1

FIVE IMPORTANT CONCEPTS OF MATHEMATICS FOR JEE MAIN

1. EQUIVALENCE RELATION

A relation R on a set A is an equivalence relation if and only if(a) R is reflexive, i.e, aRa a A(b) R is symmetric, i.e., aRb bRa(c) R is transitive, i.e., aRb and bRc aRcPartial order relation :A relation R on a set A is a partial order relation if and only if.(a) R is reflexive, i.e. aRa a A(b) R is antisymmetric i.e., aRb and bRa a = b(c) R is transitive, i.e., aRb and bRc aRc.Relation of congruence modulo m :Let m be a fixed positive integer. Two integers a and b are said to be congruent modulo m if a b is divisible by m.We write a b (mod m)Thus. a b (mod m) [Read as a is congruent to b modulo m]iff a b is divisible by m; a, b .For example :(i) 25 5 (mod 4) because 255 = 20 is divisible by 4.(ii) 23 2 (mod 3) because 23 2 = 21 is divisible by 3(iii) 20 / 3 (mod 5) because 20 3 = 17 is not divisible by 5

The relation congruence modulo m is an equivalence relation on I.SOME THEOREMS ON EQUIVALENCE RELATION1. If R and S are two equivalence relations on a set A, then R S is also an equivalence relation on A.2. If R and S are two equivalence relations on a set A, then R S is not necessarily an equivalence relation3. If R is an equivalence relation on a set A, then R1 is also an equivalence relation on A.

2. DIFFERENT METHODS OF SOLVING TRIGONOMETRIC EQUATIONS

SOLUTION OF EQUATIONS BY FACTORINGExample :

Solve : 2 cos x cos 2x = cos x.Solution :

The given equation is equivalent to the equationcos x (2 cos 2x 1) = 0.This equation is equivalent to the collection of equations

,21x2cos

0xcos

.kk6

x.,e.i,k23

x2

,n,n2

x

I

I

Answer : )k,n(k,n I62

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2SOLUTION OF EQUATIONS REDUCIBLE TO QUADRATIC EQUATIONSExample : Solve the equation 3 cos2x 10 cos x + 3 = 0.Solution : Assume cos x = y. The given equation assumes the form 3y2 10 y +3 = 0.

Solving it, we find that .3y,31y 21

The value y2 = 3 does not satisfy the condition since| cos x | 1.

Consequently, ;31xcos .n,narccosx I23

1

Answer : )n(narccos I23

1

SOLUTION OF HOMOGENEOUS EQUATIONS AND EQUATIONS REDUCIBLE TO HOMOGENEOUSFORMSEquations of the form

a0 sinnx +a1 sin

n1 x cos x + a2 sinn2 x cos2 x +.....+an1 sin x cos

n1 x + an cos n x = 0.

(where a0, a1, ......, an are real numbers), are said to be homogeneous with respect to sin x and cos x. Such an equationis solved by dividing the equation by cosn x.Example 1 : Solve the equation cos 3 x + sin 3 x = 0Solution : sin3x = cos 3x.

Since the values of x for which cos3x is equal to zero cannot serve as roots for the given equation, we divide bothsides of the initial equation by cos3x and obtain an equation, equivalent to the initial equation

.1x3tanor,x3cosx3cos

x3cosx3sin

Hence, .n,nx;nx I3124

3

Answer : In,n312

Example 2 :Solve : 6 sin2 x sin x cos x cos2 x = 3

Solution :6sin2 x sin x cos x cos2 x 3 (sin2 x + cos2 x) = 0.Removing the brackets and collecting like terms, we get 3sin2 x sin x cos x 4 cos2 x = 0.

