IIT-JEE Super Course in Mathematics - Vol 3 Calculus

473 Pages·2012·15.774 MB·English

by  Trishna Knowledge Systems

Super Course in Mathematics
CALCULUS
for the IIT-JEE
Volume 3
Trishna Knowledge Systems
A division of
Triumphant Institute of Management Education Pvt. Ltd
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Contents
Preface iv
Chapter 1 Functions and Graphs 1.1—1.94
STUDY MATERIAL
• Set Theory • Cartesian Product of Two Sets • Relations • Functions
• Composition of Functions • Inverse of a Function • Even and Odd
Functions • Periodic Functions • Some Real Valued Functions • Parametric form of
Representation of a Function • Graphs of Conic Sections • Graphs of a Few Composite
Functions • Transformation of Functions • Some Special Curves
Chapter 2 Differential Calculus 2.1—2.196
STUDY MATERIAL
• Introduction • Limit of a Function • Laws on Limits • Standard
Limits • Continuity of a Function • Types of Discontinuities of a Function • Concept
of Derivative—Differentiation • Differentiability of Functions • Derivatives of
Elementary Functions • Differentiation Rules • Concept of Differential • Successive
Differentiation • Higher Order Derivatives • Tangents and Normals • Mean Value
Theorem and its Applications • Rolle’s Theorem • L’ Hospital’s Rule • Extension
of the Mean Value Theorem • Increasing and Decreasing Functions • Maxima and
Minima of functions • Convexity and Concavity of a curve
Chapter 3 Integral Calculus 3.1—3.178
STUDY MATERIAL
• Introduction • Definite Integral as the Limit of a Sum • Anti-Derivatives
• Indefinite Integrals of Rational Functions • Integrals of the form ∫ dx ,
a+bcosx
dx acosx +bsinx
∫ , ∫ dx • Integration By Parts Method • Integrals of the
a+bsinx ccosx +dsin x
form ∫ ax2 +bx +c dx • Evaluation of Definite Integrals • Properties of Definite
Integrals • Improper Integrals • Differential Equations • Formation of a Differential
Equation • Solutions of First Order First Degree Differential Equations
Preface
The IIT-JEE, the most challenging amongst national level engineering entrance examinations, remains on the top of the
priority list of several lakhs of students every year. The brand value of the IITs attracts more and more students every year,
but the challenge posed by the IIT-JEE ensures that only the best of the aspirants get into the IITs. Students require thorough
understanding of the fundamental concepts, reasoning skills, ability to comprehend the presented situation and exceptional
problem-solving skills to come on top in this highly demanding entrance examination.
The pattern of the IIT-JEE has been changing over the years. Hence an aspiring student requires a step-by-step study
plan to master the fundamentals and to get adequate practice in the various types of questions that have appeared in the
IIT-JEE over the last several years. Irrespective of the branch of engineering study the student chooses later, it is important
to have a sound conceptual grounding in Mathematics, Physics and Chemistry. A lack of proper understanding of these
subjects limits the capacity of students to solve complex problems thereby lessening his/her chances of making it to the top-
notch institutes which provide quality training.
This series of books serves as a source of learning that goes beyond the school curriculum of Class XI and Class XII
and is intended to form the backbone of the preparation of an aspiring student. These books have been designed with the
objective of guiding an aspirant to his/her goal in a clearly defined step-by-step approach.
• Master the Concepts and Concept Strands!
This series covers all the concepts in the latest IIT-JEE syllabus by segregating them into appropriate units. The theories
are explained in detail and are illustrated using solved examples detailing the different applications of the concepts.
• Let us First Solve the Examples—Concept Connectors!
At the end of the theory content in each unit, a good number of “Solved Examples” are provided and they are designed
to give the aspirant a comprehensive exposure to the application of the concepts at the problem-solving level.
• Do Your Exercise—Daily!
Over 200 unsolved problems are presented for practice at the end of every chapter. Hints and solutions for the same are
also provided. These problems are designed to sharpen the aspirant’s problem-solving skills in a step-by-step manner.
