Formulae of Mathematics
Formulae of Mathematics
Formulae of Mathematics
GURU
JEEMAIN.GURU
FORMULAE OF MATHEMATICS
Author
Prof. R. K. Malik
Director of Newton Classes
606, 6th Floor, Hariom Tower, Ranchi
Malik Publication
177, Rajdhani Enclave, Near Rani Bagh, Pitampura
Delhi - 110034
Ph : 011-27193845
JEEMAIN.GURU
Dedicated
To
My
Sincere Students
2007 by Malik Publication. All rights reserved. No part of this book may be reproduced in
any form, by mimeograph or any other means, without permission in writing from the
publisher.
JEEMAIN.GURU
Therefore, at the outset I would like to dedicate my collection of definitions and formulae
to the anxious aspirant outside the examination hall, whose mind is meandering in the sea of
mathematical formulae needing a last minute refreshing accurate and shortcut backup of all
hard and sincere toil.
I have tried my best to give the students a very simple, short crystal clear list of intriguing
Calculus, confusing Trigonometry, vague Probability formulae. Also Two-Dimensional and
Three-Dimensional Geometry have been added along with many other important and relevant
topics.
Without going into any more description I would like to acknowledge sincere efforts and
unfailing dedication of my wife Mrs. Vandana Malik, my children Master Saksham Malik and
Master Parth Malik, my colleagues, our desktop designers Mr. Kamal Kumar and others
without which this effort would not have been in the present beautiful form and lucid style.
Though best efforts have been put in, any more suggestions for improvement are highly
welcome.
Hoping that my effort will not go in vain and fulfill the aim.
Author
Date Prof. R. K. Malik
01/06/2008
JEEMAIN.GURU
CONTENTS
Set Theory 3
(iii) n ( A − B )= n ( A ) − n ( A ∩ B ) i.e., n ( A − B ) + n ( A ∩ B ) =
n ( A)
(iv) n ( A ∆ B ) = n ( A ) + n ( B ) − 2n ( A ∩ B )
(v) n( A∪ B ∪C)
= n ( A) + n ( B ) + n ( C ) − n ( A ∩ B ) − n ( B ∩ C ) −n ( A ∩ C ) + n ( A ∩ B ∩ C )
(vi) No. of elements in exactly two of the sets A, B, C
= n ( A ∩ B ) + n ( B ∩ C ) + n ( C ∩ A ) − 3n ( A ∩ B ∩ C )
(vii) No. of elements in exactly one of the sets A, B, C
= n ( A ) + n ( B ) + n ( C ) − 2n ( A ∩ B ) − 2n ( B ∩ C ) −2n ( A ∩ C ) + 3n ( A ∩ B ∩ C )
(viii) n ( A′ ∪ B′ )= n ( A ∪ B )′ = n (U ) − n ( A ∩ B )
(ix) n ( A′ ∩ B′ )= n ( A ∪ B )′ = n (U ) − n ( A ∪ B ) .
9. Cartesian product of two or more sets
The Cartesian product of n ( ≥ 2 ) sets A1 , A2 ,....., An is defined as the set of all ordered n-tuples
( a1 , a2 ,....., an ) where ai ∈ Ai (1 ≤ i ≤ n ) . The Cartesian product of A1 , A2 ,....., An is denoted by
n
A1 × A2 × ....... × An or by ∏ A.
i =1
i
n
=
Symbolically ∏ Ai {( a , a ,......, a ) : a ∈ A ,1 ≤ i ≤ n}. If at least one of
1 2 n i i A1 , A2 ,....., An is empty set,
i =1
then A1 , A2 ,....., An is defined as empty set. If A1 , A2 ,....., An are finite sets, then
n ( A1 , A2 ,....., An ) = n ( A1 ) × n ( A2 ) × ........ × n ( An ) .
10. Important results
(i) A× B ≠ B × A (ii) A×φ = φ × A = φ
(iii) If A and B are finite sets, then n ( A × B ) = n ( A ) × n ( B ) = n ( B × A )
(iv) If A ⊆ B, then A × C ⊆ B × C
(v) (a) A × ( B ∪ C ) = ( A × B ) ∪ ( A × C ) (b) A × ( B ∩ C ) = ( A × B) ∩ ( A × C )
(c) A× ( B − C ) = ( A× B) − ( A× C )
(vi) A ⊆ B, C ⊆ D ⇒ A × C ⊆ B × D
(a) ( A × B ) ∪ (C × D ) ⊆ ( A ∪ C ) × ( B ∪ D ) (b) ( A × B ) ∩ (C × D ) = ( A ∩ C ) × ( B ∩ D ).
JEEMAIN.GURU
= =
( a + ib )( c − id ) ( ac + bd ) + i ( bc −=
ad ) ac + bd bc − ad
+i 2
( c + id )( c − id ) c +d
2 2
c +d
2 2
c + d2
GEOMETRICAL REPRESENTATION OF COMPLEX NUMBERS
A complex number z= x + iy can be represented by a point P whose Y
P ( x + iy )
Cartesian co-ordinates are ( x, y ) referred to rectangular axes Ox and Oy.
Imaginary axis
Here the x-axis and y-axis are usually called the real and imaginary axes
respectively. The plane is called the Argand plane, complex plane or Gaussian OP = z
y arg ( z ) = θ
plane. The point P ( x, y ) is called the image of the complex number z and z
θ x
is said to be the affix or complex co-ordinate of point P . X
O M
We have OP = x +y =
2 2
z . Real axis
Complex Numbers 5
Imaginary axis
Let z= a + ib be a complex number. We define conjugate of z ,
denoted by z to be the complex number a − ib .
That is, if z= a + ib, then z= a − ib.
θ
O −θ Real axis
where P ( z ) = a0 + a1 z + a2 z 2 + ... + an z n
P(z)
(xii) If R ( z ) = where P ( z ) and Q ( z ) are polynomials in z , and Q ( z ) ≠ 0, then
Q(z)
P(z ) z + 3z 2 z + 3z 2
R( z) = , e.g. =
Q(z ) z − 1 z −1
a1 a2 a3 a1 a2 a3
(xiii) If z = b1 b2 b3 , then z = b1 b2 b3
c1 c2 c2 c1 c2 c3
where ai , bi , ci ( i = 1, 2, 3) are complex numbers.
Geometrically z represents the distance of point P from the origin, i.e. z = OP.
If z = 1 the corresponding complex number is known as unimodular
complex number. Clearly z lies on a circle of unit radius having centre ( 0, 0 ) .
X
O M
Note that z ≥ 0 ∀ z ∈ C
JEEMAIN.GURU
PROPERTIES OF MODULUS
If z is a complex number, then
(i) z =0⇔ z =0
(ii) z =z =− z =− z
(iii) − z ≤ Re ( z ) ≤ z
(iv) − z ≤ Im ( z ) ≤ z
z z= z
2
(v)
If z1 , z2 are two complex numbers, then
(vi) z1 z2 = z1 z2
z1 z
(vii) = 1 , if z2 ≠ 0
z2 z2
(viii) z1 + z2 = z1 + z2 + z1 z2 + z1 z2
2 2 2
= z1 + z2 + 2 Re ( z1 z2 ) = z1 + z2 + 2 z1 z2 cos θ
2 2 2 2
z1 − z2 = z1 + z2 − z1 z2 − z1 z2
2 2 2
(ix)
= z1 + z2 − 2 Re ( z1 z2 ) = z1 + z2 − 2 z1 z2 cos θ = θ arg z2 − arg z1
2 2 2 2
where
(x) z1 + z2
2
+ z1 − z2= 2 z1 + z2
2
( 2 2
)
(xi) If a and b are real numbers and z1 , z2 are complex numbers, then
az1 + bz2
2
+ bz1 − az2
2
(
= a 2 + b2 )( z 1
2
+ z2
2
)
z1
(xii) If z1 , z2 ≠ 0, then z1 + z2 = z1 + z2 ⇔
2 2 2
is purely imaginary.
z2
(xiii) Triangle Inequality. If z1 and z2 are two complex numbers, then z1 + z2 ≤ z1 + z2
(xiv) z1 − z2 ≤ z1 + z2
(xv) z1 − z2 ≤ z1 + z2
(xvi) z1 − z2 ≥ z1 − z2
POLAR FORM OF A COMPLEX NUMBER
Let z be a non-zero complex number, then we can write
= z r ( cos θ + i sin θ ) where r = z and θ = arg ( z ) .
EULERIAN REPRESENTATION (EXPONENTIAL FORM)
Since we have, = eiθ cos θ + i sin θ
and thus z can be expressed as= z reiθ , =
where and θ
z r= arg ( z )
ARGUMENT OF A COMPLEX NUMBER
If z is a non – zero complex numbers represented by P in the complex plane, then argument of z is the angle
which OP makes with positive direction of real axis.
Note : Argument of a complex number is not unique, since if θ is a value of the argument, then 2nπ + θ
where n ∈ I , are also values of the argument of z. The value θ of the argument which satisfies the
inequality −π < θ ≤ π is called the principal value of the argument or principal argument.
JEEMAIN.GURU
Complex Numbers 7
affix of C − affix of A
or angle between AC and AB = arg
affix of B − affix of A
z -z
Interpretation of arg 3 1
z 2 - z1 y
If z1 , z2 , z3 are the vertices of a triangle ABC described in the counter-clockwise A ( z1 )
sense, then
α
AC z −z
(i) arg = arg 3 1 = ∠ BAC =α (say), and
AB z2 − z1 ( z2 ) C ( z3 )
z3 − z1 CA B
(ii)= ( cos α + i sin α )
z2 − z1 BA x
Corollary : The points z1 , z2 , z3 will be collinear if and only if angle α = 0 or π ,
z −z
i.e., if and only if 3 1 is purely real.
z2 − z1
z -z Y
D z4
Interpretation of arg 1 2
z3 - z4 z2
B
Let z1 , z2 , z3 and z4 be four complex numbers. Then the line joining z4 and z3
z1 − z2
θ = arg
is inclined to the line joining z2 and z1 at the following angle : z1 z3 − z4
A C z3
CD z −z
arg = arg 4 3
AB z2 − z1 O X
This can be proved analogously to the above result.
Corollary : The line joining z4 and z3 is inclined at 90° to the line joining z2 Y
and z1 if z3 z2
z −z π 90°
arg 1 2 = ± z1 − z2 is purely
z3 − z4 2 z3 − z4 imaginary
O X
SOME IMPORTANT GEOMETRICAL RESULTS AND EQUATIONS
1. Distance Formula Q ( z2 )
The distance between two points P ( z1 ) and Q ( z2 ) is given by
PQ = z2 − z1 = affix of Q − affix of P
2. Section Formula
P ( z1 )
If R(z) divides the line segment joining P ( z1 ) and Q ( z2 ) in the
ratio m : n ( m, n > 0 ) then
Q ( z2 )
mz + nz1 n
(i) For internal division : z= 2
m+n m R(z)
mz − nz1
(ii) For external division : z= 2
m−n P ( z1 )
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Complex Numbers 9
3. Condition (s) for four points A (z 1 ), B(z 2 ), C(z 3 ) and D(z 4 ) to represent vertices of a
(i) Parallelogram
The diagonals AC and BD must bisect each other
1 1
⇔ ( z1 + z3 )= ( z2 + z4 ) ⇔ z1 + z3 = z2 + z4
2 2
(ii) Rhombus
(a) the diagonals AC and BD bisect each other ⇔ z1 + z3 = z2 + z4 , and
(b) a pair of two adjacent sides are equal, for instance, AD = AB ⇔ z4 − z1 = z2 − z1
z −z π
(c) arg 3 1 = ±
z4 − z2 2
(iii) Rectangle
(a) the diagonals AC and BD bisect each other ⇔ z1 + z3 = z2 + z4
(b) the diagonals AC and BD are equal ⇔ z3 − z1 = z4 − z2 .
(iv) Square
(a) the diagonals AC and BD bisect each other ⇔ z1 + z3 = z2 + z4
(b) a pair of adjacent sides are equal; for instance AD = AB ⇔ z4 − z1 = z2 − z1
(c) the two diagonals are equal, that is, AC = BD ⇔ z3 − z1 = z4 − z2
4. Centroid, Incentre, Orthocentre and Circum-centre of a Triangle
A ( z1 )
Let ABC be a triangle with vertices A ( z1 ) , B ( z2 ) and C ( z3 ) ,
(i) Centroid G ( z ) of the ∆ ABC is given by
1 G (z)
z= ( z1 + z2 + z3 )
3
( z2 ) B C ( z3 )
A ( z1 )
(ii) Incentre I ( z ) of the ∆ ABC is given by
az1 + bz2 + cz3
z=
a+b+c c b
I (z)
( z2 ) B a C ( z3 )
(iii) Circumcentre S ( z ) of the ∆ ABC is given by
z1
2
z1 1 A ( z1 )
2
z2 z2 1
2
( z − z ) z + ( z3 − z1 ) z2 + ( z1 − z2 ) z3
2 2 2
z3 z3 1 F E
z= 2 3 1 ⇔ z= S (z)
z1 ( z2 − z3 ) + z2 ( z3 − z1 ) + z3 ( z1 − z2 ) z1 z1 1
z2 z2 1 ( z2 ) B D C ( z3 )
z3 z3 1
z1 ( sin 2 A ) + z2 ( sin 2 B ) + z3 ( sin 2C )
Also z=
sin 2 A + sin 2 B + sin 2C
JEEMAIN.GURU
z2 z2 1 ( z2 ) B D C ( z3 )
z3 z3 1
or z=
( tan A) z1 + ( tan B ) z2 + ( tan C ) z3 =
( a sec A) z1 + ( b sec B ) z2 + ( c sec C ) z3
tan A + tan B + tan C a sec A + b sec B + c sec C
Euler’s Line
The centroid G of a triangle lies on the segment joining the A ( z1 )
orthocentre H and the circumcentre S of the triangle. G
divides the join of H and S in the ratio 2 : 1.
2 1
H
H 2 G 1 S G S
(orthocentre) (centroid) (circumcentre) 90°
90°
B ( z2 ) L D C ( z3 )
5. Area of a Triangle
Area of ∆ ABC with vertices A ( z1 ) , B ( z2 ) and C ( z3 ) is given by
z1 z1 1 z1 z1 1
1 1
∆ = mod of i z2 z2 1 = | z2 z2 1|
4 z z 1 4
3 3 z3 z3 1
6. Condition for Triangle to be Equilateral
Triangle ABC with vertices A ( z1 ) , B ( z2 ) and C ( z3 ) is equilateral if and only if
1 1 1
+ + =
0
z2 − z3 z3 − z1 z1 − z2
⇔ z12 + z22 + z32 = z2 z3 + z3 z1 + z1 z2
1 z2 z3
⇔ 1 z3 z1 = 0
1 z1 z2
7. Equation of a Straight Line
(i) Non-parametric form
Equation of a straight line joining the points having affixes z 1 and z 2 is
z z 1
z
z1 z1 1 = 0
z2
z2 z2 1
z1
z − z1 z − z1
or =
z2 − z1 z2 − z1
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Complex Numbers 11
or z ( z1 − z2 ) − z ( z1 − z2 ) + z1 z2 − z2 z1 =
0
(ii) Parametric form
Equation of a straight line joining the points having affixes z1 and z2 is
z = tz1 + (1 − t ) z2
where t is a real parameter.
(iii) General Equation of a Straight Line
The general equation of a straight line is
az + az + b = 0
where a is a complex number and b is a real number.
(iv) Equation of the perpendicular bisector P(z1)
If p ( z1 ) and Q ( z2 ) are two fixed points and R ( z ) is a moving point such R (z)
that it is always at equal distance from P ( z1 ) and Q ( z2 ) then locus of R ( z )
is perpendicular bisector of PQ
i.e. PR = QR Q(z2)
or | z − z1 | = | z − z2 | ⇒ | z − z1 |2 = | z − z2 |2
( )
after solving z z1 − z 2 + z ( z1 − z2 ) = z1 − z2 .
2 2
⇔
( z1 − z3 )( z2 − z4 ) is purely real. But z1 − z3 is non-real.
( z1 − z4 )( z2 − z3 ) z1 − z4
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Complex Numbers 13
α
X
O
then arg ( z − z0 ) =
α z
z
(iii) If z1 and z2 are two fixed points, z − z1 z − z2
then z − z1 = z − z2
represents perpendicular bisector of the segment
joining A ( z1 ) and B ( z2 ) . A ( z1 ) B ( z2 )
(iv) If z1 and z2 are two fixed points, and k > 0, k ≠ 1 is a real number, then
z − z1
=k
z − z2
represents a circle.
For k = 1, it represents perpendicular bisector of the segment joining A ( z1 ) and B ( z2 ) .
(v) Let z1 and z2 be two fixed points and k be a positive real number.
(a) If k > z1 − z2 , then
z − z1 + z − z2 =
k
represents an ellipse with foci at A ( z1 )
C D
A ( z1 ) B ( z2 ) CD = k
(b) If =
k z1 − z2 , then z − z1 − z − z2 =k
represents the straight line joining A ( z1 ) and B ( z2 ) but excluding the segment AB.
A ( z1 ) B ( z2 )
(viii) Let z1 and z2 be two fixed points, and α be a real number such 0 ≤ α ≤ π .
P( z)
(a) If 0 < α < π
z − z1 α
arg =α
z − z 2
B ( z2 ) A ( z1 )
represents a segment of the circle passing through A ( z1 ) and B ( z2 ) .
P′ ( z )
(c) If α = 0, then
z − z1
arg = α=( 0 )
z − z2
represents the straight line joining A ( z1 ) and B ( z2 ) but excluding the segment AB.
A ( z1 ) B ( z2 )
(d) If α = π , then
z − z1
arg = α=( π )
z − z2 B ( z2 )
A ( z1 )
Complex Numbers 15
(b) ( cos θ1 + i sin θ1 )( cos θ 2 + i sin θ 2 )( cos θ3 + i sin θ3 ) ... ( cos θ n + i sin θ n )
= cos (θ1 + θ 2 + θ3 + ... + θ n ) + i sin (θ1 + θ 2 + θ3 + ... + θ n ) where θ1 , θ 2 , θ3 ...θ n ∈ R.
(c) De Moivre’s Theorem for rational index.
If n is a rational, then one of the value of ( cos θ + i sin θ ) is cos ( nθ ) + i sin ( n θ ) . In fact, if
n
n = p / q where p, q ∈ I , q > 0 and p, q have no factors in common, then ( cos θ + i sin θ ) has q
n
(d) If n is irrational then ( cos θ + sin θ ) has infinite number of values one of which is cos nθ + i sin nθ
n
∴ 1 + α + α 2 + ... + α n −1 = 0
Note
1 1 1 1 nx n −1
+ + + ... + =
x −1 x − α x − α 2 x − α n −1 x n − 1
Product of n th roots of unity is ( −1)
n−1
Complex Numbers 17
( )(
(vi) a 2 + b 2 + c 2 − bc − ca − ab = a + bω + cω 2 a + bω 2 + cω )
(vii) a 3 + b3 + c 3 − 3abc = ( a + b + c ) ( a + bω + cω 2 )( a + bω 2 + cω )
(viii) x 3 + 1 = ( x + 1)( x + ω ) ( x + ω 2 )
(ix) a 3 + b3 = ( a + b )( a + bω ) ( a + bω 2 )
(x) a 4 + a 2b 2 + b 4 = ( a + bw ) ( a + bw2 ) ( a − bw ) ( a − bw2 )
(xi) Cube roots of real number a are 3 a , 3 a w, 3 a w2
To obtain cube roots of a, we write x3 = a as y 3 = 1 where y = x / a1/ 3 .
