MATHEMATICS FUNDAE

•A game of numbers.• It is also a tool used in day to day life.

•Needed in Physics and Chemistry also.

•Helps in getting a good rank.

What is Maths?

HOW TO ENJOY MATHEMATICSFocus On Basics .Try To Understand The Formulae.Predict Its Application.Match With The Applications As In

Books.Think WHY? Write The Concepts Which You Are

Going To Use On The Paper Before Starting The Solution Of Problem.

Important Chapters??

TRIGONOMETRY

Trigonometric Functions and EquationsoWidely used in other chapters of mathematics.oRequires lot of practice.oIt has only formulae.oTry to see some pattern in the formulae and then learned.

Inverse Trigonometric Functions

Very conceptualRequires the use of graphsShould be studied after studying

functions.

• Topics are interlinked

• Visualization of problems

• Follow ‘read & draw strategy’

• Some formulae should be learned

CO – ORDINATE GEOMETRY

STRAIGHT LINENormal Form :x cos α + y sin α = p

Parametric Form : x – x1 x – x2 r

cos θ sin θ

Equation of angle bisectors a1x + b1y +c1 a2x + b2y + c2 √a1

2 + b12 √a2

2 + b22If c1, c2 > 0, then

bisector containing origin is given by +ve signACUTE & OBTUSE ANGLE BISECTOR

Use of Reflection

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 Angle between the two lines :

θ = tan-1 2 √(h2 – ab) │a + b│

Point of intersection of 2 lines :

a h g h b f g f c

ax + hy + g = 0 hx + by + f = 0

PAIR OF STRAIGHT LINES

CIRCLE (x – a)2 + (y – b)2 = r2

Equation of the tangent : •Point Form•Parametric Form•Slope Form Replace x2 xx1 y2 yy1

2x (x + x1) 2y (y + y1) 2xy (xy1 + x1y)

• Only valid for a two degree equation

Normal To A Circle

Equation for a 2nd order conic : ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

• Normal always passes through the centre of the circle.

Then normal is x – x1 y – y1

ax1 + hy1 + g hx1 + by1 + f

a h g h b f g f c

ax1 + hy1 + g , hx1 + by1 + f

Radical Axis

S – S’ = 0Perpendicular to the line joining the centres.Bisects the direct common tangents.For 3 circles, taking 2 at a time, they are concurrent.

The curve y = x3 – 3x + 2 and x + 3y = 2 intersect in points (x1,y1), (x2,y2) and (x3,y3). Then the point P(A,B) where A = Σxi and B = Σyi lies on the line

(A) x – 3y = 5 (B) x + y = 1

(C) 3x – 7 = y (D) 2x + y = 2

Conic Section

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

How a conic section is formed

PARABOLA

Focal Chord

t1t2 = -1

Length (PQ) = a*(t2 – t1)2

Tangents at P & Q will be perpendicular to each other

Length of Latus Rectum : 4 * PS * QS PS + QS

Tangents Slope Form :

Point of intersection of tangents:

( at1t2 , a(t1 + t2) )

y = mx + a m

Remembering Method :

G O A (GOA rule)

GM of at12 & at2

2 , AM of 2at1 & 2at2

i.e. at1t2 i.e. a(t1 + t2)

Equation of tangent to parabolas of different form :

Co – Normal Points Properties :

m1 + m2 + m3 = 0 y1 + y2 + y3 = 0 For the normals to be real h > 2a For the normals to be real and distinct 27ak2 < 4(h-2a)3

ELLIPSE

Tangent in slope form :

y = mx + √ (a2 m2 + b2)

- Normal in slope form :

y = mx - m (a2 – b2)

+ √ (a2 + b2m2)

Co – Normal Points

4 Normals can be drawn from any point to an ellipse.

α + β + λ + δ = (2n + 1)π

Properties:

Sin (α + β) + Sin (β + λ) + Sin (λ + α) = 0 Co – Normal Points lie on a fixed curve

called Apollonian Rectangular Hyperbola A.R.H (a2 – b2)xy + b2kx – a2hy = 0

Director Circle : Locus of the points from which perpendicular tangents can be drawn

x2 + y2 = a2 + b2

Reflection Property :

Ray passing through a focus, passes through the other focus after reflection.

HYPERBOLAAsymptotes Tangent to the hyperbola at infinity

Properties :oDifference between hyperbola and pair of asymptotes is constant.oHyperbola and its conjugate hyperbola have the same asymptotes.

x2 - y2 = 0

a2 b2

o Angle between two asymptotes is 2tan-1(b/a)o Asymptotes pass through the centre of the hyperbola.o Co – Ordinate axes are angular

bisector of the two asymptotes.o Hyperbola, Asymptotes and Conjugate Hyperbola are in A.P i.e. C + H = 2A

If normal at P (2, 1.5√3) meets the major axis

of the ellipse x2 + y2 = 1 at Q. S and S’ are

16 9

foci of given ellipse, then SQ:S’Q is(A) 8 – √7 8 + √7

(B) 4 + √7 4 – √7

(C) 8 + √7 8 – √7

(D) 4 – √7 4 + √7

DIFFERENTIAL CALCULUS

FUNCTIONSMost Important Is Concept Of DOMAIN And RANGE .