• Remember, Practice Makes You Perfect!
We recommend you work out ALL the problems on your own – both solved and unsolved – to enhance the effective-
ness of your preparation.
A distinct feature of this series is that unlike most other reference books in the market, this is not authored by an in-
dividual. It is put together by a team of highly qualified faculty members that includes IITians, PhDs etc from some of the
best institutes in India and abroad. This team of academic experts has vast experience in teaching the fundamentals and
their application and in developing high quality study material for IIT-JEE at T.I.M.E. (Triumphant Institute of Manage-
ment Education Pvt. Ltd), the number 1 coaching institute in India. The essence of the combined knowledge of such an
experienced team is what is presented in this self-preparatory series. While the contents of these books have been organized
keeping in mind the specific requirements of IIT-JEE, we are sure that you will find these useful in your preparation for
various other engineering entrance exams also.
We wish you the very best!
c h a p t e r functions and
graphs
1
nnn Chapter Outline
Preview
sTUDY MATERIAL TOPIC GRIP
• subjective Questions (10)
set Theory
• straight Objective Type Questions (5)
• Concept strands (1-8)
• Assertion–Reason Type Questions (5)
Cartesian Product of Two sets
• Linked Comprehension Type Questions (6)
Relations • Multiple Correct Objective Type Questions (3)
Functions • Matrix-Match Type Question (1)
Composition of Functions IIT AssIGnMEnT ExERCIsE
• Concept strands (9-10)
• straight Objective Type Questions (80)
Inverse of a Function • Assertion–Reason Type Questions (3)
• Concept strand (11) • Linked Comprehension Type Questions (3)
Even and Odd Functions • Multiple Correct Objective Type Questions (3)
• Matrix-Match Type Question (1)
Periodic Functions
some Real Valued Functions ADDITIOnAL PRACTICE ExERCIsE
Parametric form of Representation of a Function • subjective Questions (10)
• straight Objective Type Questions (40)
Graphs of Conic sections
• Assertion–Reason Type Questions (10)
Graphs of a Few Composite Functions
• Linked Comprehension Type Questions (9)
• Concept strands (12-13)
• Multiple Correct Objective Type Questions (8)
Transformation of Functions • Matrix-Match Type Questions (3)
some special Curves
COnCEPT COnnECTORs
• 35 Connectors
1.2 Functions and Graphs
Set theOry
The concept of a set is usually the starting point in the de- If the set contains only one element, it is called a sin-
velopment of basic Mathematics and its applications. gleton set. A set that contains no element is called a ‘null
A set is a well-defined collection of objects. Objects form- set’ and is denoted by f or { }. If the number of elements in
ing part of a set are called its ‘elements’. a set is not finite, it is called an infinite set.
The following are some examples of sets: The following examples of sets illustrate the above defi-
nitions clearly.
(i) Set of people living in a particular town
(ii) Set of English alphabets (i) Set R is an infinite set.
(iii) Set of students in a school whose weights are less than (ii) X = {x, a real number between 10 and 50} also written
45 kg as X = {x ∈ R/10< x < 50} is an infinite set.
(iv) Set of cities in India whose population is greater than (iii) A = {x ∈ N/1 ≤ x ≤ 10} is a finite set and n(A) = 10.
10 lakhs. (iv) {1}, {f} are singleton sets.
(v) If A = {x/x is a prime number and 3 < x < 5} is a null set
We require that the collection of objects, which forms
⇒A = f.
the elements of the set, be well defined. This means that we
should be able to decide without ambiguity whether an ele-
ment is or is not in a given set. Subset
Sets are usually denoted by capital letters A, B, C, X,
Y etc. If x belongs to a set A, we write x ∈ A. If x does not If every element of a set B is also an element belonging to an-
belong to a set A, we write x ∉ A. other set A, then B is said to be a subset of A and is written as
B ⊆ A. If there exists atleast one element in A not in B, then
we write B ⊂ A.
representation of sets
For example, Q ⊂ R, Z ⊂ R, N ⊂ Q.