Solution of y 3 = 1 are 1, ω , ω 2 .
∴ x = 3 a , 3 a w, 3 a w2
nth Roots of a Complex Number
Let z ≠ 0 be a complex number. We can write z in the polar form as follows :
z = r ( cos θ + i sin θ )
where r = z and θ = Pr arg ( z ) . Recall −π < θ ≤ π .
The n th root of z has n values one of which is equal to
arg z arg z
1
=z0 n z cos + i sin and is called as the principal value of z n
. To obtain other value of
n n
1
z n , we write z as
JEEMAIN.GURU
= n
r cos + i sin cos + i sin
n n n n
= z0 α=k
where k 0, 1, 2, ..., n − 1
2π 2π θ θ
and α = cos + i sin =
is a complex n th root of unity and z0 n
r cos + i sin .
n n n n
1
n
Thus, all the n n th roots of z can be obtained by multiplying the principal value of z by different roots
of unity.
Rational Power of a Complex Number
If z is a complex number and m / n is a rational number such that m and n are relatively prime integers
and n > 0.
Thus, z m / n has n distinct values which are given by
( )
m
m m
zm/n = n z cos n (θ + 2kπ ) + i sin n (θ + 2kπ ) =
where k 0, 1, 2, ..., n − 1.
Logarithm of a Complex Number
Let z ≠ 0 be a complex number. We may write z in the polar form as follows :
z = r ( cos θ + i sin θ ) =
reiθ
We define
log= z log r + iθ
= log z + i Pr arg ( z )
z= x + iy, then
log ( x 2 + y 2 ) + i Pr arg ( z ) .
1
=
log z
2
JEEMAIN.GURU
19
EQUATION
An equation is a statement of equality which is not true for all values of the variable present in the equation.
ROOTS OF AN EQUATION
The values of the variable which satisfies the given polynomial equation is called its roots i.e. if f ( x ) = 0 is a
polynomial equation and f (α ) = 0, then α is known as root of the equation f ( x ) = 0.
ROOT OF THE QUADRATIC EQUATION
A quadratic equation ax 2 + bx + c = 0 has two and only two roots. The roots of ax 2 + bx + c =0 are
−b + b 2 − 4ac −b − b 2 − 4ac
and .
2a 2a
If α and β are roots of ax 2 + bx + c = 0 , then
−b coefficient of x
1. the sum of roots = α +β = = −
a coefficient of x 2
c constant term
2. the product of roots= αβ= =
a coefficient of x 2
Also if f ( x )= ax 2 + bx + c= 0, then f ( x ) =a ( x − α )( x − β ) .
DISCRIMINANT OF A QUADRATIC EQUATION
If ax 2 + bx + c =0, a, b, c,∈ R and a ≠ 0, be a quadratic equation then the quantity b 2 − 4ac is called
discriminant of the equation and is generally denoted by D .
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∴ D= b 2 − 4ac .
NATURE OF ROOTS OF THE QUADRATIC EQUATION:
For a quadratic equation ax 2 + bx + c = 0
If a, b, c,∈ R and a ≠ 0, then
1. if D > 0 roots are real and distinct.
2. if D = 0 roots will be real and equal.
3. if D < 0 roots are imaginary and conjugate of each other.
4. if D is perfect square a, b, c ∈ Q, then roots will be rational.
5. if D is not a perfect square roots will be irrational and conjugate of each other. If one root is
p + q then other root be p − q .
6. if ax 2 + bx + c = 0 has more than two roots (complex number or real number) then this is an identity
i.e. a= b= c= 0 .
GRAPH OF A QUADRATIC EXPRESSION
We have, y = f ( x ) = ax 2 + bx + c where a, b, c,∈ R, a ≠ 0.
1. The shape of the curve y = f ( x ) is a parabola
2. The axis of the parabola is parallel to y-axis.
3. If a > 0, then the parabola opens upwards.
4. If a < 0, then the parabola opens downwards
5. For D > 0, parabola cuts x-axis in two distinct points
a < 0, D > 0
x - axis
x - axis
a > 0, D =
0
−b − D
The co-ordinates of vertex of parabola in all these cases is ,
2a 4a
7. For D < 0, parabola does not cut x-axis.
a < 0, D < 0
x - axis
x - axis
a > 0, D < 0
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Quadratic Equations 21
a > 0, D > 0
2. If α , β are real and equal ( i.e. D = 0 ) roots of the corresponding quadratic equation then the sign of
y = ax 2 + bx + c, x ∈ R is as follows:
y= ax 2 + bx + c ≥ 0 if a > 0 and y= ax 2 + bx + c ≤ 0 if a < 0
a < 0, D =
0
x-axis
x-axis
a > 0, D =
0
3. If α and β are imaginary ( i.e. D < 0 ) , then
y= ax 2 + bx + c > 0 if a > 0 and y= ax 2 + bx + c < 0 if a < 0.
a < 0, D < 0
x-axis
x-axis
a > 0, D < 0
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a<0
Note that D > 0 is surely true if f ( k ) > 0 y = ax 2 + bx + c
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Quadratic Equations 23
Combining, we may say k lies between the roots of f ( x )= ax 2 + bx + c= 0 if and only if af ( k ) < 0
CONDITIONS FOR EXACTLY ONE ROOT OF A QUADRATIC EQUATION TO LIE IN THE
INTERVAL (k 1 , k 2 ) WHERE k 1 < k 2
Case I If a > 0, then exactly one root of f ( x )= ax 2 + bx + c= 0 lies in the interval ( k1 , k2 ) if and only if
f ( k1 ) > 0 and f ( k2 ) < 0 or f ( k1 ) < 0 and f ( k2 ) > 0.
Thus, if a > 0, exactly one root of f ( x )= ax 2 + bx + c= 0 lies in the interval ( k1 , k2 ) if and only if
f ( k1 ) f ( k2 ) < 0.
a>0 a>0
f ( x ) = ax 2 + bx + c f ( x ) = ax 2 + bx + c
f ( k2 )
f ( k1 )
α k2 β α k1 β
x x
k1 f ( k1 ) k2
f ( k2 )
Case II Similarly, if a < 0, exactly one of the roots of f ( x )= ax 2 + bx + c= 0 lies in the interval ( k1 , k2 ) if
f ( k1 ) f ( k2 ) < 0.
f ( k2 ) f ( k1 )
k1 k2
x x
α k2 β α k1 β
f ( k1 ) f ( k2 )
a<0 a<0
f ( x ) = ax 2 + bx + c f ( x ) = ax 2 + bx + c
α
a>0 x
−b / 2a
f (=
x) a ( x −α )
2
a<0
−b / 2a
f (=
x) a ( x −α )
2
x
α
f (α ) 0,=
= f ′ (α ) 0 f (α ) 0,=
= f ′ (α ) 0
α2 α 1
= = .
b1c2 − b2 c1 c1a2 − c2 a1 a1b2 − a2b1 y = a2 x 2 + b2 x + c2
Eliminating α we get
( a1b2 − a2b1 )( b1c2 − b2c1 ) = ( c1a2 − c2 a1 )
2
α
x
This is the required condition for two quadratic equation to have
a common root.
y = a1 x 2 + b1 x + c1
CONDITION FOR TWO QUADRATIC EQUATIONS TO HAVE THE SAME ROOTS (i.e. Both root
common)
a1 b1 c1
= = .
a2 b2 c2
In this case both the equations have the same roots.
Note :
(a) To find the common root of two equations, make the coefficient of second degree terms in two
equations equal and subtract. The value of x so obtained is the required common root.
(b) If two quadratic equations with real coefficients have an imaginary root common, then both roots
a1 b1 c1
will be common and the two equations will be identical. ∴ = = .
a2 b2 c2
(c) If two quadratic equations with rational coefficients have an irrational root common, then both roots
a1 b1 c1
will be common and the two equations will be identical. ∴ = = .
a2 b2 c2
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Quadratic Equations 25
(d) If α is a repeated root of the quadratic equation f ( x )= ax 2 + bx + c= 0, then α is also a root of the
equation f ' ( x ) = 0 .
HIGHER DEGREE EQUATIONS
The equation f ( x=
) a0 x n + a1 x n−1 + ... + an−1 x + a=n 0 … (1)
Where the coefficient a0 , a1........an ∈ C and a0 ≠ 0 is called an equation of n th
degree, which has exactly n
roots α1 , α 2 ......α n ∈ C , Then we can write
{
p ( x ) =a0 ( x − α1 )( x − α 2 ) ..... ( x − α n ) = a0 x n − ( ∑ α1 ) x n −1 + ( ∑ α1α 2 ) x n − 2 − ...... + ( −1) α1α 2 .....α n
n
} … (2)
−a1 a2
Comparing (1) and (2) ∑α 1 = α1 + α 2 + ... + α n =
a0
; ∑α α 1 2 = α1α 2 + ... + α n −1α n =
a0
an
And so on and α1α 2 ...α n = ( −1)
n
a0
CUBIC EQUATION
For n = 3 , the equation is a cubic of the form ax3 + bx 2 + cx + d = 0 and we have in this case
b c d
α + β + γ =− ; αβ + βγ + γα = ; α βγ =−
a a a
BIQUADRATIC EQUATION
If α , β , γ , δ are roots of the biquadratic equation ax 4 + bx3 + cx 2 + dx + e = 0 , then
b c
σ 1 =α + β + γ + δ =− , σ 2 = αβ + αγ + αδ + βγ + βδ + γδ =
a a
d e
σ3 = αβγ + αβδ + αγδ + βγδ = − , σ 4 αβγδ
= = .
a a
TRANSFORMATION OF EQUATIONS
Rules to form an equation whose roots are given in terms of the roots of another equation. Let given
equation be
a0 x n + a1 x n −1 + ... + an −1 x + an = 0 (1)
Rule 1 : To form an equation whose roots are k ( ≠ 0 ) times roots of the equations in (1), replace x by
x / k in (1).
Rule 2 : To form an equation whose roots are the negatives of the roots in equation (1), replace x by − x
in (1). Alternatively, change the sign of the coefficients of x n −1 , x n −3 , x n −5 ,... etc. in (1).
Rule 3 : To form an equation whose roots are k more than the roots of equation in (1), replace x by x − k
in (1).
Rule 4 : To form an equation whose roots are reciprocals of the roots in equation (1), replace x by 1/ x in
(1) and then multiply both the sides by x n .
Rule 5 : To form an equation whose roots are square of the roots of the equation in (1) proceed as follows :
Step 1 Replace x by x in (1)
Step 2 Collect all the terms involving x on one side.
Step 3 Square both the sides and simplify.
For instance, to form an equation whose roots are squares of the roots of x3 − 2 x 2 − x + 2 = 0,
replace x by x to obtain
x x − 2x − x + 2 =0 ⇒ x ( x − 1)= 2 ( x − 1)
JEEMAIN.GURU
Squaring we get
x ( x − 1) = 4 ( x − 1)
2 2
or ( x − 4 ) ( x 2 − 2 x + 1) =
0 or x3 − 6 x 2 + 9 x − 4 =0
Rule 6 : To form an equation whose roots are cubes of the roots of the equation in (1) proceed as follows :
Step 1 Replace x by x1/ 3 .
Step 2 Collect all the terms involving x1/ 3 and x 2 / 3 on one side.
Step 3 Cube both the sides and simplify
DESCARTES RULE OF SIGNS FOR THE ROOTS OF A POLYNOMIAL
Rule 1 : The maximum number of positive real roots of a polynomial equation
f ( x=
) a0 x n + a1 x n−1 + a2 x n−2 + ... + an−1 x + a=n 0
is the number of changes of the signs of coefficients from positive to negative and negative to
positive. For instance, in the equation x 3 + 3 x 2 + 7 x − 11 =
0 the signs of coefficients are + + + −
As there is just one change of sign, the number of positive roots of x 3 + 3 x 2 + 7 x − 11 =0 is at
most 1.
Rule 2 : The maximum number of negative roots of the polynomial equation f ( x ) = 0 is the number of
changes from positive to negative and negative to positive in the signs of coefficients of the
equation f ( − x ) = 0.
RESULT ON ROOTS OF AN EQUATION
Let f ( x ) = 0 is polynomial equation of degree n then
1. Number of real roots of the equation is less than or equal to the degree of the equation.
2. Total number of roots of the equation are :
(i) Number of positive roots
(ii) Number of negative roots
(iii) Number of imaginary roots.
(iv) Number of roots which are zero
∴ Total number of roots = Number of positive roots + Number of negative roots + Number of complex
roots + number of roots which are zero.
Note:
(a) If f ( x ) = 0 be an equation and a and b are two real numbers such that f ( a ) f ( b ) < 0, then
equation f ( x ) = 0 has at least one real root or an odd number of real roots (may be repeated)
between a and b .
(b) If f ( a ) < 0. f ( b ) < 0 or f ( a ) > 0, f ( b ) > 0, then either no real root or an even number of real roots
(may be repeated) of f ( x ) = 0 lies between the numbers a and b.
CONDITION FOR THE GENERAL EQUATION OF SECOND DEGREE TO RESOLVE INTO TWO
LINEAR FACTORS
The general quadratic expression ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c in x and y may be resolved into two linear
factors if abc + 2 fgh − af 2 − bg 2 − ch 2 =
0
a h g
i.e. h b f = 0.
g f c
Case I If h 2 − ab > 0 then the given expression can be resolved into two real linear factors.
JEEMAIN.GURU
Quadratic Equations 27
Say f ( x ) = = > 0, p ( x ) ≠ 0
p ( x ) ( x − b1 )m1 ( x − b2 )m2 ...... ( x − bt )mt
<0
≥0
≤0
i = 1, 2,3.....k
(i) where ai ,b j are real number such that ai ≠ b j
j = 1, 2,3.....t
(ii) p1 , p2 ... pk , m1 , m2 ... ...mt are natural numbers
(iii) a1 , < a2 < a3 ... < ak and b1 < b2 < b3 ... < bt
Steps to find the sign of the function in an interval
(a) All zeroes of g ( x ) and p ( x ) are marked on the number line.
(b) All zeroes of p ( x ) are known as points of discontinuities
(c) g ( x ) and p ( x ) should not contain common zeroes
(d) Even and odd powers of all factors of g ( x ) and p ( x ) should be noted.
(e) Now after marking of the numbers on the number line put positive sign ( +ve ) in the right of the
biggest of these number.
(f) When passing through the next point the polynomial changes its sign if its power is odd and
polynomial have the same sign if its power is even and continue the process by the same rule.
(g) The solution of f ( x ) > 0 is the union of all those, intervals in which we have put + ve sign.
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(h) The solution of f ( x ) <0 is the union of all those intervals in which we have put negative
(minus) sign.
( x − 3) ( x + 2 ) ( x − 4 )
1 1 1
=
Example : (i) Let f ( x ) >0
( x + 1) ( x − 6 )
1 1
+ + +
− − −
−2 −1 3 4 6
In our problem each and every factor occurs odd time so interval gets alternatively + and –sign
∴ f ( x ) > 0∀x ∈ ( −2, −1) ∪ ( 3, 4 ) ∪ ( 6, ∞ )
f ( x ) < 0∀x ∈ ( −∞, −2 ) ∪ ( −1,3) ∪ ( 4, 6 )
( x − 2 ) ( x + 1) ( x − 1/ 2 ) ( x + 8)
998 249 87 6
+ + + + +
− − −
−8 −2 −1 0 1/ 2 2 3
( f ( x ))
n
we put = y and solve a0 y 2 + a1 y + a2 =
0 to obtain its roots y1 and y2 .
Finally, to obtain solutions of (1) we solve,
( f ( x ))
n
= y1
( f ( x ))
n
and = y2
3. An equation of the form
( )( ) (
ax 2 + bx + c1 ax 2 + bx + c2 ... ax 2 + bx + cn =
A )
where c1 , c2 ,..., cn , A ∈ R, can be solved by putting ax 2 + bx =
y.
4. An equation of the form
JEEMAIN.GURU
Quadratic Equations 29
( x − a )( x − b )( x − c )( x − d ) =
Ax 2
ab
where ab = cd , can be reduced to a product of two quadratic polynomials by putting y= x + .
x
5. An equation of the form
( x − a )( x − b )( x − c )( x − d ) =
A
where a < b < c < d , b − a = d − c can be solved by change of variable
( x − a) + ( x − b) + ( x − c) + ( x − d ) 1
= x − (a + b + c + d )
y =
4 4
6. A polynomial f ( x, y ) is said to be symmetric if f= ( x, y ) f ( y, x ) ∀ x, y.
A symmetric polynomials can be represented as a function of x + y and xy.
USE OF CONTINUITY AND DIFFERENTIABILITY TO FIND ROOTS OF A POLYNOMIAL
EQUATION
Let f ( x=
) a0 x n + a1 x n−1 + a2 x n−2 + ... + an−1 x + an , then f is continuous on R
Since f is continuous on R, we may use the intermediate value theorem to find whether or not f has a real
root. If there exists a and b such that a < b and f ( a ) f ( b ) < 0, then there exists at least one c ∈ ( a, b ) such
that f ( c ) = 0. Also, if f ( −∞ ) and f ( a ) are of opposite signs, then at least one root of f ( x ) = 0 lies in
( −∞, a ) . Also, if f ( a ) and f ( ∞ ) are of opposite signs, then at least one root of f ( x ) = 0 lies in ( a, ∞ ) .
Result 1 If f ( x ) = 0 has a root α of multiplicity r (where r > 1 ), then we can write
f ( x=
) ( x −α ) g ( x)
r
where g (α ) ≠ 0.
Also, f ′ ( x ) = 0 has α as a root with multiplicity r − 1.
Result 2 If f ( x ) = 0 has n real roots, then f ′ ( x ) = 0 has at least ( n − 1) real roots.
It follows immediately using Result 1 and Rolle’s Theorem.
Result 3 If f ( x ) = 0 has n distinct real roots, we can write
f ( x ) =a0 ( x − α1 )( x − α 2 ) ... ( x − α n )
where α1 , α 2 ,....α n are n distinct roots of f ( x ) = 0.
f ′( x) n
1
We can also write =∑
f ( x) k =1 x − αk
JEEMAIN.GURU
Determinants Chapter 5
DEFINITION
Any n × n ( = n 2 ) numbers arranged in a square form along horizontal rows and vertical columns having n
elements in each row and column enclosed between two vertical bars is called a determinant of order n .
EVALUATION OF DETERMINANTS
(a ∈ C∀i, j )
a11 a12
A determinant of order two is written as ij
a21 a22
and is equal to a 11 a 22 – a 12 a 21 .
A determinant of order three is written as
a11 a12 a13
a21 a22 a23 (a ij ∈ C∀i, j )
a31 a32 a33
and is equal to
a a23 a a23 a a22
a11 22 − a12 21 + a13 21
a32 a33 a31 a33 a31 a32
= a11 ( a22 a33 − a23 a32 ) − a12 ( a21a33 − a23 a31 ) + a13 ( a21a32 − a22 a31 )
= a11a22 a33 + a12 a23 a31 + a13 a21a32 − a31a22 a13 − a32 a23 a11 − a12 a21a33
A determinant of order 3 can also be evaluated by using the following diagram, due to Sarrus:
a11 a12 a13 a11 a12
Determinants 31
PROPERTIES OF DETERMINANTS
1. Reflection property : The determinant remains unaltered if its rows are changed into columns and the
columns into rows.