Knowledge About Some Important Functions Like LOGARATHMIC, TRIGONOMETRIC, GREATEST INTEGER etc.

Focus On Domain And Range Of These Functions.

Give Attention To COMPOSITE FUNCTIONS .

For Finding PERIOD, Be Careful About CONSTANT FUNCTION.

HOW TO SOLVE PROBLEMS IN EXAMS ?

Identify The Nature Of Function .Coordinate With Its Domain And Range.

Think Practically. Go What The Question Says…….Get The Knowledge Of Symmetry About Any Point Or Line.

Find the period of y = logCos(x) Sin(x)

Answer : 2π

LIMITCondition For EXISTANCE OF LIMIT

Remember Some Important Expansions.

L’HOSPITAL RULE Try To Simplify The Question If You Get A Hard And Tough Looking Problem. 95% Of Such Problems Becomes Easy After Simplification

In Case When x Tends To Infinity In Algebraic /Algebraic Function Be Careful About Constants If Given In Question.

CONTINUITY AND DIFFERENTIABILITYConcept Of LIMIT Should Be Clear Before Attempting The Questions Of Continuity.

Remember The Approach Of Continuity At End Points.

For Differentiability Of Function, Careful Where To Use Basic Funda And Where Direct Differntiation.

Differentiability At End Points In A Closed Or Open Interval.

Differentiability Implies Contuinity.

APPLICATIONS OF dy/dxROLLE’S THEOREM.LAGRANGE’S MEAN VALUE THEOREM.

Use Of dy/dx In Deciding The Nature Of Curve.

Try To Co-Relate Max. And Min. With The Help Of Graph.

Use nth Derivative Test.Concentrate On Maxima And Minima Of Discontinuous Function.

INTEGRAL CALULUS

INDEFINITE INTEGRALGet Some Important Results.Use By Parts Method Whenever There Is Any Scope.

Rearrangement Is Your Motto.

Get The Approach Of Some Important Forms.

DEFINITE INTEGRALSAn Easy Version Of Indefinite Integrals.

Use Of Properties Are Very Useful.No Need Of Gamma Function , Walli’s Function etc.

Maximum And Minimum Value of Integral Are Very Helpful During Exams.

Newton Method Of Differentiation Of Integrals.

For Area , First Draw The Curve.

Find Symmetrical Parts.Be Careful About Sign Of Integration.

(A) f(2012) + f(-2012) (B) f(2012) – f(-2012)

(C) 0 (D) 2012

DIFFERENTIAL EQUATIONOrder And Degree.Degree Is Defined For Polynomial Equation Only.

First Step Is To Check If Equation Can Be Solved By Rearranging.

If Equation Is Of Form f(ax+by+c) , Then Solve It By Taking ax+by+c=t.

For Homogeneous Equation , Use y=vx.

Concept Of Exact Equations.Linear Equation And Conversion Into Linear Equations.

ALGEBRA

COMPLEX NUMBERDe Moivre’s Theorem

1.(Cos θ + i Sin θ)n = Cos nθ + i Sin nθ2.If z = (Cos θ1 + i Sin θ1) (Cos θ2 + i Sin θ2) (Cos θ3 + i Sin θ3)……………..(Cos θn + I Sin θn) then z = Cos (θ1+θ2 +θ3+………+θn) + i Sin (θ1+θ2+θ3+………+θn)3.If z = r(Cos θ + i Sin θ) and n is a positive integer, then then z1/n = r1/n Cos (2kπ + θ) + i Sin (2kπ + θ) n n where k = 0,1,2,3………,(n-1)

Coni Method (Rotation Theorem)

z3 – z1 OQ ( Cos α + i Sin α) z2 – z1 OP CA . eiα BA │z3 – z1│ . eiα

│z2 – z1│

or arg z3 – z1 α z2 – z1

Co – Ordinate in terms of Complex

Equation of Straight Line :

z – z1 = z – z1

z2 – z1 z2 – z1

Circle : zz + az + az + b = 0 , Centre is ‘-a’ radius = √aa - b

SEQUENCE & SERIESIdentifying whether the sequence is A.P, G.P, H.P If, a – b a A.P b – c a a – b a G.P b – c b a – b a H.P b – c c Arithmetic Mean A = a + b 2 G2 = AHGeometric Mean G = √ab A > G > HHarmonic Mean H = 2ab a + b

Some tips : If first common difference is in A.P take

the General Term as ‘ax2 + bx +c’ and determine a, b, c by solving for known values.

Sn = 1 + 2 + 4 + 7 + 11 + ……. Tn = an2 + bn + c , a = 0.5 , b = -0.5 , c = 1 Sn = Σ Tn

If second common difference is in A.P then take the cubic expression as the General Term and solve for constants.

