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2

PREPARATION TIPSWEIGHTAGE ANALYSISMATHEMATICS

In EAMCET out of 160 questions first 80

questions are of Mathematics, therefore Mathe-

matics is a vital key to the success in EAMCET.

Approximately 40 questions from first year syll-

abus and 40 questions from 2nd year syllabus

will be given. Do not neglect any topic, prepare

all the basics in every topic (at least IPE level).

Out of these 80 questions 65 questions are of

easy and moderate level. Every student should

spend minimum 80 minutes out of 3hrs. Study

each question carefully and answer it quickly.

Remember not to cross your optimum speed

because as you go beyond optimum speed silly

mistakes will occur. Along with speed accuracy

is also important, so you need tips and tricks to

solve mathematics. In the last stages of prepara-

tion I suggest "Lilliput View of EAMCET" for

complete and quick revision in short time.

Identify lengthy and difficult questions and

do not spend more time on any of the problem.

During this period spend at least 2hrs every

day to prepare Formulae and tips and tricks. Ev-

en while preparing for EAMCET students sh-

ould allot half of the time for mathematics

preparation.

� To get more marks in Mathematics. Solving

more number of problems is necessary

� Identify what concepts are the most importa-

nt, set priorities and study the most important

concepts first.

� Accuracy and time management is very

important

� Where ever necessary use Tips and Shortcuts

� After attaining command over concepts time

bound work is needed

� More number of questions may be given from

Calculus, Algebra, co-ordinate geometry (2D,

3D)

� Matrix, Determinants, Binomial theorem, Tri-

gonometry should be solved in a Systematic

manner.

� In complex numbers, modulus, Amplitude,

cube root of unity nth root of unity related

problems are, Locus and maxima and minima

values are important.

� Set a realistic study schedule and begin

studying early

� Short study sessions spread out over time are

more efficient and effective than a single

period of condensed study. Begin your study

sessions with a quick review of the material

you've previously studied, so that this

previous material stays fresh even though you

studied it in detail weeks before the test.

� Actively summarize: For each major conce-

pt, integrate information from your lecture

notes, lecture presentations, text in the printed

guide, and required readings onto a summary

sheet by diagramming, charting, outlining,

categorizing in tables, or writing paragraph

summaries of the information. Your studying

should also focus on defining, explaining, and

applying terms.

� Study with other well-prepared students The-

se study sessions will give you the opportuni-

ty to ask questions and further your understa-

nding of the course material.

Preparation Tips� Memorizing land mark problems (remember-

ing standard formulae, concepts so that you

can apply them directly) being strong in men-

tal calculations (never use the calculator duri-

ng your entire EAMCET preparation), try to

do first and second level of calculations with-

out pen and paper.

� Speed is familiarity, more familiarity with

concepts and formulae leads to more speed.

� Read each question carefully, sometimes you

can identify the range of the answer to that

question and only one option will be in that

range so you can answer the question without

solving the problem. E.g: answer is positive

and only one option is positive.

� While preparing for EAMCET mathematics

students should practise substitution methods

and verification method and also tips and

tricks given in all EAMCET materials. You

cannot rely on only concepts because time is

the main factor for EAMCET.

� EAMCET exam students should maintain a

calm and cool state of mind as this exam de-

mands speed and accuracy. You must be very

confident, don't panic, its not difficult and to-

ugh. You need to learn some special tips and

tricks to solve the EAMCET questions to get

the top rank.

� Do not spend more time on any question (mo-

re than 1½ min) if you do so you will lose the

time for another problem which may be easy

for you.

� Identification of the problem to be attempted

or not to be attempted plays a major role in

your success, So while preparing for EAM-

CET you should concentrate more on this

factor.

� Do not listen to your friends and other class-

mates, the topics which are easy or difficult

need not be same for two students. Do not ch-

ange your preparation methods at this stage.

� Don't try to touch new topics as they will take

time, you will also lose your confidence on

the topics that you have already prepared. But

prepare up to IPE level (Basic concepts) in

each and every topic. As there is no negative

marking you can go for answering every

question.

� Don't try to attempt 100% unless you are

100% confident: It is not necessary to attempt

the entire question paper, don't try if you are

not sure and confident as negative marking is

there. If you are confident in 60% questions,

that will be enough to get a good rank.

� Never answers question blindly. Be wise, pre-

planning is very important.

� There are mainly three difficulty levels, simp-

le, tough and average. First try to finish all the

simple questions to boost your confidence.

� Don't forget to prepare EAMCET previous

year question papers before the examination.

� As you prepare for the board examination,

you should also prepare and solve the last ye-

ar question papers for EAMCET. You also ne-

ed to set the 3 hour time for each and every

previous year paper, it will help you to judge

yourself, and this will let you know your we-

ak and strong areas. You will gradually beco-

me confident.

� You need to cover your entire syllabus but do-

n't try to touch any new topic if the examinat-

ions are closeby.

� Most of the questions in EAMCET are not

difficult but they are just IPE level. They req-

uire simple basic concepts in each topic. You

will notice only 10 to 15 percent of questions

are difficult and lengthy ones. As time is the

major factor you take care of time managem-

ent even during preparation time. You should

not think about time when you are revising

concepts.

� It is very important to understand what you

have to attempt and what you have to omit.

There is a limit to which you can improve

your speed and strike rate beyond which what

becomes very important is your selection of

question. So success depends upon how judi-

ciously one is able to select the questions. To

optimize your performance you should quick-

ly scan for easy questions and come back to

the difficult ones later.

� Try to ensure that in the initial 2 hours of the

paper the focus should be clearly on easy and

average questions, After two hours you can

decide whether you want to move to difficult

questions or revise the ones attempted to

ensure a high strike rate.

A. BHANUKUMARSenior Faculty,

Sri Chaitanya EducationalInstitutions, Hyderabad.

Prepared by

Set a realistic study schedule..Area wise questions weightage analysis

EAMCET-2009 ANALYSIS

2005

2006

2007

2008

2009

2005

2006

2007

2008

2009

2005

2006

2007

2008

2009

2005

2006

2007

2008

2009

Vector Algebra

Probability

Probability

Vector Algebra

ALGEBRA

CALCULUS

TRIGONOMETRY

COORDINATE GEOMETRY

VECTOR ALGEBRA

PROBABILITY

TRIGONOMETRY

CORDINATE GEOMETRY

CALCULUS

ALGEBRA


3

TRIGONOMETRYVECTOR ALGEBRAMATHEMATICS

Vector equation of a line passing...VECTOR ALGEBRA & PROBABILITY

SYNOPSISEven though students feel probability a difficult

chapter the questions which have come in the

previous years are easy and moderate. The ques-

tions from random variable and distributors are

easy to score .Rarely they give one or two diffic-

ult questions in probability (Application of Ba-

ye's Theorem). Vector algebra is the scoring

area, but student should be thorough with all

basic concepts and formulae.

In probability and vector algebra around

12% of questions will appear in EAMCET. All

questions are on basic concepts. Those who are

thorough with TELUGU academy text book

(IPE) can score more, but students will feel

these topics are difficult to score as these topics

need some more analytical ability than what

students have normally at this age.

Probability

� Two events A, B in a sample space S are said

tobe disjoint or mutually exclusive if A∩B=φ.

� The events A1, A2,....An in a sample space S

are said to be mutually exclusive or pair wise

disjoint if every pair of the events A1, A2,....An

are disjoint.


to be exhaustive if A∪B=S

� The events A1, A2,....An in a sample space S

are said to be exhaustive if A1∪A2∪...∪An=S


to be complementary if A∪B=S, A∩B=φ� Let A be an event in a sample space S. An

event B in S is said to be complement of A if

A, B are complementary in S. The

complement B of A is denoted by .

� If A is an event in a sample space S, then

� Let A, B be two events in a sample space S. If

then .

� Let S be a sample space containing n sample

points. If E is an elementary event in S, then

.

� Let S be a sample space containing n sample

points. If A is an event in S containing m

sample points, then Probability

� If A, B, C are three events in a sample space

S, then

� If A, B are two events in a sample space then

the event of happening B after the event A

happening is called conditional event. It is

denoted by B|A.

� Multiple theorem of probability: Let A, B

be two events in a sample space S such that

, . Then

i)

ii)


to be independent if

� Two events A, B in a sample space S are

independent if

� If A1, A2 are two mutually exclusive and

exhaustive events and E is any event then

� If p, q are the probabilities of success, failure

of a game in which A, B play then

i) probability of A's win =

ii) probability of B's win

� If p, q are the probabilities of success, failure

of a game in which A, B, C play then

i) probability of A's win

ii) probability of B's win

iii) probability of C's win

� For probability distribution if x=xi with range

(x1,x2,x3 ----) and P(x=xi) are their probabi-

lities then mean Variance

= Standard deviation

=

� If n be positive integer p be a real number

such that a random variable X with

range (0,1,2,-----n) is said to follows binomial

distribution.

For a Binomial distribution of

i) probability of occurrence = p

ii) probability of non occurrence = q

iii) p + q = 1

iv) probability of 'x' successes

v) Mean =

vi) Variance = npq

vii) Standard deviation =

� If number of trials are large and probability of

success is very small then Poisson distributi-

on is used and given as

Vector Algebra

� If ABCDEF is regular hexagon with center 'G'

then AB+AC+AD+AE+AF=3AD=6AG.� Vector equation of sphere with center at

and radius a is or

� are ends of diameter then equation of

sphere

� where

i) is acute

ii) is obtuse

iii) two vectors are

to each other.

�

� Vector equation. of a line passing through the

point A with P.V. and parallel to 'b' is

� Vector equation of a line passing through

is r =(1-t) +t

� Vector equation. of line passing through &

to is

� Vector equation. of plane passing through a pt

and- parallel to non-collinear vectors

is . s,t ∈ R and also

given as

� Vector equation. of a plane passing through

three non-collinear

Points. is

i.e =

� Vector equation. of a plane passing throughpts and parallel to is

or

= 0

� Perpendicular distance from origin to plane

passing through a, b, c is

� If ABC is a triangle such that

then the vector area of ∆ABC is 1/2( )

and scalar area 1/2[ ] is

� If ABCD is a parallelogram and

then the vector area of ABCD is

� The volume of the tetrahedron ABCD is ±1/6

� If a,b,c are three conterminous edges of a

tetrahedron then the volume of thetetrahedron = ±1/6

Model Questions

1. The median AD of the triangle ABC is

bisected at E BE meets AC in F, then AF : AC

1) 3:1 2) 1:3 3) 1:2 4) 2:1

2. If are orthonormal vectors and is a

vector then =

1) 2) 3) 4)

3. O, A, B, C are vertices of a tetrahedron G1,

G2, G3 are the centroids of triangles OBC,

OCA and OAB. OG1, OG2, OG3 are conterm-

inal edges of a parallelopiped with volume

V1. If V2 is the volume of the tetrahedron

OABC then V2/V1

1) 4/9 2) 9/4 3) 9/2 4) None

4. Three of six vertices of a regular hexagon are

chosen at random,the probability that a

triangle with these vertices is an equilateral

triangle

1) 1/15 2) 2/15 3) 4/15 4) 1/10

5. The key for a door is in a bunch of 10 keys. A

man attempts to open the door by trying keys

at random descending the wrong key. The

probability that the door is opened in the fifth

trial is

1) 0.1 2) 0.2 3) 0.5 4) 0.6

6. Let X = {1,2 ....50}. A subset A of x is chosen

at random. The set X is reconstructed by

replacing the elements of A and another

subset B of X at random the probability that

A∩B contain exactly 5 elements

1) 2)

3) 4)

KEY: 1) 2, 2) 1, 3) 2, 4) 4, 5) 1, 6) 3

TRIGONOMETRY

SYNOPSISIn trigonometry, students usually find it difficult

to memorize the vast number of formulae. Und-

erstand how to derive formulae and then apply

them to solving problems. The more you practi-

se, the more ingrained in your brain these form-

ulae will be, enabling you to recall them in any

situation. Direct questions from trigonometry

are usually less in number, but the use of trigo-

nometric concepts in coordinate geometry and

calculus is very profuse.

In two years Intermediate TRIGONOME-

TRY around 14% of questions will appear in

EAMCET. All questions are simple and most of

the questions can be solved by substitution

method, but students should be careful about the

domains and ranges of trigonometric and inve-

rse trigonometric functions. Those who are thor-

ough with TELUGU academy text book (IPE)

can score more.