2. All-zero property : If all the elements of a row (column) are zero, then the determinant is zero.
3. Proportionality [Repetition] Property : If the elements of a row (columns) are proportional [identical]
to the element of some other row (columns), then the determinant is zero.
4. Switching Property : The interchange of any two rows (columns) of the determinant changes its sign.
5. Scalar Multiple property : If all the elements of a row (column) of a determinant are multiplied by a
non-zero constant, then the determinant gets multiplied by the same constant.
a1 + b1 c1 d1 a1 c1 d1 b1 c1 d1
6. Sum Property : a2 + b2 c2 d2 = a2 c2 d 2 + b2 c2 d2
a3 + b3 c3 d3 a3 c3 d3 b3 c3 d3
7. Property of Invariance :
a1 b1 c1 a1 + α b1 + β c1 b1 c1 a1 + α a2 + β a3 b1 + α b2 + β b3 c1 + α c2 + β c3
a2 b2 c2 = a2 + α b2 + β c2 b2 c2 = a2 b2 c2
a3 b3 c3 a3 + α b3 + β c3 b3 c3 a3 b3 c3
That is , a determinant remains unaltered under an operation of the form
Ci → Ci + α C j + β Ck , where j , k ≠ i, or an operation of the form
R i → Ri + α R j + β Rk , where j , k ≠ i,
8. Factor property : If a determinant ∆ becomes zero when we put x = α , then ( x − α ) is a factor of ∆ .
9. Triangle Property : If all the elements of a determinant above or below the main diagonal consists of
zeros, then the determinant is equal to the products of diagonal elements. That is,
a1 a2 a3 a1 0 0
=
0 b2 b3 =
a2 b2 0 a1 b2 c3
0 0 c3 a3 b3 c3
10. Product of Two Determinants :
a1 b1 c1 α1 β1 γ 1 a1α1 + b1β1 + c1γ 1 a1α 2 + b1β 2 + c1γ 2 a1α 3 + b1β3 + c1γ 3
a2 b2 c2 × α 2 β 2 γ 2 = a2α1 + b2 β1 + c2γ 1 a2α 2 + b2 β 2 + c2γ 2 a2α 3 + b2 β3 + c2γ 3
a3 b3 c3 α 3 β3 γ 3 a3α1 + b3 β1 + c3γ 1 a3α 2 + b3 β 2 + c3γ 2 a3α 3 + b3 β3 + c3γ 3
Here we have multiplied rows by rows. We can also multiply rows by columns, or columns by rows, or
columns by columns.
11. Conjugate of a Determinant : If ai , bi , ci ∈C ( i =1, 2,3) ,
a1 b1 c1 a1 b1 c1
=
and Z =
a2 b2 c2 then Z a2 b2 c2
a3 b3 c3 a3 b3 c3
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f (r ) a l
SUMMATION OF DETERMINANTS : Let ∆ r =g ( r ) b m where a, b, c, l , m and n are constants.
h (r ) c n
n
∑ f (r )
r =1
a l
n n
Then r ∑∆
=r 1 =r 1
=∑ g ( r ) b m . Here function of r can be elements of only one row or one column.
n
∑ h (r )
r =1
c n
DIFFERENTIATION OF A DETERMINANT
1. Let ∆ ( x ) be a determinant of order two. If we write ∆ x =C1 C2 , where C1 and C2 denote the 1st and
2nd columns, then
∆′ (=
x) C1' C2 + C1 C2'
where Ci' denotes the column which contains the derivative of all the functions in the i th column Ci .
R R' R
In a similar fashion, if we write ∆ ( x ) =1 , then ∆′ ( x ) = 1 + 1' .
R2 R2 R2
2. Let ∆ ( x ) be a determinant of order three. If we write ∆ ( x ) =C1 C2 C3 , then
∆ '( x)
= C '
1 C2 C3 + C1 C '
2 C3 + C1 C2 C '
3
R1 R1' R1 R1
and similarly if we consider ∆ ( x ) =R2 . Then ∆′ ( x ) = R2 + R2' + R2 .
R3 R3 R3 R3'
3. If only one row (or column) consists functions of x and other rows (or columns) are constant, viz. Let
f1 ( x ) f 2 ( x ) f3 ( x ) f1′( x ) f 2′ ( x ) f3′ ( x )
∆ ( x ) =b1 b2 b3 . Then ∆′ ( x ) = b1 b2 b3
c1 c2 c3 c1 c2 c3
f1n ( x ) f 2n ( x ) f3n ( x )
and in general ∆ n ( x ) = b1 b2 b3 , where n is any positive integer and f n ( x ) denotes the
c1 c2 c3
nth derivative of f ( x ) .
INTEGRATION OF A DETERMINANT
f ( x) g ( x) h ( x)
Let ∆ ( x ) =a b c , where a, b, c, l , m and n are constants.
l m n
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Determinants 33
∫ f ( x ) dx ∫ g ( x ) dx ∫ h ( x ) dx
b b b
a a a
∫ ∆ ( x ) dx =
b
⇒ a b c
a
l m n
Note :
If the elements of more than one column or rows are functions of x then the integration can be done only
after evaluation/expansion of the determinant.
Note :
a11 a12 a13 C11 C12 C13
Result; Suppose ∆ = a21 a22 a23 then ∆1 = C21 C22 C23 = ∆ 2
a31 a32 a33 C31 C32 C33
where C ij denotes the co-factor of the element a ij in ∆ .
SOME SPECIAL DETERMINANTS
1. Symmetric determinant
A determinant is called symmetric determinant if for its every elements
a h g
a=
ij a ji ∀ i, j e.g ., h b f
g f c
2. Skew-symmetric determinant : A determinant is called skew symmetric determinant if for its every
0 3 −1
element aij =
−a ji ∀ i, j , e.g . −3 0 5
1 −5 0
Note: Every diagonal element of a skew symmetric determinant is always zero.
The value of a skew symmetric determinant of even order is always a perfect square and that
of odd order is always zero.
3. Cyclic order : If elements of the rows (or columns) are in cyclic order i.e.,
1 a a2
(i) 1 b b2 = ( a − b )( b − c )( c − a )
2
1 c c
1 1 1
(ii) a b c = ( a − b )( b − c )( c − a )( a + b + c )
a 3 b3 c3
1 1 1
(iii) a2 b2 c 2 = ( a − b )( b − c )( c − a )( ab + bc + ca ) .
a3 b3 c3
a b c
(iv) b c a = (
− a 3 + b3 + c 3 − 3abc )
c a b
JEEMAIN.GURU
CRAMER’S RULE
a1 b1 c1
=If ∆ a2 b2 c2 ≠ 0 then the solution of the system of liner equations
a3 b3 c3
a1 x + b1 y + c1 z =
d1
a2 x + b2 y + c2 z =d2
a3 x + b3 y + c3 z =
d3
is given by
∆1 ∆2 ∆3
=x = ,y = ,z where
∆ ∆ ∆
d1 b1 c1 a1 d1 c1 a1 b1 d1
=∆1 d2 =
b2 c2 , ∆ 2 a2 d2 =
c2 and ∆ 3 a2 b2 d2
d3 b3 c3 a3 d3 c3 a3 b3 d3
CONDITIONS FOR CONSISTENCY
The following cases may arise :
1. If ∆ ≠ 0, then the system is consistent and has a unique solution, which is given by Cramer’s rule :
∆1 ∆2 ∆3
= x = , y = ,z .
∆ ∆ ∆
2. If ∆ =0 and atleast one of the determinants ∆1 , ∆ 2 , ∆ 3 is non-zero, the given system is inconsistent i.e. it
has no solution.
3. If ∆ =0 and ∆1 =∆ 2 =∆ 3 =0, then the system is consistent and dependent, and has infinitely many
solutions.
HOMOGENEOUS AND NON-HOMOGENEOUS SYSTEM
The system of linear homogeneous equations
a1 x + b1 y + c1 z = 0
a2 x + b2 y + c2 z = 0
a3 x + b3 y + c3 z = 0
has a non-trivial solution (i.e., at least one of the x, y, z is different from zero)
a1 b1 c1
=
if ∆ a= 2 b2 c2 0
a3 b3 c3
If ∆ ≠ 0 , then the only solution of the above system of equations is x = 0, y = 0 and z = 0.
Corollary If at least one of x, y, z, is non-zero and x, y and z are connected by the three given
a1 b1 c1
equations, then the elimination of x, y and z leads to the relation a2 b2 c2 = 0 .
a3 b3 c3
JEEMAIN.GURU
35
Matrices Chapter 6
DEFINITION
A rectangular arrangement of numbers (which may be real or complex numbers) in rows and columns, is
called a matrix. This arrangement is enclosed by small ( ) or big [ ] brackets. The numbers are called the
elements of the matrix or entries in the matrix.
ORDER OF A MATRIX
A matrix having m rows and n columns is called a matrix of order m × n or simply m × n matrix (read as an m
by n matrix). A matrix A of order m × n is usually written in the following manner
a11 a12 a13 ...a1 j ...a1n
a
21 a22 a23 ...a2 j ...a2 n
... ... ... ... ... i = 1, 2,... m
A= or A = aij m×n , where
ai1 ai 2 ai 3 ...aij ain j = 1, 2,... n
... ... ... ... ...
am1 am 2 am 3 ...amj ...amn
Here aij denotes the element of i th row and j th column.
3 −1 5
Example : order of matrix is 2 × 3 .
6 2 −7
A matrix of order m × n contains mn elements. Every row of such a matrix contains n elements and every
column contains m elements.
EQUALITY OF MATRICES
Two matrices A and B are said to be equal matrix if they are of same order and their corresponding elements are
equal.
TYPES OF MATRICES
1. Row matrix : A matrix is said to be a row matrix or row vector if it has only one row and any number of
columns.
Example : [5 0 3] is a row matrix of order 1× 3 and [2] is a row matrix of order 1×1 .
2. Column matrix : A matrix is said to be a column matrix or column vector if it has only one column and
any number of rows.
2
Example : 3 is a column matrix of order 3 ×1 and [ 2] is a column matrix of order 1×1 . Observe that
−6
[ 2] is both a row matrix as well as a column matrix.
3. Singleton matrix : If in a matrix there is only one element then it is called singleton matrix. Thus,
A = aij is a singleton matrix, if m= n= 1 .
m×n
0 0 0 0 0
Example : [ 0] , , 0 0 0 , [ 0 0] are all zero matrices, but of different orders.
0 0
5. Square matrix : If number of rows and number of columns in a matrix are equal, then it is called a square
matrix. Thus A = aij is a square matrix if m = n .
m×n
Matrices 37
Lower triangular matrix : A square matrix aij is called the lower triangular matrix, if
(ii)
=aij 0 when i < j.
1 0 0
Example : 2 3 0 is a lower triangular matrix of order 3 × 3.
4 5 2
A triangular matrix is said to be strictly triangular if aii = 0 for 1 ≤ i ≤ n.
ELEMENTARY TRANSFORMATION OF ELEMENTARY OPERATIONS OF A MATRIX
The following three operations applied on the rows (columns) of a matrix are called elementary row (column)
transformations.
1. Interchange of any two rows (columns)
It ith row (column) of a matrix is interchanged with the jth row (column), it will denoted by
Ri ↔ R j ( Ci ↔ C j ) .
2 1 3 2 1 3
Example : A= −1 2 1 , then by applying R2 ↔ R3 , we get B = 3 2 4
3 2 4 −1 2 1
2. Multiplying all elements of a row (column) of a matrix by a non-zero scalar.
If the elements of ith row (column) are multiplied by non-zero scalar k, it will be denoted by
Ri → Ri ( k ) Ci → Ci ( k ) or Ri → k Ri [Ci → k Ci ] .
3 2 −1 3 2 −1
Example : A = 0 1 2 , then by applying R2 → 3R2 , we obtain B = 0 3 6
−1 2 −3 −1 2 −3
3. Adding to the elements of a row (column), the corresponding elements of any other row (column)
multiplied by any scalar k.
If k times the elements of jth row (column) are added to the corresponding elements of the ith row
(column), it will be denoted by Ri → Ri + k R j ( Ci → Ci + k C j ) .
TRACE OF A MATRIX
The sum of diagonal elements of a square matrix A is called the trace of matrix A, which is denoted by tr A.
n
tr A = ∑ aii = a11 + a22 + ... ann
i =1
MULTIPLICATION OF MATRICES
Two matrices A and B are conformable for the product AB if the number of columns in A (pre-multiplier) is
Thus if A [ =
same as the number of rows in B (post multiplier).= air ]m×n and B brj are two matrices of
n× p
order m × n and n × p respectively, then their product AB is of order m × p and is defined as AB = Cij
m× p
r =1
Now we define the product of a row matrix and a column matrix.
JEEMAIN.GURU
b1
b
Let A = [ a1b1...an ] be a row matrix and B = ..2 be a column matrix.
.
.
bn
Then AB= [ a1b1 + a2 b2 + ... + an bn ]
PROPERTIES OF MATRIX MULTIPLICATION
If A . B. and C are three matrices such that their product is defined, then
1. AB ≠ BA ( Generally not commutative )
2. ( AB ) C = A ( BC ) ( Associative Law )
3. IA= A= AI , where I is identity matrix for multiplication.
4. A ( B + C ) = AB + AC , ( Distributive Law )
5. = AC ⇒ B = C ,
If AB ( Cancellation law is not applicable )
6. If AB = 0 it does not mean that A= 0 or B = 0, again product of two non zero matrix may be a zero
matrix.
7. If AB = BA then matrices A and B are called commutative matrices
8. If AB = − BA then A and B are called anti-commutative matrices.
POSITIVE INTEGRAL POWERS OF A MATRIX
The positive integral power of a matrix A are defined only when A is a square matrix.
Also then=A2 A. A= , A3 A=. A. A A2 A. also for any positive integers m and n,
(A=
) (A )
n m
(i) Am An = Am + n (ii) m
A=
mn n
Matrices 39
a h g
Example : h b f
g f c
2. Skew –symmetric matrix : A square matrix A, A = aij is called skew-symmetric matrix if aij = −a ji
for all i, j or AT = − A.
0 h g
f
Example : −h 0
− g − f 0
All principle diagonal elements of a skew-symmetric matrix are always zero because for any diagonal
element.
aii =− aii ⇒ aii = 0
PROPERTIES OF SYMMETRIC AND SKEW -SYMMETRIC MATRICES
1. If A is a square matrix, then A + AT , AAT , AT A are symmetric, while A − AT is skew – symmetric matrix.
2. If A is symmetric matrix, then − A, KA, AT , An , A−1 , BT AB are symmetric matrices where n ∈ N , K ∈ R
and B is a square matrix of order that of A .
3. If A is a skew – symmetric matrix, then
(i) A2n is a symmetric matrix for n ∈ N
(ii) A2 n +1 is a skew – symmetric matrix for n ∈ N
(iii) kA is also skew – symmetric matrix, where k ∈ R. .
(iv) BT AB is also skew-symmetric matrix where B is a square matrix order that of A .
4. If A, B are two symmetric matrices, then
(i) A ± B, AB + BA are also symmetric matrices,
(ii) AB − BA is a skew – symmetric matrix,
(iii) AB is also skew – symmetric when AB = BA .
5. If A, B are two skew – symmetric matrices, then
(i) A ± B , AB − BA, are skew symmetric matrices, then
(ii) AB + BA is a symmetric matrix.
6. If A is a skew – symmetric matrix and C is a column matrix, then C T AC is a zero matrix.
7. Every square matrix A can uniquely be expressed as sum of a symmetric and skew-symmetric matrix
1 1
( )
i.e., A = A + AT + A − AT .
2 2
(
)
SINGULAR AND NON SINGULAR MATRIX
Any square matrix A is said to be singular if A = 0, and a square matrix A is said to be non singular if
A ≠ 0, Here A means corresponding determinant of square matrix A.
2 3 2 3
Example : A= then A = =10 − 12 =
−2 ⇒ A is a non-singular matrix.
4 5 4 5
HERMITIAN AND SKEW-HERMITIAN MATRIX
A square matrix A = Aij is said to be Hermitian Matrix
( )
T
if aij =
a ji ; ∀i, j i.e., A =
A .
JEEMAIN.GURU
3 3 − 4i 5 + 2i
a b + ic
Example : b − ic , 3 + 4i
5 −2 + i
d
5 − 2i − 2 − i 2
If A is a Hermitian matrix then a= ii a ii ⇒ aii is real ∀i , thus every diagonal element of a Hermitian matrix
must be real.
A square matrix, A = aij is said to be Skew – Hermitian if a ji = −a ij .∀i, j i. e. AT =− A. If A is skew-
Hermitian matrix, then aii =−a ii ⇒ aii + a ii must be purely imaginary or zero.
3i −3 + 2i −1 − i
0 −2 + i
Example : 2 − i , 3 + 2i −2i −2 − 4i
0
1 − i 2 − 4i 0
ORTHOGONAL MATRIX
A square matrix A is called orthogonal if AAT= I= AT A e.i.if A−1= AT
cos α − sin α cos α sin α
Example : sin α cos α is orthogonal = because A−1 = A
T
− sin α cos α
In fact every unit matrix is orthogonal. Determinant of orthogonal matrix is –1 or 1
IDEMPOTENT MATRIX
A square matrix A is called an idempotent matrix if A2 = A
1 1 1 + 1 1 +1 1 1
Example : 2 2 is an idempotent matrix, because 2 4
= A
4 = 4 4 = 2 2
A
1 1 1 + 1 1 +1 1 1
2 2 4 4 4 4 2 2
1 0 0 0
= Also A = and B are idempotent matrices because = A2 A=or B 2 B
0 0 0 1
In fact every unit matrix is idempotent
INVOLUTORY MATRIX
A square matrix A is called an involutory matrix = if A2 I= or A−1 A
1 0 1 0
Example : A= is an involutory matrix because = A2 = I
0 1 0 1
In fact every unit matrix is involutory.
NILPOTENT MATRIX
A square matrix A is called a nilpotent matrix if there exists a p ∈ N such that A p = 0 .
0 0 0 0
Example : A= is a nilpotent matrix because= A2 = 0 (Here P = 2)
1 0 0 0
Determinant of every nilpotent matrix is 0.
UNITARY MATRIX
A square matrix is said to unitary, if
=A ' A I=
since A ' A=
and A ' A =
A ' A therefore =
if A' A I we have A ' A 1.
Thus he determinant of unitary matrix is of unit modulus. For a matrix to be unitary it must be non-singular.
Hence A ' A = I ⇒ A A'= I
JEEMAIN.GURU
Matrices 41
PERIODIC MATRIX
A matrix A will be called a periodic matrix if there exists a positive integer k such that Ak +1 = A . If however k
is the least positive integer for which Ak +1 = A , then k is said to be the period of A.
