Krishna's

CHEM STRY

for

By

Alok Bariyar
Ph.D. (CSIR fellow), M.Sc.(IIT Delhi), G.A.T.E., N.E.T.

Ex. Nuclear Scientist
Bhabha Atomic Research Centre,
Mumbai, India

Prakashan Media Pvt. Ltd.
KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India


CHEMISTRY for

Syllabus

PHYSICAL CHEMISTRY
1.

Basic Mathematical Concepts: Functions; maxima and minima; integrals; ordinary differential equations; vectors and
matrices; determinants; elementary statistics and probability theory.

2.

Atomic and Molecular Structure: Fundamental particles; Bohr's theory of hydrogen-like atom; wave-particle duality;
uncertainty principle; Schrodinger's wave equation; quantum numbers; shapes of orbitals; Hund's rule and Pauli's
exclusion principle; electronic configuration of simple homonuclear diatomic molecules.

3.

Theory of Gases: Equation of state for ideal and non-ideal (van der Waals) gases; Kinetic theory of gases; MaxwellBoltzmann distribution law; equipartition of energy.

4.

Solid State: Crystals and crystal systems; X-rays; NaCl and KCl structures; close packing; atomic and ionic radii; radius ratio
rules; lattice energy; Born-Haber cycle; isomorphism; heat capacity of solids.

5.

Chemical Thermodynamics: Reversible and irreversible processes; first law and its application to ideal and nonideal gases;
thermochemistry; second law; entropy and free energy; criteria for spontaneity.

6.

Chemical and Phase Equilibria: Law of mass action; Kp, Kc, Kx and Kn; effect of temperature on K; ionic equilibria in
solutions; pH and buffer solutions; hydrolysis; solubility product; phase equilibria-phase rule and its application to onecomponent and two-component systems; colligative properties.

7.

Electrochemistry: Conductance and its applications; transport number; galvanic cells; EMF and free energy; concentration
cells with and without transport; polarography; concentration cells with and without transport; Debey-Huckel-Onsagar
theory of strong electrolytes.

8.


Chemical Kinetics: Reactions of various order; Arrhenius equation; collision theory; transition state theory; chain reactions normal and branched; enzyme kinetics; photochemical processes; catalysis.

9.

Adsorption: Gibbs adsorption equation; adsorption isotherm; types of adsorption; surface area of adsorbents; surface films
on liquids.

10. Spectroscopy: Beer-Lambert law; fundamental concepts of rotational, vibrational, electronic and magnetic resonance
spectroscopy.

ORGANIC CHEMISTRY
11. Basic Concepts in Organic Chemistry and Stereochemistry: Electronic effects (resonance, inductive, hyperconjugation)
and steric effects and its applications (acid/base property); optical isomerism in compounds with and without any
stereocenters (allenes, biphenyls); conformation of acyclic systems (substituted ethane/n-propane/n-butane) and cyclic
systems (mono- and di-substituted cyclohexanes).
12. Organic Reaction Mechanism and Synthetic Applications: Chemistry of reactive intermediates (carbocations, carbanions,
free radicals, carbenes, nitrenes, benzynes etc ...); Hofmann-Curtius-Lossen rearrangement, Wolff rearrangement,
Simmons-Smith reaction, Reimer-Tiemann reaction, Michael reaction, Darzens reaction, Wittig reaction and McMurry
reaction; Pinacol-pinacolone, Favorskii, benzilic acid rearrangement, dienone-phenol rearrangement, Baeyer-Villeger
reaction; oxidation and reduction reactions in organic chemistry; organometallic reagents in organic synthesis (Grignard,
organolithium and organocopper); Diels-Alder, electrocyclic and sigmatropic reactions; functional group inter-conversions
and structural problems using chemical reactions.
13. Qualitative Organic Analysis: Identification of functional groups by chemical tests; elementary UV, IR and 1H NMR
spectroscopic techniques as tools for structural elucidation.
14. Natural Products Chemistry: Chemistry of alkaloids, steroids, terpenes, carbohydrates, amino acids, peptides and nucleic
acids.
(iii)



15. Aromatic and Heterocyclic Chemistry: Monocyclic, bicyclic and tricyclic aromatic hydrocarbons, and monocyclic
compounds with one hetero atom: synthesis, reactivity and properties.
INORGANIC CHEMISTRY
16. Periodic Table: Periodic classification of elements and periodicity in properties; general methods of isolation and
purification of elements.
17. Chemical Bonding and Shapes of Compounds: Types of bonding; VSEPR theory and shapes of molecules;hybridization;
dipole moment; ionic solids; structure of NaCl, CsCl, diamond and graphite; lattice energy.
18. Main Group Elements (s and p blocks): General concepts on group relationships and gradation in properties; structure of
electron deficient compounds involving main group elements.
19. Transition Metals (d block): Characteristics of 3d elements; oxide, hydroxide and salts of first row metals; coordination
complexes: structure, isomerism, reaction mechanism and electronic spectra; VB, MO and Crystal Field theoretical
approaches for structure, color and magnetic properties of metal complexes; organometallic compounds having ligands
with back bonding capabilities such as metal carbonyls, carbenes, nitrosyls and metallocenes; homogenous catalysis.
20. Bioinorganic Chemistry: Essentials and trace elements of life; basic reactions in the biological systems and the role of metal
ions, especially Fe2+, Fe3+, Cu2+ and Zn2+; structure and function of hemoglobin and myoglobin and carbonic anhydrase.
21. Instrumental Methods of Analysis: Basic principles; instrumentations and simple applications of conductometry,
potentiometry and UV-vis spectrophotometry; analysis of water, air and soil samples.
22. Analytical Chemistry: Principles of qualitative and quantitative analysis; acid-base, oxidation-reduction and complexometric
titrations using EDTA; precipitation reactions; use of indicators; use of organic reagents in inorganic analysis; radioactivity;
nuclear reactions; applications of isotopes.

Weightage of Inorganic, Organic & Physical Chemistry
Questions Asked in Last Three Years

JAM 2016

JAM 2018

JAM 2017


INORGANIC

INORGANIC
INORGANIC

17 Q

PHYSICAL

10 Q

13 Q
27 Q

23 Q

PHYSICAL

20 Q

20 Q

28 Q

ORGANIC

PHYSICAL

22 Q


ORGANIC
ORGANIC

Physical Chemistry : 23 Questions

Physical Chemistry : 27 Questions

Physical Chemistry : 28 Questions

Organic Chemistry : 20 Questions

Organic Chemistry : 20 Questions

Organic Chemistry : 22 Questions

Inorganic Chemistry : 17 Questions

Inorganic Chemistry : 13 Questions

Inorganic Chemistry : 10 Questions

(iv)


Preface
DEDICATED TO LORD KRISHNA

– Publishers & Author

I am happy to present this book entitled "Chemistry for IIT JAM". It has been written according to the Latest syllabus

to fulfil the requirement of students.
The book is written with the following special features:
1.

It is written in a simple language so that all the students may understand it easily.

2.

It has an extensive and intensive coverage of all topics.

3.

The complete syllabus has been divided into 3 sections having 22 Chapters.

4.

Sufficient Numerical Problems, Subjective Questions and Objective type questions with Hints & Solutions given at the end of
each chapter will enable students to understand the concept.

5.

Sufficient Mock Test Papers have also been added for the practice of students.

First of all I want to express my sincere gratitude to Dr. S.B.P. Sinha, Prof. J.C. Ahluwalia, Prof. N.K. Jha for their
invaluable guidance, immense interest and constant encouragement for the successful completion of the work. I am
also thankful to Bandana Bariyar, Abhishek Bariyar, Purnima Sinha & family for their kind help at many occasions.
I am extremely grateful to my respected and beloved parents whose incessant inspiration guided us to accomplish
this work. I also express gratitude to my respected family members for their moral support.
I am immensely thankful to Mr. S.K. Rastogi, Managing Director, Mr. Sugam Rastogi, Executive Director,
Mrs. Kanupriya Rastogi, Director and entire team of Krishna Prakashan Media (P) Ltd., for taking keen interest in

getting the book published.
The originality of the ideas is not claimed and criticism and suggestions are invited from the students, teaching
community and other readers.
Author

(v)


Exam Pattern
Paper Specific Instructions
1.

The examination is of 3 hours duration. There are a total of 60 questions carrying 100 marks. The entire paper is divided
into three sections, A, B and C. All sections are compulsory. Questions in each section are of different types.

2.

Section - A contains a total of 30 Multiple Choice Questions (MCQ). Each MCQ type question has four choices out of which
only one choice is the correct answer. Questions Q.1-Q.30 belong to this section and carry a total of 50 marks. Q.1-Q.10
carry 1 mark each and Questions Q.11 - Q.30 carry 2 marks each.

3.

Section - B contains a total of 10 Multiple select Questions (MSQ). Each MSQ type question is similar to MCQ but with a
difference that there may be one or more than one choice(s) that are correct out of the four given choices. The candidate
gets full credit if he/she selects all the correct answers only and no wrong answers. Questions Q.31-Q.40 belong to this
section and carry 2 marks each with a total of 20 marks.

4.


Section - C contains a total of 20 Numerical Answer Type (NAT) questions. For these NAT type questions, the answer is a
real number which needs to be entered using the virtual keyboard on the monitor No choices will be shown for these type of
questions. Questions Q.41-Q.60 belong to this section and carry a total of 30 marks. Q.41-Q.50 carry 1 mark each and
Questions Q.51-Q.60 carry 2 marks each.

5.

In all sections, questions not attempted will result in zero mark. In Section-A (MCQ), wrong answer will result in NEGATIVE
marks. For all 1 mark questions, 1/3 marks will be deducted for each wrong answer. For all 2 marks questions, 2/3 marks will
be deducted for each wrong answer. In Section - B (MSQ), there is NO NEGATIVE and NO PARTIAL marking provisions.
There is NO NEGATIVE marking in Sections-C (NAT) as well.