Important Statements

�

� 3/2

�

� 1/4

� 1/4

�

�

�

� If A+B =450 or 2250 then

� If A+B =1350 or 3150 then

� If A+B+C = 1800 then ,

� If A+B+C = 900 then ,

�

�

� If then

3/4

� If then

3/4

�

3/4

�

�

�

�2 2 2osaC bSin C aSin bCos a b cθ θ θ θ+ = ⇒ − = ± + −

( )Cos A B C CosACosBCosC SinASinBCosC+ + = − ∑( )Sin A B C SinACosBCosC SinASinBSinC+ + = −∑

0 0(60 ) (120 ) 3 3Tan Tan Tan Tanθ θ θ θ+ + + + =3Cos θ

3 3 0 3 0(120 ) (120 )Cos Cos Cosθ θ θ+ − − + =

2 2Sin A Sin B SinASinB+ =m

060A B± =

2 2Cos A Cos B CosACosB+ =m

060A B± =2 2CotA TanA Cot A− =2 sec 2CotA TanA Co A+ =

CotA CotA= Π∑1TanATanB =∑

1CotACotB =∑TanA TanA= Π∑

(1 )(1 ) 2;(1 )(1 t ) 2TanA TanB CotA Co B− − = + + =

(1 )(1 ) 2;(1 )(1 cot ) 2TanA TanB CotA B+ + = − − =

2 2( ). ( )Cos A B Cos A B Cos A Sin B+ − = −

2 2( ). ( )Sin A B Sin A B Sin A Sin B+ − = −

0 0. (60 ). (60 ) 3Tan Tan Tan Tanθ θ θ θ− + =3Cos θ0 0. (60 ). (60 )Cos Cos Cosθ θ θ− + =

3Sin θ. (60 ). (60 )Sin Sin Sinθ θ θ− + =(120 ) (120 ) 0Sin Sin Sinθ θ θ+ + − − =

2 2 0 2 0(120 ) (120 )Cos Cos Cosθ θ θ+ − + + =

0 0(120 ) (120 ) 0Cos Cos Cosθ θ θ+ − + + =

50 45450

.3

4

C50 45550

.3

4

C

( )45503

50

. 3

4

C504

504

C

cbar

( ),r a a∑r, ,a b c

a b c

AB AC AD uuur uuur uuur

a b×BC b=uuur ,AB a B=

uuur ua b×a b×

,AB a AC b= =uuur uuur

abc

b c c a a b

× + × + ×

, ,or r a b a c − −

( )r a s b a tc= + − +, , 0AP AB c =

( )C c( ) ( )A a B b

, , 0r a b a c a − − − = ( )1 s t a sb sc− − + +( ) ( )r a s b a t c a= + − + −

0AB AC AP = ( ) ( ) ( ), ,A a B b C c

r a bc r bc abc − = =

r a sb tc= + +&b c

( )A a

( )r a t b c= + ×,b cr⊥a

ba( ) ( ),A a B b

r a tb= +a

( ) ( ) ( )2 2 22 2; 2a a i a j a k a× + × + × =

( ) ( ) ( )2 2 2. . .a i a j a k a+ + =

r⊥. 0 90a b θ= ⇒ = ° ⇒. 0 90 180a b θ θ< ⇒ ° < < ° ⇒. 0 0 90a b θ θ> ⇒ < < ° ⇒

0 180θ° ≤ ≤ °. cosa b a b θ=

( )( ). 0r a r b− − =

,a b

2 2 22 .r r c c a− + =( )2r c−

c

( )ke

P x kk

λλ−

= =

npq

npµ =( ) n x x

i xP x x nC q p−= =

( )nq p+

0 1p≤ ≤

variance

2 2 2( )i ix P x xσ µ= = −∑

( )i ix P x xµ = =∑

2

31

q p

q=

−

31

qp

q=

−

31

p

q=

−

21

qp

q=

−

21

p

q−

1 1 2 2( ) ( ) ( | ) ( ) ( | )P E P A P E A P A P E A= +

( ) ( ) ( )P A B P A P B∩ =

( ) ( )P B A P B=

( ) ( ) ( )P A B P B P A B∩ =( ) ( ) ( )P A B P A P B A∩ =

( ) 0P B ≠( ) 0P A ≠

( ) ( ) ( )P B C P C A P A B C− ∩ − ∩ + ∪ ∪( ) ( ) ( ) ( )P A P B P C P A B+ + − ∩

( )P A B C P∪ ∪ =

( ) /P A m n=

( ) 1/P E n=

( ) ( )P A P B≤A B⊆

( ) 1 ( )P A P A= −

A


4

COORDINATE GEOMETRYTRIGONOMETRYMATHEMATICS

If the angle of elevation of...� Period of Sin x,Cos x, Sec x and Cosec x is 2π� Period of Tan x, Cot x is π� Period of x-[x] is '1'

� Period of ax-[ax] is 1/a

� Principal value of value θ for function sin θ

lies between

� Principal value of value θ for function sin θ

lies between

� Principal value of value θ for function sin θlies between

� General solution of Sin θ is nπ+(-1)n a if a

� General solution of Tan θ is nπ+a if a

� General solution of Cos θ is 2nπ ± a if a

� General solution of θ if Sin θ = 0 is θ = nπ� General solution of θ if Cosθ =0 is (2n+1)π/2� If Sin2θ = Sin2a

Cos2θ = Cos2a

Tan2θ = Tan2a

then general solution is θ = nπ ± a� For aCosθ + bSin θ = c then solution exists if

� tanθ=0⇒θ=nπ� sinθ=1⇒θ=(4n+1)π/2� sinθ=−1⇒θ=(4n-1)π/2� cosθ=1⇒θ=2nπ� cosθ=−1⇒θ=(2n+1)

� sinθ = sina and cosθ = cos a ⇒ θ = 2n+a

� sin-1(sinθ)= θ if and only if and

sin(sin-1 x) = x where -1 ≤ x ≤ 1� cosec-1 (cosecθ)= θ if and only if

or and

cosec (cosec-1x) where or

� tan-1(tanθ)= θ if and only if

and tan(tan-1 x)= x where � cos-1(cosθ)= θ if and only if and

cos(cos-1 x)=x where � sec-1(secθ)= θ if and only if

or and

sec(sec-1x)=x where or � cos-1(cosθ)= θ if and only if 0< θ < π

and cot(cot-1 x)=x where

� , ,

�

�

�

�

�

� = rs = 2R2 sin A

sin B sin C =

�

� If I,I1I2,I3 are the incenter and excenters of a

triangle ABC then

� AI= ,BI= ,CI=

� If H is ortho center of a triangle ABC then

AH=2RcosA, BH=2RcosB, CH=2RcosC

� If a2 + b2 +c2 =8R2 then it is a right angled

triangle

� If r1-r=2R then it is a right angled triangled at

A

� If r2-r =2R then it is a right angled triangled at

B

� If r: R : r1 = 2 : 5 :12 then A = 900 because r1-

r =2R, 12-2=10=2.5

� If then A= 900

� If rr1 = r2r3 then A=900

� If r1r2r3are in H..P. then a,b,c are in A.P.

� If Then

�

� If

� Sinhx =

� Tan hx =

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

� If the angle of elevation of cloud from an

observation at a height 'd' from water level is

and angle of depression

of its image in lake is βthen height of cloud is

� From three points on level ground angle of

elevations of top of tower

found to bα, 2α, 3α then

height of tower is

Model Questions

1. If secθ - tanθ = 4/3, then sin θ=

1) 7/25 2) -7/25 3) 24/25 4) -24/25

2. If then =

1) 2)

3) 4)

3. The value of =

1) 2)

3) 4)

4. then x =

1) 2) 3) 4)

5. If the base angles of a triangle are

and , then the base and height are

in the ratio

1) 1:2 2) 1:3 3) 3:1 4) 2:1

6. The area of the triangle on the argand diagram

formed by the complex number z,iz and z-iz is

1) 2) 3) 4) 2|z|2

7. If z = i log then cos z =

1) -2 2) 2i 3) -2i 4) 2

8. The minimum value of 4tan2θ+ 9cot2θ is

1) 6 2) 12 3) 4 4) 9

9. A man standing on a level plane observes the

elevation of the top of a pole to be α. He then

walks a distance equal to the double the

height of the pole and find the elevation is

now 2α then the value of α is

1) 2) 3) 4)

KEY: 1) 2, 2) 4, 3) 1, 4) 1, 5) 4,

6) 3, 7) 4, 8) 2, 9) 1

COORDINATE GEOMETRY

SYNOPSISThis section is usually considered easier than

trigonometry. There are many common concepts

and formulae (such as equations of tangent and

normal to a curve) in II year Geometry (circle,

parabola, ellipse, hyperbola). Pay attention to

Locus and related topics, as the understanding

of these makes coordinate geometry easy.

In Co-ordinate Geometry 3D- Geometry is

one area where students can score more with ba-

sic concepts and formulae, some of the questio-

ns in conic section are tricky. Over all co-ordin-

ate Geometry requires formulae support so stud-

ents should practise all theorem statements, note

points, formulae and standard problems from

Telugu Academy. Students should spend more

time in these areas while preparing as most of

the questions are the reach of the students.

Those who are tho-rough with TELUGU

academy text book (IPE) can score more.

In two years of Intermediate I year Geomet-

ry around 12 (15% of total) questions will appe-

ar in EAMCET. All questions are on basic conc-

epts sometimes one or two questions are little

bit lengthy but they can be solved with verificat-

ion or substitution methods, those who are thor-

ough with TELUGU academy text book (IPE)

can score more.

In two years of Intermediate II year Geo-

metry around 10 (12% of total) questions will

appear in EAMCET. All questions are on basic

concepts. One or two questions in conic section

may be tricky and lengthy.

3D Coordinate System

� Area of ∆le formed by origin and A(x1,y1,z1),

B (x2,y2,z2) is

� Distance of P(x, y, z) from xy plane is |z|, yz

plane is |x|, xz plane is |y|

� Distance of P(x, y, z) from x- axis is ,

y - axis is z - axis is

� Centroid of tetrahedron =

� Centroid 'G' of tetrahedron ABCD divides the

line joining any vertex to centroid of its

opposite face in 3 :1 ratio

� If α, β, γ are angles made by a line with +ve

1 2 3 4 1 2 3 4 1 2 3 4, ,4 4 4

x x x x y y y y z z z z+ + + + + + + + +

2 2y x+2 2 ,x z+

2 2y z+

( ) ( ) ( )2 2 21 2 2 1 1 2 2 1 1 2 1 2

1

2x y x y y z y z z x x z− + − + −

3

π5

12

π6

π12

π

( )2 3−

21

2z21

4z

2z

012

11201

222

2

7

2

2 7

3

7

3

2 7

1 1sin sin 23

x xπ− −+ =

2 cos16

π2cos

8

π

2cos16

π2cos

32

π

2 2 2 2+ + +

2cos2

A2cos

2

A−

2sin2

A−2sin

2

A

1 sin 1 sinA A− − +02782

A=

( )(3 )2

aa b b a

b+ −

(tan tan )

tan tan

d α βα β

−+

5 55cos sinn nC θ θ−

1 31 3sin cos sin cos sinn n n nn C Cθ θ θ θ θ− −= − +

2 4 42 4cos cos cos cos sinn n n n nn C Cθ θ θ θ θ− −= − +

tanh tan tan tanhix i x ix i x= =cosh cos ,cos coshix x ix x= =

( )sinh sin ,sin sinhix i x ix i x= =

1 1 1tanh log

2 1

xx

x− + = −

1 2sinh log 1x x x− = + +

1 2cosh log 1x x x− = + −

3cosh 3 4cosh 3coshx x x= −

3sinh 3 3sinh 4sinhx x x= +

2

2 tanhsinh 2 2sinh .cosh

1 tanh

xx x x

x= =

−

( )cosh cosh cosh sinh sinhx y x y x y± = ±( )sinh sinh cosh cosh sinhx y x y x y± = ±

22

2

1 tanh1 2sinh

1 tanh

xx

x

++ =

−

2 2 2cosh sinh cosh 2 2cosh 1x x x x+ = = − =

2 2cosh sinh 1x x− =

,cothx x x x

x x x x

e e e ex

e e e e

− −

− −− +

=+ −

,cosh2 2

x x x xe e e ex

− −− +=

2 21log log arg

2z x iy z x y i θ= + ⇒ = + +

( )( )( )2 2a b c a b c a b cω ω ω ω+ + + + + +

3 3 3 3a b c abc+ + − =

2 2 32 2 0,

23 3 ( ), 3 3 ( )

