if every element of A is an element
of B and B has at least one element which is not an element of A and is denoted by A sub B (ii) Improper Subset The set A itself and the empty set is known as improper subset For example if X = x1 x2 xn
then total number of proper sub-sets = 2n - 2 (excluding itself and the null set) The statement A sube B can be written as B supe A then B is called the super set of A
2012 POweR SetS
The collection of all subsets of set A is called the power set of A and is denoted by P(A) ie P(A) =
X X is a subset of A If A = x1 x2 x3 xn then n(P(A)) = 2n n(P(P(A)) = 22n
Thus X isin P(A) hArr X sube A ie the elements of P(A) are the subset of A
2013 DISjOINt SetS
Sets A and B are said to be disjoint iff A and B have no common element or A cap B = f If A cap B ne f then A and B are said to be intersecting or overlapping sets Eg
(i) If A = 1 2 3 B = 4 5 6 and C= 4 7 9 then A and B are disjoint set where B and C are intersecting sets
(ii) Set of even natural numbers and odd natural numbers are disjoint sets
2014 uNIVeRSAL SetS
It is a set which includes all the sets under considerations To explain this it is a super set of each of the given set Thus a set that contains all sets in a given context is called the universal set It is denoted by U For example if A = 1 2 3 B = 2 4 5 6 and C = 1 3 5 7 then U = 1 2 3 4 5 6 7 can be taken as the universal set
2015 COMPLeMeNt Set OF A gIVeN Set
Complement set of a set A is a set containing all those elements of universal set which are not in A It is denoted by cA A or Aprime So Ac = x x isin U but x notin A For example if set A = 1 2 3 4 5 and universal set
U = 1 2 3 4 50 then A = 6 7 50
Sets and Relation 20215
2016 COMPLeMeNtRy Set OF A gIVeN SetS
Two sets A and B are said to be complementry sets if A B and B A= = To explain this if elements of A are removed from universal set U we get the elements of set B and if elements of A are removed from U we get elements of set B
Remarks (i) Two disjoint sets need not be complementry eg if U = 1 2 3 4 5 A = 1 3 B = 2 4 then A
and B are disjoint but Ac = 2 4 5 ne B and Bc = 1 3 5 ne A
(ii) Two complementary sets are always disjoint
2017 COMPARAbLe SetS
Two set A and B are said to be comparable if either A sub B or B sub A or A = B If neither (A sub B or B sub A) nor A = B then A and B are said to be incomparable
2018 VeNN (euLeR) DIAgRAMS
Here we represent the universal set U as the set of all points within rectangle and the subset A of the set U is represented by the interior of a circle If a set A is a subset of a set B then the circle representing A is drawn inside the circle representing B If A and B are not equal but they have some common elements then we represent A and B by two intersecting circles
2019 OPeRAtIONS ON SetS
20191 Union of Two SetsThe union of two sets A and B is the set of all those elements which are either in A or in B or in both This set is denoted by A cup B (read as lsquoArsquo union Brsquo)
Symbolically A cup B = x x isin A or x isin Bor A cup B = x x isin A (cup v denotes OR) x isin BClearly x isin A hArr x isin A or x isin B and x notin A cup B hArr x notin A and x notin BThe union of two sets can be represented by a Venn diagram as shown in the following figures
The shaded region represents A cup B
20216 Mathematics at a Glance
20192 Intersection of Two SetsThe intersection of two sets A and B is the set of all those elements which are common in A and B This set is denoted by A cap B (read as lsquoA intersection Brsquo)
Symbolically A cap B = x x isin A and x isin Bor A cap B = x x isin A cap x isin B [cap denotes lsquoandrsquo]Clearly x isin A cap B hArr x isin A and x isin BBut x notin A cap B hArr x notin A or x notin B ie x is not found in atleast one of A and BThe intersection of two sets can be represented by a Venn diagram as shown in above figure The
shaded region represents A cap B
20193 Difference of Two SetsThe difference of two sets A and B in this order (also called lsquorelative complementrsquo of B in A) is the set of all those elements of A which are not elements of B It is denoted by A ndash B and is read as lsquoA minus Brsquo
Symbolically A ndash B = x x isin A and x notin BThus x isin A ndash B hArr x isin A and x notin BSimilarly B ndash A = x x isin B and x notin A Thus x isin B ndash A hArr x isin B and x notin AA ndash B can be represented by Venn diagram as shown in the given figure The shaded region represents A ndash B
20194 Symmetric Difference of Two SetsSet of those elements which are obtained by taking the union of the difference of A and B ie (A - B) and the difference of B and A ie (B - A) is known as the symmetric differerence of two sets A and B and it is denoted by (A D B) Thus A D B = (A - B) cup (B - A) = x x isin (A cup B) ndash (A cap B)
Representation through the Venn diagram is given in the figure here
Sets and Relation 20217
20195 Complement of a Set
The complement of a set A (also called lsquoabsolute complementrsquo of A) is the set of all those elements of the universal set S which are not elements of A It is denoted by Aprime or Ac
Clearly Aprime or Ac = S ndash ASymbolically Aprime or Ac = x x isin S and x notin AThus x isin Aprime hArr x notin AComplement of a set can be represented by a Venn diagram as shown in the figure here The shaded