(a) Roster form: the elements of the set are listed inside It is the usual convention that null set and the set itself
set brackets. are subsets of a given set.
For example, A = {1, 2, 3, 4, 5, 6, 7}
(b) Set builder form: the elements of the set are represent-
Number line and Intervals
ed by a variable satisfying certain well-defined condi-
tions, for example, The elements of the set R of real numbers can be repre-
{ x/x is a counting number less than 5} sented by points on a line called the Number line.
universal set (cid:237)(cid:20) (cid:50)(cid:3) (cid:20)(cid:3)
The set of all elements that are of interest in a study is called The point O on this line represents the number zero.
Universal set and is denoted by S. For example, if we are All positive numbers are represented by points on the line
discussing about certain books in a library, then the Uni- to the right of O and all negative numbers are represented
versal set is the collection of all books available in that li- by points on the line to the left of O.
brary. It may be noted that we can move to the right of O
indefinitely without end. We say that there are infinite
Finite and infinite sets number of positive numbers (denoted by ∞). Similarly, we
can move to the left of 0 indefinitely without end and we
say that there are infinite number of negative numbers (de-
A set consisting of a definite number of elements is a finite
noted by -∞).
set.
The set R of real numbers may be represented by
If A is a finite set, the number of elements in A is called
(-∞, ∞).
the cardinal number of A denoted by n(A).
For example A = {3, 7, 9, 11, 13, 19} is a finite set. For
this set, n(A) = 6. (cid:68)(cid:3) (cid:50) (cid:69)
Functions and Graphs 1.3
Let a and b represent two elements of R. The set of set is represented by points inside a rectangle and its sub-
points (or the set of real numbers) lying between a and b, sets are represented by points inside closed curves.
inclusive of the two extreme points a and b, may be repre-
sented by [a, b] (called the closed interval).
algebra of sets
If the set of points does not include the extreme points
a and b, we represent this by (a, b) (called the open interval).
(i) Union of sets
[a, b) – set includes a but does not include b.
If A and B are any two sets, the set of all elements that be-
(a, b] – set includes b but does not include a.
long to either A or B is called the union of A and B and is
The set of points to the left of a may be represented by denoted by A ∪ B.
(-∞, a] or (-∞, a), according as this set includes or does not
include a. The set of points to the right of b may be repre- (cid:54)(cid:3)
sented by [b, ∞) or (b, ∞), according as this set includes or
does not include b. (cid:36) (cid:3)(cid:3)(cid:37)(cid:3)
power set of a set
(cid:36)(cid:3)(cid:137)(cid:3)(cid:37)(cid:16)(cid:86)(cid:75)(cid:68)(cid:71)(cid:72)(cid:71)(cid:3)(cid:85)(cid:72)(cid:74)(cid:76)(cid:82)(cid:81)
The set of all subsets of a given finite set A is called the Power
Fig.1.2
set of A, denoted by P(A).
The number of subsets of a given finite set A is 2n, where,
A ∪ B = { x/x ∈ A or x ∈ B}
n is the cardinal number of A. Therefore, n((P(A)) = 2n.
For example, if A = {1, 2}, then P(A) = {f, {1}, {2}, A}. Consider the following examples:
Note that n(P(A)) = 22.
(i) Let A = {a, b, c, d, e}; B = {b, x, c, d, y, f} Then A ∪ B =
{a, b, c, d, e, x, y, f}.
equal sets
(ii) Let A represent the set of points {x/–1 ≤ x ≤ 5} i.e., x
lies in the closed interval [–1, 5]; B represents the set
Two finite sets A and B are said to be equal and we write
of points {x/–3 < x < 4} i.e., x lies in the open interval
A = B if every element in A is in B and every element in B is
(–3, 4). Then, A ∪ B = {x/–3 < x ≤ 5}
in A i.e., if A ⊆ B and B ⊆ A.