Solution : Sn = cn2

Tn = Sn – Sn-1

= cn2 – c(n-1)2 = c(2n – 1)Tn

2 = c2(2n – 1)2

Sn = Σ Tn

*Shortcut Method : Put n = 1 in the question

If the sum of first n terms of an A.P is cn2 , then the sum of squares of these n terms is (A) n (4n2 – 1)c2

6 (B) n(4n2 + 1)c2

3

(C) n(4n2 – 1)c2

3(D) n(4n2 + 1)c2

6

QUADRATIC EQUATION

ax2 + bx +c = 0 Conditions For A Common Root : ax2 + bx + c = 0 , a’x2 + b’x +c’ = 0 a = b = c

a’ b’ c’

Note : To find the common root between the two equations, make the same coefficient of x2 in both the equations and then subtract the 2 equations.

Graph of Quadratic Expression

f(x) = ax2 + bx + c

Location Of Roots1. If both the roots are less than k (i) D >= 0, (ii) a*f(k) > 0, (iii) k > -b 2a

2. If both the roots are greater than k (i) D >= 0, (ii) a*f(k) > 0, (iii) k < -b

2a

3. If k lies between the roots (i) D > 0 , (ii) a*f(k) < 0

4. If one of the roots lie in the interval (k1, k2) (i) D > 0,(ii) f(k1)*f(k2) < 0

5. If both the roots lie in the interval (k1,k2) (i) D >= 0

6. If k1,k2 lie between the roots (i) D > 0 (ii) a*f(k1) > 0

(iii) a*f(k2) > 0

(iv) k1 < -b < k2

2a

(iii) a*f(k2) > 0

(ii) a*f(k1) > 0

PERMUTATION & COMBINATION1. Permutation of n different things taking r at a time = nPr

2. Permutation of n things taken all at a time, p are alike of one kind, q are alike of

2nd kind, r are alike of 3rd kind, rest are different n! p! q! r!3. Number of permutations of n different things taken r at a time, when each thing

may be repeated any no. of times nr

Circular Permutation1. When anticlockwise and clockwise are treated as different :

(n – 1)!

2. When anticlockwise and clockwise are treated as same :

(n – 1)! 2

COMBINATION1. Combination of n different things taking r at a time : nCr

2. Combination of n different things taking r at a time, when k particular objects occur is:

n-kCr-k When k particular objects never occur : n-kCr

3. Combination of n different things selecting at least one of them :

nC1 + nC2 + nC3 + …………. + nCn = 2n – 1

4. If out of (p+q+r+t) things, p are alike of one kind , q are alike of 2nd kind, r are alike of 3rd kind, and t are different, then the total number of combinations is :

(p+1)(q+1)(r+1)*2t – 1

5. Number of ways in which n different things can be arranged into r different groups is :

n+r-1Pn

PROBABILITY

Different types of events like Mixed Event, Independent Events, Complimentary Events.

Learn some formulas like P(A U B) = P(A) + P(B) – P( A ∩ B) Conditional Probability. Baye’s Theorem or Inverse Probability. Binomial Theorem. Multinomial Theorem.

VECTORS AND 3D GEOMETRY

VECTORSYou Must Give More Concentration On Vectors As It Is Also Required In Physics.

You All Are Aware Of Simple Applications.

Linearly Dependent And Linearly Independent Vectors.

Combination Of DOT And CROSS In Problems.

For Solving Vector Equation, Just Try To Simplify It And Use The Conditions Given.

If Sol. Requires r In Form Of Two Vectors a & b, Take

r =λa + µb + σ(a×b)

In Case Of Three Non-Coplanar Vectors a, b And c,

r =λa + µb + σc And Then Use/Apply The Conditions Given.

3D-GEOMETRYProjection Of Segment Joining Two Points On Line.

Angle Between Two Lines.Different Forms Of Straight Line Including Vector Form.

Perpendicular Distance Of A Point From Line.

Be Careful From Skew Lines.

Shortest Distance Between Two Skew Lines(Its Better To Use Detailed Approach).

Equation Of Plane In Different Forms.

Angle Between Line And Plane.Angle Between Two Planes.

Just Try To Visualize Problem And Use Examination Hall To Put The Conditions In That Frame.

Derive The Following Yourself:

Equation Of Plane Containing The Given Lines.

Shortest Distance Between Two Lines In All Three Forms.

Condition For Lines To Intersect.

TIPS SECTION

TIPS FOR EXAMINATIONFirst Of All , Keep Faith In Yourself.

On Entering Examination Hall, Your Confidence Level Is Like That You Are JEE-2012 Topper But Don’t Let This Confidence To Become Over-Confidence.

During Examination, First Make A Quick View On Q. Paper.

Then Select The Question Which You Think ,You Will Solve Easily.

Don’t Lose Your Confidence When You Are Not Able To Solve Any Problem.

Keep In Mind, Some Questions Of JEE Are Not Given For Solving But Are Given To Leave . You Must Develop A Sense About Selection Of Question.

CONTACTS

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THANK YOU

PHONE NO : 08927482599EMAIL – ID : ashnil.anu@gmail.comFacebook.com : Ashnil Kumar