region represents Aprime
2020 LAwS FOLLOweD by Set OPeRAtIONS cup cap AND D
(i) Idempotent Operation For any set A we have(a) A cup A = A and (b) A cap A = A
(ii) Existence of identity element wrt set operationFor any set A we have(a) A cup f = A and (b) A cap U = A(c) A ndash f = A (d) A D f = AThat is f and U are identity elements for (union difference symmetric difference) and intersection respectively
(iii) Commutativity For any set A and B we have(a) A cup B = B cup A (b) A cap B = B cap A(c) A D B = B D AThat is union and intersection and symmetric difference are commutative Note that A ndash B ne B ndash A
(iv) AssociativityIf A B and C are any three sets then(a) (A cup B) cup C = A cup (B cup C) (b) (A cap B) cap C = A cap (B cap C)(c) (A D B) D C = A D (B D C)ie union and intersection are associativeNote that (A ndash B) ndash C ne A ndash (B ndash C) eg for A = 2 3 4 5 6 7 8 B = 6 7 8 9 10 C = 4 5 6 7 10 12 (A ndash B) ndash C = 2 3 A ndash (B ndash C) = 2 3 4 5 6 7
(v) Divisibility If A B and C are any three sets then(a) A cup (B cap C) = (A cup B) cap (A cup C) (b) A cap (B cup C) = (A cap B) cup (A cap C)(c) A cup (B cup C) = (A cup B) cup (A cup C) (d) A cap (B cap C) = (A cap B) cap (A cap C)ie union and intersection are distributive over intersection and union and on themselves
(vi) Complement law(a) A cup Aprime = cup (Universal set) (b) A cap Aprime = f(c) (Aprime)prime = A (d) fprime = cup and cupprime = f
2021 De-MORgANrsquoS PRINCIPLe
If A and B are any two sets then
(i) (A cup B)prime = Aprime cap Bprime (ii) (A cap B)prime = Aprime cup Bprime
20218 Mathematics at a Glance
2022 INCLuSIVe-exCLuSIVe PRINCIPLe
(i) For set A and B n(A cup B) = n(A) + n(B) ndash n(A cap B) (ii) For sets A B and C n(A cup B cup C) = n(A) + n(B) + n(C) ndash n(A cap B) ndash n(B cap C) ndash n(C cap A) +
n(A cap B cap C)
2023 SOMe ReSuLtS ON CARDINAL NuMbeRS
(i) max n(A) + n(B) ndash n(S) 0 le n (A cap B) le min n(A) n(B) (ii) max n(A) n(B) le n (A cup B) le min n(A) + n(B) n(S) (iii) n(Ac) = n(U) ndash n(A)
20231 Cartesian Product of Two SetsCartesian product of two sets A and B is a set containing the ordered pairs (a b) such that a isin A and b isin B It is denoted by
A times B ie A times B = (a b) a isin A and b isin B If set A = a1 a2 a3 and B = b1 b2 thenA times B = (a1 b1) (a1 b2) (a2 b1) (a2 b2) (a3 b1) (a3 b2) andB times A = (b1 a1) (b1 a2) (b1 a3) (b2 a1) (b2 a2) (b2 a3)Clearly A times B ne B times A until A and B are equal
Remarks
1 Since A times B has elements as ordered pairs therefore it can be geometrically located on X ndash Y plane by considering set A on X-axis and set B on Y-axis
2 Cartesian product of n sets A1 A2 A3An is denoted by A1 x A2 x A3 x x An and is the set of n ordered tuples ie A1 x A2 x A3 x x An = (a1 a2 a3 an) ai isin Ai i = 1 2 3 n Cartesian product of n sets represents n-dimensional space
3 A times B times C and (A times B) times C are not same
A times B times C = (a b c) a isin A b isin B c isin C whereas
(A times B) times C = (a b c) a isin A b isin B c isin C
20232 Number of Elements in Cartesian Product A times B
If number of elements in A denoted by n(A) = m and number of elements in B denoted by n(B) = n then number of elements in (A times B) = m times n ie n(A times B) = n(A) times n(B)
Since A times B contains all such ordered pairs of the type (a b) such that a isin A and b isin B that means it includes all possibilities in which the elements of set A can be related with the elements of set B Therefore A times B contains n(A) times n(B) number of elements
20233 Properties and Laws of Cartesian Product
202331 Distributive laws
1 (a) Cartesian product distributes over union and intersection of sets That is A times (B cup C) = (A times B) cup (A times C) and A times (B cap C) = (A times B) cap (A times C) for every group
of sets A B and C
Sets and Relation 20219
(b) Cartesian product distributes over subtraction of sets That is A times (B ndash C) = (A times B) ndash (A times C) 2 Cartesian Product is not Associative Cartesian product of sets is not associative in nature
That is A times (B x C) ne (A times B) times C As the elements of A times (B times C) are of the type (a (b c)) whereas the elements of (A times B) times C are of
the type ((a b) c) a isin A b isin B c isin C 3 Cartesian Product is not Commutative Cartesian product of sets is not commutative in nature That is A times B ne B times A until A = B 4 Cardinality of Cartesian Product (a) If A and B are two sets then n(A times B) = n(A) times n(B) (b) If A and B are sets having k number of common elements ie n(A cap B) = k then the number
of elements common to A times B and B times A = k2 5 Intersection of cross product is equal to cross product of intersection That is for sets A B S and T (A times B) cap (S times T) = (A cap S) times (B cap T) 6 For subset A of B and C of D We have (a) (A times C) sube (B cap C) for every set C (b) (A times C) sube (B cap D) (c) (A times A) sube (A times B) cap (B times A) 7 For complementary sets B and C of sets B and C (a) A times (B cup C) = (A times B) cap (A times C) (b) A times (B cap C) = (A times B) cup (A times C) 8 A times (B D C) = (A times B) D (A times C)