If the numbers of elements of two sets are equal, then
the sets are called equivalent sets. (ii) Intersection of sets
For example, if A = {1, 2, 3, 4} and B = {2, 3, 4, 5}, then
If A and B are any two sets, the set of all elements that be-
A and B are equivalent sets but A ≠ B.
long to both A and B is called intersection of A and B and
is denoted by A ∩ B.
Venn diagrams
S
Sets and operations on sets can be geometrically illustrated
by means of Venn diagrams. In such diagrams, Universal
A B
(cid:54) A ∩ B-shaded region
(cid:36)(cid:3) (cid:37)(cid:3) (cid:39)(cid:3) Fig.1.3
A ∩ B = { x | x ∈ A and x ∈ B}
If A ∩ B = f, i.e., there are no elements common to
both A and B, we say that A and B are disjoint sets.
(cid:38)(cid:3)
Examples are
Fig.1.1 (i) Let A = {1, 2, 3, 4, 5} , B = {3, 4, 5, 6, 7, 8}, then A ∩ B
= {3, 4, 5}
1.4 Functions and Graphs
(ii) The set of rational numbers and the set of irrational (v) Symmetric difference of two sets
numbers are disjoint sets.
(iii) Let A = {x ∈ Z/x ≥ 0} and B is the set of natural Let A and B be any two sets. The symmetric difference of A
numbers. Then A ∩ B = B. and B is the set of elements that belong only to A or only to
B and is denoted by A D B.
(iii) Complement of a set
S
If A is any subset of the universal set S, the set consisting
A B
of the elements in S that do not belong to A is called the
complement of A and is denoted by A’ or AC.
S A ∆ B-shaded
region
Fig.1.6
A
A D B = (A – B) ∪ (B – A) or (A\B) ∪ (B\A)
A’-shaded region For example, if A = {a, b, c, d, e, f} and B = {c, d, e, f, g, h}
Fig.1.4 then A – B = {a, b} and B – A = {g, h}.\A D B = {a, b, g, h}
Also note that A D B = A ∪ B – A ∩ B.
For example, if S = {1, 2, 3, 4, 5, 6, 7, 8, 9} and
A = {3, 7, 9}, then A’ = {1, 2, 4, 5, 6, 8} (vi) Fundamental laws of set operation
(iv) Difference of two sets (i) Identity law: A ∪ f = A; A ∩ f = f; A ∪ S = S;
A ∩ S = A.
If A and B are any two sets, the set A – B (or A\B) is the set (ii) Complement law: A ∪ A’ = S; A ∩ A’ = f; (A’)’ = A.
of all elements that belong to A but not to B. (iii) Idempotent law: A ∪ A = A; A ∩ A = A.
(iv) Commutative law: A ∪ B = B ∪ A, A ∩ B = B ∩ A.
S
(v) Associative law: (A ∪ B) ∪ C = A ∪ (B ∪ C);
A B (A ∩ B) ∩ C = A ∩ (B ∩ C).
(vi) Distributive law: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C);
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
A / B-shaded region (vii) De Morgan’s laws : (A ∪ B) ’ = A’ ∩ B’;
Fig.1.5 (A ∩ B) ’ = A’ ∪ B’
(viii) n (A ∪ B) = n(A) + n(B) – n(A ∩ B)
A – B (or A\B) = {x | x ∈ A, x ∉ B} (ix) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) –
For example, if A = {x ∈ R/1 ≤ x ≤ 2} = [1, 2] and B = n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
{x ∈ R/1 < x < 2} = (1, 2), then A - B = {1, 2} and B – A = f. (x) n(A’) = n(S) – n(A)
ConCept StrandS
Concept Strand 1 Given n(A) = 400, n(B) = 12000, n(A ∩ B) = 350 and
n(S) = 40000.
A town has total population of 40000 out of which 400
We have to find n(A’ ∩ B’) or n[(A ∪ B)’].
people own cars, 12000 people own motorcycles and
Now, n (A ∪ B) = 400 + 12000 – 350 = 12050.