(2 ) 3, (2 ) 0,

n nCos Sin Cos Sin

Cos Cos Sin Sin

Cos Sin

α α α α

α α β γ α α β γ

α β γ α β γ

= = = =

= + + = + +

− − = − − =

∑ ∑ ∑ ∑∑ ∑∑ ∑

2 2 0Cos Sinα α= =∑ ∑0If Cos Sinα α= =∑ ∑

1 2

2 3

1 1 2r r

r r

− − =

1 3 cos2

CAI r ec=

2 2 cos ,2

BBI r ec=1 1 cos ,

2

AAI r ec=

cos2

Cr eccos

2

Br eccos

2

Ar ec

2 21 2 2 3 3 1 1, 4 sin

2

Ar r r r r r s r r R+ + = − =

2 1 34 cos , 4 cos ,R B r r r r R C= + + − =2 3 1 1 3 24 , 4 cos ,R r r r r R A r r r r+ + − = + + − =

21 2 3 1 2 3

1 2 3

1 1 1 1, ,r r r r r r r r

r r r r= ∆ + + = + + − =

sin( )2

cos2

A Ba b

Cc

−− =

cos( )2 ,

sin2

A Ba b

Cc

−+ =

1sin

2bc A

( )( )( )s s a s b s c∆ = − − −1x >

1 1 12 2

2 22 tan sin tan ,

1 1

x xx

x xπ π− − −= − = +

+ −

1 1 12 2

2 22 tan sin tan , 1

1 1

x xx x

x x− − −= = <

+ −

21 1

2

12 tan cos , 0

1

xx x

x− − −

= ≥+

1

1

tan 0, 0, 11

tan , 0, 0, 11

0, 0, 12

x yx y xy

xy

x yx y xy

xy

x y xy

π

π

−

−

+ ≥ ≥ < −+ + > > > −

> > =

1 1tan tanx y− −+ =0 0x y≥ ≥

1 1 1tan tan tan1

x yx y

xy− − − −− =

+

1 1sec cos2

x ec xπ− −+ =

1 1tan cot2

x xπ− −+ =1 1sin cos

2x x

π− −+ =

x−∞ < < ∞

1 x≤ < ∞1x− ∞ < ≤ −2

π θ π< ≤02

πθ≤ <

1 1x− ≤ ≤0 θ π≤ <

x−∞ < < ∞2 2

π πθ− < <

1 x≤ ≤ ∞1x−∞ < ≤ −

02

πθ≤ ≤02

π θ− ≤ ≤

2 2

π πθ− ≤ ≤

2 2c a b≤ +

[ ]0,π

,2 2

π π−

,2 2

π π−


5

GEOMETRY IInd YEAR3D COORDINATE SYSTEMMATHEMATICS

ax2+2hxy+ by2+2gx+2fy+c=0 to.. direction of axes ;

� (i) If the d.c's (l,m,n) of two lines are

connected by the relations al + bm + cn = 0

and fmn + gnl + hlm=0

Then two lines are perpendicular if

And two lines are parallel

(ii) If the d.c's (l,m,n) of two lines are conn-

ected by the relations al + bm + cn =0 and

Then two lines are perpendicular if

And two lines are parallel

� Harmonic conjugate points: If P and Q

divide AB internally and externally in the

same ratio, then P is called as harmonic

conjugate of Q and Q is called as harmonic

conjugate of P, also P, Q are a pair of

conjugate points w.r.t A and B.

� If P, Q are harmonic conjugate points w.r.t A,

B then A, B are harmonic conjugate points

w.r.t P, Q

� If P, Q divide AB in the ratio l : m internally

and externally then A, B divide PQ in the

ratio (l - m) : (l + m)

� If D is midpoint of BC of triangle ABC then

AB2+AC2=2(AD2+BD2)

� In a triangle ABC if BC is the largest side

then

AB2+AC2=BC2 then triangle ABC is right

angled

AB2+AC2 > BC2 then triangle ABC is Acute

angled triangle

AB2+AC2 < BC2 then triangle ABC is Obtuse

angled triangle

Note: In the above 3 cases if AB = AC then

the triangle is isosceles also

� Internal angular bisector of angle A of ∆ABC

divides the opposite side BC in the ratio AB :

AC

� If a, b, c are lengths of sides BC, CA, AB of

∆ABC and if I is in-centre then I divides the

Internal angular bisector of angle A in the

ratio AI : ID= b + c : a

� Circumcentre of the right angled triangle

PQR, right angled at P is

where P(x1,y1), Q(x2,y2), R(x3,y3) are the

vertices and Orthocentre is P

� The orthocenter of the triangle with vertices

(0,0), (x1,y1), and (x2,y2) (k(y2-y1),k(x1-x2))

where

� If P,Q,R be the mid-points of the sides BC,

CA, AB respectively of �ABC, then

1 The Centroid of ∆ ABC=Centroid of ∆ PQR

2 The Circumcenter of ∆ ABC=Orthocentre

of ∆ PQR

� If H is orthocenter of triangle ABC then each

of the points A, B, C, H is the orthocentre of

the triangle formed by the remaining points.

� In ∆ ABC, ON : NG : GS = 3 : 1 : 2 Here G =

centroid, O = orthocentre, S = circum centre

of ∆ ABC N = centre of nine points circle

� The locus of a point is the path traced out by

the point under certain geometrical condition

/ conditions.

� Equation of locus is an equation in x and y,

which is satisfied by the co-ordinates of any

point on the locus.

� The curve represented by S ≅ ax2+by2+2hxy

+2gx+2fy+c=0 is

� a pair of parallel lines if h2 = ab, ∆ = 0

� a parabola if h2 = ab, ∆ ≠ 0

� an ellipse if h2 < ab, ∆ ≠ 0

� a circle of a = b, h = 0, g2 + f2 - ac ≥ 0

� a pair of intersecting lines if h2 > ab ∆ = 0

� a hyperbola if h2 > ab and ∆ ≠ 0 where

∆=abc + 2fgh - af2 - bg2 - ch2

� a rectangle hyperbola if h2> ab, a+b=0,

∆ ≠ 0.

� Translation of Axes: If we shift the origin to

(h, k) without changing the direction of axes,

the relation between original coordinate (x, y)

and new coordinate (X,Y) is given by x=X+h

& y=Y+k or equivalently X=x -h and Y=y -k,

� The first degree terms are removed from the

equation ax2+2hxy+by2+2gx+2fy+c=0, by

translation of axes to the point

In this case, the transformed equation is a

aX2+2hXY+bY2+(gx1+fy1+c)=0

� To remove the first degree terms from

ax2+by2+2gx+2fy+c=0 the origin is shifted to

the point in this case the transfor-

med equation is

� To remove the first degree terms from 2hxy +

2gx + 2fy + c = 0, the origin is shifted to the

point in this case, the transformed

equation is 2hXY + c = 0.

� Rotation of Axes: When the axes are rotated

through an angle θ without changing the orig-

in, the relations between original coordinates

(x, y) and new coordinates

(X, Y) are given by

x=X cosθ - Y sinθ Y=X sinθ - Y cosθX=x cosθ + ysinθ Y=-xsinθ +y cosθ

� The xy term is removed in ax2 + 2hxy + by2

+2gx + 2fy + c = 0, by rotation of axes

through an angle if a ≠ b

and if a=b.

� The area of the triangle formed by the line ax

+ by + c = 0 with the coordinate axes is

� The perpendicular distance of the line ax + by

+ c = 0 from the origin is

� The ratio that the line joining the two points

(x1,y1)and(x2,y2) is divided by the line L=0 is

(i) lie on the same side if L11 and L22 are of

same sign.

(ii) lie on the opposite sides if they are

opposite signs.

� The area of the rhombus formed by

ax±by±c = 0 is

� The foot of the perpendicular from (x1,y1) to

the line ax+by+c=0 is (h,k), then

� The image of the point (x1,y1) w.r.t the line

ax + by + c=0 is (h, k) then

The Reflection of the point (x1, y1) w.r.t to

1 x-axis is the point (x1, -y1)

2 y-axis is the point (-x1, y1)

3 The origin is (-x1, -y1)

4 The line y = x is the point (y1, s1)

5 The line y = -x is the point (-y1, -x1)

� If the lines a1x+b1y+c1=0, a2x+b2y+c2=0 and

a3x+b3y+c3=0 are concurrent, then

� The distance of the point P(x1,y1) from the

line ax+by+c=0 in the direction of a line

making an angle 'θ ' with the x-axis is

� Orthocentre of the ∆le formed by lx + my + n= 0 and pair of lines =0 is

� If represents a pair of parallel lines then

The distance between the parallel lines

� The product of perpendicular from (α,β) to

the pair of lines S= 0 is .

� The area of the ∆le formed by ax2+2hxy+by2

= 0, lx+my+n=0 is sq. units.

� If 'θ ' is an acute angle between

ax2+2hxy+by2 = 0 then

� If the equation ax2+2hxy+by2= 0 represents

two sides and (l,m) is orthocenter of a triangle

then third side is (a+b) (lx+my) =

am2- 2hlm+ bl2

� If ax2+2hxy+by2=0 be two sides of a

parallegram and lx+my+n=0 is one diagonal

then eq of the other diagonal is

� If the pair of lines S = 0 intersect the x-axis at

P &Q then the length PQ is called x-intercept.

x - intercept, PQ =.

Similarly y- intercept =

� If G is the centroid of the ∆le and D is the

midpoint of the 3rd side then D =

Model Questions

1. If the straight line x+y+1 = 0 is changed into

the form x cosα+y sinα = p , then α =

1) π/4 2) 3π/4 3) 5π/4 4) 7π/42. A straight line passing through Q(2,3) makes

an angle of π/4 with x- axis in +ve direction.If

this straight line intersects x+y-7=0 at p, then

1) 2) 3) 4)

3. If a+b+c=0, the straight line 2ax+3by+4x=0

passes through the fixed point

1) 2) 3)

4) No such fixed point

4. The difference of the slopes of the lines 3x2-

8xy-3y2= 0 is

1) 3/10 2) -3/10 3) 10/3 4) -10/3

5. Perpendicular distance of (1,4,3) to x-axis is

1. 2 . 3. 4. 56. The equation of the plane which makes inter-

cepts - 4, 5 , 6 on co-ordinate axes of X,Y,Z

respectively is

1) 2)

3) 4)

KEY: 1) 3, 2) 1, 3) 1, 4) 3, 5) 4, 6) 3

Geometry - IInd Year

� The conditions that the equation ax2+2hxy+by2+2gx+2fy+c=0 to represent a circle are

(i) a=b (ii) h=0 (iii) g2+f2-ac≥ 0� If 'C' is centre, r is radius of a circle S=0,

P (x1,y1) is any point then CP2-r2 is called

power of the point P with respect to S=0.

� The power of the point P (x1, y1) with respect

to the circle S=0 is S11

i) If p lie outside the circle then power S11 >0

ii) If p lie inside the circle then power S11 <0

iii) If p lie on the circle then power S11 =0

� The equation of circle having the line segme-

nt joining the points A(x1,y1), B(x2,y2) as

diameter is (x-x1)(x-x2)+(y-y1)(y-y2)=0

� If S=0 is a circle, L=0 is a straight line. Then

equation of the circle passing through point of

1 04 5 6

x y z+ + + =1

4 5 6

x y z− − =

14 5 6

x y z− − =1

4 5 6

x y z+ + =

171026

4 4,

3 3

( )2,24

2,3

7 25 23 22

3.

2

G

22 f bc

b

−

22 g ac

a

−

x y

bl hm am hl=

− −

2 2

2 2

2 2;

( ) 4

h ab h abTam Sin

a b a b hθ θ− −

= =+ − +

2 2;

( ) 4

a bCos

a b hθ

+=

− +

2 2

2 22

n h ab

am hlm bl

−− +

11

2 2( ) 4

S

a b h− +

( ) ( )2 2

2 2g ac f bc

a a b b a b

− −= =

+ +

2 2 2, .h ab af bg= =

2 22 2 2 0S ax hxy by gx fy c≡ + + + + =

( ) 2 2

( ),

2

n a bkl km k

am hlm bl

− +=

− +

2 22ax hxy by+ +

1 1

cos sin

ax by c

a bθ θ+ +

+

1 1 1

2 2 2

3 3 3

0

a b c

a b c

a b c

=

( )1 11 12 2

2.

ax by ch x k y

a b a b

+ +− −= =

+

1 1 1 12 2

.h x k y ax by c

a b a b

− − + += =

+

22.

c

ab

11

22

L

L

−

2 2.

c

a b+

2

. .2

csq units

ab

=4

πθ

11 2= tan

2

h

a bθ −

−

,f g

h h

− −

2 22 2 0

g fax by c

a b

−+ + − + =

,g f

a b

− −

2 2,

hf bg gh af

ab h ab h

− − − −

1 2 1 2

1 2 2 1

x x y yk

x y x y

+= −

2 3 2 3,2 2

x x y y+ +

2 2 2

0a b c

u v w+ + =

2 2 2( ) ( ) ( ) 0a v w b u w c u v+ + + + + =

2 2 2 0ul vm wn+ + =

0af bg ch± ± =

0f g h

a b c+ + =

2 2 2sin sin sin 2α β γ+ + =

2 2 2cos cos cos 1α β γ+ + =


6

intersection S=0, L=0 is S+λL=0 where λ isa parameter.