2024 ReLAtIONS
A relation R from set X to Y (R X rarr Y) is a correspondence between set X to set Y by which none one or more elements of X are associated with none one or more elements of Y Therefore a relation (or binary relation) R from a non-empty set X to another non-empty set Y is a subset of X times Y That is R X rarr Y is nothing but subset of A times B For example consider set X and Y as set of all males and females members of a royal family of the Ayodhya kingdom
20220 Mathematics at a Glance
X = Dashrath Ram Bharat Laxman Shatrughan and Y = Koshaliya Kaikai Sumitra Sita Mandavi Urmila Shrutkirti and a relation R is defined as was husband of from set X to set Y
Then R = (Dashrath Koshaliya) (Ram Sita) (Bharat Mandavi) (Laxman Urmila) (Shatrughan Shrutkirti) (Dashrath Kaikai ) (Dashrath Sumitra)
2025 DOMAIN CO-DOMAIN AND RANge OF ReLAtION
Domain Domain of a relation R from set A to set B is the collection of elements of set A which are participating in the correspondence ie it is set of all pre-images under the relation R For example domain of R = (1 5) (2 10) (3 6) is
DR = 1 2 3 where R is a relation from set A = 1 2 3 4 to set B = 5 6 7 8 9 10
Co-domain Co-domain of a relation R from set A to set B is set B itself irrespective of the fact whether an element of set B is related with any element of A or not For example B =5 6 7 8 9 10 is co-domain of above relation R
Range Range of a relation R from set A to set B is the set of those elements of set B which are participating in the correspondence ie set of all images under the relation R For the above relation range is given by the set RR = 5610
2026 uNIVeRSAL ReLAtION FROM Set A tO Set b
Since A times B contains all possible ordered pairs which relate each element of A to every element of B therefore (A times B) is largest possible relation defined from set A to set B and hence also known as Universal relation from A to B
2027 NuMbeR OF ReLAtIONS FROM Set A tO Set b
Since each relation from A to B is a subset of Cartesian product A times B therefore number of relations that can be defined from set A to set B is equal to the number of subsets of A times B Thus the number of relations from A to B = 2n(A times B) = 2n(A) x n(B)
2028 ReLAtION ON A Set
A relation R from set A to itself is called relation on set AFor example let A = 1234916 Define a relation from set A to itself as a R b if b is square of a
but a ne b thenR = (2 4) (3 9)(4 16) Here domain = 2 3 4 co-domain = A range = 4 9 16
2029 RePReSeNtAtION OF ReLAtION IN DIFFeReNt FORMS
(i) By representing the relation as a set of ordered pairs (Roster form)In this method we represent the relation by a set containing ordered pairs (a b) where a isin A and b isin B such that aRb as shown for the relation R from A = 1 2 3 4 to set B = 2 3 4 5 6 7 when b isin B is to be related to a isin A here such that b = 2a + 1 R = (13) (2 5) (3 7)
Sets and Relation 20221
(ii) Analytical method or set builder from In this method we represent the relation as R = (a b) a isin A b isin B ahellipb where the dots are replaced by an equation connecting image b with its pre-image a For example let R be a relation from set A = 1 2 3 4 to set B = 2 3 4 5 6 7 given by R = (13)(25)(37) then it can be represented by R = (x y) x isin A y isin B x R y iff y = 2x + 1
(iii) Graphical representation or representation by lattice In this method we take set X along x-axis and set B along y-axis then plot the points (a b) isin R in x y plane For example in the above illustration the relation can be represented as shown in the diagram given below
(iv) By arrow diagram In this method we represent set A and set B by two circles or by two ellipses and join the images and their pre-images by using arrows as shown below for above illustration
(v) Tabular form In this form of representation of a relation R from set A to set B elements of A and B are written in the first column and first row respectivelyIf (a b) isin ℝ then we write lsquo1rsquo in the row containing a and the column containing b and if (a b) notin ℝ then we write lsquo0rsquo in the row containing a and the column containing bFor example for the relation R = (1 3) (2 5) (3 7) from set A = 1 2 3 4 to set B = 2 3 4 5 6 7 we have the following tabular representation
R 2 3 4 5 6 7
1 0 1 0 0 0 02 0 0 0 1 0 03 0 0 0 0 0 14 0 0 0 0 0 0
20222 Mathematics at a Glance
2030 CLASSIFICAtION OF ReLAtIONS
One-one or Injective Relation
If different elements of set X are related with different elements of set Y ie no two different elements of domain are related to same element of set Y then R is said to be one-one relation or injective relation from set X to set Y
Many-one Relation
When there exists at least one group having more than one element of set X which are related with same element of set Y then R is said to be many one relation from set X to set Y
One-many Relation
Relation R from set X to set Y is said to be one-many if there exists an element in set X which is related with more than one element of set Y
Many-many Relation
Relation R from set X to set Y is said to be many-many if it is many-one as well as one-many