350 people own both cars and motorcycles. How many in
\ n [(A ∪ B)’] = 40000 – 12050 = 27950.
the town do not own either?
Solution Concept Strand 2
Let A = people owning cars; B = people owning motor Let A = {1, 2, 3, 4, …., 20}. Find the number of subsets of A
cycles; S = people in the town which contain 5, 6, 7, 8, 9 and 10.
Functions and Graphs 1.5
Solution We are given:
(1) + (2) + (3) + (4) + (5) + (6) + (7) + (8) = 3000
Since 5, 6, 7, 8, 9 and 10 are to be there in all the subsets,
(2) + (3) + (5) + (6) = 2270;
the number of subsets is clearly the number of subsets that
(2) + (5) = 750; (4) + (5) = 450;
can be formed with the remaining 14 elements of A.
(5) = 400; (5) + (6) = 1000;
The answer is 214.
(1) = 250; (7) = 200
Answer (i) = (1) + (2) + (4) + (5) = 250 + 750 + 450
Concept Strand 3
– 400 = 1050.
Answer (ii) = (8) = 3000 – (2270 + 450 – 400 + 200
In an examination, 75% students passed in English, 65%
+ 250) = 230.
passed in Hindi and 10% failed in both. Find the percent-
Answer (iii) = (3) = 2270 – 1000 – 750 + 400 = 920.
age of students who passed in both subjects.
Solution
Concept Strand 5
Let A = Set of students who passed in English;
In a survey of 100 students in a music school, the num-
B = Set of students who passed in Hindi
ber of students learning different musical instruments was
Given n (A) = 75, n (B) = 65, n (A’ ∩ B’) = 10.
found to be: Guitar: 28, Veena: 30, Flute: 42, Guitar and
⇒ n(A ∪ B) = 90
We have, 90 = 75 + 65 – n (A ∩ B) or n (A ∩ B) = 50 or Veena: 8, Guitar and Flute: 10, Veena and Flute: 5, All mu-
sical instruments: 3
The required answer is 50%.
(i) How many students were learning none of these three
Concept Strand 4 musical instruments?
(ii) How many students were learning only the flute?
In a competitive examination consisting of three tests viz.,
general knowledge, arithmetic and English, the number of
participants was 3000. Only 2270 participants were able to Solution
get through the arithmetic test, 750 through both arith-
(i) n(G) = 28, n(V) = 30, n(F) = 42
metic and general knowledge, 450 through both general
⇒ n (G ∪ V ∪ F) = 28 + 30 + 42 – 8 – 10 – 5 + 3 = 80
knowledge and English, 400 through all the three tests
and 1000 through arithmetic and English. There were 250 Hence n (G ’ ∩ V ’ ∩ F ’) = 100 – 80 = 20.
participants who got through general knowledge alone but (ii) n(Flute only) = 42 – 10 – 5 + 3 = 30
not through the other two and 200 passed in English only.
(i) How many were in a position to get through general Concept Strand 6
knowledge?
(ii) How many participants failed in all the three subjects? Out of 500 students who appeared at a competitive exami-
(iii) How many got through arithmetic only? nation from a centre, 140 failed in Mathematics, 155 failed
in Physics and 142 failed in Chemistry. Those who failed
in both Mathematics and Physics were 98, in Physics and
Solution
Chemistry were 105, and in Mathematics and Chemistry
Set G : Candidates who got through general knowledge. 100. The number of students who failed in all the three
Set A : Candidates who got through arithmetic subjects was 85. Assuming that each student appeared in
Set E : Candidates who got through English all the 3 subjects, find
(i) the number of students who failed in at least one of
the three subjects.
G A
(ii) the number of students who passed in all the subjects.
1 2 3 (iii) the number of students who failed in Mathematics
only.
5
4 6
7 8 Solution
E
Let M: failed in Mathematics; P: failed in Physics; C: failed
Fig.1.7
in Chemistry