Source : jam.iitb.ac.in

(vi)


Common Mistakes Done By Students
1.

Avoiding the Basic Concepts : Students generally start their study with reference books and other advanced study
materials. They forget that to fight a battle in the field it's important to learn basic concepts of the fight, which is applicable
to IIT JAM as well. Not working on conceptual terms tend to make students puzzled when the question is slightly twisted.

2. Tomorrow Never Comes : It's a natural tendency of humans to say, “I'll do it from tomorrow onwards”. Making a schedule
alone is not enough to make student win this battle, you must work on it with full dedication. To create an effective timeschedule, save the important dates for the exam and plan up your activities accordingly.
3.

Self Confidence : Self confidence is the key to success, you can lose even a small fight if you have doubts about yourself.
On the other hand, self-confidence is the only key that can lead you to win over the battles. Giving a little space to doubts

leads to hesitation and lack of confidence. Have trust in yourself and make a positive mind set about whatever you have
prepared is more than enough for exams. This will increase your confidence and you will perform well in the exam.

4. Mock Tests : Mock Tests are practice tests used as a mirror to your skill sets. They help you in finding out the true analysis of
your performance & knowledge level. Helping you in increasing your speed, they also improve your answer giving skills
with greater accuracy. One should never skip these tests as they are the most important part of preparation.
5. Skipping Revision : Taking too much stress or no stress, both can be harmful to a student. Many students panic a lot which
worsens their performance in the exams and many of them are so relaxed that they do not bother even about the revision.
Students must understand that the preparation they have done might be enough but revision is of equal importance.

Revision Tips
1.

Start Revising Early: Always make a time table for revision well before the exams. A time table can include which topics to
revise to score good marks. Divide time accordingly for all the subjects.

2.

Make Summary Notes: By making summary notes students can focus on key points of important topics.

3.

Take Short Breaks: During revision take short breaks after every hour not every 10 minutes so that you feel energetic.

4.

Have a Proper Sleep: Having less sleep could impact your next day revision and could lead to less concentration for
revision. Research has shown that lack of sleep leads to clogging of brain and makes us forget sooner.

5.


Revision Room and Sitting Conditions: Revise in a room that is quiet and you don't get distracted from outside noises. Use
proper chair and table for revision. Don't try to revise in bed. Wash your face and eyes with water to avoid sleeping state.
Lastly, be confident before exams with all the revision you have done. Remember revision is a skill that leads to
improvement of memory and enahnces learning ability of students. Practice makes one perfect but effective Revision
Techniques put excellence in ones perfectness.

(vii)


Contents
SECTION - A
Chapter 1

:

Basic Mathematical Concepts......................................................................................................(A-01-A-40)

Chapter 2

:

Atomic and Molecular Structure....................................................................................................(B-01-B-24)

Chapter 3

:

Theory of Gases..........................................................................................................................(C-01-C-16)


Chapter 4

:

Solid State .................................................................................................................................(D-01-D-16)

Chapter 5

:

Chemical Thermodynamics.........................................................................................................(E-01-E-20)

Chapter 6

:

Chemical and Phase Equilibria......................................................................................................(F-01-F-28)

Chapter 7

:

Electrochemistry.........................................................................................................................(G-01-G-20)

Chapter 8

:

Chemical Kinetics .......................................................................................................................(H-01-H-28)


Chapter 9

:

Adsorption ....................................................................................................................................(I-01-I-08)

Chapter 10 :

Spectroscopy................................................................................................................................(J-01-J-16)

SECTION - B
Chapter 1

:

Basic Concepts in Organic Chemistry and Stereochemistry............................................................(A-01-A-32)

Chapter 2

:

Organic Reaction Mechanism and Synthetic Applications..............................................................(B-01-B-44)

Chapter 3

:

Qualitative Organic Analysis.......................................................................................................(C-01-C-24)

Chapter 4


:

Natural Products Chemistry.........................................................................................................(D-01-D-36)

Chapter 5

:

Aromatic and Heterocyclic Chemistry..........................................................................................(E-01-E-40)

SECTION - C
Chapter 1

:

Periodic Table.............................................................................................................................. (A-01-A-24)

Chapter 2

:

Chemical Bonding and Shapes of Compounds ........................................................................... (B-01-B-20)

Chapter 3

:

Main Group Elements (s and p blocks)..........................................................................................(C-01-C-60)


Chapter 4

:

Transition Metals (d Block)...........................................................................................................(D-01-D-32)

Chapter 5

:

Bioinorganic Chemistry................................................................................................................(E-01-E-08)

Chapter 6

:

Instrumental Methods of Analysis.................................................................................................(F-01-F-12)

Chapter 7

:

Analytical Chemistry...................................................................................................................(G-01-G-40)

Mock Test Papers................................................................................................................................................................
Latest Examination Paper..................................................................................................................................................

(viii)



Section

A
Chapter 1: Basic Mathematical Concepts
Chapter 2: Atomic and Molecular Structure
Chapter 3: Theory of Gases
Chapter 4: Solid State
Chapter 5: Chemical Thermodynamics
Chapter 6: Chemical and Phase Equilibria
Chapter 7: Electrochemistry
Chapter 8: Chemical Kinetics
Chapter 9: Adsorption
Chapter 10: Spectroscopy


A-3

C HAPTER

Section-A

1

Basic Mathematical Concepts
Syllabus

• Functions • Maxima and minima • Integrals • Ordinary differential equations • Vectors and matrices • Determinants • Elementary statistics
and probability theory.

INTRODUCTION TO FUNCTION

Let A & B be two sets & let there exist a rule or manner or correspondence ’f ’ which associates to each element of A, a unique element in B. Then f is
called a function or mapping from A to B.It is denoted by the symbol f : A → B or A f B which reads “f ” is a function from A to B or f maps A′ toB.

So it can be concluded that

f:A→B
Input

Output
y = f(x)

One many ⇒ Not a function
Many one ⇒ Function.

x

y

A

B

INTERVALS
The set of numbers between any two real numbers is called interval.
Types of intervals :

A relation R from a set A to a set B is called a function if

1.


(i)

Representation : [a, b] i. e., a ≤ x ≤ b

Each element of A is associated with some element of B.

(ii)
Each element of A has unique image in B.
Some important points :
To every x 1 there is one & only one y. i.e.
1
2
3
4

a
b
c

2.

Closed intervals :
Open intervals :

Representation : ] a, b [ or (a, b) i. e., a < x < b
3.

Semiclosed or semi open :
[a, b] → a ≤ x < b


Not a function

[a, b] → a < x ≤ b)
Example 1: Solve

a
1
2

1
2
3

1
2
3
4

b

Not a function

c

a
b
c

a
b

c
d

Is a function

3 − 2x
x −4
7− x
+
≥ 1 − x and
≥ 1.
2
−3
−3

Solution :
6x − 9 + 2x − 8
7−x
≥1
⇒
≥ 1− x &
−6
−3
7−x
≥ −3
⇒
8x − 17 ≤ 6x − 6 &

or


2x ≤ 11
11
x≤
2
11
x ε  −∞, 

2

11/2

& − x ≤ − 10
& x ≥ 10
& x ε [10, ∞)

10

Is a function

Hence, no common value so x ε φ.


A-4

EXERCISE A

+

1.


Solve −6 ≤ 2 (1 − x ) ≤ 8

2.

Solve −3 ≤ 3 − 2x ≤ 7 & −4 ≤ 1 − 5x < 6

+

–∞

–1

WAVY CURVE METHOD
If a, b = 0 ⇒ either a = 0 or b = 0. But if a ⋅ b > 0 ⇒ if we conclude
a > 0 or b > 0 it is wrong.
Similarly, if (x − 1) (x − 2) > 0, then this doesn’t mean (x − 1) > 0 or
(x − 2) > 0.
These wrong interpretation can be corrected by wavy curve
method.

x ε [1, 4]

Example 3: Solve

( x + 1)2 ( x − 2)3 ( x − 4)4
( x − 3)2 ( x − 7)5

–1

f (x ) =


n2

...... (x − a k )

m1

m2

...... (x − bk )mp

(x − a1) (x − a 2)
(x − b1)

(x − b2)

(≥ 0, ≤ 0)

Step I : First arrange all values of x at which either numerator or
denominator becomes zero.
Step II : Values of x at which numerator becomes zero should be
marked with dark circle.

So wave will not change the direction at these points.
+

So

Step IV : From right to left, draw a wavy curve (Beginning above
the number line in case of value of f (x ) is positive in step III otherwise

form below the number line). Passing throughly all the marked
points so that when passes through a points (exponent or power
whose factor is odd) intersect the number line & when passing
thoroughly a point (exponent whose corresponds factor is even) the
curve doesn’t intersect the real line & remain on the same side of real
line.
Step V : The appropriate intervals are chosen in accordance with
the sign of inequality (the function f (x ) is positive wherever the
curve is above the number line. It is negative if the curve is found
below the number line). Their union represents the solution of
inequality.
x =1
x=3

3

2

–

+∞

Example 2: Solve 2x 2 + x − 1< x 2 + 4x + 3 or x 2 − 3x − 4 < 0
x 2 − 3x − 4 < 0
x = 4 & x = −1

–

4


+

–

∞

7

–∞

0

+
–

1

2

–

+
∞

4

3

x ε [0, 1] ∪ [3, ∞]


Example 5: Solve

( x − 1)3 ( x − 2)5 ( x − 7)
( x − 3)2 ( x − 1)5 x 2

≤0

Solution : Don’t cancel common factor, instead, add their power. Here
for x − 1, 3 + 5 = 8 i.e., even wave will not change the direction.
+
–∞

+
0

+

+
2

1

–

3

–

7


∞

EXERCISE B
Solve
1.