Onto Relation (Surjective Relation)
A relation R X rarr Y is said to be onto or surjective relation if there is no such element y isin Y which is not related with any x isin X ie for each y isin Y there exist at least one element x in X which is related with y In such a relation
Range (RR) = co-domain ie range of onto relation is nothing but the co-domain of the relation
RemarkIn onto relation all elements of set X may or may not participate in relation but all elements of co-domain set Y participate in relation
2031 INtO ReLAtION
A relation R X rarr Y is said to be into iff there exist at least one y isin Y which is not related with any x isin X
Sets and Relation 20223
That is if range (RR) sub co-domain that is range of relation is a proper subset of co-domain
That is R6 (x1 y1) (x1 y2) (x2 y3)Clearly under relation R6 y4 has no pre-image in X
20311 One-One-Onto Relation (Bijective Relation)
A relation R X rarr Y is said to be bijective relation iff it is both onendashone as well as onto relation
For example R7 (x1 y2) (x2 y1) (x3 y3) where X = x1 x2 x3 x4 and Y = y1 y2 y3)
2032 tyPeS OF ReLAtIONS
20321 Reflexive Relation
R X rarr Y is said to be reflexive iff x R x x isin X That is every element of X must be related to itselfTherefore if for each x isin X (x x) isin R then relation R is called reflexive relation
RemarkIf R X rarr Y is a reflexive relation then its domain is X For example if R is a relation on set of integers (ℤ) defined by ldquoxRy iff x divides yrdquo then it is reflexive and hence its domain set is ℤ
20322 Identity RelationA relation R X rarr Y is said to be an identity relation if each element of X is related to itself only For example if X = x1 x2 x3 and Y = x1 x2 x3 x4 then the relation R = (x1 x1) (x2 x2) (x3 x3) is an identity relation from set X to set Y
Remarks 1 Every identity relation from set X to set Y is reflexive relation from set X to set Y but converse is
not true That is every reflexive relation need not be identity For example R X rarr Y where X = x1 x2 x3 and Y = x1 x2 x3 x4 then the relation R = (x1 x1) (x2 x2) (x3 x3) (x1 x2) is reflexive but not identity relation from set X to set Y because x1 R x1 as well as x1Rx2
2 If R is a relation from set X to itself then the relation is called relation on set X
(a) R is said to be reflexive on set X if xRx x isin X
(b) R is said to be identity relation on set X if x R x x isin X and x is not related to any other element and it is denoted by Ix
3 Symmetric Relation R X rarr Y is said to be symmetric iff (x y) isin R rArr (y x) isin R
That is x R y rArr y R x For example perpendicularity of lines in a plane is symmetric relation
20323 Transitive Relation
R X rarr Y is said to be transitive iff (x y) isin R and (y z) isin R rArr (x z) isin RThat is x R y and yR zrArr x R z For example the relation ldquobeing sister ofrdquo among the members of a family is always transitive
20224 Mathematics at a Glance
Notes (i) Every null relation is a transitive relation
(ii) Every singleton relation is a transitive relation
(iii) Universal and identity relations are reflexive as well as transitive
20324 Anti-symmetric RelationA relation R from set X to set Y is said to be an anti-symmetric relation iff (a b) isin R and (b a) isin R rArr a = b
That is for two different elements x isin X and y isin Y the relation R does not contain the ordered pairs (x y) and (y x) simultaneously
For example relations ldquobeing subset of rdquo ldquois greater than or equal tordquo and ldquoidentity relationrdquo are anti-symmetric relations
RemarkA relation R from set X to set Y may be both symmetric as well as anti-symmetric any one or not bothFor example let X = 1 2 3 4 and Y = 1 2 3 4 5 6
203241 Consider the relations
(i) R1 = (11) (22) (ii) R2 = (1 2) (2 1) (2 3) (3 2) (iii) R3 = (1 1) (2 2) (3 4) (iv) R4 = (1 2) (2 1) (3 4)
1 R1 is symmetric as whenever ordered pair (x y) isin R1 rArr (y x) isin R1Also R1 is anti-symmetric as for no two different elements x y the ordered pairs (x y) and (y x) occur in R1
2 R2 is symmetric but not anti-symmetric as (1 2) (2 3) isin R2 rArr (2 1) (3 2) isin R2 but 1 ne 2 and 2 ne 3 3 R3 is anti-symmetric but not symmetric as (3 4) isin R3 but (4 3) notin R3 4 R4 is neither symmetric nor anti-symmetric as (3 4) isin R3 but (4 3) notin R3 and (1 2) (2 1) both are
in R3 but 1 ne 2
20325 Equivalence RelationA relation R from a set X to set Y (R X rarr Y) is said to be an equivalence relation iff it is reflexive symmetric as well as transitive The equivalence relation is denoted by ~ For example relation ldquois equal tordquo Equality Similarity and congruency of triangles parallelism of lines are equivalence relations
2033 COMPOSItION OF ReLAtIONS
Let R and S be two relations from set A to B and B to C respectively Then we can define a relation SoR from A to C such that (a c) isin SoR hArr exist b isin B such that (a b) isin R and (b c) isin S
This relation is called the composition of R and S Diagrammatically it is shown in the following figure
Sets and Relation 20225
2034 INVeRSe OF A ReLAtION
Let A B be two sets and let R be a relation from a set A to B Then the inverse of R denoted by R-1 is a relation from B to A and is defined by R-1 = (b a) (a b)isinR Clearly (a b) isin R hArr (b a) isin R-1