Since the values n2

x are not roots of the equation and cos x 0, we divide both sides of the equation by cos2 x :

3 tan2 x tan x 4 = 0,

Whence tan x = 1, .n,nx I4

and Ik,karctanx,xtan3

4

3

4

Answer : )k,n(k34arctan,n

4I

SOLVING EQUATIONS BY INTRODUCING AN AUXILIARY ARGUMENT

Equations of the form a sin x + b cos x = c is solved by dividing the two sides of equation by 22 ba and substituting

2 2 2 2

a bcos and sina b a b

3Example :

Solve the equation sin x + cos x = 2Solution :

2xcos2

1xsin2

12

,14

sinxcos4

cosxsin or ,14

xsin

)n(nx I224

)n(nx I242

Answer : )n(n I24

SOLVING EQUATIONS BY TRANSFORMING A SUM OF TRIGONOMETRIC FUNCTIONS INTO APRODUCT

We take the help of the formulae sin C + sin D = ......etc.Example :

Solve the equation cos 3x + sin 2x sin 4x = 0Solution :

cos 3x + (sin 2x sin 4x) = 0Transforming the expression in brackets into product form we obtaincos 3x + (2 sin x cos 3x) = 0, or cos 3x (1 2 sin x) = 0The last equation is equivalent to the collection of equations

cos 3x = 0 or 21xsin

Consequently, )k,n(k)(x,nx k I6

136

.

The set of solutions )k(k)(x k I6

1 belongs entirely to the set of solutions ).n(3

n6

x I

Therefore this set alone remains as a set of solutions.

Answer : ).n(n I36

SOLVING EQUATIONS BY TRANSFORMING A PRODUCT OF TRIGONOMETRIC FUNCTIONS INTOA SUM

We use the set of formulae 2 sin A sin B = ..........etc.Example :

Solve : sin 5x cos 3x = sin 6x cos 2x.Solution :

We apply formula of 2 sin A cos B to both sides of the equation :

),x4sinx8(sin21)x2sinx8(sin

21

042 xsinxsin 0x3cosxsin2

0x3cos0xsin

I

I

k,kx,kx

n,nx

3623

4

Answer : )k,n(k,n I36

SOLVING EQUATIONS WITH THE USE OF FORMULAS FOR LOWERING A DEGREEExample :

Solve the trigonometric equation sin2 x + sin2 2x = 1Solution :

We have 12

x4cos12

x2cos1

0x4cosx2cos 0xcos.x3cos2 .

The last equation is equivalent to the collection of two equations

(a) cos 3x = 0, ,n2

x3 ,3n

6x n I.

(b) cos x = 0, ,k2

x n I.

The set of the solutions of equation (b) is a subset of the set of solutions of (a) and, therefore in the answer wewrite only roots of equation (a).

Answer : )n(n I36

SOLVING EQUATIONS WITH THE USE OF HALF ANGLE FORMULASExample :

Solve : 02xsin2xcos 2

Solution :

cos x (1 cos x) = 0 01xcos2

)n(nxxcos I232

1

Answer : )n(n I23

SOLVING EQUATIONS WITH THE USE OF FORMULAS FOR DOUBLE AND TRIPLE ARGUMENTSExample 1 :

Solve : .xcos2x2sinSolution :

Using formula, we obtain

xcos2xcosxsin2

It is impossible to divide both sides of the equation by cos x since it would lead to the loss of solution which areroots of the equation cos x = 0.

We transfer xcos2 to the left - hand side and get

0xcos2xcosxsin2 0)1xsin2(xcos2

1xsin2

0xcos

I

I

k,k)(x

n,nx

k4

1

2

5

Answer : )k,n(k4

)1(,n2

k I

Example 2 :Solve the equation

xsin2xsin2xcos

2xsin2 22 xsinxcos 22

Solution :The given equation is rewritten as

.xsinxcos)xsinx(cos2xsin2 2222

Replacing the expression x2cosbyxsinxcos 22 , we get

x2cosx2cos2xsin2 or 0x2cosx2cos

2xsin2

012

22xsinxcos

2

1

2

02

xsin

xcos

.k,k)(x

n,nx

k I

I

23

1

24

Answer : )k,n(k)(,n k I23

124

SOLVING EQUATIONS BY A CHANGE OF VARIABLE(a) Equations of the form P (sin x cos x, sin x cos x) = 0

Where P (y, z) is a polynomial, can be solved by the substitution.cos x sin x = t 1 2 sin x cos x = t2.Let us consider an example .