(x − 1) (x − 2)
≥0
(x − 3) (x − 4)

2.

p2 ≥ 16

3.

f (x ) = (x 2 − 4) (x 2 − 3x + 2) (x 2 − 1) (x 2) (x 2 + 3x + 2) for
which f (x ) is positive.

x ε [1, 3]

(x − 4) (x + 1) < 0 or

–

2

Solution :

+

1

+

Example 4: Solve ( x − 1)3 ( x − 2)2 ( x − 3)2015 x 3 ( x − 4)2020 ≥ 0

Example 1: Solve ( x − 1) (3 − x ) ≥ 0

Solution :

+

x ε [−∞, 1] ∪ [−1, 2] ∪ (7, ∞)

Step III : Values of x at which denominator becomes zero should
be marked with empty circle. Check the value of f (x ) for any real
number greater than the right most marked number or the number
line.

–

7

∵ Exponents of factor of −1, 3, & 4 are even,

Method

–∞

4


Check f (x) for x > 7, f(8) > 0

–∞ –1

3 − x = 0 or

3

2

nk

W h e r e n1, n 2 ...... n k ; m1, m2, ...... mp a r e r e a l n u m b e r s &
a1, a 2, ...... a k , b1, b2 ...... bk are any real numbers such that a1 ≠ b j
where i = 1, 2, 3 ..... & j = 1, 2, 3 ...... p.

Solution : Roots are x − 1 = 0 or

≥0

Solution : Mark points with dark circle for which numerator is zero &
point of discontinuity (at which denominator becomes zero) with empty
circle.

In order to solve the inequalities of the form
n1

+∞


4

–

4.

y < y (1 − y)

5.

2x 2 + x − 6 ≥ 0

Answers:
1.

x ε [−∞ 1] ∪ [2, 3] ∪ [4, ∞]

2.

p ε [−∞, 4] ∪ [4, ∞]

3.

x ε R  { −2, − 1, 2, 0}

4.

yεφ



A-5

5.
x ε [−∞, − 2] ∪ [3 / 2, ∞]
Some basics for functions

Answers :

f : A→ B
1
2
3
4

1
4
9
16
25
36

A

B

i.e.,

(1, 1)

f (2) = 4


i.e.,

(2, 4)

f (3) = 9

i.e.,

(3, 9)

f (1) = 1

1.

D f = [−1, 1]

2.

D f = (2, 3) ∪ (3, ∞)

MODULUS FUNCTION
 x , if x ≥ 0
|x|= 
 − x , if x < 0
Now in case (I) | x | ≥ 0
in case (II) | x | ≥ x
|x| > x
x ε (–∞, 0)
|x| = x

x ε (0, ∞)

A → Domain set
B → Co-domain {1, 4, 9, 16, 25, 36, }
Range → Set of images {1, 4, 9, 16}

But | x | ≠ less than x.
Example 1: Solve for x : | x − 1| + 2 ( x − 2) = x + 4

Range is subset of codomain

Solution :

i.e., Range ≤ Codomain

Case I : If x ≥ 1

Domain : Set of values of x for which the (input) function is defined
(Not zero in the denominator) & real (Not negative in square root).
Range : Set of collection of values of y (output) for which f (x ) is
defined & real.

x − 1 + 2 (x − 2) = x + 4
x − 1 + 2x − 4 = x + 4
or

or

2x − 9 = 0


x=

9
2

i.e., x ≥ 1 (acceptable)

f:X→Y

Case II : If x < 1
x

y

X

Y

− (x − 1) + 2 (x − 2) = x + 4
− x + 1 + 2x − 4 = x + 4
or

Example 1: Find the domain of following functions.
1.

f (x ) =

1
x −1


Solution : Let | x − 1 | = y
or

x ≠1

(y − 4) (y + 3) = 0

D f = (−∞, 1) ∪ (1, ∞)

y=4

3
x −1

+ 5 4− x +

x 2 − 3x + 2
x2 − 4

Solution : In given question
x −1> 0 4− x ≥ 0
x2 − 4 ≠ 0
x>1
x≤4
(x − 2) (x + 2) ≠ 0

FIND DOMAIN

2.


y= −3

Now | x − 1 | = 4 & | x − 1 | = − 3 (Can’t possible)
x −1 = ± 4
or

x = 5 & x = −3

Example 3:

x 2 − 5x + 6
≤0
|x|+3

So

x 2 − 5x + 6 ≤ 0
(x − 3) (x − 2) ≤ 0
+

f (x ) = 1 − 1 − 1 − x 2
f (x ) =

&

Solution : Clearly | x | + 3 is positive value.

D f = (1, 2) ∪ (2,4)

EXERCISE C


1.

y 2 − y − 12 = 0

D f = R − {1}

Example 2: f ( x ) =

or

−3 = 4 (impossible)
9
Overall answer x =
2
Example 2: | x − 1| 2 − | x − 1| − 12 = 0

Solution : ∵ x − 1 ≠ 0

or

x−3= x + 4

x−2
x 2 − 5x + 6

+
2

–


3

x ε [2, 3]


A-6

EXERCISE D

−1 ≤ | 2x − 1 | − 3 ≤ 1
2 ≤ | 2x − 1 |≤ 4

SOLVE FOR x
| x 2 + 4x + 3 | + 2x + 5 = 0

2.

2| x | + |2 x −1 − 1 | = 2 x −1 + 1

3.

| x − 2 | + 2 | x − 3 | = 5 (x − 4)
x = {−4, − 1 −

2.

x = [0, 1]

3.


x = 6 only

| 2x − 1 |≤ 4

Using property II
or

2x − 1 ≤ − 2
−1
x≤
2

Answers :
1.

&

| 2x − 1 |≥ 2

1.

or

2x − 1 ≥ 2
3
x≥
2

Using property I


3}

−3
5
≤x≤
2
2
−3 −1   3 5 

Df ε
,
∪
,
 2 2   2 2 
−4 ≤ 2x − 1 ≤ 4 or

FIND DOMAIN OF FOLLOWING FUNCTION
1

Example 1: f ( x ) =

x −|x |

–∞

–3/2

–1/2


3/2

5/2

∞

Solution : x − | x |> 0
or

| x |< x, clearly x ε φ

EXERCISE F

1− | x |
|x |−2

Example 2: f ( x ) :

2

1 − |x |
Solution :
≥0
|x |− 2
or

|x |− 1
≤0
|x |− 2


x − 3x − 1

1.

Solve for x

2.

Solve for x & find domain

x2 + x + 1

x 2 − 5x + 4

Roots of numerator

x2 − 4

x = ±1

Roots of denomenator x = ± 2
+
–

≤1

Answers :

+
–2


<3

–1

+
1

–

2

1.

x ε (−∞, − 2) ∪ (−1, ∞)

2.

x ε (−∞, − 2) ∪ (0, 2) ∪ (5 / 2, ∞)
x ε (−2, 8 / 5) ∪ (2, ∞)

So, Df = (2, –1) ∪ (1, 2)

EXERCISE E

D f ε (0, 8 / 5) ∪ (5 / 2, ∞)

PROPERTIES OF MODULUS

FIND DOMAIN


1.

1.

f (x ) =

x − |x |

2.

2.

f (x ) =

− |x |

3.

|x + 3 | + x
f (x ) =
−1
x+2

Answers :

3.

| x 1, x 2, x 3 , ...., x n | = | x 1 |.| x 0 |...| x n |
x

|x|
=
y
| y|
| x ± y|≤ | x | + | y|

GREATEST INTEGER FUNCTION OR STEP UP FUNCTION
The function y = f (x ) = [x ] is called the greatest integer function
where [x ] denotes the greatest integer less than or equal to x.

1.

D f = {0, ∞}

2.

Df = φ

3.

D f = (−5, − 2) ∪ (−1, ∞)

INEQUALITY OF MODULUS
Property I : For any positive value of K if | f (x )| ≤ K

f (x ) : [x ] = Integer part of x just less than or equal to x.
 1
2
e.g.,
2 ⋅ 3 = 2, [Π ] = 9, [−7 ⋅ 1] = − 8



[3] = 3, [−5] = − 5

then

− K ≤ f (x ) ≤ K

i.e.,

| x + 3| ≤ 1

1.

−1 < x + 3 ≤ + 1

2.

x − 1 < [x ] ≤ x

−4 < x ≤ − 2

3.

[x ] = x if x ∈ I

Properties :

x ε [−4, − 2]
Property II : If | f (x )| ≥ K then f (x ) ≤ − K or f (x ) ≥ K

Example 1: | | 2x − 1| − 3 | ≤ 1
Solution : Using property I

4.

[x ] gives integer only

[x ] < x if x ∉ I
 −1 [x ] , if x ∉ I
[− x ] = 
, if x ∈ I
− x


A-7

5.

 0, if x ∈ I
[x ] + [− x ] = 
 −1, if x ∉ I

6.

[x ± I ] = [x ] ± I

So

2


x − x + 1 always positive. So not possible
+

+

e.g., [x + 2] = [x ] + 2

0

–

1

Also [2x ] ≠ 2[x ]
Graphical representation

D f = (−∞, 0] ∪ [1, ∞)

EXERCISE G

+3

–3

–2

+2

FIND DOMAIN


+1

1.

f (x ) =

2.

–1

3.

f (x ) = ln (2 [x + 2] −14)
1
f (x ) =
2
[x ] − [x ] − 6

–2

4.

f (x ) = [x ] − x

–3

Answers :

–1


1

2

3

EQUALITY & INEQUALITY OF [x]
1.