Also Dom (R) = Range (R-1) and range (R) = Dom (R-1)For example let A = 1 2 3 4 and B = 2 3 4 5Define a relation R from A to B as xRy iff y = x + 1 then R = (1 2)
(2 3) (3 4) (4 5)rArr Rndash1 = (2 1) (3 2) (4 3) (5 4)Thus we can define Rndash1 a relation from B to A as xRy iff y = x ndash 1
The arrow diagram represents the relations R and Rndash1
Remark(SoR)ndash1 = Rndash1oSndash1 where R is a relation from A to B and S is a relation from B to C
Tips and Tricks
If number of elements in A n(A) = m and n(B) = n then number of elements in (A times B) = m times n A times B is termed as the largest possible relation defined from set A to set B it is also known as the
universal relation from A to B If A sube B then (A times B) cap (B times A) = A2 = A times A If A has m elements and B has n elements then number of relations that can be defined
from A to B = 2m times n If A is a set containing n elements then the number of relations that can be defined
on set ( )2nA 2=
If A and B are two non-empty sets having n elements in common then A times B and B times A have n2 elements in common
If A is related to B then symbolically it is written as (aRb) where a is pre-image and b is image If A is not related to B then symbolically it is written as a R b All identity relations are reflexive but all reflexive relations are not identity Every null relation is a transitive relation Every singleton relation is a transitive relation Universal and identity relations are reflexive as well as transitive Identity relation is symmetric as well as anti-symmetric or both Union of two reflexive (or symmetric) relations on a set A also reflexive (or symmetric) on set A Union of two transitive relations need not be transitive on set A Union of two equivalence relations need not be equivalence 1 If R and S are two equivalence relations on a set A then R cap S is also an equivalence
relation on A 2 The inverse of an equivalence relation is an equivalence relation 3 The set (a a)a isin A = D is called the diagonal line of A times A Then lsquothe relation R in a is
antisymmetric iff R cap Rndash1 sube D
Chapter 21FunCtions
211 Definition of function
Let X and Y be two non-empty sets Then a function lsquof rsquo from set X to set Y is denoted as f X rarr Y or y = f(x) x isin X and y isin Y A function f(x) from X (domain) to Y (co-domain) is defined as a relation f from set X to set Y such that each and every element of X is related with exactly one element of set Y
Image and Pre-image Let f be a function from set X to set Y ie f X rarr Y and let an element x of set X be associated to the element y of set Y through the rule lsquof rsquo then (x y) isin f ie f(x) = y then y is called lsquoimage of x under f rsquo and x is called lsquopre-image of y under f rsquo
Natural Domain The natural domain of a function is the largest set of real number inputs that give real number outputs of the function
Co-domain Set Y is called co-domain of function f
Range of Function If f Df (sube X) rarr Y is a function with domain Df then the set of images y (output isinY) generated corresponding to input x isin Df is called range of function and it is denoted by Rf
ie Rf = f(x) xisinDf sube Y
Remarks
(i) Every function is a relation but every relation read not be a function
(ii) A relation R A rarr B is a function if its domain = A and it is not one-many ie either one-one or many-many
Functions 21227
(iii) To find domain of function we need to know when does a function become undefined and when it is defined
ie we need to find those values of x where f(x) is finite and real and those values of x where f(x) is either infinite or imaginary
(iv) When its value tends to infinity (infin)
eg =minus2
1y
x 1 at x = plusmn1 f(x) is not defined at x = plusmn1 and defined forall x isin ℝ except for plusmn1 therefore
domain of f(x) = ℝ ~ 1 ndash1
(v) When it takes imaginary value eg = minusy x 1 at x isin (ndashinfin 1) f(x) is not defined on (ndashinfin 1) and
defined on [1 infin) therefore domain of f(x) = [1 infin)
(vi) When it takes indeterminate form ie becomes of the form infininfininfin infin minus infin
infin0 00
1 0 etc0
212 RepResentation of a function
A fanction can be represented analytcally as orduced pass parametrically wita arrow diagram praphibly
Remarks All function cannot be represented by all the above methods
(i) The Drichlet-Function which is defined as f(x) =
0 when x is rational
1 when x is irrational cannot be graphed since there exist
infinite number of rationals as well as irrationals between any two real numbers
(ii) Consider the Eulerrsquos totient or Eulerrsquos phi function f(n) = Number of positive integers less than or equal to n and co-prime to n where n is a natural number
The domain of f is the set of positive integers Its range is the set of positive integers 1 2 3 hellip
We cannot represent this function analytically A portion of the graph of f(n) as shown here for understanding of the function
(iii) Consider another function called prime number function defined by f(x) = number of prime numbers less than or equals to x where x is non-negative real number