Example :Solve the equationsin x + cos x = 1 + sin x cos x

Solution :We introduce the substitution sin x + cos x = t.Then (sin x + cos x)2 = t2,

1 + 2 sin x cos x = t2 sin x cos x 2

1t2

The initial equation in new variable is

01t2tor2

1t1t 22

(t 1)2 = 0 or t = 1,

i.e., sin x + cos x = 1, 1xcos21xsin

212

21xsin

4sinxcos

4cos

22

4xcos

In,nx 244

, In,nx 244

Answer: )n(n,n I222

(b) Equation of the from a sin x + b cos x + d = 0,Where a, b, and d are real numbers, and a, b 0, can be solved by the change

6

2xtan1

2xtan2

xsin,

2xtan1

2xtan1

xcos22

2

, x + 2 n (n I).

Note that such equations can be solved by dividing by 22 ba as well.

Example : Solve the equation 3 cos x + 4 sin x = 5

Solution : 5

2xtan1

2xtan2

4

2xtan1

2xtan1

322

2

,2xtan55

2xtan8

2xtan33 22

2 x x4 tan 4 tan 1 02 2

012

22xtan

In,ntanxxtan 22

12

2

1

21

Answer : In,ntan 22

12 1

(c) Many equation can be solved by introducting a new variable :Example :

Solve the equation

,x2cosx2sinx2cosx2sin 44

Solution :The given equation can be written as

2 2 2 2 2(sin 2x cos 2x) 2sin 2x cos 2x sin 2x cos 2x ,

.01x2cosx2sinx2cosx2sin2 22

We introduce the substitution sin 2x cos 2x = y.

The last equation assumes the form 2y2 + y 1 = 0, or 021y)1y(2

We pass to the variable x and obtain

(1) sin 2x cos 2x = 1 2 sin 2x cos 2x = 2 sin 4x = 2 x

(2) sin 2x cos 2 x =21

sin 4x = 1

In,nx 22

4 In,nx28

Answer : )n(,n I28

7

3. MULTINOMIAL THEOREM

Use of expansion to evaluate number of combinationRESULT-1 : The total number of ways of dividing 'n' identical things among 'r' persons, each of whom can

receive 0,1,2 are more ( n ) things is 1r1rn C (Alternatively, n identical balls to be distributed in r boxes, when

blank box is allowed).Let us consider 'r' brackets corresponding to 'r' persons.We take in each bracket an experssion given by 1 + x + x2 + .... + xn , where the various powers of x; namely 0, 1, 2,...., n correspond to the number of things each persons can have in the distibution.The required number of ways

= coefft. of xn in the produt (1 + x + x2 ..... + xn) ( ) ( )...... repeated r times

= coefft xn in

r1n

x1x1

= coefft of xn in

rr1n )x1()x1( coefft of xn in (1 x)r

= 1r1rn C

)1r.....(3.2.1)1rn).......(2n)(1n(

RESULT-2 : The total number of ways to divide 'n' identical things among 'r' persons when each gets at least one

thing is 1r1n C = coeff of xn in ( x + x2 + ....+ xn-r+1)r

= coeff of xn r in (1 x)r

RESULT-3 : The total number of selections of 'r' things from 'n' things where each thing can be repeated as many

times as one can is r1rn C .

The required number of ways = coefficient of xr in(1 + x + x2 + .........)n

= coefficient of xr in n)x1(1

coefft. of xr in (1 x)n

= r1rn C

)!1n()1nr).....(2r)(1r(

RESULT-4 : Number of ways of dividing n identical things into r groups such that no group contains less than mthings and more than k (m < k) things is coefficient of xn in the expansion of (xm + xm + 1 + ..... + xk)r

RESULT-5 : The number of ways of selecting r things out of n things of which p are alike and are of one kind, q arealike and are of second, s are alike and are of third kind and so on, is

= coefficient of xr in [(1 + x + x2 + ..... + xp) (1 + x + x2 + ..... + xq) (1 + x + x2 + ..... + xs) .....]