� The equation of the circum circle of the trian-

gle formed by co-ordinate axes and the line

ax+by+c = 0 is ab(x2+y2)+ c(bx + ay) = 0.

� If a1x+b1y+c1=0, a2x+b2y+c2=0 cuts the co-

ordinate axes at four distinct points then those

points are concyclic⇔ a1a2=b1b2. The equat-

ion of the circle passing through these points

is

(a1x+b1y+c1)(a2x+b2y+c2)-xy(a1b2+a2b1)=0

and centre of circle is

� If the circle x2+y2+2gx+2fy+ c = 0 touches

both the coordinate axes the g2=f2=c

� The equation of the circle of radius 'a' which

touches the co-ordinate axes is

x2+y2 ± 2ax ± 2ay ± a2= 0.

� Length of tangent drawn from P(x1,y1) to the

circle S = 0 is .

� If lx+my+n=0 is a polar of the point (x1,y1)

with respect to x2+y2+2gx+2fy+c = 0 then

= = .

� If P(x,y) is a point on the circle with centre

(h,k) and r is radius, then x=h+r cosθ,

y=k+r sinθ (0≤ θ<2π ).

� If P, Q are conjugate point w.r.to S=0, l1,l2 are

lengths of tangent from P and Q then P

Q2=l12+l2

2.

� The vertices of the square whose sides are

parallel to the axes and inscribed the circle

S=0 are

� The simplest form of a coaxal system of circ-

les is x2+y2+2λx+c=0 where λ is parameter

and c is a constant. Line of centres is X- axis.

The radical axis is Y-axis If c<0, then the

system is intersecting system and intersecting

at .If c>0, then the system is non

touching system.

� Limiting points (Def): The point circles of a

coaxal system of circles are called as limiting

points of that coaxal system.

For intersecting system there are no limiting

points

For touching system the point of contact is

the only limiting point.

For non-touching, non-intersecting coaxal

system there are two limiting points

Limiting points lie on the line of centres

Common radical axis is the, perpendicular bi-

sector of the segment joining limiting points.

Every circle passing through the limiting poi-

nts cut every member of the coaxal system

orthogonally.

Limiting points are a pair of inverse points

with respect to every member of the coaxal

system.

The polar of one limiting point with respect to

any member of the coaxal system passes

through other limiting point.

Any common tangent to two circles of a coa-

xal system subtends a right angle at either of

the limiting points

If x2+y2+2λx+c=0 is a coaxal system of

circles then the limiting points are

∴c>0⇒Two limiting points.

c = 0⇒One limiting point, it is origin

c< 0⇒No limiting point.

� If origin is one limiting point of a coaxal syst-

em, of which the circle x2+y2+2gx+2fy+c=0

is a member, then the other

limiting point is

� The locus of a point which moves in a plane

such that its distance from a fixed point is at

a constant ratio to its distance from the fixed

line is called a conic

The fixed point is called focus.

The fixed line is called the directrix

The fixed ratio is called eccentricity(e).

If e = 1, the conic is a parabola

If e < 1, the conic is an ellipse

If e > 1, the conic is a hyperbola

Axis is a line perpendicular to the directrix

and passing through the focus.

Vertex: It is a point where the conic cuts the

axis.

Focal chord: Chord of the conic passing

through the focus.

Latus rectum: It is the focal chord

perpendicular to the axis.

Focal distance: It is the distance from the

focus to any point on a conic.

� Parametric point on the parabola y2=4ax is

(at2,2at)

� Equation of the chord joining t1 and t2 on the

parabola y2=4ax is y (t1+t2) = 2x+2at1t2.

� Condition for the line y=mx+c to touch the

parabola y2=4ax is c=a/m and point of

contact is (a/m2, -2a/m)

� Condition for the line y=mx+c to touch

x2=4ay is c = -am2.

� The condition for the line lx+my+n=0 to

touch i) y2=4ax is am2=ln

ii) x2=4ay is al2+mn=0

� From an external point (x1,y1) two tangents

can be drawn to the parabola y2=4ax.

� If m1 and m2 are slopes, then

m1+m2 = , m1m2 =

� If PQ is the focal chord of the parabola then

the tangents at P and Q on the parabola are

perpendicular and intersect on directrix. Also

the tangent at P is parallel to the normal at Q.

� Pole of the line lx + my + n=0 w.r.t

i) y2=4ax is and

ii) x2=4ay is .

� The point of intersection of tangents at 't1' and

't2' on y2=4ax is [at1t2 , a(t1+t2)].

� Through a given point three normals can be

drawn to the parabola. The sum of the slopes

of these three normals is 0.

� If two tangents are drawn to the parabola

y2=4ax, then the angle between the pair of

tangents is.

� Area of the triangle formed by the tangents

drawn from (x1,y1) to the parabola S=0 and

the chord of contact is

� The locus of the point whose ratio from the

fixed point to the fixed line bears a constant

ratio less than 1 is called an ellipse.

The fixed point is called the focus

The fixed line is called the directrix

The fixed ratio is called the eccentricity (e)<1

Equation of the ellipse in the standard form

Another def. of ellipse: The locus of a point

whose sum of the distances from two fixed

points is always a constant is an ellipse

� Ellipse is a closed curve and its area is πab.

� If S and S' are the foci and P is any point on

the ellipse, then

SP + S'P =2a (if a>b) SP + S'P=2b(if a<b)

� Equation of the normal at (x1,y1) on the

ellipse m is a2-b2.

� The normal at P of the ellipse S=0 is the inte-

rnal bisector of SPS' where S and S' are the

foci.

� The product of perpendiculars from the foci

on any tangent to the ellipse S=0 is b2.

� The locus of point of intersection of perpen-

dicular tangents of an ellipse

is called a director circle and its

equation is x2+y2 = a2+b2.

� The locus of the foot of the perpendicular fr-

om the foci on any tangent to the ellipse

is called an auxiliary circle and

its equation is

x2+y2=a2 if a>b x2+y2=b2 if b>a

� Equation of the chord joining α and β of the

ellipse is

� Equation of the tangent at 'θ' on the ellipse

is

� Four normals can be drawn from any point to

the ellipse. The sum of the eccentric angles of

their feet is an odd multiple of ππ.

� A conic section is said to be a hyperbola of it's

eccentricity is greater than 1. The equation of

a hyperbola in standard form is x2/a2-y2/b2=1.

� A point (x1,y1) is said to be an

1) external point to the hyperbola

= 1 if < 1.

2) internal point to the hyperbola

= 1 if > 1.

� A hyperbola is said to be a rectangular hype-

rbola if the length of it's transverse axis is eq-

ual to the length of it's conjugate axis. The ec-

centricity of a rectangular hyperbola is .

� If e1,e2 are eccentricities of two conjugate

hyperbolas then e12+e2

2=e12e2

2.

� The condition that the line y = mx + c may be

a tangent to the hyperbola

= 1 is c2=a2m2-b2.

� The equation of the chord joining two points

α and β one the hyperbola x2/a2-y2/b2 = 1 is

� The equation of the tangent at P(θ) on the

hyperbola

= 1 is

� The equation of a rectangular hyperbola

whose asymptotes are the coordinate axes is

xy =c2.

The parametric equations of xy = c2 are

x = ct, y = c/t.

The eccentricity of xy = c2 is .� Area of ∆le PQR=

r1r2 sin (θθ11−−θθ22)) ++ r2r3 sin (θθ22−−θθ33)) ++ r3r1 sin(θθ33−−θθ11))

� Equation of circle with centre (C,α) and

radius 'a' is r2-2rc Cos (θ −α) = a2-c2

� If ZSX is initial line Equation of conic

Directrix

� If SZX is initial line...

Equation of conic

Directrix

� Equation of the line ax + by + c = 0 in polar

form is given by

⇒a(rCosθ) + b(rSinθ) + c = 0

⇒a Cos θ+ b Sinθ =

� Equation of the line 11 to ax + by + c = 0 is

given by

a Cosθ+ b Sinθ = [ Since || line differ byK

r

c

r

−

cosl

er

θ

1 cosl

er

θ= +

cosl

er

θ= −1 cosl

er

θ= −

1

2

2

sec tan 1x y

a bθ θ− =

2 2

2 21

x y

a b− =

cos sin cos2 2 2

x y

a b

α β α β α β− + +− =

2 2

2 21

x y

a b− =

2

2 21 12 2

x y

a b−

2 2

2 21

x y

a b− =

2 21 12 2

x y

a b−

2 2

2 21

x y

a b− =

cos sin1

x y

a b

θ θ+ =

2 2

2 21

x y

a b+ =

cos sin cos2 2 2

x y

a b

α β α β α β+ + −+ =

2 2

2 21

x y

a b+ =

2 2

2 21

x y

a b+ =

2 2

2 21

x y

a b+ =

2 2

1 1

a x b y

x y− =

2 2

2 21

x y

a b+ =

2 2

2 21

x y

a b+ =

3/211

2

S

a

111

1

tanS

x a−

+

2,

al n

m m

−

2,

n am

l l

−

1

a

x1

1

y

x

2 2 2 2

-gc -fc,

g f g f

+ +

( ),0c±

(0, )c± −

,2 2

r rg f

− ± − ±

2

g

r

l mf n+ −1y f

m

+1x g

l

+

11S

Sumof x - intercept Sumof y - intercept,

2 2

LIMITING POINTSGEOMETRY IInd YEARMATHEMATICS

A point (x1,y1) is said to be an..


7

L-HOSPITAL'S RULECALCULUSMATHEMATICS

If f is continuous at x=a and g is..by constant]

� Equation of the line .to ax + by + c = 0

is of the form bx - ay = k

⇒b(rCosθ) - a(rSinθ) = K

⇒ b Sin + a Cos =

∴ a Cos + bSin =

MODEL QUESTIONS

1. If the points (1, -6), (k, 0), (5, 2), (-1, -4) are

concyclic then k =

1) 7 2) -7 3) 4)

2. If the lines 2x+3y+19=0 and 9x+ky-17=0

cut the coordinate axes in concyclic points

then k =

1) 6 2) -6 3) 2 4) 1/6

3. The angle between the two circles, each

passing through the centre of the other

1) 2) 3) 4)

4. The equation of the straight line meeting the

circle x2+y2=a2 in two points equal distance

'd' from a point (x1, y1) on the circumference

is xx1+yy1=K then k =

1) a-ad2 2)

3) 4) 0

5. If a normal subtends a right angle at the ver-

tex of the parabola y2=4ax then its length

1) 2) 3) 4)

6. The angle between asymptotes of the hyper-

bola 27x2-9y2=24 is

1) 2) 3) 4)

7. The product of the lengths of the perpen-

diculars from any point of the hyperbola

x2 - y2 = 8 to its asymptotes is

1) 2 2) 3 3) 4 4) 8

8. A chord PQ of a conic subtends a right angle

at the focus S then

1) 2) 3) 4)

KEY:1) 1, 2) 1, 3) 4, 4) 3, 5) 3, 6) 2, 7) 4, 8) 2

CALCULUS

SYNOPSISCalculus includes concept-based problems whi-

ch require analytical skills. Functions are the ba-

ckbone of this section. Be thorough with prope-

rties of all types of functions, such as trigonom-

etric, algebraic, inverse trigonometric, logarith-

mic, exponential, and signum. Approximating

sketches and graphical interpretations will help

you solve problems faster. Practical application

of derivatives is a very vast area, but if you und-

erstand the basic concepts involved, it is very

easy to score.

In Calculus differentiability and continuity

questions are conceptual while practising and

solving these problems students should carefu-

lly observe the nature of the function in left and

right neighborhoods. In "Limits" topic most of

the questions can be solved using L-Hospitals

Rule while solving the problem of limits obser-

ves the existence .Differentiation is scoring. Ap-

plication of derivatives need more practice befo-

re final Eamcet. For second year integral calcul-

us students should practise, the methods of inte-

gration, formulae (repeated reading of formulae

is needed), and properties of definite integrals,

modulus and step function problems in definite

integration and graphs for Areas Topic. Differe-

ntial equation is scoring for the students. They

should be through with IPE concepts.