Then domain of f(x) is (0 infin) and range is the set of non-negative integers ie 0 1 2 3 hellip
The graph of function is shown here
As x increases the function f(x) remains constant until x reaches a prime at which the graph of function jumps by 1Therefore the graph of f consists of horizontal line segments This is an example of a class of function called step functions
(iv) Another function which is so complicated that it is impossible to draw its graph
h(x) = minus
x if x is rational
x is irrationalif x2
As we know that between any two real numbers there lie infinitely many relations and irrational number so it is impossible to draw its graph
21228 Mathematics at a Glance
213
s
oM
e s
tan
Da
RD
fu
nc
tio
n
S
NO
St
anda
rd
Func
tion
Bas
ic D
efini
tion
D
omai
nR
ange
Form
of C
urve
Fun
ctio
n
1C
onst
ant
func
tion
y =
c c
isin ℝ
is a
fine
d re
ad n
umbe
rR
c
2lin
e as
fu
nctio
nsy
= ax
2 + b
a b
isin ℝ
a ne
0R
R
3Q
uade
atic
fu
nctio
ny
= ax
2 + b
x +
c a
b c
isin ℝ
a ne
0R
D
4a
minus
infin
whe
re
D =
b2 ndash
4ac
4Cu
bic
func
tion
y =
ax3 +
bx2 +
cx +
d a
b c
d isin
ℝ
a ne
0R
R
Functions 21229
5Bi
quad
ratic
fu
nctio
ny
= ax
4 + b
x3 + cx
2 + d
x +
e a
b c
d e
isin
ℝ a
ne 0
R[f(
k) infin
] fo
r a gt
0
whe
re K
is th
e po
int o
f loc
al
min
ima
havi
ng
leas
t im
age
and
(ndashinfin
f(k
)] f
or
a lt
0 w
here
k
is th
e po
int o
f lo
cal m
axim
a ha
ving
gre
ates
t im
age
6Po
lyno
mia
l fu
nctio
n of
nt
h de
gree
y =
a 0xn + a
1xnndash1 +
a2xnndash
2 + hellip
+ a
nndash1 x
+
a n ai isin
ℝ a
0 ne 0
n isin
ℕR
= ℝ
if n
is o
dd
= [f(
k) infin
] for
a 0 gt
0 k
is p
oint
of
loca
l min
ima
havi
ng le
ast
imag
e if
n is
even
=
(ndashinfin
f(k
)] fo
r a 0 gt
0 k
is p
oint
of
loca
l max
ima
havi
ng g
reat
est
imag
e if
n is
even
21230 Mathematics at a Glance
7M
odul
us
func
tion
xx
0f(
x)x
xx
0ge
=
= minus
lt
R[0
infin)
8Si
gnum
fu
nctio
n1f
orx
0f(
x)sg
n(x)
0fo
rx0
1for
x0
minuslt
=
=
gt
Rndash
1 0
1
9G
reat
est
inte
ger
func
tion
xif
xf(
x)[x
]k
ifk
xk
1k
isin
==
ltlt
+
isin
ie
[x] =
gre
ates
t am
ong
the
inte
gers
le
ss th
an o
r equ
al to
x
Rℤ
= se
t of a
ll in
tege
rs
10
Leas
t int
eger
fu
nctio
n or
ce
iling
of x
xif
xf(
x)x
k1i
fkx
k1
isin
==
+lt
lt+
ie
|x| =
leas
t am
ong
the
inte
gers
gr
eate
r tha
n or
equ
al to
x
Rℤ
= se
t of a
ll in
tege
rs
Functions 21231
11
Nea
rest
in
tege
r fu
nctio
n
11
xif
kx
k2
2f(
x)(x
)1
3k
1ifk
xk
22
minus
lelt
+
==
++
lelt
+
ie
(x) =
inte
ger n
eare
st to
x a
nd if
x
is of
the
form
1
kk
2+
isin
the
n
(x) =
k +
1
Rℤ
= Se
t of
inte
gers
12
Frac
tiona
l pa
rt fu
nctio
n0i
fxf(
x)x
fi
fxk
fan
df
(01
)k
isin
=
==
+
isinisin
R[0
1)
13
Iden
tity
fu
nctio
nf(x
) = x
RR
14
Expo
nent
ial
func
tion
f(x) =
ax a
gt 0
a ne
1 a
is fi
xed
and
x va
ries o
ver s
et o
f rea
d nu
mbe
rsR
(0 infin
)
21232 Mathematics at a Glance
15
Loga
rithm
ic
func
tion
f(x) =
log ax
a gt
0 a
ne 1
and
a is
fixe
d re
al n
umbe
r x v
orie
s ove
r set
of r
eal
num
bers
(0 infin
)R
16
Reci
proc
al
func
tion
or
rect
angu
lar
hype
rbol
a
1y
x=
ℝ ndash
0
ℝ ndash
0
Functions 21233
17
Pour
fu
nctio
ny
= xk k
isin ℝ
(i)
k =
(2n
+ 1)
n isin
ℕR
R
(ii)
k =
2n
n isin
ℕR
[0 infin
)
(iii)
k =
ndash(2
x +
1) n
isin ℕ
ℝ ndash
0
ℝ ndash
0
21234 Mathematics at a Glance
(iv)
k =
ndash(2
n) n
isin ℕ
ℝ ndash
0
(0 infin
)
(v)
1
kn
2n1
=isin
+
RR
(vi)
1
kn
2n=
isin
[0
infin)
[0 infin
)
Functions 21235
(vii)
1
kn
(2n
1)=minus
isin+
ℝ ndash
0
ℝ ndash
0
(viii
)
1k
n2n
=minus
isin
(0 infin
)(0
infin)
(ix)
2n
kn
m2n
(2m
1)=
isin+
minus
R[0
infin)
(x)
2n
kn
m2n
(2m
1)=
isinminus
minus
R[0
infin)
21236 Mathematics at a Glance
(xi)
2n
1k
nm
2m
1minus
=isin
minus
2n1
nm
(01
)2m
1minus
ltisin
minus
RR
(xii)
2n1
kn
m2m
1minus
=isin
minus
and
n gt
m k
gt 1
RR
(xiii
)
(2n
1)k
nm
2mminus
=minus
isin
(0 infin
)(0
infin)
(xiv
)
(2n
1)k
nm
(2n
1)minus
=minus
isinminus
ℝ ndash
0
ℝ ndash
0
Functions 21237
(xv)
2n
kn
m(2
m1)
=minus
isinminus
R ndash
0
(0 infin
)
(xvi
) k
= a
rega
live
irrat
iona
l num
ber
(0 infin
)(0
infin)
(xvi
ii)
k =
a ne
galiv
e irr
atio
nal
num
ber
(0 infin
)(0
infin)
18
Trig
onom
etry
fu
nctio
nsy
= f(s
in x
cos
x t
an x
cot
x s
ec x
co
sec x
) eg
f(x
) = si
n x
+ co
s xf(x
) = 1
ndash co
s x +
sec2 x
Com
mon
dom
ain
of tr
igon
omet
ric
func
tions
in
volv
ed
Can
be
foun
d us
ing
prop
ertie
s fu
nctio
ns li
ke
cont
inui
ty
mon
oton
icity
bo
unde
d he
re
etc
Dep
ends
upo
n th
e tr
igon
omet
ric
ratio
n in
volv
ed
21238 Mathematics at a Glance
19
Alg
ebra
ic
func
tions
Func
tions
cons
istin
g of
fini
te n
umbe
r of
tern
s inv
olvi
ng p
ower
s and
lots
of
inde
pend
ent v
aria
ble
and
the
four
fu
ndam
enta
l ope
ratio
ns (+
ndash times
divide)
Dep
ends
on
func
tion
eg
32
x1
f(x)
xminus
= h
as
its d
omai
n [1
infin)
Dep
ends
on
func
tion
and
can
be fo
und
usin
g ca
lcul
us
Dep
ends
upo
n th
e fu
nctio
n
20
Tran
scnd
ien-
tal f
unct
ion
The
func
tions
whi
ch a
re n
ot a
lgeb
raic
eg
1f(
x)nx
sinx
minus=
minus
etc
Dep
ends
on
func
-tio
nca
n be
foun
d us
ing
calc
ulus
Dep
ends
upo
n th
e fu
nctio
n
21
Ratio
nal
func
tion
P(x)
yf(
x)
Q(x
)=
=
P(x)
and
Q(x
) are
pol
ynom
ial
func
tion
ℝ ndash
x
Q(x
) = 0
Ex
pres
s x in
te
rms o
f y
and
by th
e kn
owle
dge
of q
uadi
atic
eq
uatio
n th
ose
valu
es o
f y fo
r w
hich
x is
real
an
d be
long
to
dom
ain
eg
grap
h of