DIFFERENTIATION OF A MATRIX
f ( x ) g ( x ) dA f ' ( x ) g ' ( x )
If A = then is a differentiation of matrix A.
h ( x) l ( x) dx h ' ( x ) l ' ( x )
x 2 sin x dA 2 x cos x
= If A = then
dx 2 0
Example :
2 x 2
CONJUGATE OF A MATRIX
The matrix obtained from any given matrix A containing complex number as its elements, on replacing its
elements by the corresponding conjugate complex numbers is called conjugate of A and is denoted by A
1 + 2i 2 − 3i 3 + 4i 1 − 2i 2 + 3i 3 − 4i
A =− 4 + 5i 5 − 6i 6 + 7i
Example : 4 5i 5 + 6i 6 − 7i then A =
8 7 + 8i 7 8 7 − 8i 7
PROPERTIES OF CONJUGATES
1. ( A) = A 2. ( A + B) =A + B
3. (α A) = α A, α being any number 4. ( AB ) = AB, A and B being conformable for multiplication.
TRANSPOSE CONJUGATE OF A MATRIX
The transpose of the conjugate of a matrix A is called transpose conjugate of A and is denoted by Aθ . The
conjugate of the transpose of A is the same as the transpose of the conjugate of A
i.e. (=A ') = ( )
A ' Aθ
= aij then Aθ b ji=
If A = n×m where b ji aij
m×n
T
a11 a12 a13 C11 C12 C13 C11 C21 C31
If
A = a21 a22 a= adj A = C21 C22 C23
23 , then C12 C22 C32
a31 a32 a33 C31 C32 C33 C13 C23 C33
where Cij denotes the cofactor of aij in A .
PROPERTIES OF ADJOINT MATRIX
If A, B are square matrices of order n and I n is corresponding unit matrix, then
(i) A ( adjA
= ) =
A In ( adj A) A (Thus A (adj A) is always a scalar matrix)
adj ( adj A ) = A
n −1 n−2
(ii) adj A = A (iii) A
adj ( adj A ) = A
( n −1)
( )
adj AT = ( adj A )
2
T
(iv) (v)
adj ( AB ) = ( adj B )( adj A ) adj ( A ) ( adj A )
m
(vi) (vii) = m
,m∈ N
=
(viii) adj ( kA ) k n −1 ( adj A ) , k ∈ R (ix) adj ( I n ) = I n
(x) adj ( O ) = O (xi) A is symmetric ⇒ adj A is also symmetric
(xii) A is diagonal ⇒ adj A is also diagonal.
(xiii) A is triangular ⇒ adj A is also triangular.
(xiv) A is singular ⇒ adj A = 0
Note : The adjoint of a square matrix of order 2 can be easily obtained by interchanging the diagonal
elements and changing the signs of off-diagonal (left hand side lower corner to right hand side
upper corner) elements.
INVERSE OF A MATRIX
A non -singular square matrix of order n is invertible if there exists a square matrix B of the same order such
that AB= I= n BA .
In such a case, we say that the inverse of A is B and we write A−1 = B . The inverse of A is given
1
A−1 = adj A . The necessary and sufficient condition for the existence of the inverse of a square matrix A
A
is that A ≠ 0.
PROPERTIES OF INVERSE MATRIX
If A and B are invertible matrices of the same order, then
1. Every invertible matrix possesses a unique inverse.
(A ) = A
−1
−1
2.
(A ) =(A )
−1 T
T −1
3.
4. (Reversal Law) If A and B are invertible matrices of the same order, then AB is invertible and
( AB )
−1
= B −1 A−1
In general, if A, B, C ,... are invertible matrices then
( ABC ...)
−1
= ...C −1 B −1 A−1.
(A ) = ( A ) , k ∈ N In particular ( A ) ( )
A−1
−1 k −1 2
−1
5. k 2
=
adj ( A ) = ( adj A )
−1 −1
6.
JEEMAIN.GURU
Matrices 43
−1 1 −1
7. A= = A
A
8. diag ( a1−1 a2−1...an−1 )
A = diag ( a1a2 ...an ) ⇒ A−1 =
9. A is symmetric ⇒ A−1 is also symmetric
10. A is diagonal, A ≠ 0 ⇒ A−1 is also diagonal.
11. A is a scalar matrix ⇒ A−1 is also a scalar matrix.
12. A is triangular, A ≠ 0 ⇒ A−1 is also triangular.
MINOR OF A MATRIX
Definition : Let A be a m × n matrix. If we retain any r rows and columns of A we shall have a square sub-
matrix of order r. The determinant of the square sub-matrix of order r is called a minor of A order r. Consider
1 3 4 5
any matrix A which is of the order of 3 × 4say, A = 1 2 8 7 . It is 3 × 4 matrix so we can have minor, of
1 5 0 1
order 3, 2 or 1 . Taking any three rows and three columns we get a minor of order three. Hence a minor of
1 3 4
order 3 = 1 2 8
1 5 0
Similarly we can consider any other minor of order 3. Minor of order 2 is obtained by taking any two
rows and any two columns.
1 3
Minor of order 2 = Minor of order 1 is every matrix of any element of the matrix.
1 2
RANK OF A MATRIX
The rank of a given matrix A is said to be r if
1. Every minor of A of order r + 1 is zero.
2. There is at least one minor of A of order r which does not vanish. Here we can also say that the rank of a
matrix A is r, if
(i) Every square sub matrix of order r + 1 is singular.
(ii) There is at least one square sub matrix of order r which is non-singular.
The rank r of matrix A is written as p ( A ) = r Echelon form of a matrix.
A matrix A is said to be in Echelon form if either A is the null matrix or A satisfies the following conditions :
1. Every non- zero row in A precedes every zero row.
2. The number of zeros before the first non zero element in a row is less than the number of such zeros in the
next row.
Rank of a matrix in Echelon form: The rank of a matrix in Echelon form is equal to the number of non
zero rows in that matrix.
Homogeneous and non- homogeneous systems of linear equations : A system of equations AX = B is
called a homogeneous system if B = O . If B ≠ O , it is called a non-homogeneous system of equations.
e.g., 2x + 5 y = 0
3x − 2 y = 0
is a homogeneous system of linear equations whereas the system of equations given by
e.g., 2x + 3y = 5
x+ y = 2
JEEMAIN.GURU
45
n
S= 2a + ( n − 1) d
2
n
Note :
n
(a) If S n be the sum of n terms of an A.P. whose first term is a and last term is l then=
Sn ( a + l ).
2
(b) If we are given the common difference d, number of terms n and the last term l then
n
S= 2l − ( n − 1) d
2
n
(c) t= n S n − S n −1.
PROPERTIES OF A.P.
I. If a1 , a2 , a3 ,..., an are in A.P., then
1. a1 + k , a2 + k ,..., an + k are also in A.P.
2. a1 − k , a2 − k ,..., an − k are also in A.P.
3. ka1 , ka2 ,..., kan are also in A.P.
a a a
4. 1 , 2 ,..., n , k ≠ 0 are also in A.P.
k k k
II. If a1 , a2 , a3 ,...and b1 , b2 , b3 , ... are two A.P.s, then
1. a1 + b1 , a2 + b2 , a3 + b3 ,... are also in A.P.
2 a1 − b1 , a2 − b2 , a3 − b3 ,... are also in A.P.
III. If a1 , a2 , a3 ,..., an are in A.P., then
1. a1 + an = a2 + an −1 = a3 + an − 2 = ...
ar − k + ar + k
= 2. ar , 0 ≤ k ≤ n − r.
2
3. If number of terms of any A.P is odd, then sum of the terms is equal to product of middle term and
number of terms.
4. If number of terms of any A.P. is odd then its middle terms is A.M. of first and last term.
ARITHEMETIC MEAN (A.M.)
SINGLE ARITHMETIC MEAN
A number A is said to be the single A.M. between two given numbers a and b if a, A, b are in A.P.
n ARITHMETIC MEANS
The numbers A1 , A2 ,..., An are said to be the n arithmetic means between two given numbers a and b if
a, A1 , A2 ,..., An , b are in A.P. For example, since 2,4,6,8,10,12 are in A.P., therefore 4,6,8,10 are the four
arithmetic means between 2 and 12.
INSERTING SINGLE A.M. BETWEEN TWO GIVEN NUMBERS
Let a and b be two given numbers and A be the A.M. between them. Then, a ,A, b are in A.P.
a+b
∴ A − a = b − A or 2 A = a + b, ∴ A = .
2
INSERTINS n-ARITHMETIC MEANS BETWEEN TWO GIVEN NUMBERS
Let A1 , A2 ,..., An be the n arithmetic means between two given numbers a and b.
Then a, A1 , A2 ,... An , b are in A.P. Now, =
b ( n + 2 ) th term of A.P.
b−a
b =a + ( n + 1) d ∴ d = , where d is common difference of A.P.
n +1
JEEMAIN.GURU
a1 a2 a3
2. , , ,... are also in G.P.
k k k
1 1 1
3. , , ,... are also in G.P.
a1 a2 a3
II. If a1 , a2 , a3 ,... and b1 , b2 , b3 ,... are two G.P. s, then,
1. a1b1 , a2b2 , a3b3 ,... are also in G.P.
a a a
2. 1 , 2 , 3 ,... are also in G.P.
b1 b2 b3
III. 1. In a finite G.P. the product of terms equidistant from the beginning and the end is always the same
and is equal to the product of the first and last term. If a1 , a2 , a3 , ..., an be in G.P
Then, a1a= n a2 an −=
1 a3 an −=
3 ...= ar ⋅ an − r +1
=
2. ar ar − k ar + k , 0 ≤ k ≤ n − r.
3. If the terms of a given G.P. are chosen at regular intervals. Then the new sequence so formed also
forms a G.P.
4. If a1 , a2 , a3 ....an ... is a G.P. of non-zero, non-negative terms, then log a1 , log a2 , log a3 ,..log an ,.... is
an A.P and vice-versa.
5. If first term of a G.P. of n terms is a and last term is l , then the product of all terms of the G.P. is
( al )
n/2
6. If there be n quantities in G.P. whose common ratio is r and S m denotes the sum of the first m
r
terms, then the sum of their product taken two by two is S n S n −1.
r +1
7. If a x1 , a x2 , a x3 ,...., a xn are in G.P. then x1 , x2 , x3 ,....xn will be in A.P.
GEOMETRIC MEAN (G.M) : SINGLE GEOMETRIC MEAN
A number G is said to be the single geometric mean between two given numbers a and b if a, G, b are in G.P.
For example, since 2, 4,8 are in G.P., therefore 4 is the G.M. between 2 and 8 .
n-GEOMETRIC MEANS
The numbers G1 , G2 ,..., Gn are said to be the n geometric means between two given positive numbers
a and b if a, G1 , G2 ,..., Gn , b are in G.P.
For example, since 1, 2, 4,8,16 are in G.P., therefore 2,4, 8 are the three geometric means between 1
and 16 .
INSERTING SINGLE G.M. BETWEEN TWO GIVEN NUMBERS
Let a and b be two given positive numbers and G be the G.M. between them. Then a, G, b are in G.P.
G b
∴ = or G 2 = ab, ∴ G = ab
a G
INSERTING n GEOMETRIC MEANS BETWEEN TWO GIVEN NUMBERS
Let G1 , G2 , G3 ,..., Gn , be the n geometric means between two given numbers a and b.
Then, a, G1 , G2 , G3 ,...., Gn , b are in G.P. Now , =
b ( n + 2 ) th term of G.P. = ar n +1 , where r is the common
ratio
1 1 2 n
b
n +1 b n +1
b n +1
b n +1
b n +1
∴r = or r= ∴ G1 =ar =a =
, G2 = a
ar 2
=
, ..., Gn = a
arn
Note :
1. The product of n geometric means between two given numbers is nth power of the single G.M. between
them i.e. if a and b are two given numbers and G1 , G2 ,...., Gn are n geometric means between them,
( )
n
then G1 G2 G3 ... Gn = ab .
2. If A and G are respectively arithmetic and geometric means between two positive numbers a and b then
(a) A > G
(b) The quadratic equation having a, b as its roots is x 2 − 2 Ax + G 2 =
0
(c) the two positive numbers are A ± A2 − G 2 .
HARMONIC PROGRESSION
A sequence of non –zero numbers a1 , a2 , a3 ,... is said to be a harmonic progression if the sequence
1 1 1
, , ,... is an A.P.
a1 a2 a3
n TH TERM OF AN H.P.
1
nth term of H.P. =
nth term of the correspondingA.P.
HARMONIC MEAN (H.M)
SINGLE HARMONIC MEAN
A number H is said to be the single harmonic mean between two given numbers a and b if a, H, b are in
1 1 1 1 1 1
H.P. For example, since , , are in H.P., therefore, is the H.M. between and
2 3 4 3 2 4
n-HARMONHIC MEANS
The number H1 , H 2 ,..., H n are said to be the n harmonic means between two given numbers a and b if
1 1 1 1 1
a, H1 ,H 2 ,...,H n , b are in H.P. i.e. , , ,..., , are in A.P.
a H1 H 2 Hn b
INSERTING SINGLE H.M. BETWEEN TWO GIVEN NUMBERS
Let a and b be two given numbers and H be the H.M .between them .Then, a, H, b are in H.P.
2ab
∴H = .
a+b
INSERTING n– HARMONIC MEANS BETWEEN TWO GIVEN NUMBERS
Let H1 ,H 2 ,...H n be the n harmonic means between two given numbers a and b.
Then, a,H1 ,H 2 ,...,H n , b are in H.P.
1 1 1 1 1
∴ , , , ... , are in A.P.
a H1 H 2 Hn b
a −b
d= . where d is common difference of the corresponding A.P.
ab ( n + 1)
ab ( n + 1) ab ( n + 1) ab ( n + 1)
H1 = , H2 = , ..., Hn = .
bn + a 2a + ( n − 1) b na + b
Note :
1 1 1 1 1 1 2ac
1. Three numbers a, b, c are in H.P. if and only if , , are in A.P. i.e. =+ 2. =i .e . b .
a b c a c b a+c
2. No term of H.P. can be zero
JEEMAIN.GURU
2. If A1 , A2 be twoA.M ' s G1 , G2 be two G.M.’s and H 1, H 2 be two H.M;s between two numbers a and b,
G1G2 A + A2
then, = 1 .
H1 H 2 H1 + H 2
3. Recognization of A.P., G.P., H.P., : If a, b, c are thee successive terms of a sequence.
a −b a
If = , then a, b c, are in A. P.
b−c a
a −b a
If = , then a, b, c are in G. P.
b−c b
a −b a
If = , then a, b, c are in H. P.
b−c c
4. The sum of the fourth power of first n natural numbers
= ∑ n 4 = 14 + 24 + 34 + ........ n 4
=
( )
n ( n + 1)( 2n + 1) 3n 2 + 3n − 1
30
n ( n + 1)
2
Sn =
ab bdr 1 − r
+
( n −1
−
)
a + ( n − 1) d br n
1− r (1 − r ) 1− r
2
If the difference of the successive terms of a series in Arithmetic series or Geometric series we can find
nth term of series by the following steps :
Step I : Denote the nth term by Tn and the sum of the series upto n terms by S n
Step II : Rewrite the given series with each term shifted by one place to the right.
Step III : By subtracting the later series from the former, find Tn .
Step IV : From Tn , S n can be found by appropriate summation.
Lagrange’s identity
(x )( y ) −(x y + x y + ..... + xn yn )
2 n
2
1 + x22 + x32 + ... + xn2 2
1 + y22 + y32 + ..... + yn 1 1 2 2
( x1 y2 − x2 y1 ) ( x1 y3 − x3 y1 ) ( x1 y4 − x4 y1 ) ( x1 yn − xn y1 )
2 2 2 2
= + + +....... +
( x2 y3 − x3 y2 ) ( x2 y4 − x4 y2 ) +....... + ( x2 yn − xn y2 )
2 2 2
+ +
( x3 y4 − x4 y3 ) +....... + ( x3 yn − xn y3 )
2 2
+
+ ( xn −1 yn − xn yn −1 )
2
.........................
Note : For a odd positive integer to be perfect square. It is necessary that it should be of the form 8k + 1.
JEEMAIN.GURU
53
Inequalities Chapter 8
1. ELEMENTARY RESULTS
(i) If a > b and b > c, then a > c. In general, If a1 > a2 , a2 > a3 , ... , an −1 > an , then a1 > an .
(ii) If a > b, then a ± c > b ± c, for each c ∈ R.
a b
(iii) If a > b and c > 0, then ac > bc and > .
c c
a b
(iv) If a > b and c < 0, then ac < bc and < .
c c
1 1
(v) If a > b > 0, then < .
a b
(vi) = If ai > bi > 0 for i 1, 2,....., n, then a1a2 .....an > b1b2 ....bn .
1
(vii) If 0 < a < 1 and r is a positive real number, then 0 < a r < 1 < r .
a
1
(viii) If a > 1 and r is a positive real number, then a r > 1 > r .
a
1 1
(ix) If 0 < a < b and r is positive real number, then a r < b r and r > r .
a b
r r
a a b
Also, 0 < < 1, 0 < < 1 < .
b b a
(x) Triangle Inequality a + b ≤ a + b , a, b ∈ R More generally,
a1 + a2 + .... + an ≤ a1 + a2 + .... + an .
a, if a ≥ 0
(xi) For a real number ‘a’, a =
−a,if a < 0
∴ a = max .{a, −a} .
(xii) − a ≤ a ≤ a , ∀a ∈ R.
(xiii) If b ≥ 0, then x − a ≤ b ⇔ a − b ≤ x ≤ a + b.
(xiv) If a > 1 and x > y > 0, then log a x > log a y.
(xv) If 0 < a < 1 and x > y > 0, then log a x < log a y.
1
( a1 + a2 + ..... + an ) , G = ( a1a2 ...., an ) n
1
2. For n positive numbers a1 , a2 ,...., an , A=
n
n
H= are called the Arithmetic Mean, the Geometric Mean and the Harmonic Mean
1 1 1
+ + .... +
a1 a2 an
respectively. Note. A ≥ G ≥ H The equality holds only when all the numbers are equal.
Also, A, G and H lie between the least and the greatest of a1 , a2 ,...., an .
3. For n positive real number a1 , a2 ,...., an and n positive real numbers w1 , w2 ,...., wn
JEEMAIN.GURU
w1a1 + .... + wn an 1
Aw =
w1 + .... + wn
w1
and Gw a1 ....an ( wn w +....+ w
1 n
)are called weighted A.M and weighted G.M. of
4. CAUCHY-SCHWARZ INEQUALITY
For real number a1 , a2 ,...., an and
b1 , b2 ,...., bn ( a1b1 + a2b2 + .... + anbn ) ≤ ( a12 + a22 + .... + an2 )( b12 + b22 + .... + bn2 )
2
a1 a2 a
The equality holds iff = = ..... = n.
b1 b2 bn
5. TCHEBYCHEF INEQUALITY
For any real numbers a1 , a2 ,...., an and b1 , b2 ,...., bn such that a1 ≤ a2 ≤ ..... ≤ an and b1 ≤ b2 ≤ ..... ≤ bn ,
n ( a1b1 + a2b2 + .... + anbn ) ≥ ( a1 + a2 + .... + an )( b1 + b2 + .... + bn )
The equality holds if either all a’s or all b’s are equal. Moreover,
a1b1 + a2b2 + .... + anbn a1 + a2 + .... + an b1 + b2 + ....bn
≥ × .
n n n
6. WEIERSTRASS’ INEQUALITY’
If a1 , a2 , an are n positive real numbers, then for n ≥ 2
(1 + a1 )(1 + a2 ) (1 + an ) > 1 + a1 + a2 + + an and
if a1 , a2 , an are n positive real numbers less then 1, then
(1 − a1 )(1 − a2 ) (1 − an ) > 1 − a1 − a2 − − an
Arithmetic Mean of mth Power :
Let a1 , a2 ,......, an be n positive real numbers (not all equal) and let m be a real number, then
a1m + a2m + ..... + anm a1 + a2 + ..... + an
m
> if m ∈ R − [ 0, 1]
n n
a1m + a2m + ..... + anm a1 + a2 + ..... + an
m
LOGARITHM
1
1. log a β n = log a n
β
2. a=
log c b
blogc a , ( a, b, c > 0 and c ≠ 1)
3. Usually log a n = n log a but if n is even log a n = n log a
JEEMAIN.GURU
Inequalities 55
LOGARITHMIC INEQUALITIES
a > mb , if m > 1
8. If log m a > b ⇒
a < m , if 0 < m < 1
b
a < mb , if m > 1
9. If log m a < b ⇒
a > m , if 0 < m < 1
b
(ii) Each combination corresponds to many permutations. For example, the six permutations ABC,
ACB, BCA, BAC, CBA and CAB correspond to the same combination ABC.