1
(x − 2)

If [x ] = K then x ∈ [K, K + 1]

1.

D f = (−∞, 2) ∪ (3, ∞)

2.

D f = (6, ∞)

3.

D f = (−∞, − 2) ∪ (4, ∞)

4.

Df = x ∈ I


i.e., [x ] = 2, x ∈[ 2, 3]

Fractional part of x { x } :

2.

[x ] ≥ K i. e., x ≥ K

3.

[x ] > K then x ≥ K + 1

∵

4.

[x ] ≤ K then x < K + 1

5.

[x ] < K then x < K

x = [x ] + { x }
Definition f (x ) = { x } = x − [x ] gives the fractional part of x &
x ∈(0, 1)

Example 1: Find the domain of following functions.
(a) f (x) = x − [x]

e.g.,


{ 27
. } = 07
.
{ 3} = 0

Solution : x − [x] ≥ 0
(b)

2.3 = 2 + 0.3

{ −3} = 0
2 − 2 2

α α
 −3 ⋅  =
7
7
7


x ≥ [x] or x ∈ R
1
f (x ) =
:
x − [x ]

=1−

Solution : For domain x − [x] > 0

or

[x ] < x

So

x ∉I

So
(c)

2 5
=
7 7

{ ∩} = ∩ − [∩] = ∩ − 3
Properties :

Df = R — I
sin x + cos x
:
[x / 3]

1.

f (x ) =

2.

x

Solution : For domain   ≠ 0
 3

3.

{ x } = 0 when x = I
1
1
{ x } = when x = I +
2
2
1 − { x } , x ∉ I
{x} = 
, x ∈I
0

or

 x  < 0 or x < 0 or x < 0
 3 
3

or

 x  > 0 or x ≥ 1 or x ≥ 3
 
 3
3

4.


0 , x ∈ I
{x} + {− x} = 
1 , x ∉ I

D f = (−∞, 0) ∪ (3, ∞)

5.

{x ± I} = {x }

(d)

−1

f (x) = sec

i.e., { x + 2} = { x }

2

[x − x + 1] :

Solution : For domain
2

[x − x + 1] ≤ − 1
2

Example 1: Solve 4 { x } = x + [ x ]

2

or [x − x + 1] ≥ 1
2

x − x + 1 < − 1 + 1 or x − x + 1 ≥ 1

Here

x2 − x + 1< 0

or x 2 − x + 1 ≥ 1

D< 0& a> 0

x2 − x ≥ 0

Solution : ∵ x = [x] + {x}
Now

4 {x} = x + [x]
4 {x} = [x] + {x} + [x]
3 {x} = 2 [x]


A-8
or

{x} =


2
[x ]
3

(ii) x = 2m & x = 3n or x = 6k
(iii) No solution

Now ∵ 0 ≤ { x} < 1
2 [x ]
or
0≤
<1
3
3
0 ≤ [x ] <
2

EXPONENTIAL & LOGARITHMIC FUNCTIONS
Exponential Function:
The function f (x ) = a x = e x where

4 (x − [x]) = x + [x]

a > 1 & a ≠ 1 is called an exponential function. The inverse of this
 a > 0
function is called logarithmic function i.e., f (x ) = a x 

 a ≠ 1

4x − 4 [x] = x + [x]


gives the positive value

or

…(1)

[x] = 0, 1

Also

4 {x} = x + [x]

3x = 5 [x]
5 [x ]
x=
3

or

…(2)

if

a>1

i.e.,

2x , 3x then graph


Put (1) in (2)
5× 0
3
5
x = 0,
3

or

x=

ax a > 1

5
×1
3

1

Example 2: Find domain for f ( x ) =

{ x} 2 − { x} +
{x}2 − {x} +

Solution :
Let

or

3

≠0
16

16y 2 − 16y + 3 ≠ 0
(4y − 1) (4y − 3) ≠ 0
1
3
or y ≠
y≠
4
4
1
3
{x} ≠ or {x} ≠
4
4
1
3
or {x} ≠ n +
x ≠ n+
4
4
3
1

D f = R − n + , n + ; n ∈ I 
4
4




f (x ) = { x } − 1
f (x ) =

2x = 8 → x = log 2 8
Definition : a x = b ⇔ x = log ba
Here b > 0, a > 0 & a ≠ 1
e.g., log 2(1 − x ) = 1 ⇔ 1 − x = 2′
Properties of logarithm
b

alog a = b

4 {x}2 − 1

2.

a 0 = 1 ⇔ log a 1 = 0

3.

a1 = a ⇔ log a a = 1

4.

log a (m × n) ⇔ log a m + log a n
m
log a = log a m + log a n
n


x 2 − 4 + | x 2 − 3x + 2| + (x − 2)2 + { x } = 0
 x   x  5x
 2 +  3 = 6
   

(iii)

[1 + sin x ] + [1 − cos x ] = sin x

5.

Answers :

6.

log a m n = n log a m

7.

log a b =

8.

log ba =

9.

log ba . log ab = log bb = 1

(i) D f = φ

(ii) D f − R / n +

2.

It is reverse of exponential function.

1.

(ii)

1.

LOGARITHMIC FUNCTION

1

Q.2. Solve for x
(i)

then graph
a < x <1

Q.1. Find domain of following :

(ii)

0< a < 1

3
≠0

16

EXERCISE H
(i)

if

{x} = y
y2 − y +

or

3
16

(i) x = 2

1
n ∈I
2

log m b
log m a
1

log ab


A-9


m
log a b
n

10.

log a n bm =

11.

alog c = blog c

b

or

x (2x − 7) < 0
+

a

+

–∞

0

Example 1: Find the value of:
(a)


–

∞

7/2

27loga 36
2

3 log 2 6 2
3

Here, x ∈(0, 7 / 2)

log3 6

⇒

3

⇒ (33 ) 2

or

33 log3 6

⇒ 3log3 6

But
3


⇒

63 = 216

7
Taking common x ∈  3, 
 2

Example 2: Find the value of x:

Case II:

x log4 3 + 2.3log4 x = 27

⇒ 3

log4 x

log4 x

+ 2.3

logn

3

x − 3> 1
x>4


x log4 3 + 2.3log4 x = 27

Solution :

x ∈(3, 4)

then

= 27

(x − 3)2 x

2

−7 x

2x 2 − 7 x > 0

3

(1 + 2) = 3

4

log4 x

x (2x − 7) > 0

⋅ 3 = 33


3

+

3log4 x = 32
or

> (x − 3)0

+

–∞

log 4 x = 2

0

2

–

7/2

∞

x = 4 = 16
x ∈ (−∞, 0) ∪ (7 / 2, ∞)

EXERCISE I
Example 1: Find the value of :

(i)

0.8 [1 + 9log3 8] log5 5

(ii)

log 2 3 log 3 4 log 4 5 log 5 6 log 6 7 log7 8

(iii)

8(log2

3

4

(ii)

3

(iii)

242

121 + 1 / 3)

 1
 
 2


x ∈ (4, ∞)
or

x ∈ (4, ∞)

7
Taking union x ∈  3,  ∪ (4, ∞)
 2
Example 2: Find domain of
log e (x + 3)
(i)
x 2 + 3x + 2

Example 2: Solve for x

(i)

x>4

So

7
Now we have x ∈  3, 
 2

Answers :
(i)

But


Solution : Here x + 3 > 0

x3−x

x> −3

>1
x 3 −x

 1
> 
 2

–∞

x (x − 1) < 0

(ii)

x (x − 1) (x + 1) < 0
+

So

x ∈ (−∞, − 1) ∪ (0, 1)

(ii)

(x − 3)2x


− 7x

+
0

–

1

3< x < 4
2

−7 x

> (x − 3)0

or

2x 2 − 7 x < 0

f (x ) =

–1

∞

 5x − x 2 
log1 / 2 

4 



5x − x 2  1 
≤ 
 2
4

Case I: If 0 < x −3 < 1
(x − 3)2 x

–2

 5x − x 2 
5x − x 2
Solution : For domain log1 / 2 
>0
 ≥ 0&
4 
4


>1

then

–3

D f = (−1, ∞)

[∵ 0 < a < 1]


2

–1

x ≠ −1

0

x3 − x < 0

–

x 2 + 3x + 2 ≠ 0

(x + 2) (x + 1) ≠ 0
&

x ≠ −2

 1
Solution : Here,  
 2

2

&

⇒


0

& 5x − x 2 > 0

5x − x 2
≤1
4

& x 2 − 5x < 0

5x − x 2 ≤ 4

& x 2 − 5x < 0

5x − x 2 − 4 ≤ 4

& x (x − 5) < 0

(x − 4) (x − 1) ≥ 0 & x(x − 5) < 0


A-10
+

+

+

+


&
1

4

–

0

–

5

Find domain of g(y). It will be range for f (x).
Example 1: Find range of function:
(i)

x ε (–∞,1) ∪ [4, ∞]

f (x ):

x ε (0, 5)

x2 + 1
x2

Solution : (i) Let

So common value D f = (−∞, 1) ∪ [0, 5)


EXERCISE J

2

x y = 1+ x

Example 1: Find domain of

x=±

2

(i)

f (x) = log (1 − log (x − 5x + 16)

(ii)


 x3 
f (x) = sin −1  log 2   
 3  


(iii)

f (x) = log10 sin (x − 3) +

(iv)


log x − 2 (3 − x)

: (D f = R − {0}

y = f (x ) =
2

1+ x2
x2

2

or

x (y − 1) = 1

1
y −1

for this to be defined

1
≥0
y −1

16 − x 2

+

Answers :


–

(i)

D f = (2, 3)

(ii)

D f ∈ [− 6 , − 3 / 2] ∪ [ 3 / 2 , 6]