In two years of Intermediate CALCULUS

around 19(25% of total questions) questions

will appear in EAMCET out of which 8 will be

from first year and 8 from second year. Only 4

or 5 questions will be tricky. All the remaining

questions are on basic concepts, while preparing

topics like continuity, differentiability and num-

eric integration students should think about the

depth of the subject, in all other topics. Students

should be thorough with TELUGU academy

text book (IPE).

� Indeterminate Forms: For some value ofx, say x = a if the function f(x) takes any ofthe forms

∞-∞, 0 x ∞, 1∞ ∞∞ Then f(x) is said

to be indeterminate at x = a.� L-Hospital's Rule: Let φ (x) and ψ (x) be

two functions, such that

then =

� Some Important Limits:

1) = 1, Where θ is measured in

radians,

2)

3)

4)

5) = e,

6) = e,

7) = ea,

8) = e

9)

10)

11)

12) Let and if

f(x), g(x) ∈S then

i)

ii)

iii)

iv)

v)

vi) If a1a2a3 are in A.P with commondifference 'd' then

� If

then

� Right hand limit of f(x) at x= a, is f(x) = f(a + h) (h>0 How ever small)

� Left hand limit of f(x) at x = a, is f(x)

=f(a-h) (h>0 How ever small)

� If f(x)= f(x)=l (ie.L.H.L.=

R.H.L. then

� If f(x)≠ f(x) (ie. L.H.L ≠R.H.L. then is does not exist.)

� A function f(x) is said to be continuous at

x= a if

� If then f(x) is discontin-

uous function at x= a.

� If then f(x) is right con-

tinuous function at x= a.

� If then f(x) is left conti-

nuous function at x= a.

� If ≠ f(a) then f(x) is removable

discontinuous at x= a.

Standard Statements

� If f is continuous at x=a and g is continuousat f(a),then gof is continuous at x=a.

� Every constant function is continuous on R.

� The identity function is continuous on R.

� Every polynomial function is continuous on R.� The functions Sin x,Cos x are continuous

on R. � The functions tan x and sec x are

continuous on where n is

any integer� The functions cot x and cosec x are contin-

uous on R-{nπ} where n is any integer� The function f(x) = |x| is continuous on R.� The functions f(x) =ex and f(x) = ax (a>o)

are continuous on R.� The function f(x) =[x] is continuous at all

non integral values and discontinuous at allintegral values.

� A function f(x) is said to have a derivativeat x= a if f is defined in a - δ<x<a+δ and

Exists, where.

� Thus if f '(a) exists, then

f '(a) =

� Left hand derivative of f =L f '(a)=

, a -δ < x < a

� Right hand derivative of f = R f '(a) =

, a< x < a +δ.

� f'(a) exists if L f '(a), R f '(a) exist and areequal

� If a function is differentiable at a point it isnecessarily continuous at that point. Theconverse of above theorem may not hold.E.g. The function f(x) = , is continuousat x= 0, but not differentiable at x = 0

Important Points

�

�

�

� v(x)≠0

� f(x) ≠ 0

� u = f(x)

� (log a), a being a positive

constant, a ≠1

�

� a being apositive

constant, a ≠1

�

� [u(x) v(x) w(x)] = u(x) v(x).

(w(x))+u(x) (v(x))w(x) + (u(x)).

v(x). w(x)

� .dy dy du

dx du dx=

d

dx

d

dx

d

dx

d

dx

1log .

d duu

dx u dx=

1log log ,a a

d duu e

dx u dx=

( )u ud de e u

dx dx=

,du

dxu ud

a adx

=

1( ) ( ) ,n nd duu n u

dx dx−=

'

2

( )

( ( ))

f x

f x

−1

( )f x

d

dx

2

( ). ( ) ( ). ( )( )

( ) ( ( ))

d dv x u x u x v xd u x dx dx

dx v x v x

− =

[ ( ) ( )] ( ) ( ) ( ) ( )d d d

u x v x v x u x u x v xdx dx dx

= +

[ ( )] ( )d d

kf x k f xdx dx

=

[ ( ) ( )] '( ) '( )d

f x g x f x g xdx

± = ±

x

( ) ( )lim

x a

f x f a

x a+→

−−

( ) ( )lim

x a

f x f a

x a−→

−−

( ) ( )limx a

f x f a

x a→

−−

0 x a δ< − <( ) ( )f x f a

x a

−−lim

x a→

( )2 1 .2

R nπ − +

lim ( )x a

f x l→

=

( ) ( )x a

Lt f x f a−→

=

( ) ( )x a

Lt f x f a+→

=

( ) ( )x aLt f x f a→

≠

( ) ( )x aLt f x f a→

=

lim ( )x a

f x→

limx a−→

limx a+→

lim ( )x a

f x l→

=

limx a−→

limx a+→

0limh→

limx a−→

0limh→

limx a+→

( )( ) ( ) 1( )( ) x aLt g x f x

g x

x aLt f x e →

−

→=

( )( ) 1g x

x aLt f x ∞

→=

1

1

.a d=

1 2 2 3 3 4

1 1 1....

. . .nLt nterms

a a a a a a→∞

+ + +

1 2....nna a a

1/

1 2

0

....lim

xx x xn

x

a a a

n→

+ + +=

( ) ( )[ ]

2

2

tan sin

2( )0lim

n n n

n

ax ax na

f xx

+

+

−=

→

( ) ( )2 2cos cos

20lim

ax bx b a

f cx g dx cdx

− −=→

( ) ( )21 cos

20lim

ax a

f cx g dx cdx

− =→

0

( )lim

( )x

f mx m

g nx n→=

1 1 1 1sin , tan ,sinh , tanh }x x x x− − − −{ ,sin , tan ,sinh , tanh ,S x x x x x=

{ }2lim2x

ax ax b x

→∞+ + − =

( )0

1lim log 0

x

ex

aa a

x→

−= >

0

1lim 1

x

x

e

x→

−=

( )1

1 xx+0limx→

1n

a

n

+ limn→∞

11

n

n

+ lim

n→−∞

11

n

n

+ limn→∞

0

tan1

x

xlim

x→=

m mm n

n nx a

x a mlim a

nx a−

→−

=−

0

sinx

axlim a

x→=

sinθθ0

limθ →

'( )

'( )

x

x

φψ

limx a→

( )

( )

x

x

φψ

limx a→

( ) 0,

( ) 0

aor

a

φψ

∞=∞

0,

0

∞∞

e

l2

2e

l2

2

e

l

e

l

221 1 1 1

SP SQ

− + − = l l

5

6

π2

π2

3

π6

π

7 3 a6 3 a3 5 a5 a

2 21

2a d

−

2 21

2a d

+

2

3

π2

π6

π4

π

1

7−

1

7

K

r2

π θ + 2

π θ +

K

r2

π θ + 2

π θ +

lar⊥


8

IMPORTANT FORMULAECALCULUSMATHEMATICS

The area between the parabolas y2=4ax and..�

�

�. a; a, b, n being

real numbers and x∈R

� (c) =0, c being a constant

� , c being a constant

�

�

� sin x=cos x∀x∈R; sin u= cos u

� cos x =-sin x ∀x∈R and

cos u= -sin u

� tan x=sec2x, x∈R except odd multiples

of π/2

� cot x=-cosec2 x tan x, x∈R expect even

multiples of π/2

� sec x = sec x tan x x∈R except even

multiples of π/2

� cosec x = -cosec x cot x, x∈R except

even multiples of π/2

�

�

�

�

�

�

�

(ax) = ax loge a, a ≠1

�

�

�

� If 's' be the distance of the particle mea-sured form a fixed point O on the line attime 't', then the velocity of the movingparticle at any time t is

υ = and acceleration a or

f =

� A differentiable real function f(x) is increa-

sing in an interval I if f '(x)>0 for all x in I.A differentiable real function f(x) is decre-asing in an interval I if f '(x)<0 for all x inI.

� If y=f(x) and ∆x is small change in x thenerror in y = ∆y or δy = f(x+ ∆x) -f(x)Approximate change in y=dy=f '(x).δx or f'(x)(where dy is approximate value of δy)

Relative error in y = and Percentage

error in y = or

� i) Slope of tangent at P (x0,y0) to the curve

y=f(x) is m=f '(x0)=

ii) Equation of tangent at (x0,y0) is:y-y0= f '(x0) (x-x0)

iii) Equation of normal at (x0,y0) is:

y-y0= (x-x0)

� i) Length of tangent =,

where m=

ii) Length of Normal =

iii) Length of sub tangent =

iv) Length of sub normal = � Let m1, m2 are slopes of the tangents to the

curve at the common point of intersection.i) Two curves touch each other at theirpoint of intersection if m1= m2.ii) Two curves intersect orthogonally attheir point of intersection if m1m2= -1.

� For Max.or Min. value of y = f(x) at x= a.

I) = 0 and if < 0 at x = a, then

f(a) is local maxima.

II) = 0 and if > 0 at x = a, then

f(a) is local minima.� Euler's Theorem:If u=f (x,y)is homoge-

nous function of nth degree then

i)

ii)

iii)

iv)

v) for trigonometric function 'u '

vi) For exponential function ' u '

� Leibniz's theorem: If f(x), g(x) arefunctions having nth derivatives then

Integration - Important Formulae

�

(for n not equal to -1)

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

� For n N, if In =






� For n N, if Im, n=

� For n N, If In =

� The area of the triangle formed by thetangent and normal at P(x1,y1) and x-axis(m is the slope of tangent) is

sq. units.

� The area of the triangle formed by the tang-ent and normal at and y-axis (m is the

slope of tangent) is sq. units.

� The area of the ellipse is πab

sq. units.� The area between the parabolas y2=4ax

and x2=4by is 16ab/3 sq. units.� The area between the parabola y2=4ax and

the line y= mx is 8a2/3m3 sq. units.� The area between the parabola y2=4ax and

its latusrectum is 8a2/3 sq. units.� The area enclosed by the curve

is 3πab/8 sq. units.

� The area of the asteroid x2/3+y2/3=a2/3 is3πa2/8 sq. units.

� The area bounded by and thecoordinate axes is a2/6 sq. units.

� The area enclosed between one arc of thecycloid, x=a(θ+sinθ), y=a(1-cosθ) and itsbase is 3πa2 sq. units.

� Trapezoidal Rule: Let y=f(x) be givenfunction and for equally spaced (n+1)arguments and x =a, a+h, a+2h,….+a+(n-1)h, a+nh=b and y0= f(x0), y1=f(x1),....,yn-

1= f(xn-1)in [a, b] then

� Simpson's Rule: Let y=f(x) be givenfunction and for equally spaced (n+1)arguments x=a, a+h, a+2h, ...a+(n-1)h,a+nh = b and y0=f(x0), y1=f(x1),... yn-1 =f(xn-1), yn=f(xn) here [a, b] is divided into nsubintervals where n is even then

� Differential Equation:An equation containing an independentvariable, dependent variable and differe-ntial coefficient of dependent variable withrespect to independent variable is called adifferential equation

Eg:1. 2.

� Ordinary differential equation: Adifferential equation involving derivativeswith respect to single independent variableis called ordinary differential equation.

� Eg:

� Partial differential equation: A different-ial equation involving at least 2 independe-nt variables and partial derivatives with re-spect to either of these independent variab-les is called a partial differential equation.

Eg: . 2∂ ∂+ =∂ ∂u u

x y ux y

3 sin= +dyx x

dx

.∂ ∂+ =∂ ∂u u

x y ux y

22

25 6+ + =d y dy

y xdx dx

( )2 4 22 ... ny y y −+ + + +

( ) ( )0 1 3 14 ...3

b

n na

hydx y y y y y −= + + + + +∫

( ) ( )0 1 2 12 ...2

b

n na

hydx y y y y y − = + + + + + ∫

x y a+ =

2 23 3

1 + =

x y

a b

2 2

2 21+ =x y

a b

21

1 1

2+y m

m

( )2 21 11

2

+y m

m

( ) ( ) 1log , then log .n n

n nx dx I x x n I −= −∫

1 1

2,

1.

m n

m n

Sin xCos x mI

m n m n

− +

−−= − +

+ +

1 1

, 2

1.

m n

m n

Sin xCos x nI

m n m n

+ −

−−= +

+ +

, thenm nnSin xCos x dx I∫

2

2

sec 2.

1 1

n

n

Co x Cotx nI

n n

−

−− −= +

− −

sec , thennnCo x dx I∫

2

2

tan 2.