Number of Combinations without repetition
The number of combinations (selections or groups) that can be formed from n different objects taken
n!
r ( 0 ≤ r ≤ n ) at a time is nCr = .
r !( n − r ) !
SOME USEFUL RESULTS OF PERMUTATIONS
n!
1. n
Pr =
( n − r )!
= n ( n − 1)( n − 2 ) ... ( n − r + 1) , 0 ≤ r ≤ n.
2. Number of Permutations of n different things taken all at a time n Pn = n !
3. The number of permutations of n things, taken all at a time, out of which p are alike and are of one type,
n!
q are alike and are of second type and rest are all different = .
p !q !
4. The number of permutations of n different things taken r at a time when each thing may be repeated any
number of times is n r .
Conditional Permutations
1. Number of permutations of n dissimilar things taken r at a time when p particular things always occur
= n − pCr − p ⋅ r !.
2. Number of permutations of n dissimilar things taken r at a time when p particular things never occur
= n − pCr r !.
3. The total number of permutations of n different things taken not more than r at a time, when each thing
n ( n r − 1)
may be repeated any number of times, is .
n −1
4. Number of permutations of n different things, taken all at a time, when m specified things always come
together is m ! × ( n − m + 1) !.
5. Number of permutations of n different things, taken all at a time, when m specified things never come
together is n !− m ! × ( n − m + 1) !.
6. Let there be n objects, of which m objects are alike of one kind, and the remaining (n – m) objects are
alike of another kind. Then, the total number of mutually distinguishable permutations that can be formed
n!
from these objects is .
( m !) × ( n − m )!
The above theorem can be extended further i.e., if there are n objects, of which p 1 are alike of one kind;
p 2 are alike of another kind; p 3 are alike of 3rd kind; ….; p r are alike of rth kind such that
n!
p1 + p2 + .... + pn =n; then the number of permutations of these n objects is .
( p1 !) × ( p2 !) × .... × ( pr !)
Circular Permutations
In circular permutations, what really matters is the position of an object relative to the others.
Thus, in circular permutations, we fix the position of the one of the objects and then arrange the other
objects in all possible ways.
There are two types of circular permutations :
JEEMAIN.GURU
(i) The circular permutations in which clockwise and the anticlockwise arrangements give rise to
different permutations, e.g. seating arrangements of persons round a table.
(ii) The circular permutations in which clockwise and the anticlockwise arrangements give rise to same
permutations, e.g. arranging some beads to form a necklace.
Difference between Clockwise and Anti-Clockwise Arrangement
If anti-clockwise and clockwise order of arrangement are not distinct e.g., arrangement of beads in a
necklace, arrangement of flowers in garland etc. then the number of circular permutations of n distinct
items is
( n − 1)! .
2
Note :
(a) The number of circular permutations of n different objects is ( n – 1)!.
(b) The number of ways in which n persons can be seated round a table is ( n – 1) !.
1
(c) The number of ways in which n different beads can be arranged to form a necklace, is ( n − 1) !.
2
(d) Number of circular permutations of n different things, taken r at a time, when clockwise and
n
P
anticlockwise orders are taken as different is r .
r
(e) Number of circular permutations of n different things, taken r at a time, when clockwise and
n
pr
anticlockwise orders are not different is .
2r
PROPERTIES OF n Cr
1. n
Cr = nCn − r 2. =
n
C0 n
=
Cn 1, n=
C1 n
n +1
3. If nC x = nC y then either x = y or x + y =n. 4. n
Cr + nCr −1 = Cr
5. n n −1Cr −1
r ⋅ nCr =⋅ 6. n ⋅ n −1Cr −1 = ( n − r + 1) nCr −1
n
Cr n − r +1
7. n
=
Cr −1 r
8. If n is even then the greatest value of nCr is nCn / 2 .
9. If n is odd then the greatest value of nCr is nC n +1 or nC n −1 .
2 2
n = (n 1 + n 2 + ….) Things, when n 1 are alike of one kind, n 2 are alike of second kind, and so on
is {( n1 + 1)( n2 + 1) ....} − 1.
(iv) The number of selections of r objects out of n identical objects is 1
(v) Total number of selections of zero or more objects from n identical objects is n + 1.
(vi) The number of selections taking at least one out of a1 + a2 + a3 + .... + an + k objects, where a 1 are
alike (of one kind),a 2 are alike (of second kind ) and ….a n are alike (of nth kind) and k are distinct
=( a1 + 1)( a2 + 1)( a3 + 1) ..... ( an + 1) 2k − 1.
Conditional Combinations
1. The number of ways in which r objects can be selected from n different objects if k particular objects are
(i) Always included = n − k Cr − k (ii) Never included = n − k Cr
2. The number of combinations of n objects, of which p are identical, taken r at a time is
n− p
Cr + n − p Cr −1 + n − p Cr − 2 + .... + n − p C0 , if r ≤ p and
n− p
Cr + n − p Cr −1 + n − p Cr − 2 + .... + n − p Cr − p , if r > p.
Division into groups
Case I : 1. The number of ways in which n different things can be arranged into r different groups is
n + r −1
pn or n ! n −1Cr −1 according as blank group are or not admissible.
2. The number of ways in which n different things can be distributed into r different groups is
r n − r C1 ( r − 1) + r C2 ( r − 2 ) − .... + ( −1) Cr −1 or Coefficient of x n is n !( e x − 1) .
n n r −1 r r
4. Number of ways in which m × n different objects can be distributed equally among n persons
(or numbered groups) = (number of ways of dividing into groups) × (number of groups) !
= =
( mn )!n ! ( mn )! .
( m !) n ! ( m !)
n n
5. If N = a p b q c r ..... where a, b, c are distinct primes and p, q, r... are any positive integers
then number of all positive integers which are less than N and are prime to it are
1 1 1
N 1 − 1 − 1 − ..... It is called Euler’s Totient function.
a b c
Case II : The number of ways in which ( m + n ) different things can be divided into two groups which
m+n
contain m and n things respectively is, = Cm .n Cn
( m + n )! , m ≠ n.
m !n !
Corollary : If m = n , then the groups are equal size. Division of these groups can be given by two types.
Type I : If order of group is not important : The number of ways in which 2n different things can be
=
( 2n )! ×=2!
2n !
.
2!( n !) ( n !)
2 2
2. The number of ways in which (m + n + p) different things can be divided into three groups
Type II : If order of group is important : The number of ways in which 3p different things can be divided
(i) If order of group is not important : The number of ways in which m n different things can be
mn !
divided equally into n groups is = .
( m !) n !
n
(ii) If order of group is important : The number of ways in which m n different things can be
DERANGEMENT
Any change in the given order of the things is called a derangement. If n things form an arrangement in a row,
the number of ways in which they can be deranged so that no one of them occupies its original place is
1 1 1 n 1
n !1 − + − + .... + ( −1) . .
1! 2! 3! n!
NUMBER OF RECTANGLES AND SQUARES
n
(a) Number of rectangles of any size in a square of size n × n is = ∑r
r =1
3
n
and number of squares of any size is = ∑r .r =1
2
np
(b) Number of rectangles of any size in a rectangle of size n × p ( n < p ) is = ( n + 1)( p + 1)
4
n
and number of squares of any size=
is ∑ ( n + 1 − r )( p + 1 − r ) .
r =1
EXPONENT OF PRIME p in n!
Exponent of a prime p in n ! is denoted by E p ( n !) and is given by
n n n n
E p ( n !=
) + 2 + 3 + ... + k .
p p p p
n n
where p k < n < p k +1 and denotes the greatest integer less than or equal to .
p p
For example, exponent of 3 in (100 )! is
JEEMAIN.GURU
= ( n − 1)!( a1 + a2 + ... + an )
(10 n
) . and if one of terms is zero digit then
−1
9
10n − 1 10n −1 − 1
sum = ( n − 1) !( a1 + a2 + ... + an ) − ( n − 2 ) !( a1 + a2 + .... + a )
n
a a
( n !)! is divisible by ( n !)(
n −1)!
7.
( kn )! is divisible by ( n !)
k
8.
9. n
Cr is divisible by n if n is prime ( n ≠ r )
Multinomial Theorem
Let x1 , x2 ,...., xm be integers. Then number of solutions to the equation
x1 + x2 + .... + xm = n … (i)
Subject to the condition
a1 ≤ x1 ≤ b1 , a2 ≤ x2 ≤ b2 ,...., am ≤ xm ≤ bm … (ii)
n
is equal to the coefficient of x in
(
( xa1 + xa1 +1 + .... + xb1 )( xa2 + xa2 +1 + .... + xb2 ) .... xam + xam +1 + .... + xbm ) … (iii)
This is because the number of ways, in which sum of m integers in (i) equals n, is the same as the number of
times x n comes in the expansion of (iii).
JEEMAIN.GURU
Use of solution of linear and coefficient of a power in expansions to find the number of ways of
distribution :
(i) The number of integral solutions of x1 + x2 + x3 + .... + xr = n where x1 ≥ 0, x2 ≥ 0,....xr ≥ 0 is the
same as the number of ways to distribute n identical things among r persons.
This is also equal to the coefficient of x n in the expansion of ( x 0 + x1 + x 2 + x3 + ....)
r
r
1
n
= coefficient of x in
1− x
= coefficient of x n in (1 − x ) = 1 + r C1 x + r +1C2 x 2 + r + 2C3 x 3 + .....
−r
r + n −1 n + r −1
= = Cn Cr −1
n −1
(ii) Similarly total number natural solutions of x1 + x2 + .... + xr =
n are Cr −1
(iii) The number of possible arrangements permutations of P object out of n1 identical objects of kind
1, n2 identical objects of kind 2 and so on is
x2 x n1 x2 x nk
= P ! × coefficient of x P in the expansion of 1 + x + + ... + ..... 1 + x + + ... +
2! n1 ! 2! nk !
Number of Divisors
Let N = p1α1 . p2α 2 . p3α3 .... pkα k , where p1 , p2 , p3 ,.... pk are different prime and α1 , α 2 , α 3 ,...., α k are natural number
then :
(i) The total number of divisors of N including 1 and N is = (α1 + 1)(α 2 + 1)(α 3 + 1) .... (α k + 1) .
(ii) The total number of divisors of N excluding 1 and N is = (α1 + 1)(α 2 + 1)(α 3 + 1) .... (α k + 1) − 2.
(iii) The sum of these divisors is =
( )
( p10 + p11 + p12 + .... + p1α1 )( p20 + p12 + p22 + .... + p2α2 ) .... pn0 + p1n + pn2 + .... + pnαn
(iv) The number of ways is which N can be resolved as a product of factors is
1
2 (α1 + 1)(α 2 + 1) .... (α k + 1) , if N is not a perfect square
1 (α + 1)(α + 1) .... (α + 1) + 1 , if N is a perfect square
2 1 2 k
(v) The number of ways in which a composite number N can be resolved into two factors which are
relatively prime (or co-prime) to each other is equal to 2n−1 where n is the number of different
factors in N.
The Pigeon-Hole Principle (PHP)
If more than n objects are distributed into n compartments. Some compartments must receive more than one
object.
Formally, the pigeon hole principle states the following :
If rn + 1 pigeons ( r ≥ 1) are distributed among n pigeon holes, one of the pigeon-holes will contain at least
r + 1 pigeons.
Alternatively the pigeon hole principle states the following :
n − 1
If n pigeons are placed in m pigeon-holes, then at least one pigeon hole will contain more than pigeons,
m
where [ x ] denotes the greatest integer ≤ x
JEEMAIN.GURU
65
∑ n Cr . x n − r . y r
i.e., ( x + y ) =
n
… (i)
r =0
n!
=
Here n C0 ,n C1 ,n C2 ,......n Cn are called binomial coefficients and n Cr , for 0 ≤ r ≤ n.
r !( n − r ) !
SOME IMPORTANT EXPANSIONS
1. Replacing y by − y in (i), we get
( x −=
y) C0 x n −0 y 0 − n C1 x n −1 y1 + n C2 x n − 2 y 2 − ..... + ( −1) nCr x n − r y r + .... + ( −1)
n n r n n
Cn x o y n
n
( x − y) = ∑ ( −1) nCr x n − r y r
n r
i.e.
r =0
The terms in the expansion of ( x − y ) are alternatively positive and negative, the last term is positive or
n
r =0
n
i.e. (1 − x ) = ∑ ( −1)
n r n
Cr x r
r =0
( x + y ) + (=
x − y) 2 n C0 x n y 0 + nC2 x n − 2 y 2 + nC4 x n − 4 y 4 + ..... and
n n
4.
( x + y) − ( x − y) = 2 nC1 x n −1 y1 + nC3 x n −3 y 3 + nC5 x n −5 y 5 + .....
n n
GENERAL TERM
The general term of the expansion is ( r + 1) th term usually denoted by Tr +1 and
Tr +1 = nCr x n − r y r
Note :
In the binomial expansion of ( x − y ) , Tr +1 =
( −1) nCr x n−r y r
n r
(a)
JEEMAIN.GURU
n
one middle term i.e. + 1 th term is the middle term
2
T n = Cn / 2 x n / 2 y n / 2
n
2 +1
When n is odd, then total number of terms in the expansion of ( x + y ) is n + 1 (even). So, there are two
n
2.
n +1 n+3
middle terms i.e. th and th are two middle terms.
2 2
n +1 n −1 n −1 n +1
T n +1 = C n−1 x
n 2
y 2
and T n +3 = C n +1 x
n 2
y 2
2 2 2 2
Note :
(a) When there are two middle terms in the expansion then their binomial coefficients are equal.
(b) Binomial coefficient of middle term is the greatest binomial coefficient.
PROPERTIES OF BINOMIAL COEFFICIENTS
In the binomial expansion of (1 + x )
n
(1 + x )
n
= n
C0 + nC1 x + nC2 x 2 + .... + nCr x r + .... + nCn x n
Where nC0 , nC1 , nC2 ,......., nCn are the coefficients of various powers of x and called binomial coefficients,
and they may be written as C0 , C1 , C2 ,.....Cn .
(1 + x )
n
Hence, =C0 + C1 x + C2 x 2 + .... +Cr x r + .... + Cn x n … (i)
The sum of binomial coefficients in the expansion of (1 + x ) is 2n
n
1.
Putting x = 1 in (i), we get, 2n = C0 + C1 + C2 + .... + Cn … (ii)
2. Sum of binomial coefficients with alternate signs is 0
Putting x = −1 in (i), we get, C0 − C1 + C2 − C3 + ..... =0 … (iii)
Sum of the coefficient of the odd terms in the expansion of (1 + x ) is equal to sum of the coefficients of
n
3.
even terms and each is equal to 2n−1
i.e. C0 + C2 + C4 + ... = C1 + C3 + C5 + ... = 2n −1
n n −1 n n − 1 n−2
4. n
Cr = Cr −1= ⋅ Cr − 2 and so on.
r r r −1
5. Sum of product of coefficients in the expansion
2n !
C0Cr + C1Cr +1 + ..... + Cn − r .C= 2n
Cn += … (iv)
n r
( n − r )!( n + r )!
6. Sum of squares of coefficients
JEEMAIN.GURU
Binomial Theorem 67
14.
( 2n ) !
C02 + C12 + C22 ...... + Cn2 = 2
( n !)
0, if n is odd
15. C02 − C12 + C22 − C32 + ....... =
( −1) . Cn / 2 . if n is even
n/2 n
Note :
n!
(a) The general term in the above expansion is x1r1 .x2r2 ...xkrk
r1 !r2 !...rk !
(b) The total number of terms in the above expansion is
= number of non – negative integral solutions of the equation r1 + r2 + ... + rk =n
n + k −1 n + k −1
= Cn or Ck −1
Coefficient of x1r1 .x2r2 ...xkrk in the expansion of ( a1 x1 + a2 x 2 + ... + ak xk )
n
(c)
n!
= a1r1 .a2r2 ...akrk
r1 !r2 !...rk !
Greatest coefficient in the expansion of ( x1 + x2 + ... + xk )
n
(d)
n!
= where q is the quotient and r the remainder when n is divided by k.
( q !) . ( q + 1) !
k −r r
(e) Sum of all the coefficient is obtained by putting all the variables xi equal to 1 and is n m .
In the above expansion, the first term must be unity. In the expansion of ( a + x ) , where n is either a
n
1.
negative integer or a fraction or an irrational number, we proceed as follows :
x n ( n − 1) x
n
x
n 2
n x
( a + x ) = a 1 + = a 1 + = a 1 + n. +
n
+ ...
n
a a a 2! a
x
and the expansion is valid when < 1 i.e. x < a .
a
There are infinite number of terms in the expansion of (1 + x ) , when n is a negative integer or a
n
2.
fraction.
3. If x is so small that its square and higher powers may be neglected, then approximate value of
(1 + x )
n
=
1 + nx.
GENERAL TERM IN THE EXPANSION (1 + x )
n
n ( n − 1)( n − 2 ) ... ( n − r + 1)
Tr +1 = xr
r!
SOME IMPORTANT DEDUCTIONS
1. Replacing n by −n in (1), we get
n ( n + 1) 2 n ( n + 1)( n + 2 ) 3 r n ( n + 1)( n + 2 ) ... ( n + r − 1) r
(1 + x ) =1 − nx + x + ... + ( −1)
−n
x − x + ...
2! 3! r!
r n ( n + 1)( n + 2 ) ... ( n + r − 1) r
General Term : Tr +1 = ( −1) x
r!
2. Replacing x by − x in (1), we get
n ( n − 1) 2 n ( n − 1)( n − 2 ) 3 r n ( n − 1)( n − 2 ) ... ( n − r + 1) r
(1 − x ) =1 − nx + x + ... + ( −1)
n
x − x + ...
2! 3! r!
r n ( n − 1)( n − 2 ) ... ( n − r + 1) r
General Term : Tr +1 = ( −1) x
r!
3. Replacing x by − x and n by − n in (1) , we get
n ( n + 1) n ( n + 1)( n + 2 ) n ( n + 1)( n + 2 ) ... ( n + r − 1)
(1 − x )
−n
=+
1 nx + x2 + x3 + ... + x r + ...
2! 3! r!
n ( n + 1)( n + 2 ) ... ( n + r − 1)
General Term : Tr +1 = xr
r!