RESULT-6 : The number of ways of selecting r things out of n things of which p are alike and are of one kind, q arealike and are of second kind and rest (n p q) things are all different is :

= coefficient of xr in [(1 + x + x2 + ..... + xp) (1 + x + x2 + ..... + xq) (1 + x)n p q]

RESULT-7 : The number of ways of selecting r things out of n things of which p are alike and are of one kind, q arealike and are of second kind, s are alike and are of third kind when each thing is taken at least once :

= coefficient of xr in[(x + x2 + ..... + xp) (x + x2 + ..... + xq) (x + x2 + ..... + xs) .....]

RESULT-8 : The number of non-negative integral solutions of the equation x1 + x2 + ..... + xr = n is n + r 1Cr.

RESULT-9 : The number of terms in the expansion of(a1 + a2 + a3 + .... + an)

r is n + r 1Cr.

8

An important Expansion : ...xC...xCxCxC1)x1( rr1rn332n221n1nn , 1|x|andn N

Product of two infinite power series :

......)xbxbb(....)xaxaa( 22102

210 ......xcxcc2

210

where 0n2n21n1n0n ba......bababac for all 0n .

4. PROPERTIES OF BINOMIAL COEFFICIENTS :

We have n

nn1n

1nn

rr

n22

n1

n0

nn

xCxC

....xC....xCxCC)x1(

Also,

nn

1nnrn

rn

2n2

n1n1

nn0

nn

CxC....xC

....xCxCxC)x1(

Let us denote nn

2n

1n

0n C.....,,C,C,C by C0, C1, C2,...., Cn respectively..

Then the above expressions become nnr

r2

210n xC.....xC....xCxCC)x1(

And n1

1n2n

21n

1n

0n CxC....xCxCxC)x1(

C0, C1, C2,....,Cn are called the binomial coefficient and have the following properties:1. In the expansion of (1 + x)n the coefficient of terms equidistant from the beginning and end are equal.

The coefficient of (r + 1)th term from the beginning is nCr. The (r + 1)th term from the end is (n r + 1)th termfrom the beginning. Therefore, its coefficinet is nCn r.But nCr =

nCn rHence the coefficient of terms equidistant from the beginning and end are equal.

2. The sum of the binomial coefficient in the expansion of (1 + x)n is 2n.Putting x = 1 in(1 + x)n = C0 + C1 x + C2 x

2 + ..... + Cr xr + .... + Cn x

n, we get

C0 + C1 + C2 + ..... + Cn = 2n or

n

0r

nr

n 2C .

3. The sum of the coefficient of the odd terms is equal to the sum of the coefficient of the even terms and each isequal to 2n 1 i.e., C0 + C2 + C4 + ...... = C1 + C3 + C5 + ..... = 2

n 1

Putting x = 1 and 1 respectively in the expansion.n

n1n

1n3

32

210n xCxC....xCxCxCC)x1( ,

we get C0 + C1 + C2 + C3 + ..... + Cn 1 + Cn = 2n

and C0 C1 + C2 C3 + .... + (1)n Cn = 0

Adding and subtracting these two equations, we get C0 + C2 + C4 + .... = C1 + C3 + C5 + .... = 2n 1.

4. 1 . C1 + 2 . C2 + 3 . C3 + .... + n . Cn = n . 2n 1 or

n

0r

1nr

n 2.nC.r

we have nn1n1n332210n xCxC....xCxCxCC)x1(

Differentiating both sides w.r. to x and putting x = 1, we get n3211n Cn.....C3C2C2.n

EXTRA IMPORTANT RESULTS

1. r1rn

CC

1rn

rn

2. 1r1n

rn C

rnC that is, 1r

1nr

n C.nC.r and 1n

C1r

C 1r1n

rn

9

3. 1rn

1r1n C)1rn(C.n

4. r1n

rn

1rn CCC

5. rnn

rn CC

6. nyxoryxCC yn

xn

7. nnn

2n

1n

0n 2C.....CCC

8. 1n3n

1n

2n

0n 2........CC.....CC

9. n2n1n2

11n2

01n2 2C.....CC

Example : If the sum of the coefficients in the expansion of 512 )1x2x( vanishes then =.