(iii)

D f = (3 − 2 π, 3 − π) ∪ (3, 4]

(iv)

D f (2, 3)

y ∈ (1, ∞) or

1

R f = (1, ∞)

2

(ii)

RANGE OF FUNCTION


f (x ):

x − x +1
x2 + x + 1

Let

Set of collection of values of y for which f (x ) is defined & real.

y=

; Df = R :

x2 − x + 1
x2 + x + 1

yx 2 + yx + y = x 2 − x + 1

f:X→Y

x 2 (y − 1) + (y + 1) x + y − 1 = 0
y

x

Taking y as constant
y − 1 ≠ 0 or y ≠ 1

X


Y

Also D ≥ 0

or (y + 1)2 − 4 (y − 1) (y − 1) ≥ 0
or (3y − 1) (y − 3) ≤ 0

Method of find range :

+

Method I: When D f = Finite element
= { a, a 2, a 3 ...... a n}
then

1/3

R f = { f (a1), f (a 2), ...... f (a n)}

f : A → R, f (x ) = 2x + 1, find R f .
Solution : Clearly here domain A = {1, 2, 3}
Now

f(1) = 2 × 1 + 1 = 3

1
So, range ∈  , 3
 3 


Example 2: f ( x ) = x 2 + 3

f(3) = 2 × 3 + 1 = 7

Solution : Range = [3, ∞)

R f = {3, 5, 7}
Example 2: Find range for f ( x ) :
Solution : For domain
7 − x ≥ 0 Also x − 3 ≥ 0
x≤7
x≥3

7 −x

Px − 3

and 7 − x ≥ x − 3
x≤5

Example 3: f ( x ) =

3
2 cos x + 5



3
3
3 

Solution : Range = 
,
 or R f =  7 , 1
2
×
−
1
+
5
2
×
1
+
5





D f = [3, 4, 5]

EXERCISE K

R( f ) = {1, 2, 3}

Example 1: Find range of function if Df = R—{Some finite
points}

Putting domain in f (x)
Method II: When D f = R or

then let

3

Method III: By thinking i.e., use of definition

f(2) = 2 × 2 + 1 = 5

D f = [3, 5] or

–

1
y ∈  , 3
 3 

Example 1: Let A = {1, 2, 3}

So

+

R — {Some finite point}

y = f (x )

Express x in terms of y i.e., x = g(y).

1.


f (x ) =

x 2 − 5x + 6
x 2 − 3x + 2


A-11
2.

2x − 5
f (x ):
5x − 3

3.

5 + x − [x ]
f (x ) =
3

Solution : ∵ f (x) is bijective i.e., one-one and onto
So

2.
3.

f ′(x ′) ≤ 0

3x + 2 (a + 2) x + 3a ≥ 0 or 3x 2 + 2 (a + 2) + 3a ≤ 0
(Can’t possible)


Answers :
1.

or

f ′(x) ≥ 0

2

Here a > 0,

Range : R − {1, − 1}
2
R f = (−∞, ∞) −  
 5

or
or

So, D < 0

(2 (a + 2))2 − 4 × 3 × 34 < 0
(a − 1) (a − 4) < 0
a ∈(0, 4)

5
R f =  , 2
 3 

+


+

Example 2: If x , [ x ] & { x } are in GP, find valve of x

Answers : 0,

1

1+ 5 1− 5
,
2
2

INCREASING & DECREASING FUNCTION
Increasing
function

x,

Decreasing
function

1
,
x

ex ,
e


−x

lnx,
log a x ,
(0 < a < 1)

VALUE OF A FUNCTION
sin−1 x ,

tan−1 x ,

−1

−1

cos x ,

cot

x,

sec −1 x
−1

cosec x

Example

1:


Let

CLASSIFICATION OF FUNCTION
I. One-one function (injective):

or

It can be defined as different x ⇒ diferently.
II. Many one function:
It can be defined as more than one value of x = same y.

HOW TO IDENTIFY ONE-ONE & MANY ONE FUNCTION
Method I: By inspection :
If domain R → R, many one
If domain ∈ N one-one
Method II: By graphical method :
If a line parallel to x-axis cut the graph exactly at one point, then
function is one-one otherwise many one if D f → R → R.
Method III: By increasing or decreasing function for one one
function f ′(x ) ≥ 0 or f ′(x ) ≤ 0 i.e., function must be strictly increasing
or decreasing.

Method V: INTO function
Here in this function, at least one element extra in the co-domain.
Method VI : If a function is injective & surjective then that function
is called bijective.
How to test into function.
First of all find range.

Example 1: Let f : R → R

f (x ) : x 3 + (a + 2) x 2 + 3ax + 5 is a bijective function, then find
zone of ‘a’.

then

prove

that

 1 + 2x 



2x 

1+ x2 
f
=
log

1+ x2 
 1 − 2x 



1+ x2 
1+ x
= 2 log
= 2 f (x )
1− x

4x
4x + 2

then prove that

1.
2.

f (x) + f (1 − x) = 1
Hence and otherwise find the value of
1 
 2  + f  3  + ...... f  2011 
f 
 + f 





 2012 
 2012 
 2010 
 2012 
ψ
Solution : 1. Let 4x = y
f (x ) =
y+ 2
Now
So


f (1 − x) =

41 − x
4x + 2

4

+2

=

4 / 4x
4 / 4x + 2

=

2
2
=
2 + 4x y + 2

f (x) + f (1 − x)
y
2
+
=1
y+ 2 y + 2

2.


2 
 2  + f  3  + ...... f  2011 
f 
 + f 





 2012 
 2012 
 2012 
 2012 

Let

x=

1
2011
then 1 − x =
2012
2012

Using previous result
  1 
  2011     2 
 2010  
 f  2012   + f   2012   +  f  2012  + f  2012  
1006 

+ ...... 1005 times + f 

 2012 

R f = co-domain

then it is onto elso into

1+ x
1− x

1+ x
1− x

Example 2: Let f ( x ) =

Method IV: ONTO function (surjective)
Here in this function, no extra element in co-domain i.e., R f =
Co-domain.

f ( x ) = log

 2x 
f 
 = 2f ( x ).
 1+ x 2 
Solution : ∵ f (x) = log

If


4

–

1
(1 + 1 + 1 + ...... 1005 times) + f  
 2
1005 +

41 / 2
1/ 2

4

+ 2

= 1005 +

1
= 1005.5
2


A-12

EXERCISE L
Example 1: Let y = f ( x ) =

INVERSE OF FUNCTION


1− x
, then find f (2x ) in terms of
1+ x

Only bijective functions are invertible.
f:X→Y

f ( x ).
3 f (x ) − 1
3 − f (x )

Solution : f (x) =

Example 2: If f ( x + 2y, x − 2y ) = xy then f ( x , y ) will be
Solution : f (x, y) =

x 2 − y2
8

1
4
9
16

1
4
9
16

3x 2 1

2
Solution : f (x) =
− −
5
5 5x 2

f : X→ Y
x

Inverse of a function is unique.

Example 1: If f : (2, ∞ ) → R and f ( x ) : − 1 + ln ( x − 2). Find f −1( x ).
g : Y→ Z

y

X

Def II: If f (g (x )) = g ( f (x )) = x = I each one is inverse of other.
If we are given whether a function is invertibile or not, we will check
that y = f (x ) is one-one and onto.

COMPOSITION OF A FUNCTION

Solution : ∵ f (x) : − 1 + ln (x − 2) is bijective
z

g (x)

Y


g (y)
g f(x)
g of

Let

y = f (x) = − 1 + ln (x − 2)
y + 1 = ln (x − 2)
ey + 1 = x − 2

Z

Def :gof : X → Z , gof (x ) = g ( f (x )) exist.

ey + 1 + 2 = x
or

Only when R f ≤ D g

f −1 ( x ) = 2 + e x + 1

Note : ( fog)−1 = g −1 of −1

fog = g (g (x )) Z → X
Rg ≤ D f

EXERCISE N

Example 1: Let f ( x ) : x 2 − 1 and g ( x ) : sin x + cos x then 18

find
(i) fog

Example 1: Assuming bijective f ( x ) =
(ii) gof

(iii) gog

(iv) fof
Solution : f −1 (x) :

Solution : (i) fog : f (g(x)) : f (sin x + cos x)
2

: (sin x + cos x) − 1 = 1 + 2 sin x cos x − 1 = sin 2x
(ii)

gof : g( f (x)) : g(x 2 − 1) = sin(x 2 − 1) + cos (x 2 − 1)

(iii)

gog : g(g(x)) : g(sin x + cos x) = sin (sin x + cos x)

(iv)

1
2
3
4


Def I: If y = f (x ) ⇔ x = f −1(y)

 1
Example 3: If 3f ( x ) + 2f   = x 2 − 1, then find f ( x )
x

Similarly

Def : f : Y → X

1
2
3
4

2

2

2

4

fof : f f (x) : f (x − 1) : (x − 1) − 1 : x − 2x

2

10x − 10− x

find f −1( x ).