1 1

n

n

Sec x x nI

n n

−

−−= +

− −

, thennnSecx x dx I∫

1

2

tan

1

n

n

xI

n

−

−= −−

tan , thennnx dx I∫

1

2

1n

n

Cos xSinx nI

n n

−

−−+

, thennnCos x dx I∫

1

2

1n

n

Sin xCosx nI

n n

−

−− −= +

, thennnSin x dx I∫

1, .ax

n ax nn n

e nx e dx then I x I

a a −= −∫

1 1 2 2. hx x

Cosh dx xCos x a ca a

− − = − − + ∫

1 1 2 2. nhx x

Sinh dx xSi x a ca a

− − = − + + ∫

1 1 2cos 1Cosh x dx x h x x c− −= − − +∫

1 1 2. 1Sinh x dx xSinh x x c− −= − − +∫

1 1 1sec . secCo x dx xCo x Cosh x c− − −= + +∫

1 1 1.Sec x dx xSec x Cosh x c− − −= − +∫

( )1 1 21. log 1

2Cot x dx xCot x x c− −= + + +∫

1 1 2. 1Cos x dx xCos x x c− −= − − +∫

1 1 2. 1Sin x dx xSin x x c− −= + − +∫( ) ( ) ( )log cos sina bx c b bx c k+ + + +

( )( )2 2log

xx a

a Cos bx c dxa b

+ =+∫

[ ](log ). ( ) ( )a Sin bx c bCos bx c K+ − + +

2 2( )

(log )

xx a

a Sin bx c dxa b

+ =+∫

[ ]( ) (aCos bx c bSin bx c K+ + + +

2 2( )

axax e

e Cos bx c dxa b

+ =+∫

[ ]( ) (aSin bx c bCos bx c K+ − + +

2 2( )

axax e

e Sin bx c dxa b

+ =+∫

[ ]'( ) ( ) ( )xf x f x dx xf x c+ = +∫

'( ) ( )( )

axax f x e f x

e f x dx ca a

+ = + ∫

[ ]( ) '( ) ( )x xe f x f x dx e f x c+ = +∫

1 1log log

1 1

nn x

x xdx x cn n

+ ⋅ = − + + + ∫

( ) ( ) ( ) ( ) ( ) ( )0

. .n

n r rnrn

r

f x g x C f x g x−

== ∑

.1u u

x y nx y

∂ ∂+ =∂ ∂

u u nux y

x y u

∂ ∂+ =′∂ ∂

( )2 2

21

u u ux y n

x y yy

∂ ∂ ∂+ = −∂ ∂ ∂∂

( )2 2

22

1u u u

x y nx y xx

∂ ∂ ∂+ = −∂ ∂ ∂∂

( )2 2 2

2 22 2

2 1u u u

x xy y n n ux yx y

∂ ∂ ∂+ + = −∂ ∂∂ ∂

u ux y nu

x y

∂ ∂+ =∂ ∂

2

2

d y

dx

dy

dx

2

2

d y

dx

dy

dx

0y m

0y

m

20 1y m+

( )0 0,x y

dy

dx

20 1y m

m

+

0

1

'( )f x

−

( )0 0,x y

dy

dx

100dy

y×100

y

y

δ ×

y

y

δ

2

2.

dv d s dvv

dt dsdt= =

ds

dt

( ) 1log , 0e

dx x

dx x= ≠

( ) 1log log ,a 1a a

dx e

dx x= ≠

( ) x=exde

dx

d

dx

-1

2

1cosec x = , 1

1

dx

dx x x

−>

−

-1

2

1sec x = , 1

1

dx

dx x x>

−

-12

1cot x = ,

1

dx R

dx x

−∈

+

-12

1tan x = ,

1

dx R

dx x∈

+

-1

2

1cos x = , 1

1

dx

dx x

−<

−

-1

2

1sin x = , 1

1

dx

dx x<

−

d

dx

d

dx

d

dx

d

dx

du

dx

d

dx

d

dx

du

dx

d

dx

d

dx

| || | , x 0

d xx

dx x= ≠

' '1 2 1 2( ( ) ( ) ...) ( ) ( ) ....

df x f x f x f x

dx± ± = ± ±

( ( )) ( )d d

cf x c f xdx dx

=

d

dx

1( ) ( )n ndax b n ax b

dx−+ = +

n n-1(x ) = n x , (x) = 1 x Rd d

dx dx∀ ∈

/, 0

/

dy dy dt dx

dx dx dt dt= ≠


9

ALGEBRACALCULUSMATHEMATICS

If the roots of ax2+bx+c=0 are 1, c/a� Order of a differential equation: The

order of a differential equation is the orderof the highest derivative appearing in it.

� Degree of a differential equation: Thedegree of a differential equation is thehighest derivative which occurs in it, afterthe differential equation has been madefree from radicals and fractions as for asthe derivatives are concerned.

Order=1, Degree=1

Order=2, Degree=1

Order=2, Degree=3

� Solution of the differential equation: Arelation between the variables with out de-rivatives which satisfy the given differen-tial equation is called a solution of thegiven differential equation.

� Type(i): General form

. Then is the

integrating factor and the solution is

Type (ii): General Form:

Then integrating factor is . Then

solution is

� Bernoulli's differential equation:Type: 1: General form

n∈R, is called Bern-

oulli's equation in y. Dividing by yn and

substituting convert into linear

differential equation.

Type 2: General form .

n∈R is called Bernoulli equation in x.

Standard results

� The solution of the equationxdy+ydx=0 is xy=C

� The solution of the equation

is


is


is


is

� The solution of is





MODEL QUESTIONS

1.

1) 2) 3) 4)

2. The focal length of a mirror is given by

If equal errors α are made in

measuring u and v then the relative errorin f is

1) 2)

3) 4)

3. If the curves 4x2 + 3y2 = 1 and cx2 + 5y2

= 1 intersect orthogonally then c =

1) 2) 3) 4)

4.

1) tan-1x - tan-1x3 + c

2) tan-1x - tan-1(x3)+c

3) tan-1x+tan-1(x3)+c

4) tan-1(x)+ tan-1(x3)+c

5.

1) 2) 3) 4)

6.

1) 2) 3) 4)

7. The solution of the differential equation

is

1) 2)

3) 4)

KEY: 1) 4, 2) 2, 3) 2, 4) 4, 5) 3, 6) 2, 7) 1

ALGEBRA

SYNOPSISDon't use formulae to solve problems in topics

which are logic-oriented, such as permutations

and combinations, location of roots of a quadrat-

ic, geometrical application of complex numbers.

Except functions all other problems are easy

but sometimes one or two lengthy problems

where more calculations are required are given.

In two years of Intermediate Algebra (Exce-

pt probability and vector algebra) around 24%

of questions will appear in EAMCET out of wh-

ich only five or six questions will be tricky are

on basic concepts. By thorough in TELUGU

academy text book (IPE) can score more.

QUADRATIC EXPRESSIONS ANDTHEORY OF EQUATIONS

� If the roots of ax2+bx+c=0 are 1, c/a then

a+b+c=0.

� If the roots of ax2+bx+c=0 are in ratio m:n

then mnb2=(m+n)2ac

� If one root of ax2+bx+c=0 is square of the

other then ac2+a2c+b3= 3abc

� If the two roots are negative, then a, b, c will

have same sign

� If the two roots are positive, then the sign of

a, c will have different sign of 'b'

� f(x)=0 is a polynomial then the equation

whose roots are reciprocal of the roots of f(x)

= 0 is increased by 'K' is

f(x-K)=0, multiplied by K is f(x/K)=0

� Three roots of a cubical equation are A.P, they

are taken as a-d, a, a+d

� Four roots in A.P, a-3d, a-d, a+d, a+3d

� If three roots are in G.P are taken as

roots

� If four roots are in G.P are taken

as roots

� For ax3+bx2+cx+d=0

i)

ii)

iii)

iv)

v) In axn+bxn-1+cxn-2.....=0 to eliminate

second term roots are diminished by

� A polynomial cannot have more positive ro-

ots than there are changes of sign in f [x] and

cannot have more negative roots than there

are changes in f (-x).

Binomial Theorem and Partial Fractions

� Number of terms in the expansion

(x1+x2+...+xr)n is n+r-1Cr-1

� In

� For independent term is

� In above , the term containing xs is

� Coefficient of xn in (x+1) (x+2).... (x+n)=n

� Coefficient of xn-1 in (x+1) (x+2).... (x+n) is

� Sum of coefficients of even terms is equal to

� Sum of coefficients of odd terms is equal to

� For (x+y)n, if n is even then only one middle

term that is term.

� For (x+y)n, if n is odd there are two middle

terms that is term and term.

� In the expansion (x+y)nif n is even greatest

coefficient is

� In the expansion (x+y)n if n is odd greatest

coefficients are

if n is odd

�

� Partial fractions of

�

�

�

�

where A=

� In Number of Rational terms are =

Hint:

Exponential and Logarithmic Series

�

�

�

�

�

�

�

1

11

!n

en

∞

== −∑

( ) ( )0 1 2

1 1 1

! 1 ! 2 !n n n

en n n

∞ ∞ ∞

= = == = =

− −∑ ∑ ∑

( )1

0

11 1 1 1 11 ....

1! 2 ! 3! 4! 5! !

n

n

en

∞−

=

−= − + − + − ∞ = ∑

0

1 1 11 ..........

1! 2 ! !n

en

∞

== + + + ∞ = ∑

( )2 1

0

2 2sinh2 1 !

n

n

xx

n

+∞

=

= =+∑

3 5

2 ........1! 3! 5!

x x x x xe e−

− = + + ∞ =

( )2

1

2 2cosh2 !

n

n

xx

n

∞

=

= =∑

2 4

2 1 ........2! 4!

x x x xe e−

+ = + + + ∞ =

2 3

0

, 1 .......1! 2! 3! !

nx

n

x x x xx R e

n

∞

=

∀ ∈ = + + + + ∞ = ∑

241 1

5,7LCM

+ ⇒

( )2475 3 2+

( )( )( )

f cC

c a c b=

− −

( )( )( )

( )( )( )

; ;f a f b

Ba b a c b a b c

=− − − −

( )( )( )( )

f x A B C

x a x b x c x a x b x c= + +

− − − − − −

( )( ) 2 2 2 2 2 22 2 2 2

1 1 1 1

b a x a x bx a x b

= − − + ++ +

( )( ) ( ) ( )2 2

f x A B C

x a x bx a x b x a= + +

− −− − −

( )( )1 1 1 1

x a x b a b x a x b = − − − − − −

( )( )1 1 1 1

x a x b b a x a x b = − + + − + +

( )223 1

;4

n nn

+=∑

( ) ( )( )21 1 2 1; ;

2 6

n n n n nn n

+ + += =∑ ∑

1 1

2 2

,n nn nC C− +

2

nnC

3

2

thn +1

2

thn +

12

thn +

( ) ( )1 1

2

f f+ −

( ) ( )1 1

2

f f− −

( )1

2

n n +

1np s

p q

− ++

1np

p q+

+

n

p

q

bax

x

+

( ) 1 1,

n r

r

T n rx a

T r+ − ++ =

b

na

−

3 3 3 31 1 2 33 3s s s sα β γ+ + = − +

4 4 4 4 2 21 1 2 1 3 24 4 2s s s s s sα β γ+ + = − + +

2 2 2 21 22s sα β γ+ + = −

1 2 33s s s= −( )( )2 3α β αβ βγ γα α β γ αβγ∑ = + + + + − =

3

3, , ,

a aar ar

r r

, ,a

a arr

10f

x

=

cos2

x yc x

+ = sin

2

x yc x

+ = −

tan2

x yc x

+ = + tan

2

x yc x

+ = −

1cosdy

x ydx

− = +

8

642

7

647

8

6935

693

/26 5

0

sin cosx xdxπ

=∫

250 2200 2

150 2100 2

100

0

1 cos 2x dxπ

− =∫

1

3

1

3

4

6

1

1

xdx

x

+ =+∫

60

3360

7−

60

7

60

47

1 1 1

u vα −

1 1

u vα −

1 1

u vα +

2

α

1 1 2.