(a) 1 (b) 21

(c) 2 (d) 1

Sol. 5125122 ])x1[()1x2x(

)say(.....xAxAA)x1( 2210102 Put x = 1,

we get )given(0.....AAA)1( 210102

101[Note : The sum of the coefficients (not the binomial coefficients) in any expansion can be obtained by substitutingall the variables equal to one]Answer (a)

5. RESULTS RELATED TO A POINT AND A STRAIGHT LINE

(i) Position of a given point Relative to a Given Line :

O

Y

X

P(x , y )11

P(x , y )11

ABOVEBELOW

Let the given line be ax + by + c = 0 or L(x, y) = 0.

A point P (x1, y1) will lie above or below this line according as bcbyax 11 is positive or negative.

That is, if 0b

)y,x(L 11 , P lies above the line and if 0b

)y,x(L 11 , P lies below the line,

where L(x1, y1) = ax1 + by1 + c.(ii) The perpendicular distance of the point P(x1, y1) from the given line

O

Y

X

L

P(x , y )11

Q(x , y )22

N(h, k)

10

2211

2211

ba

)y,x(L

ba

cbyaxd

(iii) Coodinates of foot of perpendicular N(h, k) from the point P(x1, y1) on the line L (x, y) = 0 may be given by

221111

bacbyax

byk

axh

= 2211

ba)y,x(L

(iv) Coordinates of the image of Q(x2, y2) of the point P(x1, y1) in the line mirror L(x, y) = 0 may be given by

22111212

ba)cbyax(2

byy

axx

= 2211

ba)y,x(L2

RESULTS RELATED TO TWO POINTS AND A STRAIGHT LINE(i) The ratio in which the join of two given points P(x1, y1) and Q (x2, y2) is divided by the straight line ax + by + c

= 0 or L (x, y) = 0 is given by

O

Y

X

L

M

P(x , y )11

Q(x , y )22

)y,x(L)y,x(L

cbyaxcbyax

QMPM

22

11

22

11

If the above result comes out to be negative, it indicates an external division.(ii) The relative position of the points P and Q with respect to the line may be expressed as following :

If 0)y,x(L)y,x(L

22

11, both the points lie to the same side of the given line.

If 0)y,x(L)y,x(L

22

11 , the points lie in opposite sides of the given line.

RESULTS RELATED TO TWO STRAIGHT LINESSuppose that a1x + b1y + c1 = 0 or L1 (x, y) =0 ....(i)and a2x + b2y + c2 = 0 or L2 (x, y) = 0 ....(ii)be two straight lines.(1) The Point of Intersection : Point of intersection of the two given straight lines can be found by solving equations

(i) and (ii).

Thus, 122112211221 baba

1acac

ycbcb

x,

provided 2

1

2

11221 b

baa0baba .

That is the point of intersection of the lines is 1221

1221

1221

1221babaacac,

babacbcb

, provided 2

1

2

1bb

aa

.

If 2

1

2

1

2

1cc

bb

aa

, then the straight lines L1 and L2 are parallel and distinct.

If 2

1

2

1

2

1cc

bb

aa

, then the straight lines L1 and L2 are coincident.

11(2) Angle between two Straight Lines :

O

Y

X

L1

L2

If the slope of two straight lines be m1 and m2 then the acute angle ( ) between the lines is given by

21

21mm1mmtan .

The lines are parallel if m1 = m2.The lines are perpendicular if 1 + m1m2 = 0 1mm 21Thus, for the lines given above by equation (i) and (ii), we get

(a) angle ( ) between them 2121

1221bbaababatan

(b) the lines are parallel if

2

1

2

11221 b

baa0baba

(c) the line are perpendicular if 0bbaa 2121 .