1
x+1
log
2
x −1

Example 2: Let f : [1, ∞) → [2, ∞) f (x) : x +
Solution : f −1 (x) :

10x + 10− x

x+

1
find f −1 (x)
x

x2 − 4
2

Example 2: Let g ( x ) : 2x + 1and h ( x ) : 4x 2 + 4x + 7 then find a

Example 3: Let f : [1 / 2, ∞ ) → [3 / 4, ∞ ) & f ( x ) : x 2 − x + 1 then

function f ( x ) such that fog = h .

solve

Solution : ∵ fog = h


or

f (g (x ) = h

f (2x + 1) = 4x 2 + 4x + 7

x 2 − x + 1=

2

f (2x + 1) = 4x + 4x + 1 + 6
f (2x + 1) = (2x)2 + 2 × 2x + 1 + 6
f (2x + 1) = (2x + 1)2 + 6

EXERCISE M
Example 1: Let f ( x ) : 2x and g ( x ) = 1 / x and h ( x ) : e x then
prove that (gog ) oh = fo(goh ).
Example 2: If the domain of f ( x ) is [1 / 2, 1], then find the
domain of f (sin x ) :
π
π
Solution : x ∈ ∪  2nπ + , 2n ∩ + 
n ∈I 
6
2 


x −


3
4

Solution : x = 1

ODD FUNCTION AND EVEN FUNCTION
(i)

f (x ) = x 2 + 6

1
+
2

ODD function :
If f (− x ) = − f (x ) ⇒ f (x ) + f (− x ) = 0
then called as odd function.
e. g ., x odd power , sin x , tan x . cot x , cosec x
odd functions are symmetric about the origin or opposite
quadrant and f (o) = 0 (if defined)


A-13
Y
f(x) = x3

X

f ( x) = const.


Periodic but having no fundamental
period

sin x 2 , cos x , sin x 3

Not a periodic function

HOW TO FIND PERIOD
Rule I: If f (x ) → T then f (ax + b) =

T
. But f (x ) ± a, a. f (x ) and
| a|

f (x )
has period T only.
a

EVEN FUNCTION

Rule II: gof (x ) = g ( f (x )) is a periodic if f (x ) is a periodic function.

If f (− x ) = f (x ) i.e., f (x ) − f (− x ) = 0 then called as even function.

And period of gof (x ) = period of f (x ) but it may give fundamental
period.

e.g., x

even power


, cos x , sec x ,| x |.

Rule III: If f (x ) = f1(x ) − f 2(x ) + f 3(x ) +... f n(x ) then period of

Even functions are symmetrical about y axis y =| x |.

f (x ) = LCM (T1, T2, T3 ...... Tn). Again it may not give always
fundamental period.

y = |x|

Example 1: Find period of f ( x ) = 2 | cos ( πx + 3) | + 7
Solution : T =

π
=1
π

Example

Let f ( x ) = e 3x + 1−[ 3x + 1]

g( x ) = e

2:

3(x −[ x ])

has


a

period T1 &

has a period T2. Find relation between T1 & T2.

Solution : As f (x ) = e 3x +1−[ 3x −1] = e{ 3x +1} is a periodic function

f (x ) = 0 is even as well as odd function.
f(x)

g(x)

f(x) + g(x)

f(x) − g(x)

f(x)g(x)

f(x)
g(x)

gof(x)

odd

odd

odd


odd

even

even

odd

even

even

even

even

even

even

even

odd

even

NONE

NONE


odd

odd

even

even

odd

NONE

NONE

odd

odd

even

with period 1 / 3.
1
So
T1 =
3
Also
g (x ) = e 3( x −[ x ])
g (x ) = e 3{ x }
or

Hence

T2 = 1
3T1 = T2

Example 3: Find period for f ( x + 4) = f ( x + 6)
Solution : Replace x by x − 4
f (x − 4 + 4) = f (x − 4 + 6)

PERIODIC FUNCTION

f (x) = f (x + 2)

A function f (x ) is called periodic if there exist a positive number
T (T > 0) called the period of the function such that:

By using f (x) = f (x + T), period T = 2

f (x + T ) = f (x ) for all values of x within the domain of x.

Example 4: Check whether function is even or odd,

Note : { x } is a periodic function with the periodic = 1 sin x is a
periodic function with the period = 2π.

(i)

Fundamental period

sin x , cosecx , cos x , sec x


2π

tan x , cot x

π

|sin x |,|cos x |,|cot x |
|tan x |,|sec x |,|cosecx |

π

sinn x , cosn x , cosec n x , sec n x

n (even) = 17
n (other than even) = 2π

tann x , cotn x

π

{x}

1

1+ x + x 2

Solution : Put x = − x

LIST OF SOME STANDARD PERIODIC FUNCTION

Periodic function

f (x ) : 1− x + x 2 +

f (− x ) = 1 + x + x 2 +

1− x + x2

As f (x) = f (x) = So even function
(ii)

f (x ) :

2. x 2017 . (x + sin x)
1 + cos 2 x

It is even function.

EXERCISE O
Example 1: Check for even and odd function:
(1 + x)
(i)
f (x) = log
(1 − x)
(ii)

f (x ) = (a x + a − x ) + (a x − a − x )

(iii)


a + x 

f (x) =  log 


a − x  


2016

log (x + 1 + x 2 )
(a + a − x ) (a x − a − x )
x


A-14

Form III:

Solution : (i) Odd
(ii) None

∫a

(iii) Even
Example 2: Find period for given function:
(i)
(ii)

f (x) : 3x + 1 − [3x + 1] + 2013

x
x
f (x) : 2 cos − 3 sin
!n
!n + 1

(iii)

f (x) : sin 2 x + cos 4 x

(iv)

f (x + 1) + f (x − 1) =

x

∫e

1

∫ 1 + x 2 dx = tan
1

∫

2 f (x).

=

ex

+ c or e x + c
lne

1− x

2

−1

x + c or − cot −1 x + c

dx = sin −1 x + c or − cos−1 x + c

Form V:

∫ sin x dx = − cos x + c ∫ cos x dx = sin x + c
2
2
∫ sec dx = tan x + c
∫ cosec x dx = − cot x + c
∫ sec x tan x dx = sec x + c ∫ cosec cot x dx = − cosec x + c

(ii) 2 π (n + 1)!
(iii) π
(iv) 8
Example 3: If f ( x ) : 2 sin [a ] x + 7 has a period of π / 3, find

Form VI:

the value of a.


1.

Solution : a ∈[36, 37]

2.

INDEFINITE INTEGRATION

3.

Integration is also known as anti-derivative or primitive or integral.
d
Definition : If
( f (x ) + c) = F (x )
dx

∫ F (x ), dx = f (x ) + c

where F (x ) is called integrand

∫ cot x dx = ln | sin| + c
∫ tanx dx = ln |sec x | + c
∫ sec x dx = ln | sec x + tan x | + c
n x
or ln tan  +  + c
 4 2

4.


∫ cos ec dx = ln | cos ec x − cot x | + c

or

ln | tan x | 2| + c

f (x ) : integral

EXERCISE A

c : Constant of integration.

Example 1: Integrate the following:

Theorem I:
Integral of some function may differ but difference is only of
constant.
g1 (x) + c1

∫ f(x) dx

g2 (x) + c2

1.

∫ f1(x ) ± f2(x ) dx = ∫ f1(x ) dx + ∫ f2(x ) dx
∫ Kf (x ) dx = K ∫ f (x ) dx
Theorem III:

∫ f (x ) dx = g (x ) + c


x n+1
∫ x dx = n + 1 + c, n ≠ − 1
n

x

6.

x4

∫ 1+ x2

dx

sin 6 + cos 6 x

∫ sin 2 x cos 2 x
4

dx

1

∫

2x + 1 − 5x + 1
dx
10x
1


8.

∫ 25x 2 + 30x + 10 dx

10.

∫ 1 − sin x

12.

∫

1

∫ cos

13

∫ tan x tan 2x tan 3x dx if x + 2x + 3x = π

14.

∫ cos (x − α) dx

x dx

cos x

Solution : 1.

2.
3.

x6
+ c
6

3
2x − 3 + c
2
−3
+ c
2 (2x + 1)

dx

cos 8x − cos 7 x
dx
1 + 2 cos 5x

11.

(ax + b)
+c
a (n + 1)

Form II:
1
1.
∫ x dx = ln | x | + c

1
1
2.
∫ ax + b dx = a ln | ax + b| + c

∫ 4x 2 − 9 dx

∫ 4x 2 + 12x + 9 dx

n+1

dx =

∫

5.

n

∫ (ax + b)

2.

4.

7.

1
dx
2x − 3

3
8

dx

3

9.

Form I:

5 lnx

∫ 4x 2 + 4x + 1 dx

g (ax + b)
f (ax + b) dx =
+c
a

FORMS OF INTEGRATION

∫e

3.

Theorem II:

∫


x

Form IV:

Solution : (i) 1/3

then

ax
+c
lna


A-15
4.

2x − 3
1
ln
+ c
12
2x + 3

5.

1
2

6.


5
ln2

7.
8.

1

3
 2 ln | 2x + 3 | + 2 (2x + 3)  + c


x

 1 − 2
 
 2
ln5

n +1

n

3.

∫ f (x )

4.

∫ f (x ) n


f ′(x) dx =

f ′(x)

dx =

x

EXERCISE B

 1 + c
 
 5

Example 1: Evaluate integration of following:
x 2 tan −1 x 3
1.
∫ 1 + x 6 dx

x3
− x + tan −1 x + c
3
1
tan −1 (5x + 3) + c
5

55

2.


∫5

x

x

.55 .5x dx

tan x − cot x − 3x + c

10.

tan x + sec x + c
sin 4x 
1
3x + sin 2x +
+ c
8 
4 
1
1
sin 3x − sin 2x + c
3
2
ln | sec 3x |
− ln | sec 2x | − ln | sec x | + c
3

3.


∫

4.

∫

5.

∫

6.

x cos α − ln | sec (x − α) | sin α + c

7.

∫ tan x dx
3
5
∫ sin x cos x dx

12.
13.
14.