v u f− =

1

3−

2

2−2

3

1

3

2 20

1 1

sinxLt

x x→

− =

2 2log + =x y C

2 2 0xdx ydy

x y

+ =+

log =xC

y0

− =ydx xdy

xy

log =yC

x0

− =xdy ydx

xy

log =xy C0+ =xdy ydx

xy

1tan− =xC

y2 2 0− =+

ydx xdy

x y

1tan− =yC

x2 2 0− =+

xdy ydx

x y

=ye

Cx

20

− =y yxe dy e dx

x

=yc

x20

− =xdy ydx

x

=xc

y2 0− =ydx xdy

y

( ) ( )+ = ndxP y x Q y x

dy

1

1− =

nZ

y

( ). ( ).+ = ndyP x y Q x y

dx

( ) ( ) ( ). .∫ ∫= +∫

P y dy P y dyx e Q y e dy c

( )∫e

p y dy

( ) ( ),.dx

P y x Q ydy

+ =

( ) ( ) ( ). .∫ ∫= +∫

P x dx P x dxy e Q x e dx c

( )∫e

p x dx( ). ( )+ =dyP x y Q x

dx

5/32 2

25 5

+ =

dy d y

dx dx

22

2 2 2 sin + + =

d y dyy x

dx dx

2

2

1

1

+=+

dy y

dx x


10

PERMUTATIONSMATRICESMATHEMATICS

Sum of even divisiors=...�

�

Matrices and Determinants

� A square matrix in which every element is

equal to '0', except those of principal diagonal

of matrix is called as diagonal matrix

� A square matrix is said to be a scalar matrix if

all the elements in the principal diagonal are

equal and other elements are zero's

� A diagonal matrix A in which all the elements

in the principal diagonal are 1 and the rest '0'

is called unit matrix

� A square matrix A is said to be Idempotent

matrix if A2=A,

� A square matrix A is said to be Involuntary

matrix if A2=I

� A square matrix A is said to be Symmetric

matrix if A=AT

� A square matrix A is said to be Skew

symmetric matrix if A=-AT

� A square matrix A is said to be Nilpotent

matrix If their exists a positive integer n such

that An=0 'n' is the index of Nilpotent matrix

� If 'A' is a given matrix, every square matrix

can be expressed as a sum of symmetric and

skew symmetric matrix where

Symmetric part

Skew-symmetric part

� A square matrix 'A' is called an orthogonal

matrix if AAT=I or AT=A-1

� A square matrix 'A' is said to be a singular

matrix if det A = 0

� A square matrix 'A' is said to be non singular

matrix if det A≠ 0

� If 'A' is a square matrix then det A=det AT

� If AB=I=BA then A and B are called inverses

of each other

� (A-1)-1=A, (AB)-1=B-1A-1

� If A and AT are invertible then (AT)-1 = (A-1)T

� If A is non singular of order 3 , A is

invertible, then

� If

if ad-bc ≠ 0

� (A-1 )-1 = A, (AB)-1 = B-1 A-1, (AT)-1 = (A-1)T

(ABC)-1 = C-1 B-1 A-1

� If A and B are two non-singular matrices of

the same type then

i) Adj (AB) = (Adj B) (Adj A).

ii)|Adj (AB)|=|Adj A| |Adj B|=|Adj B| |Adj A|

� To determine rank and solution first convert

matrix into Echolon form

i.e

.

Echolon form of

No of non zero rows = n = Rank of a matrix

If the system of equations AX = B is

consistent if the coeff matrix A and

augmented matrix K are of same rank

Let AX = B be a system of equations of 'n'

unknowns and ranks of coeff matrix = r1and

rank of augmented matrix = r2

If , then AX = B is inconsistant, i.e. it

has no solution

If then AX=B is consistant, it has

unique solution

If then AX=B is consistant and it

has infinitely many number of solutions

Standard Results

� The determinant of a unit matrix is 1

�

�

�

�

�

�

�

=(a-b)(b-c)(c-a)(a+b+c)

�

= (a-b)(b-c)(c-a)(ab+bc+ca)

Permutations and Combinations

� The number of ways of dividing 'mn' things

into two groups of 'm' things and 'n' things is

� FOR a number 2a3b5c

1) Number of divisors = (a + 1) (b +1) (c +1)

2) Number of divisors excluding unity

(a +1) (b +1) (c +1) -1

3) Excluding number & unity =

(a+1) (b +1) (c +1) - 2

i.e proper divisors or non-trival solutions

4) Odd divisors = (b +1) (c +1)

5) Even divisors=total divisors - Odd divisors

6) Sum of divisors

=

7) Sum of even divisors =

sum of total divisors - sum of odd divisors

� Number of diagnols in a polygon of 'n'sides is

=

� The Number of ways of dividing '2k' things

into two equal groups is

� Number of ways of arranging A1,A2, A3, A4 ,...

A10 such that

1. A1 always before A2

2. A1 before A2, A2 before A3=

GRAPH OF SOME REAL FUNCTIONS

� Quadratic Functions:

If a, b, c are fixed real numbers, then the

quadratic function is expressed as

Which is equation of a parabola, downward if

a<0 and upward if a>0 and vertex at

The domain of f(x)=R

The range of f(x) is if a<0

and if a>0

� Modulus function (of Absolute value funct-ion): Modulus function is given by

y=f(x)=|x|, where |x| denotes the absolute

value of x, that is

Domain of f(x) =R. Range of f(x) =

� Signum function:Signum function is defined as follows

or

Symbolically, signum function is denoted by

sgn(x). Thus y=f(x) =sgn(x) Where

Domain of sgn(x)=R, Range of sgn(x)={-

1,0,1}

� Exponential functions:Exponential function is given by

y=f(x)=ax, where a=0, a�1

The graph of the function is as shown below,

which is increasing if a>1 and decreasing if

0<a<1

Properties of exponential functions:1. 2.

3. 4.

� Logarithmic functions:

A logarithmic function may be given by

y=f(x)=logaX, where a>0, a ≠ 1 and x>0

The graph of the function is as shown below,

which is increasing if a>1 and decreasing if

0<a<1

Domain of f(x) = {x R: x>0} = (0, ) Range of

f(x) =R

� The greatest integer function:

( ) yyx xy xa a a= ≠1xx

aa

− =

. ,x y x ya a a x y R+ = ∀ ∈0xa x R> ∀ ∈

( )1, 0

, 00 0

1 00, 0

x if xif x

xsgm x if x

if xif x

− < ≠ = = = >=

, 0

0, 0

xif x

x

if x

≠ =

( ) , 0

0, 0

xif x

y f x xif x

≠= =

=

[0, )+∞

, 0

, 0

x if xx

x if x

≥= − <

24,

4

ac b

a

−∞

24,

4

ac b

a

−−∞

24,

2 4

b ac b

a a

−−

2 24

2 4

b ac ba x

a a

− = + +

( ) 2 , 0y f x ax bx c a y= = + + ≠ ⇒

10!

3!⇒

10!

2!⇒

(2 )!

! !2!

k

k k

( 3)

2

n n −

1 1 12 1 3 1 5 1

2 1 3 1 5 1

a b c+ + + − − − − − −

( )!

! !

m n

m n

+

2 3 2

2 3 2

2 3 2

1

1

1

a a a a bc

b b b b ca

c c c c ab

=

3 2

3 2

3 2

1 1

1 1

1 1

a a a bc

b b b ca

c c c ab

=

2

2

2

1 1

1 1 ( )( )( )

11

a a a bc

b b b ca a b b c c a

c abc c

= = − − −

2

2 2 2 2

2

1

1 1

1

a ab ac

ab b bc a b c

ac bc c

+

+ = + + +

+

1 1 1

1 1 1 (1 1/ 1/ 1/ )

1 1 1

a

b abc a b c

c

++ = + + +

+

1

1 1

1

a b c

a b c a b c

a b c

++ = + + +

+

3 3 33

a b c

b c a abc a b c

c a b

= − − −

2 2 22

a h g

h b f abc fgh af bg ch

g f c

= + − − −

1 2r r n= <

1 2r r n= =

1 2r r≠

1 2 3 4

0

0 0

A x y z

k l

=

1 2 3 4

2 3 1 2

3 2 1 0

A

=

1 1a b d bA A

c d c aad bc− −

= ⇒ = −−

1

det

AdjAA

A− =

2

TA A−=

2

TA A+=

21 11 ....

2 ! 3! 1

aa a a e e

a

+ + + −+ + + =−

2

12

!n

en

∞

== −∑


11

Ist & IInd YearQUICK REVIEWMATHEMATICS

Minimum value of a2Sin2x+b2Cosec2x is..The greatest integer function is defined as

y=f(x)=[x],where [x] represents the greatest

integer less than or equal to x. This is for any

integer n, [x] =n if ;

Properties of logarithmic functions

�

� loga(x y)=loga x+loga y

�

� loga xy= yloga x

�

� loga x.logy a=logy x

�

� if a>1and x<ay if 0<a<1

� if a>1 and x<ay if 0<a<1

� y=f(x)=loge x(or 1n x) is a particular function

form of above.

MODEL QUESTIONS

1. The number of quadratic expressions with

the coefficients drawn from set {0,1,2,3} is

1) 27 2) 36 3) 48 4) 64

2. The number of terms in the expansion

(x + y + z)n is

1) n 2) n + 1 3) Σn 4) Σ (n+1)

3.

1) e 2) e/2 3) e/3 4) e/4

4. The reciprocal equation is

1) 2x3+4x2+2x+4= 0

2) 2x3+4x2+4x+2= 0

3) 2x3+2x2+4x+4= 0

4) 2x3+4x2+2x+2= 0

5. If a, b, c are positive and not all equal,

then is

1) < 0 2) ≤ 3) ≥ 4) > 0

KEY: 1) 3, 2) 4, 3) 2, 4) 2, 5) 1

QUICK REVIEWPROBABILITY

� If A is an event in a sample space S, then the

ratio is called the odds favour to

A and is called the odds against

to A.

� Addition theorem on probability: If A, B

are two events in a sample space S, then

� If A, B are two events in a sample space S,

then and

� If A, B are two events in a sample space S

such that , then

P(BA) =

� If n letters are put at random in the n

addressed envelopes, the probability that

i) all the letters are in right envelops = 1/n!

ii) at least one letter may be in wrongly

placed = 1-1/n!

iii) all the letters may be in wrong envelops =

VECTOR ALGEBRA

� If are unit vectors then unit vector

along bisector of ∠AOB is

or

� i) The component of is

ii) the projection of is

iii) the projection of on a vector perpendi-

cular to' ' in the plane generated by a, b is

� The perpendicular distance from a point P to

the line joining the points A, B is

� The shortest distance between the skew lines

r = a +s b and r = c+ td is

�

�

are called reciprocal system of vectors

TRIGNOMETRY

� Range of Sin x are Cos x are ± 1� Range of aCosx+bSinx+c is

� Minimum value of a2Sin2x+b2Cosec2x is 2ab

� Minimum value of a2Tan2x+b2Cot2x is 2ab

� Minimum value of a2Cos2x+b2Sec2x is 2ab

� Minimum value of a2Sec2x+b2Cosec2x is

(a+b)2

� If then

� If then

� If then

� If then

� If x r1 = yr2=z r3 then a:b:c= y+z:x+z:x+y

� If Cot A/2: Cot B/2: CotC/2 =x:y:z then

a:b:c= y+z:x+z:x+y and Cot A: Cot B: CotC

=x:y:z then a:b:c=

If Cot A/2,Cot B/2,CotC/2 are in A.P then

a,b,c are in A.P

� For equilateral triangle a=b=c=1 and

A=B=C=600,

by substituting these values general notation

problems can be solved

� Length of altitude from C to AB in ∆ABC is

� Area of polygon of n sides inscribed in a

circle with radius r is

� Radius of inscribed triangle for n sided

polygon of side 'a'is r =

�

and

3D COORDINATE SYSTEM

� If (l1,m1,n1) (l2,m2,n2) are d.c's of two lines

which include an angle θ then

i) d.c's of internal angle bisector is

ii) d.c's of external angle bisector is

� If projections of a line of length 'd'

i) on co-ordinate axis are d1,d2,d3 then

d12+d2

2+d32=d2

ii) on co-ordinate planes d1,d2,d3 then

d12+d2

2+d32=2d2

� If (x1,y1) and (x2,y2) are the vertices of base of

an isosceles triangle and the angle made by

sides with the base is θ then third vertex is

� If A= (a, b) B = (-a, b) then the locus of P such

that PA + PB = k or is

or y = b if k = 2a.

� If A = (a, b) B = (a, -b) then the locus of p

such that PA + PB = k or is

or x = a if k=2b.

� The angle of rotation of the axes so that the

equation ax+by+c=0 may be reduced to the

form

i. X = constant is

ii. Y = constant is

� Area of the parallelogram formed by a1x+

b1y+c1=0, a2x+b2y+c2=0, a1x+b1y+d1=0

and a2x+b2y+d2=0 is.