(x
1)f(
x)(x
2)(x
3)minus
=minus
minus is
show
n be
low
(i)
If ax
bf(
x)0
0cx
b+
=ne
+d
~c
minus
a~
c
22
Irra
tiona
l Fu
nctio
nTh
e al
gebr
enc f
unct
ions
hav
ing
ratio
nal (
non-
inte
ger)
pou
res o
f x
are
coul
ed ir
ratio
nal f
unct
ions
eg
3f(
x)x
1f(
x)x
=+
=
3
2
x1
x1
f(x)
xx
1
+minus
minus=
++
etc
Dep
ends
upo
n th
e fu
nctio
nC
an b
e fo
und
by
usin
g ca
lcul
usD
epen
ds u
pon
the
func
tion
eg
3
3f(
x)x
x1
=minus
+ h
as fo
llow
ing
wav
e
form
Functions 21239
214 equal oR iDentical functions
Two functions f and g are said to be equal if
1 The domain of f = the domain of g 2 The range of f = the range of g 3 f(x) = g(x) for every x belonging to their common domain eg f (x) =1x and g(x) = xx2 are identical
functions f(x) =log(x2) and g(x) =2log(x) are not-identical functions as domain of f(x) = (ndashinfin infin) ~ 0 whereas
that of g(x) = (0 infin)
RemarkGraphs of trigonometric function and inverse trigonometric functions with their domain and range are givenin the same book under corresponding topics
215 pRopeRties of GReatest inteGeR function (BRacket function)
(i) Domain of [x] ℝ Range of [x] ℤ (ii) [[x]] = [x] (iii) [x + m] = [x] + m provided m isin ℤ (iv) [x + [y + [z]]] = [x] + [y] +[z] (v) [x] gt n n isin ℤ rArr [x] isinn + 1 n + 2 n + 3 rArr x isin [n + 1 infin) (vi) [x] ge n rArr x isin [n infin) (vii) [x] lt n rArr x isin (ndashinfin n) (viii) [x] le n rArr n isin (ndashinfin n + 1)
(ix) [x] x if x
[ x]1 [x] if x
minus = minus isinminus = minus minus notin
(x) x ndash 1 lt [x] le x equality holds iff x isin ℤ (xi) [x] le x lt [x] + 1
(xii) xc
=xc
for c isin ℕ and x isin ℝ
(xiii) [x] + [y] le [x + y] le [x] + [y] + 1
(xiv) [x] = x x 12 2
+ + forall x isin ℝ
(xv) The number of positive integers less than or equal to n and divisible by m is given by nm
m
and n are positive integers
(xvi) If p is a prime number and e is the largest exponent of p such that pe divides n then k
k 1
nep=
infin =
sum
21240 Mathematics at a Glance
2151 Properties of Least Integer Function
1 The domain of the function is (-infin + infin) 2 The range is the set of all integers 3 [x] converts x = (I + f) into I while x converts it into I + 1 Eg If x = 24 then 2lt x lt3 rArr x = 3 = I + 1 4 When x is an integer [x] = x = x
5 x + n = x + n where n is an integer
2152 Properties of Fractional Part Function
(i) Domain of fractional part function = Df = ℝ Range of fractional part function = Rf = [0 1) (ii) x is periodic function with period 1 (iii) [x] = 0 (iv) [x] = 0 (v) x = x this result is true when fractional part function is applied on x on left hand side more
than or equal to twice
(vi) 0 x
x1 x x
isinminus = minus notin
(vii) [x] [y] 0 x y 1
[x y][x] [y] 1 1 x y 2
+ le + lt+ = + + le + lt
2153 Properties of Nearest Integer Function
(i) (x) =
1[x] if 0 x2
1[x] 1 if x 12
le lt + le lt
(ii) (x + n) = (x) + n if n isin ℤ
(iii)
2n 1(x) x ~ x n2( x)
2n 1(x) 1 for x n2
+ minus forall isin = isin minus = + minus + = isin
(iv)
1[x] n if n x n2(x)
1[x] 1 n 1 if n x n 12
= le lt += + = + + le lt +
Properties of Modulus of a real number 1 |x1 x2 x3 xn| = | x1 | | x2 | | x3 | | xn| forall xi isin ℝ
2 x | x |y | y |= forall x y isin ℝ and y ne 0
3 | xn | = | x |n forall n isin ℤ 4 | ndashx | = | x | forall x isin ℝ 5 | x | = d rArr x = d or x = -d
6 | x | lt d rArr x isin (- d d) and | x | gt d 7 | x - a | lt d rArr x isin (a ndash d a + d)
Functions 21241
8 | x ndash a | = d rArr x = a + d or a ndash d 9 | x ndash a | gt d rArr x gt a + d or x lt a ndash d
10 2x | x |= forall x isin ℝ 11 |x| = maxndashx x forall x isin ℝ 12 |x| = |y| hArr x2 = y2
13 |x + y| is not always equal to | x | + | y | 14 (Triangle inequality) | x + y | le | x | + | y | for all real x and y inequality holds if xy lt 0 ie x and y are
of opposite signs equality holds if xy ge 0 ie x and y are of same sign or at least one of x and y is zero 15 |x ndash y| le |x| + |y| for real x and y inequality holds if xy gt 0 ie x and y are of same sign equality holds
if xy le 0 ie x and y are of opposite sign or at least one of x and y is zero 16 ||x| ndash |y|| le |x + y| for real x and y Equality holds if x and y are of opposite signs and for same sign
inequality holds 17 ||x| ndash |y|| le |x ndash y| for real x and y Equality holds if x and y are of same sign and for opposite signs
inequality holds
21531 Methods of testing a relation to be a function
Method 1 When the relation to be tested is represented analytically A relation f X rarr Y defined as y = f(x) will be function iff x1 = x2 rArr f(x1) = f(x2) since otherwise an element of X would have two different image
Method 2 When the relation to be tested is represented as a set of ordered pairs
A relation f X rarr Y represented as a set of ordered pairs will be function from X to Y iffSet of abscissa of all ordered pairs is equal to XNo two ordered pairs should have same abscissa
RemarkBecause f is a relation from X rarr Y therefore abscissa of ordered pairs must belong to X where as ordinates of ordered pairs must belong to Y
Method 3 When the relation to be tested is represented graphically relation f X rarr Y y = f(x) is function iff all the straight line x = a forall a isin X intersect the graph of function exactly once as shown below