INTEGRATION BY SUBSTITUTION
Example 1:

∫x


cos x − cos 3 x
1 − cos 3 x

2 sin ( x 2 + 1) − sin 2x ( x 2 + 1)
2 sin ( x 2 + 1) − sin 2 ( x 2 + 1)

dx

2.
3.

dt
x dx =
2

dx

tan x
dx
sin 2 x
x + e x (sin x + cos x) + sin 2x
(x 2 + 2e x sin x − 2 cos 2 x)5

(tan −1 x 3 )2
+ c
6

1
(ln 5)


55

3

5x

x

55 5x + c

−2
5m −1 (cos 3 / 2 x) + C
3

4.

tan x + c

∫

2 sin t − sin 2t
dt
2 sin t + sin 2t

5.

−1
8


1
2

∫

2 sin t − 2 sin t cos t
dt
2 sin t + 2 sin t cot t

6.

tan 3 x
− tan x + x + c
3

or

1
2

∫

2 sin t  1 − cos t 

 dt
2 sin t  1 + cot t 

7.

sin 4 x

sin 8 x 2 sin 6 x
+
−
+ c
4
8
6

or

1
2

∫

⇒

1 log | sec t / 2 |
+ c
2
1/ 2

or

1
2

or

2 sin 2 t / 2

2 sin 2 t / 2

log sec2

or

or

dt

1
tan t / 2
2∫

2

(x + 1)
+ c
2

2

Example 2: ∫ e 2x + lnx dx :
Solution : ∫ e 2 x . e lnx dx or

∫e

2x 2

. x dx


Put

2x 2 = t

or

4x dx = dt or

or

1
e t dt
4∫

or

e2x
+ c
4

2.

dt
4

4




1
 2
 + c
x
2
x
+
2
e
sin
x
−
2
cos
x



Some shortcuts:

∫ sin

m

∫

f ′(x)
dx = 2 f (x) + c
f (x )


∫

f ′(x)
dx = ln | f (x) | + c
f (x )

x cosn x dx

m1n ∈ N

1.

If m → even and n → odd; put sin x = t

2.

If m → odd and n → even; put cos x = t

3.

m1 n is odd, put either sin x or cos x = t

4.

m1 n is even, then simplify

5.

If m1n is fraction and m + n = negative even integer put
tan x = t


NINE SPECIAL INTEGRALS
1

2

SOME SHORTCUTS
1.

x dx =

dx

4

Solution : 1.

Solution : Put x 2 + 1 = t then 2x dx = dt
or

+ c

f (x)1 − n
+ c
1− n

9.

11.


| f (x ) |
n+ 1

1.

∫

2

a −x

2

1

dx = 5m −1
1

2.

∫ a 2 + x 2 dx = a tan

3.

∫

1
2

a +x


2

−1

x
+c
a
x
+c
a

dx ln| x + a 2 + x 2 | + c


A-16

1

∫

5.

∫ x 2 − a 2 dx = 2a ln | x + a | + c

6.

x 2 − a2

sec 4 x dx


dx = ln| x + x 2 − a 2 | + c

4.

1

1

1

∫ tan 4 x + 1

| x − a|

1

Put

∫ t4 + 1

|a + x|

7.

∫

x
a − x dx =
2


8.

∫

a 2 + x 2 dx =

2

x
2

Put

a2
x
a −x +
sin −1 + c
2
a
2

dZ

a2
ln| x + a 2 + x 2 | + c
2

∫ Z2 + (


ex

2.

∫

= (a ± b)2 ∓ 2ab as per required.

3.

∫

Example 1. Integrate the following:
ex
1.
∫ a + e2x dx

4.

∫

5.

∫ x 4 + 1 dx

6.

∫ x 2 + 4x + 7 dx

or Algebric twins

2

Divide numerator and denomenator by x and use a + b

(e)2 = t

Put

dt
2
dt
1
= ln | t + 32 + t 2 | + c
32 + t 2 2

1
2

∫

=
(ii)

x

∫

x4 + x2 + 1
2


Put

x =t

Now

1
2

⇒
or
(iii)

2

e x dx =

or
or

2e x dx = dt

or

∫

1
t2 + t + 1

2

1
2
1
2

tan −1

⇒

1
2

∫

1
 3
(t + 1 / 2)2 +  
 2

Divide by cos x numerator and dinomenator

2

+c

t−1/ t

tan −1

tan x − 1 / tan x


+c

2

2

+c

a − 9x 4

dx

dx

2 sin 2x − cos x

dx

6 cos 2 x − sin x
tan x dx
1

2x + 1

7 ex − 1
1
tan −1
+ c
20

20 / 7

2.

1
3x 2
sin −1 2 + c
6
a

3.

4 sin 2 x − sin x + 5 + ln sin x −

1
+
2

sin 2 x − sin x + 5 + c

4.

tan x − 1
tan x + 1
1
1
tan −1
+
tan −1
+ c

2
2 tan x
2
2 tan x

5.

1
(x − 1 / x )
1
x + 1/ x
tan −1
−
tan −1 
 + c

2 2
2
2 2
2 

6.

ln | x 2 + 4x + 7 | −

2xdx = dt
dt

Z


tan −1

dx

1
log| t + 1 / 2 + t 2 + t + 1 | + c
2
1
log| x 2 + 1 / 2 + x 4 + x 2 + 1 | + c
2
1
∫ sin 4 x + cos4 x dx
4

x
4

Solution : 1.

1
ln | e 2x + a + e 2x | + c
2

or

∫ (t − 1 / t)2 + 2 dt

Example 1: Integrate the following:

∫ 3 − 2ex + 7 e2x


2

1 + 1 / t3

dt or

EXERCISE C
1.

x 4 ± Kx 2 + 1

1

=

cosquare.
Linear
L
Type II:
or
or L Q put L = AQ′ + B
Quadratic
Q
Type III:

∫ t2 + 1 / t8

=


Const
1
Type I:
,
, Q then make Q as a perfect
Quadratic (Q)
Q

x2 ± 1

1 + 1 / t2

⇒

2)2

x 2 − a 2 dx =

∫

sec 2 x dx = dt

1 − 1 / t 2 dt = dZ

x
a3
x 2 − a2 −
ln| x + x 2 − a 2 | + c
2
2

Some tricks in the integration:
9.

dx

t−1/ t= Z

2

a2 + x 2 +

(tan 4 x + 1)

or

or

dt

sec 2 x (tan 2 x + 1)

∫

tan x = t

t2 + 1

∫ − a 2 − x 2 dx = 2a ln | a − x | + c
2


or

2

1
x + 3
tan −1 
 + c
 3 
3

SOME OTHER TYPES OF INTEGRATIONS
Type I:

1
a ± b sin 2 x

,

1
1
,
a ± b cos2 x (a sin x + b cos x )2
1

2

2

a sin x + b cos x ± a sin x cos x

Divide numerator and denomenator by cos2 x .


A-17

Type II:
1
1
1
1
,
,
,
a ± b sin x a ± b cos x a sin x ± b cos x a sin x ± b cos x + c
2

2 tan x / 2

Rule: Use sin x =

2

1 + tan x / 2

1
ln sec x [(− cot x)] − ∫
sec x . tan x (− cot x) dx
sec x
or


− cot x ln sec x + x + c

2

, cos x =

1 − tan x / 2
2

1 + tan x / 2

or convert

into polar form.
± sin x ± cos x
Type III :
:
f (sin 2x )
Use

sin 2x = (sin x + cos x )2 − 1

or

sin 2x = 1 − (sin x − cos x )2

Example 2:

Solution : ∵ At x > 0
∴


2
π
=
=

EXERCISE D
=
Example 1: Integrate the following:
sin x
(i)
∫ sin 3x dx
1
(ii)
∫ sec x + cos ecx dx

=
=

cos x − sin x
dx
a + 16 sin 2x

=

(iii)

∫

(iv)


∫ cos x (cos x + 2) dx

1

Solution : (i)

1
2 3

log

(iii)

1
ln sin x + cos x +
4

(iv)

1
1
tan x / 2 
log | sec x + tan x | −
tan −1 
 + c

2
3
3 


 7
(sin x + cos x)2 + 

 4 

∫ (1 + x )2 (

1 + x 2 + 1)

∫e

2x

sin 3x dx

2

+ c

e2x 3
− ∫ cos 3x . e 2 x dx
2
2

I = sin 3x

e2x 3 
e2x
−  cos 3x .

+
2
2
2

I = sin 3x

e2x 3
e2x 3 × 3
− cos 3x
−
sin x e 2 x dx
2
2
2
2× 2 ∫

∫ sin 3x .

e2x 3
9
− cos 3e 2 x − I
2
4
4
9
3
3

I + I −  sin 3x − cos 3x 


4
4
2
or

I =

CONCEPT OF ILATE

e2x
(2 sin 3x − 3 cos 3x) + c
13

Shortcut

I : Inverse function
L : Logarithimic function

1.

∫e

ax

2.

∫e

ax


A : Algebric function
T : Trigonometric function
E : Exponential function

cos bx dx =
sin bx dx =

e ax
2

a + b2
e ax
2

a + b2

(a cos bx + b sin bx) + c
(a sin bx − b cos bx) + c

Two classical integerals :


∫ vdx  dx

1.
1 2 3 3 5

2.


As per concept v will be higher preference as I L A T E i.e., in

x

x

∫ e ( fx(x) + f ′(x)) dx = e f (x) + c
∫ f (x) + x f ′(x) dx = x f (x) + c

EXERCISE E

sin x . lnx , sin x will be v.
Example 1: Integrate the following :
2
1.
∫ cos ec x ln (sec x) dx

Example 1: Integrate the following :

Solution : Using ILATE

2.

ln sec x


3e 2 x
dx 
2



I = sin 3x

dx . Also f ( 0) = 0.

2|− π / 4

e2x
3e 2 x
− ∫ cos 3x .
dx
2
2

I = sin 3x .