� ax+by+c= 0 and (ax+by)2-tan2α(bx-ay)2=0

form an isosceles ∆le with equal angles as α.

1) If tan2α = 3 → equilateral triangle

2) If tan2α=1→ Right angled Isosceles

triangle

3) If tan2α < 1→ Isosceles, obtuse angled

triangle

� The condition that slopes of pair of lines

ax2+2hxy+by2=0 are in the ratio p:q is

ab(p+q)2=4h2pq .

GEOMETRY - Second year

� If l1x+m1y+n1=0, l2x+m2y+n2=0 are conju-

gate lines with respect to the circle S = 0 then

r2(l1l2+m1m2)=(l1g+m1f-n1)(l2g+m2f-n2)

� The area of the triangle formed by the two

tangents from (x1,y1) to S≡ x2+y2+2gx+2fy

+c=0and chord of contact is .

� The focal distance of the point P (x1, y1) on

the parabola

i) y2 =4ax is SP = |x1+a|

ii) x2 =4ay is SP = |y1+a|

� Equation of tangent at 't' on the parabola

y2=4ax is yt = x+at2.

� Equation of the normal at 't' on the parabola

y2=4ax is xt + y = 2at+at3.

� Equation of the tangent to y2=4ax having

slope 'm' is y =mx+a/m.

� Equation of the normal to y2=4ax having

slope 'm' is y=mx-2am-am3 and foot of the

normal is (am2,-2am).

� If t1and t2 are the ends of the focal chord of

the parabola y2=4ax then t1t2 = -1.

� The condition for the lines l1x+m1y+n1=0

and l2x+m2y+n2=0 to be conjugate w.r.t

i) y2=4ax is l1n2+l2n1=2am1m2.

ii) x2=4ay is m1n2+m2n1=2al1l2.

� If the normal at 't1' of the parabola meets the

parabola again at 't2', then

� If l1x+m1y+n1=0 and l2x+m2y+n2=0 are two

conjugate lines, then a2l1l2+b2m1m2=n1n2.

� Equation of the normal at 'θ ' on the

ellipse is

� The condition for the line lx +my +n=0

is a normal to an ellipse is

2 2 2 2 2

2 2 2

( ).

a b a b

l m n

−+ =

2 2

2 21

x y

a b+ =

2 2.cos sin

ax bya b

θ θ− = −

2 2

2 21

x y

a b+ =

2 1

1

2.t t

t= − −

3

211

211

( )r S

r S+

1 1 2 2

1 2 2 1

( ) ( )d c d c

a b a b

− −−

1 aTan

b− −

1 bTan

a−

2 2

2 2 2

4( ) 41

4

x a y

k b k

− + =−

PA PB k− =

( )22

2 2 2

441

4

y bx

k k a

−+ =

−

PA PB k− =

( ) ( )1 2 1 2 1 2 1 2tan tan,

2 2

x x y y y y x xθ θ+ ± − + −

m

1 2 1 2 1 2, ,2sin 2sin 2sin2 2 2

l l m m n nθ θ θ

− − −

1 2 1 2 1 2, ,2cos 2cos 2cos2 2 2

l l m m n nθ θ θ

+ + +

2log log2

i ii e

π π= =( )2 2iii e i e

π π−= =

cot2

a

n

π

2nr Tann

π

1 2

1 2

2r r

r r+

1 2 3 1 2 3

3,

23r r r h h h= = = = = =

3 1 1, , ,

4 2 3 3r R r∆ = = =

: :y z x z x y+ + +

2cos 1 sin 1 sin2

AA A= + + −

2sin 1 sin 1 sin2

AA A= + − −

4 2 4

Aπ π− < <

2cos 1 sin 1 sin2

AA A= − + + −

2sin 1 sin 1 sin2

AA A A= − + − −

5 7

4 2 4

Aπ π< <

2cos 1 sin 1 sin2

AA A= − + − −

2sin 1 sin 1 sin2

AA A= − + + −

3 5

4 2 4

Aπ π< <

2cos 1 sin 1 sin2

AA A= + − −

2sin 1 sin 1 sin2

AA A= + + −

3

4 2 4

Aπ π< <

2 2 2 2,c a b c a b − + + +

1 1 1, ,b c c a a b

a b cabc abc abc

× × ×= = =

2

, , , ,a b b c c a a b c × × × =

, ,a c b d

b d

− ×

AP AB

AB

×uuur uuur

uuur

( )2

.b a ab

a−

ab

( )2

.b a a

ab on a

( ).b a a

ab on a

( )ˆ

ˆ

a b

a b

+±

+

$

$

a b

a b

++

,a b

1 1 1 ( 1)1 ......

1! 2! 3! !

n

n

−− + − + +

( )

( )

p A B

p A

∩( ) 0P A ≠

( ) ( ) ( )P A B P A P A B− = − ∩( ) ( ) ( )P B A P B P A B− = − ∩

( ) ( ) ( ) ( )P A B P A P B P A B∪ = + − ∩

( ) : ( )P A P A

( ) : ( )P A P A

a b c

b c a

c a b

1 1 2 1 2 3 1 2 3 4....

2! 3! 4! 5!

+ + + + + ++ + + + =

log ya x y x a< ⇒ <

log ya x y x a> ⇒ >

log ya x y x a= ⇒ =

log logkm

aa

mx x

k=

log log loga a ax

x yy

= −

loga xa x=

[ ]1x x x− < ≤1n x n≤ < +

� ��


12

SUCCESS TIPSQUICK REVIEWMATHEMATICS

Most of the students follow 'CMP' order� The equation of the normal to the hyperbola

at P(x1,y1) is

� The tangents of a hyperbola which touch the hyperbola at

infinity are called asymptotes of the hyperbola. The equations

of the asymptotes of the hyperbola

S = 0 are

� The angle between the asymptotes of the hyperbola S = 0 is

or 2 Sec-1e

� The equation of the normal at P(θ ) on the hyperbola

is

� The condition that the line lx+my+n=0 to be a normal to the

hyperbola is

� Vertex of conic section

CALCULUS

� Maximum value of acos2x+bsin2x is 'a' and Minimum value =

b (if a>b )

� Minimumvalue of a tanx+bcotx is and attained

at x=tan-1

� Minimumvalue of a2sec2 x+b2cosec2x is (a+b)2 and is and

attained at x=tan-1

� The sum of two numbers is k .if the sum of their squares is

minimum then numbers are k/2, k/2

� The maximum value of (1/x)x is e1/e

� The maximum value of (x)1/x is e1/e

� The minimum value of (x)x is e-1/e

� The area of the greatest rectangle inscribed in

an ellipse is 2ab

� Maximum value of area of ∆le with vertices (a,0) (acosθ,

bsinθ), (acosθ, −bsinθ) is

� Two sides of a ∆le are given. If area of ∆le is maximum then ∆le

is right angled. The sum of hypotenuse and one side of right

angled triangle is given then the angle between them is if area

is π/3 maximum

�

�

�

�

�

�

�

�

� If a > 0

� If a > 0

�

�

�

�

�

ALGEBRA

� If ncr-1ncr

ncr+1 are in A.P (n-2r)2= n+2� The first negative term in the expansion of (1+x)p/qis equal

to [p/q] +3

� (i) (ii)

(iii) (iv) (v)

� If A is a n × n non- singular matrix, thena) A(AdjA)=|A|Ib) Adj A = | A| A -1c) (Adj A) -1 = = Adj (A -1)d) Adj AT = (Adj A)T

e) Det (A-1 ) = ( Det A)-1

f) |Adj A| = |A| n -1

g) lAdj (Adj A ) l= |A|(n - 1)2

h) For any scalar 'k'� If x1+x2+x3+.....+xr=n then

1. Number of non-negative solutions = (n+r-1) cr-1

2. Number of positive solutions = (n-1) cr-1

� The power raised to prime number 'p' in the number (n!)⇒[n/p]+[n/p2]+...Ex: Power of 2in 50!= [50/2]+[50/4]+[50/8]+[50/16]+ [50/32]+[50/64]=47

4

1

15!n

ne

n

∞

==∑

3

1

5!n

ne

n

∞

==∑

2

1

2!n

ne

n

∞

==∑

1

1

( 1)!n

ne

n

∞

=

− =−∑

0 !n

ne

n

∞

==∑

( )2

b b

a a

x a b xdx dx b a

b x x a

π− −= = −

− −∫ ∫

1

( )( )

b

a

dxx x a b x ab

π=

− −∫

1

( )( )

b

a

dxx a b x

π=− −∫

2( )( ) ( )8

b

a

x a b x dx b aπ

− − = −∫

22

220 0

1 sec

(sec n ) 1( 1)n

x ndx dx

x ta x nx x

∞

= =+ −+ −

∫ ∫

2 20

sinax be bx dx

a b

∞− =

+∫

2 20

cosax ae bx dx

a b

∞− =

+∫0 0

tan sin ( 2)

sec tan 1 sin 2

x x x xdx dx

x x x

π π π π −= =

+ +∫ ∫

2

20 0

sin sin

sec cos 41 cos

x x x xdx dx

x x x

π π π= =

+ +∫ ∫

2

2 2 2 20

1

2sin cosdx

aba x b x

π π=

+∫

2

2 2 2 20

1

2sin cosdx

aba x b x

π

π=

+∫

( )2

0

12 log 2 1

sin cosdx

x x

π

= ++∫

2 2

0 0

og(sin ) log(cos ) 22

l x dx x dx log

π π

π= = −∫ ∫

2

4

og(1 cot ) 28

l d log

π

π

π+ θ θ =∫

4

0

og(1 tan ) 28

l d log

π

π+ θ θ =∫

3 3

4

ab

2 2

2 21

x y

a b+ =

b

a

b

a

2 ab

1 ( ) ,1

l leCos is

r eθ α α = + − +

2 2 2 2 2

2 2 2

( )a b a b

l m n

+− =2 2

2 21

x y

a b− =

2 2

sec tan

ax bya b

θ θ+ = +

2 2

2 21

x y

a b− =

12b

Tana

−

0.x y

a b± =

2 22 2

1 1

ya x ba b

x y+ = +

2 2

2 21

x y

a b− =

� Focus on solving as many problems as you can, rather than

just reading theories, formulae, and solutions

� More than rigid reliance on rules without un-derstanding

(rule-oriented study) rely on an understanding of mathemati-

cal concepts and flexibility in problem solving (concept

oriented study)

� Master the fundamentals, as most questions are designed to

evaluate the candidates' clarity of fundamental concepts and

the ability to apply these concepts to problem solving

� Don't be in a rush to solve problems. In EAMCET. Both spe-

ed and strikerate matter. You need to be quick as well as acc-

urate to achieve high scores. High speed with low accuracy

can actually ruin your results.

� Master the fundamentals, practise a lot, and manage time

well. Most of the students follow CMP (Chemistry Maths

Physics) order

� CHEMISTRY -35 min, MATHEMATICS -75min and PHY-

SICS- 40 min remaining time for marking and left difficult

and lengthy questions. Practice rounding off method, identi-

fy the difficult and lengthy problem and round off move to

other problem because scoring is important.

SUCCESS TIPS

sinx R [-1,1] Cosx R [-1,1] tanx R – {(2n +1)? /2, n ? z} R cotx R – n ? /2 R secx R – {(2n +1)? /2, n ? z} ( , 1] [1, )−∞ − ∪ ∞ cosecx R – n? /2 ( , 1] [1, )−∞ − ∪ ∞ sin-1x [-1,1]

,2 2π π −

cos-1x [-1,1] [0,? ] tan-1x R

,2 2π π −

cot-1x R (0,? ) sec-1x ( , 1] [1, )−∞ − ∪ ∞

0, ,2 2π π ∪ π

cosec-

1x ( , 1] [1, )−∞ − ∪ ∞

, 0 0,2 2π π − ∪

ax, a>0 R ( )0, ∞

ex R ( )0, ∞

logax ( )0, ∞ R

|x| R )0,∞

[x] R Z x-[x] R )0,1

CALCULUSFunctions Domain Range

�

�

�

�

�

�

� �

Function Nth derivative 1

ax b+ ( )

( )

1

1

1 !n n

n

n a

ax b

−

+

−+

log ax b+ ( ) ( )( )1 1 !n n

n

n a

ax b

− −+

( )cos ax b+ sin

2n n

a ax bπ + +

( )sin ax b+ = cos

2n n

a ax bπ + +

( )sinaxe bx c+ ( )2 2 1sin tann

ax ba b e bx c n

a + + +

( )cosaxe bx c+ ( )2 2 1cos tann

ax ba b e bx c n

a− + + +

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