NUMERICALLY GREATEST TERM IN THE EXPANSION OF (1+x)n :
Case 1: ‘n’ is + ve integer
n +1
If
x = an integer denote it by p; then Tp +1 = Tp both are greatest.
1+ x
n +1
If x = integer + proper fraction, then Tp +1 is greatest term
1+ x p
Binomial Theorem 69
n +1
If
1 + x
x = integer + proper fraction, then Tp +1 is greatest term
p
AN IMPORTANT THEOREM
( )
n
If A+B =+
I f where I and n are positive integers, If n being odd and 0 ≤ f < 1 then
(1 + f ) ⋅ f =Kn where A − B 2 = K > 0 and A − B < 1.
( ) +( )
n n
If n is even integer then A+B A−B =1 + f + f '
Hence L. H.S. and I are integers
∴ f + f ′ is also integer ⇒ f + f ′ =1 ; ∴ f ′ =(1 − f ) ,
( I + f )(1 − f ) = (I + f ) f ′ = ( )( ) ( )
n n n
Hence A+B A−B = A − B2 = K n.
SOME IMPORTANT EXPANSIONS
n ( n − 1) 2 n ( n − 1)( n − 2 ) ..... ( n − r + 1) r
(1 + x ) =+
n
(i) 1 nx + x + ..... + x + .....
2! r!
n ( n − 1) 2 n ( n − 1)( n − 2 ) .... ( n − r + 1)
(ii) (1 − x ) =1 − nx + ( − x ) + .....
n r
x − ..... +
2! r!
n ( n + 1) 2 n ( n + 1)( n + 2 ) 3 n ( n + 1) .... ( n + r − 1) r
(iii) (1 − x ) =1 + nx +
−n
x + x + .... + x + ...
2! 3! r!
n ( n + 1) 2 n ( n + 1)( n + 2 ) 3 n ( n + 1) ..... ( n + r − 1)
(iv) (1 + x ) =1 − nx + ( − x ) + .....
−n r
x − x + .... +
2! 3! r!
(v) (1 + x ) q =
− p p
1− x +
p ( p + q ) x2 − (p p + q )( p + 2 q ) x3 + .......
q q ⋅ 2q q ⋅ 2q ⋅ 3q
−p p p ( p + q ) 2 p ( p + q )( p + 2q ) 3
(vi) (1 − x ) q = 1+ x + x + x + .....
q q ⋅ 2q q ⋅ 2q ⋅ 3q
(1 + x ) = 1 − x + x 2 − x3 + ....... + ( −1) x r + .....
−1 r
(vii)
(1 − x ) =1 + x + x 2 + x3 + .......... + x r + .......
−1
(viii)
(1 + x ) = 1 − 2 x + 3x 2 − 4 x3 + ....... + ( −1) ( r + 1) x r + ......
−2 r
(ix)
(1 − x ) =1 + 2 x + 3x 2 + 4 x3 + ...... + ( r + 1) x r + .....
−2
(x)
( r + 1)( r + 2 ) ⋅ −1 r x r + ......
(1 + x ) = 1 − 3x + 6 x 2 − 10 x3 + .............. + ( )
−3
(xi)
2
( )( r + 2 )
r + 1
(1 − x )
−3
(xii) =+
1 3 x + 6 x 2 + 10 x3 + ................ + x r + ......
2
(xiii) (1 + x ) =
−n 1− n n +1
C1 x + C2 x ..... where n ∈ N
2
(xiv) (1 − x ) =
−n 1+ n
C1 x + n +1C2 x 2 + n + 2C3 x 3 ..... where n ∈ N
JEEMAIN.GURU
X‘ X
IV quadrant
III quadrant
only cos θ
only tanθ are + ve
are + ve sec θ
cotθ
Y’
Formulae for the trigonometric ratios of sum and differences of two angles
1. sin ( A=
+ B ) sin A cos B + cos A sin B
2. sin ( A=
− B ) sin A cos B − cos A sin B
3. cos ( A
= + B ) cos A cos B − sin A sin B
4. cos ( =
A − B ) cos A cos B + sin A sin B
tan A + tan B tan A − tan B
5. tan ( A + B ) = 6. tan ( A − B ) =
1 − tan A tan B 1 + tan A tan B
cot A cot B − 1 cot A cot B + 1
7. cot ( A + B ) = 8. cot ( A − B ) =
cot B + cot A cot B − cot A
9. sin ( A + B ) sin ( A − B=
) sin A − sin B= cos B − cos A
2 2 2 2
− cos ( B + A )
14. tan A − cot B =
cos A sin B
Formulae for the trigonometric ratios of sum and differences of three angle
1. sin ( A + B + C ) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C
or sin (A + B + C ) = cos A cos B cos C (tan A + tan B + tan C − tan A. tan B. tan C )
2. cos= ( A + B + C ) cos A cos B cos C − sin A sin B cos C − sin A cos B sin C − cos A sin B sin C
=
cos ( A + B + C ) cos A cos B cos C (1 − tan A tan B − tan B tan C − tan C tan A)
tan A + tan B + tan C − tan A tan B tan C
3. tan ( A + B + C ) =
1 − tan A tan B − tan B tan C − tan C tan A
cot A cot B cot C − cot A − cot B − cot C cot A + cot B + cot C − cot A cot B cot C
4. cot ( A + B + C ) = =
cot A cot B + cot B cot C + cot C.cot A − 1 1 − cot A cot B − cot B cot C − cot C cot A
In general
5. sin ( A1 + A2 + ... + An ) = cos A1 cos A2 ...cos An ( S1 − S3 + S5 − S7 + ...)
6. ( A1 + A2 + .... + An ) cos A1 cos A2 ....cos An (1 − S2 + S4 − S6 ....)
cos=
S1 − S3 + S5 − S7 + ...
tan ( A1 + A2 + ... An ) =
7.
1 − S 2 + S 4 − S6 + ...
where, = S1 tan A1 + tan A2 + ... + tan An = The sum of the tangents of the separate angles.
=S 2 tan A1 tan A2 + tan A1= tan 3 + ... The sum of the tangents taken two at time.
= S3 tan A1 tan A2 tan A3 + tan A2 ⋅ tan A3 tan A4 + ... = Sum of tangents three at a time, and so on.
If A1= A2= ...= An= A, then S1= n tan A
=S2 n
=
C2 tan 2 A, S3 n
C3 tan 3 A,...
8. =
sin nA cos n A ( n C1 tan A − n C3 tan 3 A + n C5 tan 5 A − ...)
9. cos nA =cos n A (1 − n C2 tan 2 A + n C4 tan 4 A − ...)
n
C1 tan A − n C3 tan 3 A + n C5 tan 5 A − ...
10. tan nA =
1 − n C2 tan 2 A + n C4 tan 4 A − n C6 tan 6 A + ...
( )
11.
sin β
2
cos {α + ( n − 1) ( β )}.sin ( n β )
2 2
cos (α ) + cos (α + β ) + cos (α + 2 β ) + ... + cos (α + ( n − 1) β ) =
sin ( β )
12.
2
Formulae to transform the product into sum or difference
1. = sin ( A + B ) + sin ( A − B )
2sin A cos B
2. = sin ( A + B ) − sin ( A − B )
2 cos A sin B
3. B cos ( A + B ) + cos ( A − B )
2 cos A cos =
4. B cos ( A − B ) − cos ( A + B )
2sin A sin =
JEEMAIN.GURU
C+D C−D
Let =
A + B C and =
A− B D =
Then A = and B
2 2
Formulae to transform the sum or difference into product
C+D C−D
1. sin C + sin D = 2sin cos
2 2
C+D C−D
2. sin C − sin D = 2 cos sin
2 2
C+D C−D
3. cos C + cos D = 2 cos cos
2 2
C+D D −C C+D C−D
4. cos C − cos D = 2sin sin = −2sin sin
2 2 2 2
Trigonometric ratio of multiple of an angle
2 tan A
1.= =
sin 2 A 2sin A cos A
1 + tan 2 A
1 − tan 2 A
2. cos 2 A = 2 cos 2 A − 1 = 1 − 2sin 2 A = cos 2 A − sin 2 A =2 .
1 + tan A
2 tan A
3. tan 2 A =
1 − tan 2 A
4. sin 3 A =3sin A − 4sin 3 A =4sin ( 60o − A ) .sin A.sin ( 60o + A )
5. cos 3 A = 4 cos3 A − 3cos A = 4 cos ( 60o − A ) .cos A.cos ( 60o + A )
3 tan A − tan 3 A
6. tan 3 A = 2 =− tan ( 60o A ) .tan A.tan ( 60o + A )
1 − 3 tan A
=
7. sin 4θ 4sin θ .cos3 θ − 4 cos θ sin 3 θ
8. cos 4θ = 8cos 4 θ − 8cos 2 θ + 1
4 tan θ − 4 tan 3 θ
9. tan 4θ =
1 − 6 tan 2 θ + tan 4 θ
10. sin 5 A = 16sin 5 A − 20sin 3 A + 5sin A
11. cos 5 A = 16 cos5 A − 20 cos3 A + 5cos A
Trigonometrical values
1 1 1 1 1 1
1. sin 22= ° cos 67= ° 2− 2 2. sin 67= ° cos 22= ° 2+ 2
2 2 2 2 2 2
3 −1 3 +1
3. =
sin15 ° cos= 75° 4. =
cos15 ° sin=75°
2 2 2 2
3 −1 3 +1
5. tan15° = cot 75° = = 2− 3 6. tan 75° = cot15° = = 2+ 3
3 +1 3 −1
1 ° 1 ° 1 ° 1 °
7. tan 22 =
cot 67 = 2 −1 8. tan 67 =
cot 22 = 2 +1
2 2 2 2
5 −1 10 − 2 5
9. =
sin18 ° = cos 72° 10. sin=
36° = cos 54°
4 4
5 +1 10 + 2 5
11. sin=
54° = cos36° 12. sin=
72° = cos18°
4 4
JEEMAIN.GURU
n +1
nπ ( −1) 2 sin θ , if n is odd
7. cos +θ =
2 ( −1) 2 cos θ , if n is even
n
A −1 ± 1 + tan 2 A
8. tan =
2 tan A
JEEMAIN.GURU
n −1
sin 2n A
9. ∏
r =0
cos 2 A = n
r
2 sin A
10. tan α = cot α − 2 cot 2α , cosec 2α =
cot α − cot 2α
11. sec α .sec
= (α + β ) cosec β {tan (α + β ) − tan α }
12. (α + β )
tan α .tan= cot β {tan (α + β ) − tan α } − 1
13. (α + β )
cosec α cosec= cosec β {cot α − cot (α + β )}
14. tan α .sec
= 2α tan 2α − tan α
15. tan 2 α .tan
= 2α tan 2α − 2 tan α
sin ( A + B )
16. tan A + tan B =
cos A cos B
sin ( A − B )
17. tan A − tan B =
cos A cos B
sin ( A + B )
18. cot A + cot B =
sin A sin B
sin ( B − A )
19. cot A − cot B =
sin A sin B
cos ( A − B )
20. 1 + tan A tan B =
cos A cos B
cos ( A + B )
21. 1 − tan A tan B =
cos A cos B
cos ( A − B )
22. 1 + cot A cot B =
sin A sin B
− cos ( A + B )
23. 1 − cot A cot B =
sin A sin B
24. If α + β + γ = 0 then
α β γ
(a) sin α + sin β + sin γ =
− 4sin sin sin , (b) tan α + tan β + tan γ =
tan α tan β tan γ
2 2 2
α β γ
(c) cos α + cos β + cos γ =−1 + 4 cos cos cos
2 2 2
24. tan θ + cot θ =
2cosec2θ 25. cot θ − tan θ =
2 cot 2θ
26. sin ( A + B ) sin ( A − B )= sin 2 A − sin 2 B= cos 2 B − cos 2 A .
27. cos ( A + B ) cos ( A − B=
) cos 2 A − sin 2 B= cos 2 B − sin 2 A .
α +β β +γ γ +α
28. cos α + cos β + cos γ + cos (α + β + γ ) =4 cos cos cos
2 2 2
α +β β +γ γ +α
29. sin α + sin β + sin γ − sin (α + β + γ ) =4sin sin sin
22 2
cot α + cot β + cot γ
30. tan α + tan β + tan γ − tan
= (α + β + γ ) tan α tan β tan γ 1 −
cot (α + β + γ )
31. For any α , β and γ we have the following identities.
JEEMAIN.GURU
But it is better to remember the following results instead of using above formulae in the following cases.
sin θ = 0 θ = nπ where n ∈ I
cos θ = 0 θ nπ + π / 2 where n ∈ I
=
cos θ = −1 θ
= ( 2n + 1) π where n ∈ I
sin θ = ±1 θ
= ( 2n + 1) π / 2 where n ∈ I
cos θ = ±1 θ = nπ where n ∈ I
b
⇒ θ − α= 2nπ ± β ⇒ θ= 2nπ ± β + α , where tan α = is the general solution.
a
JEEMAIN.GURU
Y
(1, π / 2 ) ( −1, π ) Y
X X
O O (1, 0)
( −1, − π / 2 )
(iii) Graph of y = tan −1 x (iv) Graph of y = cot −1 x
Y Y
y =π /2 y =π
( 0, π / 2 )
X
O
y = −π / 2 O X
( −1, π ) Y Y
(1, π / 2 )
y =π /2
X X
O
O (1, 0) ( −1, − π / 2 )
JEEMAIN.GURU
π π
(iii) tan −1 ( tan θ ) = θ if and only if − <θ < and
2 2
π y = tan −1 ( tan x )
x, 0≤ x<
2 π /2
π
π 3π 3π / 2
tan −1 ( tan x ) = x − π , <x<
2 2 π /2 2π
3π −π / 2
x − 2π , < x ≤ 2π
2
⇒ f ( x) = tan −1 ( tan x ) is periodic with period π .
y = cot −1 ( cot x )
(iv) cot −1 ( cot θ ) = θ if and only if 0 < θ < π and
π (π , π ) ( 2π , π )
x + π , −π < x < 0
cot ( cot x ) x,
= −1
0< x <π
x − π , π < x < 2π
−π O π 2π 3π
⇒ f ( x) =
cot ( cot x ) is periodic with period π
−1
π π
(v) sec −1 ( sec θ ) = θ if and only if 0 ≤ θ < or < θ ≤ π and y = sec −1 ( sec x )
2 2
π
π
− x, −π ≤ x ≤ 0, x≠−
sec −1 ( sec x ) = 2
x, π −π −π / 2 O π /2 π 3π / 2 2π
0 < x ≤π, x≠
2
⇒ f ( x) = sec ( sec x ) is periodic with period 2π .
−1
π π
(vi) cosec −1 ( cosec θ ) = θ if and only if − ≤ θ < 0 or 0 < θ ≤ and
2 2
π
x, 0< x≤ , y = cosec −1 ( cosec x )
2 π /2
π 3π
cosec −1 ( cosec x ) = π − x, <x≤ x ≠π
2 2
3π 3π / 2 2π
x − 2π , < x < 2π π /2 π
2
⇒ f ( x) = cosec −1 ( cosec x ) is periodic with period 2π .
−π / 2
JEEMAIN.GURU
Y = sin ( sin −1 x ) Y
(vii) (a) ( )
−1
sin sin x = x iff −1 ≤ x ≤ 1
Y = cos ( cos −1 x ) ( 0, 1) (1, 1)
(b) cos ( cos x ) = x iff −1 ≤ x ≤ 1
−1
( −1, 0 )
X′ X
O
(1, 0 )
( −1, − 1) ( 0, − 1)
Y′
Y
Y = tan ( tan −1 x )
(viiii) (a) ( )
tan tan −1 x = x for all x Y = cot ( cot −1 x )
(b) cot ( cot x ) = x for all x
−1
X′ X
O
Y′
Y = cosec ( cosec −1 x ) Y
Y = sec ( sec −1 x ) ( 0, 1) (1, 1)
(ix) (a) ( )
sec sec −1 x = x iff x ≥ 1 or x ≤ −1
(b) ( )
cosec cosec −1 x = x iff x ≥ 1 or x ≤ −1
X′
( −1, 0 )
X
O
(1, 0 )
( −1, − 1) ( 0, − 1)
Y′
2. (i) sin (−x) =
−1
− sin x , −1
cos ( − x ) = π − cos x
−1 −1
−1 π
= −1
sin x + cos x 2 , for all x ∈ [ −1, 1]
−1 π
3. x + cot −1 x
tan = , for all x ∈ R
2
−1 π
sec x + cosec −1 x
= , for all x ∈ ( −∞, − 1] ∪ [1, ∞ )
2
JEEMAIN.GURU
5. Conversion property :
sin −1 x
= cos −1 1 − x 2 , 0 ≤ x ≤1
(i)
sin −1=
x − cos −1 1 − x 2 , −1 ≤ x ≤ 0
−1 1 − x2
= sin x cot −1 , 0 < x ≤1
x
(ii)
−1 1 − x2
= sin x cot −1
−π , −1 ≤ x < 0
x
x
(iii) sin −1 x = tan −1 x <1
1− x
2
=cos −1 x sin −1 1 − x 2 , 0 ≤ x ≤1
(iv)
−1
cos= x π − sin −1 1 − x 2 , −1 ≤ x ≤ 0
−1 1 − x2
= cos x tan −1 , 0 < x ≤1
x
(v)
−1 1 − x2
cos=x π + tan , −1
−1 ≤ x < 0
x
x
(vi) cos −1 x = cot −1 x <1
1− x
2
−1 1
= tan x cos −1 , x≥0
1 + x2
(vii)
tan −1 x = 1
− cos −1 , x≤0
1 + x2
JEEMAIN.GURU
−1 1
= tan x cot −1 , x>0
x
(viii)
tan −1 x = 1
cot −1 − π , x<0
x
x
(ix) tan −1 x = sin −1 ∀ x∈R
1 + x2
−1 1
= cot x sin −1 , x≥0
1 + x2
(x)
cot −1 x = 1
π − sin −1 , x<0
1 + x2
−1 1
= cot x tan −1 , x>0
x
(xi)
cot −1 x = 1
π + tan −1 , x<0
x
x
(xii) cot −1 x = cos −1 ∀x ∈ R
1+ x
2
6. General values of inverse circular functions : We know that if α is the smallest angle whose sine is x,
then all the angles whose sine is x can be written as nπ + ( −1) α , where n ∈ I . Therefore, the general
n
π π
Thus, we have Sin −1 x= nπ + ( −1) α , − 1 ≤ x ≤ 1 if sin α = x and − ≤α ≤
n
.