Then d f (1).

 dµ
∫ µ . vdx = µ ∫ µdx − ∫  dx

x − cot −1 x dx

2
π
tan −1 x −  − tan −1 x  dx
2

π ∫
2

π
tan −1 x − + tan −1 x dx
π ∫
2
4
−1
tan x dx − ∫ 1 . dx
π ∫
4
1 . tan −1 x dx − x
π ∫
4 
x
x tan −1 x − ∫
dx  − x

π 
1+ x2
4x
2
tan −1 x − log |1 + x 2 | − x + c
π
π

I = sin 3x

− cos x
sin x
1
+

−
log | sec x − π / 2 | − tan (x − π / 4) + c
2
2
2 2

x2

−1

tan −1 x + cot −1 x = π / 2

Solution :

(ii)

Solution : f(1) = log |1 +

∫ tan

Example 3:

3 + tan x
+ c
3 − tan x

Example 2: Let f ( x ) =

tan−1 x − cot −1 x


∫ tan−1 x + cot −1 x dx; x > 0

2
2 
 d ln (sec x)
∫ cos ec dx − ∫  dx ∫ cos ec x  dx

1.

3.

4x

2

∫ e .sin x dx
∫ sin(log x) dx
x sin −1 x

∫ (1 − x 2)3 / 2

dx


A-18
4.
5.

6.
7.

8.

3

∫ sec
∫

x dx

x 2 + 5x + 7
(x + 3)2

dx

 2 + sin x 

 dx
 1 + cos 2 x 

1
e4x
Solution : 1. e 4 x −
(4 cos 2x + 2 sin 2x) + c
8
40
x
2.
sin (log x) − cos (log x) + c
2
3.

4.
5.

1

3.

2

sin −1 x sec sin −1 x − log | sec sin −1 x + tan −1 x | + c
1
(sec x tan x + log | sec x + tan x |) + c
2
(x + 2)
ex
+ c
(x + 3)

put L = t 2

Q L

2

∫ sin (log x) + cos (log x) dx
x

φ (Polyn of degre 0 or 1)

2.


 log x − 1 

put L2 = t 2

L1 L2

e x dx

∫  1 + (log x)2 

∫e

1

1.

put L = 1 / t

L Q

4.

1

Q1 = ax 2 + b

Q1 Q2

Q2 = ax 2 + d


put x = 1 / t
−t dt
Again it takes form
Q1 Q2

put Q2 = z 2 and solve it.

1

Example 1:

∫ sin x − sin 2x dx :

Solution : ∫

1
dx
sin x (1 − 2 cos x)

⇒

sin x

∫ sin 2 x (1 − 2 cos x) dx

sin x

⇒


∫ (1 − cos 2 x) (1 − 2 cos x) dx

6.



1
x
 + c
2
1
+
(log
x
)



⇒

∫ (1 + cos x) (1 − cos x) (1 − 2 cos x)

7.

x sin (log x) + c

Put

8.


e x tan x + c

cos x = t or − sin x dx = dt
dt
A
B
C
−∫
=
+
+
(1 + t) (1 − t) (1 − 2t) 1 + t 1 − t 1 − 2t

sin x dx

INTEGRATION BY USING PARTIAL FRACTION

Type P(x)
q(x)

aid
is s e
b
to

4 / 3 dt
1 / 6dt
1 / 2dt
−∫
+ ∫

1+ t
1− t
1 + 2t
1
1
1
1
3
1
= ∫
=
+
dt
6 1 + t 2 ∫ 1 − t 4 ∫ 1 + 2t
1
1
2
= log |1 + t | + log |1 − t) + log |1 + 2t | + c
6
2
3
1
1
2
= log |1 + cos x | + log |1 − cos x | + log |1 + 2 cos x | + c
6
2
3
=


Improper is degree p(x) degree
of q(x)
Here no rule of partial
fraction is valid.

is said
to be

Proper if degree p(x) < degree of q(x)
H rule is valid

Type I: When denominator has only linear factor (non-reperted).
How to partiate the fraction.
2x + 1
e.g., Partiate
(x − 1) (x + 2) (x + 3)
2x + 1
A
Β
C
=
+
+
(x − 1) (x + 2) (x + 3) x − 1 x + 2 x + 3
x = 1 x = −2 x = −3
=

2×1+1
2x − 2 + 1
2 × −3 + 1

+
+
(1 + 2) ( 1 + 3) (−2 − 1) (−2 + 3) (−3 − 1) (−3 + 2)
x −1
x+2
x+3

1/ 4
1
−5 / 4
=
+
−
x −1 x + 2 x + 3
Type II: When denominator has linear as well as repeated factors.
Ax + 3
P
Q
R
=
+
+
2
2
x − 2 (x − 2)
x −1
(x _ 1) (x − 2)
Type III: Quadratic factor
Ax + B
P

Rx + T
i.e.,
=
+ 2
2
(x − 1) (x + 1) x − 1 x + 1
Type IV: Some other types

Example 2:

∫

3x 2 + 1

∫ ( x − 1) ( x 2 + 1) dx

Solution : Let ∫

3x 2 + 1
(x − 1) (x 2 + 1)

dx =

A
Bx + C
+ 2
x −1
x +1

Solving A = 2, B = 1 and C = 1

2
x+1
=∫
dx + ∫ 2
dx
x −1
x +1
1
= 2 log | x − 1 | + log | x 2 + 1 | + tan −1 x + c
2
Example 3:
Solution : ∫
Put

1
x2

∫

(x − x 3 )1 / 3
x4

dx

x (1 / x 2 − 1)1 / 3 dx
x4

− 1 = Z or

or


−1
Z dZ
2 ∫

or

−1
4

−2
x3
⇒

dx = dZ
−1 Z 3
+ c
2 2

2

 1 − 1 + c
 2

x


Example 4: ∫

(x + 3) dx

(x 2 + 3x + 3) x + 2

Solution : Put x + 2 = t 2

or

x = t3 − 2


A-19
dx = 2t dt
t 2 − 2 + 3 . 2t dt

⇒

∫ ((t 2 − 2)2 + 3 (t 2 − 2) + 3) t

⇒

2∫

or

2∫

8.

1 
1 + 3


1+ x 

1
tan −1
2

2

+ c

2

(t 2 + 1) dt
4

t + 4 − 4t 2 + 3t 2 − 6 + 3
t2 + 1

dt ⇒ 2∫

t4 − t2 + 1
1+ 1

or

2∫

or

2 tan −1 (t − 1 / t) + c


or

2 tan −1  x + 2 −


(t − 1 / t)2 + 2 − 1

1 + 1 / t3
t2 − 1 + 1 / t2

dt ⇒ 2∫

DIMENSION GEOMETRY
dt

1 + 1/ t

DIRECTION COSINE OF LINE

3

(t − 1 / t)2 + 1

dt

Just like in 2D, slope is basic for a line. In same, 3D, for a directed line
segment or vector, direction cosine is basic to represent it
Let


→

^

^

^

a=xi+yj+zk

1 
 + c
x + 2

→
a

EXERCISE F
Example 1:

z-axis

(x 2 + 1) (x 2 + 2)

1.

∫ (x 3 + 3) (x 2 + 4) dx

2.


∫ (x − 1) (x 2 − 3x + 2) dx

2x

tan x + tan 3 x

3.

∫

4.

∫ (x − 3)

5.

6.
7.
8.

x +1

y-axis

dx

1

∫ (x 2 − 5x + 6)
∫


x2 + x + 1

x + x 2 / 3 + x1 / 6

∫ x3

x + x 4/ 3

Then angle in cosine made by x-axis, y-axis and z-axis is known as
direction cosine of a vector or line.

dx

→

Also represented in l 1m1 n

dx

2.
3.

2x 4 − 2x 2 + 1

dx

4.
5.


| a|
then DC < l , m, n >

Also cos α + cos2 β + cos2 γ = 1

x + 2x + 4

tan −1

l 2 + m2 + n 2 = 1

i.e.,

^

^

^

^^

a=l i +m j+n kk

Three numbers < a, b, c > are said to be direction ratio if it is
proportional to the direction cosine.

tan x − 1 / 2
tan x − 1 / 3
2
+

tan −1
3+ 2
3 3
3/2

x+1 − 2
+ c
x+1 + 2

−1
1
7
log
+
+
3
x − 3 26

or

DIRECTION RATIO

−1
1
ln | tan x + 1 | + ln | tan 2 x − tan x + 1 |
3
6

1
log

2

→

y
→

| a|

2
x
x
tan −1
− 3 tan −1 + c
3
3
2
2
− 4 log | x − 1 | +
+ 4 log | x − 2 | + c
(x − 1)

3 3

, m = cos β =

2

2


x+

1

z

n = cos γ =

dx

+

→

| a|

x2 − 1

∫ x 2 + 2x + 2

x

l = cos α =

Solution :
1.

x
α


dx

1 + tan 3 x
1

α

DC

i.e.,

l 2 + m2 + n 2
l m n
=
= = 2
a b c
a + b2 + c 2

i.e.,

l=

a
2

2

a +b +c
3
(x − 3)2


+

7
+1
x−3

1
1
5
−
log
+
+
7
x − 2 14
6.

3 4/ 6
x
+ 6 tan −1 x 1 / 6 + c
2

7.

1
2
1
2− 2 + 4 + c
2

x
x

7

5
+
(x − 2)2 (x − 2)

2

^

^

2

2

a +b +c

thus D.R. < a, b, c >
a
DC <
,
2
a + b2 + c 2
→

b


;m =

b
2

2

a +b +c

2

2

,n =

c
2

a + b2 + c 2
c

,

2

a + b2 + c 2

^


In case of a = x i + y j + z k
dr = < x , y, z >
x
DC <
,
2
x + y2 + z 2

y
2

2

x +y +z

2

,

z
2

x + y2 + z 2

>