A relation f X rarr Y will not be a function in following two conditions 1 If for some a isin X line x = a does not cut the curve y = f(x) eg in the graph of function shown below
the line x = a does not cut the graph of function and a isin X (Df) = [a b] ie no output for input x = a
rArr f(x) is not a function from X to Y 2 If for atleast one a isin X line x = a intersects y = f(x) more than once ie there exists an input having
more than one output say at (a y1) (a y2) and (a y3) rArr For input x = a f(x) has three outputs y1 y2 as well as y3 Hence f(x) is not function
21242 Mathematics at a Glance
Method 4 When the relation to be tested is represented diagrammatically A relation f X rarr Y is a function if no input has two or more outputs in Y and no x isin X is un-related
216 classification of functions
2161 One-one (Injective) Function
f X rarr Y is called injective when different elements in set X are related with different elements of set Y ie no two elements of domain have same image in co-domain In other words we can also say that no element of co-domain is related with two or more elements of domain
217 Many-one functions
f X rarr Y is many-one when there exist at least two elements in the domain set X which are related with same element of co-domain Y
2171 Onto (Surjective) FunctionA function f X rarr Y is called surjective only when each element in the co-domain is f-image of at least one element in the domain ie f X rarr Y is onto iff y isin Y there exists x isin X such that f(x) = y ie iff Rf = co-domain (Y)
Surjective f X rarr Y reduces the co-domain set to range of function
218 MethoD of testinG foR injectivity
(a) Analytical Method A function f X rarr Y is injective (one-one) iff whenever two images are equal then it means that they are outputs of same pre-image ie f(x1) = f(x2) hArr x1 = x2 forall x1 x2 isin X Or by using contra-positive of the above condition ie x1 ne x2 hArr f(x1) ne f(x2) forall x1 x2 isin X
Notes
1 If f (x) is not one-one then it is many-one function If we go according to definition consider f(x1) = f(x2) rArr x1 is not necessarily equal to x2
ie If two f-images are equal then their pre-images may or may not be equal
2 To test injectivity of f(x) consider f(x1) = f(x2) and solve the equation and get x2 in terms of x1 If x2 = x1 is only solution then function f is injective but if other real solutions also exist then f is many-one function
Functions 21243
(b) Graphical Method For one-one every line parallel to x-axis y = k isin Rf cuts the graph of function exactly once then the function is one-one or injective
For many-one If there exists a line parallel to x-axis which cuts the graph of function at least twice then the function is many one
(c) Method of Monotonicity for one-one If a function f(x) is continuous and monotonic
(f (x) ge 0 f (x) = 0 occures at isolated points) on an interval I then it is always one-one on interval I because any straight line parallel to x-axis y = k isin I intersects the graph of such functions exactly once
For many-one
(i) If a function is continuous and non-monotonic on interval I then it must be many-one on interval I
(ii) If a function is discontinuous and monotonic on interval I then it can be one-one or many-one on I as is clear from the figures given below
(iii) Even functions and periodic functions are always many-one in their natural domains whereas they are one-one in their principal domain They can be made one-one by restricting the domain
eg cosx is many one on ℝ but is one-one on [0 p] or 02π
Similarly fraction part function
x is periodic function with period 1 It is many one on ℝ but one-one on [n n + 1) for each integer n
(iv) If a function is discontinuous and non-monotonic on an interval I then it can be one-one or many one on I It can be understood well by the graph shown as follows
21244 Mathematics at a Glance
(v) All polynomials of even degree defined in ℝ have at least one local maxima or minima and hence
are many-one in the domain ℝ Polynomials of odd degree can be one-one or many-one in ℝ (d) Hit and trial method to test many-one functions It is possible to find an element in the range of function which is f image of two or more than two
elements in the domain of function
219 into (non-suRjective) function
While defining function we have mentioned that there may exist some element in the co-domain which are not related to any element in the co-domain
f X rarr Y is into iff there exists at least one y isin Y which is not related with any x isin X
Thus the range of the into function is proper subset of the co-domain ie range sub co-domain (properly)
2110 one-one onto function (Bijective function)
If a function is both one-one as well as onto then f(x) is set to be bijective function or simply bijection
2111 testinG of a function foR suRjective
Method 1 The equality of range of function to co-domain forms the condition to test surjectivity of function For instance to test surjectivity of f [0 infin) rarr [2 infin) such that f(x) = x2 + 2
Using the analytic formula we obtain the rule of function for argument x in terms of y as shown below
∵ y = x2 + 2 x2 = y ndash 2 ie |x| = y 2minus
rArr x y 2= minus ∵ x ge 0