2 2
Similarly, general values of other inverse circular functions are given as follows :
x 2nπ ± α , − 1 ≤ x ≤ 1 ;
Cos −1= If cos α= x, 0 ≤ α ≤ π
π π
Tan −1 x =nπ + α , x ∈ R ; If tan α = x, − <α <
2 2
Cot x =nπ + α , x ∈ R
−1
If cot α= x, 0 < α < π
π
−1
Sec= x 2nπ ± α , x ≤ −1 or x ≥ 1 If sec α = x, 0 ≤ α ≤ π and α ≠
2
π π
Cosec −1 x= nπ + ( −1) α , x ≤ −1 or x ≥ 1 If cosec α = x, − ≤α ≤ and α ≠ 0
n
2 2
Note : The first letter in all above inverse Trigonometric function are CAPITAL LETTER
Formulae for sum, difference of inverse trigonometric function
(1)
sin −1 x + sin
= −1
{
y sin −1 x 1 − y 2 + y 1 − x 2 ; } x ≥ 0, y ≥ 0 and x 2 + y 2 ≤ 1
{
sin −1 x + sin −1 y =π − sin −1 x 1 − y 2 + y 1 − x 2 ;
}
x ≥ 0, y ≥ 0 and x 2 + y 2 ≥ 1
cos −1 x − cos=
(4)
−1
{
y cos −1 xy + 1 − x 2 1 − y 2 ; x ≥ 0, y ≥ 0, x ≤ y}
cos −1 x − cos −1 y =
{
− cos −1 xy + 1 − x 2 1 − y 2 ; x ≥ 0, y ≥ 0, x ≥ y }
−1 −1 −1 x + y
= tan x + tan y tan ; x ≥ 0, y ≥ 0 and xy < 1
1 − xy
−1
tan x + tan y π / 2 ;
−1
=
(5) = x > 0, y > 0 and xy 1
tan −1 x + tan −1 y =+ x+ y
π tan −1 ; x ≥ 0, y ≥ 0 and xy > 1
1 − xy
x− y
(6) tan −1 x − tan −1 y =
tan −1 ; x ≥ 0, y ≥ 0
1 + xy
=
2sin
−1
(
x sin −1 2 x 1 − x 2 ) If −
1
2
≤x≤
1
2
1.
2sin x = π − sin 2 x 1 − x
−1 −1
( 2
) If
1
2
< x ≤1
2sin x =−π + sin 2 x 1 − x
−1 −1
( 2
) If −1 ≤ x ≤
−1
2
−1
3sin= −1
(
x sin −1 3 x − 4 x3 , ) If
2
≤x≤
1
2
2. −1
(
3sin x = π − sin 3 x − 4 x ,
−1 3
) If
1
2
< x ≤1
3sin x = −π − sin 3 x − 4 x ,
−1 −1
( 3
) If −1 ≤ x < −
1
2
2 cos
=
−1
(
x cos −1 2 x 2 − 1 , ) If 0 ≤ x ≤ 1
3.
2 cos x= 2π − cos 2 x − 1 ,
−1 −1
(
2
) If − 1 ≤ x ≤ 0
=
3cos
−1
(
x cos −1 4 x 3 − 3 x , ) If
1
2
≤ x ≤1
4. 3cos x = 2π − cos 4 x − 3 x ,
−1 −1
(
2
) 1
If − ≤ x ≤ ,
2
1
2
−1
3cos = (
x 2π + cos −1 4 x 3 − 3 x , ) If − 1 ≤ x ≤ −
1
2
JEEMAIN.GURU
−1 −1 2 x
= 2 tan x tan 1 − x 2 , If − 1 < x < 1
2x
5. −1
2 tan x = π + tan −1 2
, If x > 1
1− x
−1 2 x
2 tan x = −π + tan
−1
2
, If x < −1
1− x
−1 −1 2 x
= 2 tan x sin 1 + x 2 , If − 1 ≤ x ≤ 1
2x
6. −1
2 tan x = π + sin −1 2
, If x > 1
1+ x
−1 2 x
2 tan x = −π + sin
−1
2
, If x < −1
1+ x
−1 −1 1 − x
2
= 2 tan x cos 2
, If 0 ≤ x
1+ x
7.
2 tan −1 x = −1 1 − x
2
− cos 2
, If x ≤ 0
1+ x
−1 −1 3 x − x
3
−1 1
= 3 tan x tan 2
, If <x<
1 − 3x 3 3
−1 3 x − x
3
1
8. 3 tan −1
x π
= + tan 2
, If x >
1 − 3x 3
3 tan −1 x = −π + tan −1 3x − x2 ,
3
1
If x < −
1 − 3x 3
JEEMAIN.GURU
2ca
a + b2 − c2
2 B a C
3. c 2 = a 2 + b 2 − 2ab cos C ⇒ cos C =
2ab
PROJECTION FORMULAE
1.= a b cos C + c cos B
2.= b c cos A + a cos C
3.= c a cos B + b cos A
APOLLONIUS THEOREM A
α β
THE m – n Rule
If the triangle ABC , point D divides BC in the ratio m : n , and
∠ ADC = θ , then θ
(i) ( m + n ) cot θ = m cot α − n cot β ; B m D n C
A B C
1. Formulae for sin , sin , sin
2 2 2
(i) sin
A
=
( s − b )( s − c ) (ii) sin
B
=
( s − a )( s − c ) (iii) sin
C
=
( s − a )( s − b )
2 bc 2 ca 2 ab
A B C
2. Formulae for cos , cos , cos
2 2 2
A s (s − a) B s ( s − b) C s (s − c)
(i) cos = (ii) cos = (iii) cos =
2 bc 2 ca 2 ab
A B C
3. Formulae for tan , tan , tan
2 2 2
(i) tan
A
=
( s − b )( s − c ) = ( s − b )( s − c ) (ii) tan
B
=
( s − c )( s − a ) = ( s − c )( s − a )
2 s (s − a) ∆ 2 s ( s − b) ∆
(iii) tan
C
=
( s − a )( s − b ) = ( s − a )( s − b ) = (iv) cot
A
=
s (s − a) s (s − a)
2 s (s − c) ∆ 2 ( s − b )( s − c ) ∆
B s ( s − b) s ( s − b) C s (s − c) s (s − c)
= (v) cot = = (vi) cot = .
2 ( s − a )( s − c ) ∆ 2 ( s − a )( s − b ) ∆
Note : ( a + b − c )( b + c − a )( c + a − b ) = a 2b + b 2 a + a 2 c + ac 2 + b 2 c + bc 2 − a 3 − b3 − c3
OBLIQUE TRIANGLE
The triangle which are not right-angled is called oblique triangle. We can solve a triangle if we know three of
its parts at least one of which is a side. Different cases are as follows.
Case I. The three sides are given.
Case II. Two sides and included angles are given.
Case III. Two sides and the angle opposite to one of them are given.
Case IV. One side and two angles are given.
CASE I. Given the three sides, to solve the triangle.
Proof : Let ABC be the triangle in which all the tree sides a, b, c are given. The three angle A, B, C are to
be determined. A
Since, 2s = a + b + c
∴ values of s, s − a, s − b, s − c are known.
(i) To find A c b
tan =
A ( s − b )( s − c )
2 s (s − a)
B C
A 1 a
∴ =
log tan log ( s − b ) + log ( s − c ) − log s − log ( s − a )
2 2
A A
∴ log tan and ∴ can be determined with the help of tables.
2 2
∴ A is determined.
(ii) To find B
tan
B
=
( s − a )( s − c )
2 s ( s − b)
JEEMAIN.GURU
B 1
∴ =
log tan log ( s − a ) + log ( s − c ) − log s − log ( s − b )
2 2
B B
∴ log tan and ∴ can be determined with the help of tables.
2 2
∴ B is determined.
(iii) To find C
A + B + C = 180°
∴ = 180° − A + B , is determined.
C
Thus, A, B, C being known, the triangle is solve.
CASE II : Given two sides and the included angle ; to solve the triangle.
Proof : Let ABC be a triangle, in which sides b, c (b > c) and the included angle A are given.
The side a and angles B, C are to be determined.
(i) To find B and C.
B −C b−c A
tan = cot [Napier’s Analogy]
2 b+c 2
b−c B+C A B+C B+C
= = cot 90° − =
cot tan
2
tan
b+c 2 B 2
B −C B+C
∴ log tan = log ( b − c ) + log tan − log ( b + c ) A
2 2
B −C
∴ log tan and
2 c
B −C b
∴ can be obtained with the help of tables.
2
B −C
∴ is known. …(1) B
a
C
2
B+C A
Also = 90° − [ A + B + C = 180°] …(2)
2 2
∴ from (1) and (2), by addition and subtraction B and C are known.
(ii) To find a
a b
since = [Sine formula]
sin A sin B
b sin A
∴ a=
sin B
∴ log a = log b + log sin A − log sin B
∴ log a and ∴ a can be determined with the help of the table.
Thus, B, C and a are known, the triangle is solved.
Note: If C > B , then use the formula
C − B c−b A
tan = cot .
2 c+b 2
CASE III : Given one sides and two angles ; to solve the triangle.
Proof : Let ABC be a triangle in which ‘a’ be the given side and B, C be the given angles.
Sides b, c and angle A are to be determined.
(i) To find A
A + B + C = 180°
JEEMAIN.GURU
∴ =A 180° − ( B + C )
∴ A is known.
(ii) To find b
b a
Since =
sin B sin A
a sin B
b= [Sine formula]
sin A
∴ log b = log a + log sin B − log sin A
∴ With the help of tables, log b and therefore, b is determined.
(iii) To find c
c a
Again = [Sine formula]
sin C sin A
a sin C
∴ c=
sin A
∴ log c = log a + log sin C − log sin A
∴ With the help of the tables, log c and therefore c is determined.
Thus, A, b, c being known, the triangle is solved.
CASE IV : When two sides and an angle opposite to one of them is given. (Ambiguous case)
Let the two sides say a and b and an angle A opposite to a be given.
Here we use a / sin A = b / sin B .
∴ sin B = b sin A / a …(1)
We calculate angle B from (1) and then angle C is obtained by using
∠C = 180° − ( ∠ A + ∠ B ) .
Also, to find side c, we use
a / sin A = c / sin C C
∴ c = ( a sin C ) / sin A …(2)
From relation (1), the following possibilities will arise : a
Case I : When A is an acute angle. b b sin A
(a) If a < b sin A , there is no triangle. When a < b sin A , from (1),
sin B > 1 , which is impossible. Hence no triangle is possible in this A B
case. A X
N
From the following fig., if
AC =b ; ∠CAX =A ,
then perpendicular CN = b sin A . Now taking c as centre, if we draw an arc of radius a then
if the line AX and hence no triangle ABC can be constructed in this case.
(b) If a = b sin A , then only one triangle is possible which is right C
angled at B.
When a = b sin A , then from (1),
sin B = 1 , ∴ ∠B= 90°
b a = b sinA
From fig. it is clear that
CB= a= b sin A .
Thus, in this case, only one triangle is possible which is right angled A 90°
at B. A X
B
JEEMAIN.GURU
A
3. Escribed circle of a triangle and their radii
In any ∆ABC , we have
B C A/2
a cos cos
∆ 2 2 s
(i) = r1 =
s−a cos
A s–c
B M C
2
A B
= s tan = ( s − c ) cot . s–c
2 2 E1
C a F1 r1
= ( s − b ) cot =B C
r1
2 tan + tan
2 2
I1
C A
b cos cos
∆ B A C b
(ii) =r2 = 2 2 = s tan = ( s − c ) cot = ( s − a ) cot =
s −b cos
B 2 2 2 C
tan + tan
A
2 2 2
A B
c cos cos
∆ C A B c
(iii) r3 == 2 2 =
s tan = ( s − b ) cot = ( s − a ) cot =
s−c cos
C 2 2 2 A
tan + tan
B
2 2 2
1 1 1 1
(iv) r1 + r2 + r3 − r =4 R (v) + + =
r1 r2 r3 r
1 1 1 1 a 2 + b2 + c2 1 1 1 1
(vi) + + + = (vii) + + =
r 2 r12 r22 r32 ∆2 bc ca ab 2 Rr
(viii) r1r2 + r2 r3 + r3 r1 =
s2
A B C
=
(ix) ∆ =
2R 2
sin A.sin B.sin C 4 Rr cos .cos cos
2 2 2
A B C A B C A B C
=(x) r1 4=
R sin cos cos ; r2 4=
R cos .sin .cos , r3 4 R cos cos sin
2 2 2 2 2 2 2 2 2
JEEMAIN.GURU
DISTANCE OF CIRCUMCENTRE (O) FROM THE ORTHOCENTRE (H), INCENTRE (I) AND
EXCENTRES (I 1 , I 2 , I 3 )
1. Distance between circumcentre (O) and orthocentre (H)
= R 1 − 8cos A cos B cos C
OH
2. Distance between circumcentre (O) and incentre (I)
A B C
OI = R 1 − 8sin sin sin = R R − 2r
2 2 2
3. (a) Distance between circumcentre (O) and excentre (I 1 ) of the escribed circle having opposite angle A
A B C
OI1 = R 1 + 8sin cos cos = R R + 2r1
2 2 2
(b) Distance between circucentre (O) and excentre (I 2 ) of the escribed circle having opposite angle B
A B C
OI 2 = R 1 + 8cos sin cos = R R + 2r2
2 2 2
(c) Distance between circumcentre (O) and excentre (I 3 ) of the escribed circle having opposite angle C
A B C
OI 3 = R 1 + 8cos cos sin = R R + 2r3
2 2 2
DISTANCE OF INCENTRE FROM THE VERTICES OF THE TRIANGLE
A
Let I be the In-centre. Let IP. ⊥ AB. Clearly, IP = r.∠PAI =
2
From right angled triangle IPA, A
A
A r A
sin = ⇒ AI = r cosec 2
2 AI 2 P A
2
B C I
Similarly BI = r cosec and CI = r cosec
2 2
A B C B D C
=
Thus, AI r= cosec , BI r cosec and CI = r cosec .
2 2 2
Note :
(i) The centroid of any triangle divides the join of circumcentre and orthocentre internally in the ratio 1 : 2.
(ii) If H is the orthocentre of ∆ ABC and AH produced meets BC at D and the
A
circumcircle of ∆ ABC at P, then HD = DP.
H E
= BD c=cos B and DP BD cot C
=
∴ DP c cos B
=.cot C 2 R cos B cos C B D
P
(iii) The orthocentre of an acute angled triangle is the incentre of the Pedal triangle. C
(iv) The centre of the circum circle falls inside the triangle if triangle is acute angled
but outside when it is obtuse angled. If the triangle is right angled the centre lies
on mid-point of the hypotenuse.
(v) The orthocentre falls inside the triangle if triangle is acute angled and outside
when it is obtuse angled. B A
If the triangle is right angled the orthocentre (B) lies on the triangle.
JEEMAIN.GURU
PEDAL TRIANGLE
Let the perpendicular AD, BE and CF from the vertices A, B and C on the opposite sides BC, CA
and AB of ∆ ABC respectively, meet at O. Then O is the orthocentre of the ∆ ABC . The triangle DEF is
called the pedal triangle of ∆ ABC .
Orthocentre of the triangle is the incentre of the pedal triangle. If O is
A
the orthocentre and DEF the pedal triangle of the ∆ ABC , where AD, BE , CF
are the perpendiculars drawn from A, B, C on the opposite side BC, CA, AB
respectively, then F E
= (i) OA 2= R cos A, OB 2 R cos B and OC = 2 R cos C
= (ii) OD 2= R cos B cos C , OE 2 R cos C cos A O
and OF = 2 R cos A cos B B C
1. Sides and angles of a pedal triangle D
The angles of pedal triangle DEF are : 180 − 2 A, 180 − 2 B , 180 − 2C A
and sides of pedal triangle are : EF = a cos A or R sin 2 A ;
FD = b cos B or R sin 2 B ; DE = c cos C or R sin 2C .
F
If given ∆ ABC is obtuse, then angles are represented by b cos B E c cos C
2 A, 2 B, 2C − 180° and the sides are a cos A, b cos B, − c cos C . O
H
h h
β x
α β α α β
a d Q d
a = h ( cot α − cot β ) =
h sin ( β − α ) =H x cot α tan (α + β ) A a B
sin α .sin β =a h ( cot α + cot β ) , where by
h sin α + sin β cosec ( β -α ) and
∴=
=h a sin α sin β . cosec ( α + β ) and
(6) (7) y
H
α D
a β
α β
H
a sin (α + β ) C x B
H=
sin ( β − α ) α + β
= =
If AB CD.Then, x y tan
2
P
(8) P (9)
N
N h
h
W E
W β B E α O β
α A
d d B
A
S d
S d P h=
h= cot α − cot 2 β
2
cot α + cot β
2 2
P
(10) (11)
Q
h
θ
β α
δγ
α β
A Q A B
a
a γ
=AP a sin γ .cos ec (α − γ ) ,
AQ a sin δ .cos ec ( β − δ )
B
=
= =
h AP sin α α sin α .sin γ .cos ec ( β − γ ) and apply,
If AQ= d= AP cos α= α cos α .sin γ .cosec ( β − γ ) PQ 2 = AP 2 + AQ 2 − 2 AP. AQ cos θ
JEEMAIN.GURU
Mensuration Chapter 15
AREA AND PERIMETER
1. Triangle
(a) Perimeter (2s) = a + b + c
(b) Area = s ( s − a )( s − b )( s − c ) c
b
1 height
(c) Area =× base × height where a, b, c are sides
2
a
a+b+c
of the triangle and S is semi-perimeter, S = base
2
2. Right-angled triangle
(a) Perimeter = b + p + h
1
(b) Area = × b × p p
2 h
(c) Hypotenuse (h) = b 2 + p 2
where b = base, p = perpendicular, h = hypotenuse b
3. Right-angle Isosceles triangle
(a) Hypotenuse = a2 + a2 =2a
2a
(b) Perimeter = 2a + 2a a
1 1 1
(c) Area = × base × height = × a × a = a 2
2 2 2
a
4. Equilateral triangle
(a) Perimeter = 3a
3 a a
(b) Height = a h
2
3 2
(c) Area = a a
4
5. Rectangle
(a) Perimeter = 2 (l+ b)
b
(b) Diagonal = l 2 + b2
(c) Area = l × b
l
6. Square a
(a) Perimeter = 4a
(b) Diagonal (d) = 2a a a
a
JEEMAIN.GURU
Mensuration 97
1 2
(c) Area = a 2 = d
2
7. Parallelogram
(a) Perimeter = 2 (a + b) h b
(b) Area = Base × Height
a
8. Rhombus
1
(a) Side = d12 + d 22 d1
2
(b) Perimeter = 2 d12 + d 22
d2
1
(c) Area = d1d 2
2
b
9. Trapezium
1
(a) Area = (a + b) × h c h d
2
(b) Perimeter = a + b + c + d
a
10. Quadrilateral C
c
1
(a) Area = × AC × ( h1 + h2 ) D
2 h1 b
d h2
(b) Perimeter = a + b + c + d
A a B
11. Circle
(a) Diameter (d) = 2r
(b) Circumference= 2= πr πd
πd 2
(c) Area = π r 2 =
4 r
π r2
(d) Area of semi-circle =
2
πr 2
(e) Area of quadrant =
4
12. Sector
θ 1
(a) Area of sector (A) = ×π r2 = lr r
360 2
θ r θ
(b) Length of arc (l) = × 2π r
360
13. Regular Polygon l
1
(a) Area = number of sides × radius of the inscribed circle
2
n−2
(b) Vertex angle of a regular polygon (=
θ) ×180°
n
JEEMAIN.GURU
(ii) Cube
(a) Volume = a 3
(b) Surface Area = 6a 2
a
(c) Diagonal = 3a
a
a
(iii) Right Circular Cylinder
(a) Volume = π r 2 h
(b) Curved Surface Area = 2π rh h
(c) Total Surface Area = 2π r ( r + h )
(d) Area of each end or base area = π r 2
r
(iv) Right Circular Cone
1
(a) Volume = π r 2 h
3
l
(b) Slant height = r 2 + h 2 h
(c) Curved Surface Area = π rl
(d) Base Area = π r 2 r
(e) Total Surface Area = π r ( r + l )
(v) Sphere
4 3 r
(a) Volume = πr
3
(b) Total Surface Area = 4π r 2
(vi) Hemisphere
2 3
(a) Volume = πr
3 r
(b) Total surface area = 3π r 2
(c) Curved Surface Area = 2π r 2
h
(d) Spherical Gap of Radius ‘r’ and Height ‘h’
Volume = 1/ 3π h 2 ( 3r − h ) r O
Surface area = 2 π rh
JEEMAIN.GURU
Mensuration 99
πh
(b) Volume (V) =
3
( R 2 + r 2 + Rr ) h l
A. Prism
(a) Volume of a Right Prism = Area of the base × Height
(b) Lateral Surface Area = Perimeter of the Base × Height
B. Pyramid
1 O
(a) Volume of a Pyramid = × Area of the base × height
3
(b) The whole surface area of a Pyramid is the sum of the
areas of the base and the lateral surface areas.
D
C
E
Q P
A B
JEEMAIN.GURU
Function Chapter 16
DEFINITION OF FUNCTION
Let X and Y be any two non-empty sets. “A function from X to Y is a rule or correspondence that assigns
to each element of set X, one and only one element of set Y’’. Let the correspondence be ‘f’’ then
=
mathematically we write f :X → Y where y f ( x ) , x ∈ X and y ∈ Y . We say that ‘y’ is the image of ‘x’
under f (or x is the pre image of y).
It is important to note that
(i) A mapping f : X → Y is said to be a function if each element in the set X has its image in set Y.
It is also possible that there may be few elements in set Y which are not the images of any
element in set X.
(ii) Every element in set X should have one and only one image. That means it is impossible to have
more than one image for a specific element in set X. Functions can not be multi-values
(A mapping that is multi-valued is called a relation from X and Y) e.g.
Set X Set Y Set X Set Y
a 1 a 1
b 2 b 2
c 3 c 3
Function Function
X´ O X
X´ O X X´ O X X´ O X
Y´ Y´ Y´ Y´
(i) (ii) (iii) (iv)
Figure (i) and (ii) are representing a graph while figure (iii) and (iv) are representing a function.
JEEMAIN.GURU
Function 101
NUMBER OF FUNCTIONS
Let X and Y be two finite sets having m and n elements respectively. Then each element of set X can be
associated to any one of n elements of set Y . So, total number of functions from set X to set Y is n m .
VALUE OF THE FUNCTION
If y = f ( x ) is an function then to find its values at some value of x, say x = a , we directly substitute x = a in
its given rule f ( x ) and it is denoted by f ( a ) .
e.g., If f ( x=
) x 2 + 1 , then f (1) = 12 + 1 = 2 , f ( 2 ) = 22 + 1 = 5 , f ( 0 ) = 02 + 1 = 1 etc.
DOMAIN, CO-DOMAIN AND RANGE OF FUNCTION
If a function f is defined from a set A to set B then for f : A → B set A is called the domain of function f and set
B is called the co-domain of function f. The set of all f-images of the elements of A is called the range of
function f.
If nothing is specified about domain and codomain then
Domain = All possible values of x for which f (x) exists.
And co-domain is taken to be R
Range = All possible values of f(x), x in D f .
A B A B
f
Domain Range
a p =
Domain {=
a , b, c , d } A
b q =
Co- Domain {= p, q, r , s} B
c r
d s Range = { p, q, r }
Co-Domain
INTERVALS
There are four types of interval.
1. Open interval : Let a and b be two real numbers such that a < b , then the set of a<x<b
all real numbers lying strictly between a and b is called an open interval and is
denoted by ] a, b [ or ( a, b ) . a b
Open interval
Thus, ] a, b [ or ( a, b ) = { x ∈ R : a < x < b} .
a≤ x≤b
2. Closed interval : Let a and b be two real numbers such that a < b , then the set of
all real numbers lying between a and b including a and b is called a closed interval [ ]
a b
and is denoted by [ a, b ] .
Closed interval
Thus, [ a, b] = { x ∈ R : a ≤ x ≤ b} . a< x≤b
( f ± g )( x ) = f ( x ) ± g ( x ) .
3. Multiplication of functions
(=fg )( x ) (=gf )( x ) f ( x ) g ( x ) .
4. Division of functions
f f ( x)
( x) = .
g g ( x)
SOME IMPORTANT DEFINITIONS
1. Real numbers : Real numbers are those which are either rational or irrational. The set of real numbers is
denoted by R.
2. Related quantities : When two quantities are such that the change in one is accompanied by the change
in other, i.e., if the value of one quantity depends upon the other, then they are called related quantities.
3. Variable : A variable is a symbol which can assume any value out of a given set of values.
(i) Independent variable : A variable which can take any arbitrary value, is called independent
variable.
(ii) Dependent variable : A variable whose value depends upon the independent variable is called
dependent variable.
4. Constant : A constant is a symbol which does not change its value, i.e., retains the same value throughout
a set of mathematical operation. These are generally denoted by a, b, c etc. There are two types of
constant, absolute constant and arbitrary constant.
5. Equal functions : Two function f and g are said to be equal functions, if and only if
(i) Domain of f = Domain of g .
(ii) Co-domain of f = Co-domain of g .
(iii) f ( x=
) g ( x ) ∀ x ∈their common domain.
6. Real valued function : If R , be the set of real numbers and A, B are subsets of R , then the function
f : A → B is called a real function or real-valued function.
1. One-one function (injection or injective function) : A function f : A → B is said to be a one-one
function or an injection, if different elements of A have different images in B. Thus, f : A → B is
one-one.
a ≠ b ⇒ f ( a ) ≠ f ( b ) for all a, b ∈ A
⇔ f ( a=) f ( b ) ⇒=
a b for all a, b ∈ A
e.g. let f : A → B and g : X → Y be two functions represented by the following diagrams.
A B X Y
f b1 x1 g y1
a1
b2 x2 y2
a2 b3 y3
a3 b4 x3 y4
a4 b5 x4 y5
Function 103
Y Y
(0, 1)
X´ X X´ X
O
f ( x=
) ax + b
O f ( x=
) a x ( 0 < a < 1)
Y´ Y´
(c) All even functions are many one.
(d) All polynomials of even degree defined in R have at least one local maxima or minima
and hence are may one in the domain R. Polynomials of odd degree may be one-one or
many-one.
(e) If f is a rational function then f ( x1 ) = f ( x2 ) will always be satisfied when x1 = x2 in
the domain. Hence we can write f ( x1 ) − f ( x2 ) =
( x1 − x2 ) g ( x1 , x2 ) where g ( x1 , x2 )
is some function in x1 and x2 . Now if g ( x1 , x2 ) = 0 gives some solution which is
different from x1 = x2 and which lies in the domain, then f is many-one else one-one.
(f) Draw the graph of y = f ( x ) and determine whether f ( x ) is one-one or many-one.
(ii) Number of one-one functions (injections or injective functions) : If A and B are finite sets
having m and n elements respectively, then number of one-one function from A to B
n Pm , if n ≥ m
=
0, if n < m
2. Many-one function : A function f : A → B is said to be a many-one function if two or more
elements of set A have the same image in B.
Thus, f : A → B is a many-one function if there exist x, y ∈ A such that x ≠ y but
f ( x) = f ( y) .
In other words f : A → B is a many-one function if it is not a one-one function.
A B X g Y
f
a1 b1 x1 y1
a2 b2 x2 y2
a3 b3 x3 y3
a4 b4 x4 y4
a5 b5 x5 y5
b6
Note :
JEEMAIN.GURU
(a) If function is given in the form of set of ordered pairs and the second element of atleast two
ordered pairs are same then function is many-one.
(b) If the graph of y = f ( x ) is given and any line parallel to x-axis with in the codomain cuts the
curve in its domain at more than one point then function is many-one.
Y Y
X´ X X´ X
O f ( x ) = x2 O f ( x) = x
Y´ Y´
3. Onto function (surjection) or Surjective function : A function f : A → B is onto if each element
of B has its pre-image in A. Therefore, if f −1 ( y ) ∈ A, ∀ y ∈ B then function is onto. In other words.
Range of f = Co-domain of f e.g. The following arrow-diagram shows onto function.
A B X Y
f g
a1 b1 x1 y1
x2 y2
a2 b2 x3
a3 b3 x4 y3
Number of onto functions (surjections) : If A and B are two sets having n and r elements
respectively such that r ≤ n , then number of onto functions from A to B is
r n − r C1 ( r − 1) + r C2 ( r − 2 ) ......... + ( −1)
n n r −1 r
Cr −1 .
4. Into function : A function f : A → B is an into function if there exists an element in B having
no pre-image in A.
In other words, f : A → B is an into function if it is not an onto function e.g. The fig (i), arrow-
diagram shows an into function.
A B X Y
f g
a1 b1 x1 y1
x2 y2
a2 b2
y3
a3 b3 x3 y4
Function 105
Note :
(a) An into function can be made onto by redefining the codomain as the range of the
original function.
(b) Any polynomial function f : R → R is onto if degree is odd; into if degree of f is
even.
5. One-one onto function (bijection) or Bijective Function : A function f : A → B is a bijection if it
is one-one as well as onto.
In other words, a function f : A → B is a bijection if
(i) It is one-one i.e., f ( =
x ) f ( y ) ⇒=
x y for all x, y ∈ A .
(ii) It is onto if for all y ∈ B , there exists x ∈ A such that f ( x ) = y .
A B
f
a1 b1
a2 b2
a3 b3
a4 b4
Function 107
11. Transcendental function : A function which is not algebraic is called a transcendental function
e.g., trigonometric; inverse trigonometric, exponential and logarithmic functions are all
transcendental functions.
(i) Trigonometric functions : A function is said to be a trigonometric function if it involves
circular functions (sine, cosine, tangent, cotangent, secant, cosecant) of variable angles.
(ii) Inverse trigonometric functions
Definition
Function Domain Range
of the function
y = sin −1 x
sin −1 x [ −1, 1] [ −π / 2, π / 2] ⇔x= sin y
y = cos −1 x
cos −1 x [ −1, 1] [0, π ] ⇔ x= cos y
y = tan −1 x
tan −1 x ( −∞, ∞ ) or R ( −π / 2, π / 2 ) ⇔ x= tan y
y = cot −1 x
−1
cot x ( −∞, ∞ ) or R ( 0, π ) ⇔ x= cot y
y = cosec −1 x
cosec −1 x R − ( −1, 1) [ −π / 2, π / 2] − {0} ⇔ x= cosec y
y = sec −1 x
sec −1 x R − ( −1, 1) [0, π ] − {π / 2} ⇔ x= sec y
(iii) Exponential function : Let a ≠ 1 be a positive real number. Then f : R → ( 0, ∞ ) defined
by f ( x ) = a x called exponential function. Its domain is R and range is ( 0, ∞ ) .
Y Y
a>1 0 < a <1
(0, 1) f ( x) = ax f ( x) = ax (0, 1)
X´ X X´ X
O O
Y´ Y´
=
Graph of f ( x ) a x , when a > 1 =
Graph of f ( x ) a x , when 0 < a < 1
(1, 0)
X´ X X´ X
O (1, 0) O
f ( x ) = log a x
Y´ Y´
=
Graph of f ( x ) log a x, when a > 1 =
Graph of f ( x ) log a x, when 0 < a < 1
JEEMAIN.GURU
12. Explicit and implicit functions : A function is said to be explicit if it is expressed directly in
terms of the independent variable. If the function is not expressed directly in terms of the
=
independent variable or variables, then the function is said to be implicit e.g. y sin −1 x + log x is
( a x ) ( b − y ) are implicit functions.
2 2
xy and x 3 y 2 =−
explicit function, while x 2 + y 2 =
Y
14. Identity function : The function defined by f ( x) = x for
all x ∈ R , is called the identity function on R. Clearly, the domain f ( x) = x
and range of the identity function is R.
The graph of the identity function is a straight line passing X´ X
O
through the origin and inclined at an angle of 45° with positive
direction of x-axis.
Y´
Y
15. Modulus function : The function defined by
x, when x ≥ 0
f (=
x) = x is called the modulus
− x, when x < 0 f ( x) = −x f ( x) = x
function. The domain of the modulus function is the set R of X´ X
all real numbers and the range is the set of all non-negative real O
numbers.
Y´
Y
16. Signum function : The function defined by
x 1, x > 0 (0, 1)
, x≠0
f ( x) = x or= f ( x ) =0, x 0
0, x=0 −1, x < 0 X´ X
O
is called the signum function. The domain is R and the range is the (0, –1)
set {−1, 0, 1} .
Y´
JEEMAIN.GURU
Function 109
Y
17. Reciprocal function : The function that associates each non-zero
1 f ( x ) = 1/ x
real number x to be reciprocal is called the reciprocal function.
x
The domain and range of the reciprocal function are both equal to
X´
R − {0} i.e., the set of all non-zero real numbers. The graph is as O
shown.
Y´
18. Power function : A function f : R → R defined by, f= ( x ) xα , α ∈ R is called a power
function.
EVEN AND ODD FUNCTION
1. Even function : If we put ( − x ) in place of x in the given function and if f ( − x ) =f ( x ) , ∀x ∈ domain
then function f ( x ) is called even function
e.g. f ( x ) =
e x + e− x , f ( x ) =
x2 , f ( x ) =
x sin x, f ( x ) =
cos x, f ( x ) =
x 2 cos x all are even functions.
2. Odd function : If we put ( − x ) in place of x in the given function and if f ( − x ) =− f ( x ) , ∀ x ∈ domain
then f ( x ) is called odd function. e.g., f ( x ) =
e x − e− x , f ( x ) =
sin x, f ( x ) =
x3 , f ( x ) = x cos x ,
f ( x ) = x 2 sin x all are odd functions.
PROPERTIES OF EVEN AND ODD FUNCTION
1. Every function defined in symmetric interval D (i.e., x ∈ D ⇒ − x ∈ D ) can be expressed as a sum of an
even and an odd function.
f ( x) + f (−x) f ( x) − f (−x)
= f ( x) +
2 2
f ( x) + f (−x) f ( x) − f (−x)
Let h ( x ) = and g ( x ) = . It can now easily be shown that h ( x ) is even
2 2
and g ( x ) is odd.
2. The first derivative of an even differentiable function is an odd function and vice-versa.
3. If x= 0 ∈ domain of f, then for odd function f ( x ) which is continuous at x = 0 , f ( 0 ) = 0 i.e., if for a
function, f ( 0 ) ≠ 0 , then that function can not be odd. It follows that for a differentiable even function
f ' ( 0 ) = 0 i.e., if for a differentiable function f ' ( 0 ) ≠ 0 then the function f can not be even.
4. The graph of even function is always symmetric with respect to y-axis. The graph of odd function is
always symmetric with respect to origin.
5. The product of two even functions is an even function.
6. The sum and difference of two even functions is an even function.
7. The sum and difference of two odd functions is an odd function.
8. The product of two odd functions is an even function.
9. The product of an even and an odd function is an odd function. It is not essential that every function is
even or odd. It is possible to have some functions which are neither even nor odd function.
e.g., f ( x ) = x 2 + x3 , f ( x ) =
log e x, f ( x ) =
ex .
JEEMAIN.GURU
10. The sum of even and odd function is neither even nor odd function.
11. Zero function f ( x ) = 0 is the only function which is even and odd both.
PERIODIC FUNCTION
A function f : X → Y is said to be a periodic function if there exists a positive real number T such that
f (x +T ) = f ( x ) , for all x ∈ X . The least of all such positive numbers T is called the principal period or
fundamental period of f. All periodic functions can be analysed over an interval of one period within the
domain as the same pattern shall be repetitive over the entire domain.
To test for periodicity of the function we just need to show that f ( x + T ) = f ( x ) for same T ( > 0 )
independent of x whereas to find fundamental period we are required to find a least positive number T
independent of x for which f ( x + T ) = f ( x ) is satisfied for all x.
The following points are to be remembered :
If f ( x ) is periodic with period T , then af ( x ) + b where a, b ∈ R ( a ≠ 0 ) is also periodic with period T.
T
1. If f ( x ) is periodic with period T , then f ( ax + b ) where a, b ∈ R ( a ≠ 0 ) is also period with period .
a
2. Let f ( x) has =
period T1 m / n ( m, n ∈ N and co-prime ) and g ( x) has period T2 = r / s
LCM of ( m, r )
( r , s ∈ N and co-prime ) and T be the LCM of T1 and T2 i.e., T = .
HCF of ( n, s )
Then T shall be the period of f + g provided there does not exist a positive number k ( < T ) for which
f ( k + x ) + g ( k + x=
) f ( x ) + g ( x ) else k will be the period. The same rule is applicable for any other
algebraic combination of f ( x ) and g ( x ) .
Note :
LCM of p and q always exist if p / q is a rational quantity. If p / q is irrational then algebraic
combination of f and g is non-periodic.
(a) sin n x, cos n x, cosec n x and sec n x have period 2π if n is odd and π if n is even.
(b) tan n x and cot n x have period π whether n is odd or even.
(c) A constant function is periodic but does not have a well-defined period.
(d) If g is periodic then fog will always be a periodic function. Period of fog may or may not be the
period of g.
(e) If f is periodic and g is strictly monotonic (other than linear) then fog is non-periodic.
COMPOSITE FUNCTION
If f : A → B and g : B → C are two function then the composite function of f and g .
gof A → C will be defined as =gof ( x ) g f ( x ) , ∀ x ∈ A
Properties of composition of function :
1. f is even, g is even ⇒ fog even function.
2. f is odd, g is odd ⇒ fog is odd function.
3. f is even, g is odd ⇒ fog is even function.
4. f is odd, g is even ⇒ fog is even function.
5. Composite of functions is not commutative i.e., fog ≠ gof .
6. Composite of functions is associative i.e., ( fog ) oh = fo ( goh ) .
JEEMAIN.GURU
Function 111
Limit Chapter 17
INDETERMINATE FORMS
If a function f assumes any one of the forms :
0 ∞
, , ∞ − ∞, 0 × ∞, 1∞ , 00 , ∞ 0
0 ∞
at x = a, then we say that f is indeterminate at x = a.
where 0 in every case denotes → 0 and 1 denotes → 1, ∞ denotes → ∞.
NEIGHBOURHOOD OF A POINT
Let ‘ a ’ be a real number. Then for a positive real number δ > 0, the open interval ( a − δ , a + δ ) is called a
neighbourhood of a. ( a − δ , a + δ ) − {a} , i.e. is called a deleted neighbourhood of a. The interval ( a − δ , a ) is
called a left hand neighbourhood of a and ( a, a + δ ) is called a right hand neighbourhood of a. If
x ∈ ( a, a + δ ) , we say that x approaches a from the right and we write x → a , etc. +
LIMIT OF A FUNCTION
A real number l is called the limit of the function f defined in a deleted neighbourhood of ‘ a ’ as x → a if
any given ε > 0, there exists δ > 0 such that
f ( x ) − l < ε , whenever 0 < x − a < δ i.e. l − ε < f ( x ) < l + ε , whenever x ∈ ( a − δ ) ∪ ( a + δ ) and
we write lim f ( x ) = l or f ( x ) → l as x → a.
x→a
Limit 113
(b)
x→a
( x→a
)(
lim ( fg )( x ) = lim f ( x ) lim g ( x )
x→a
)
f lim f ( x )
(c) lim ( x ) = x →a (provided lim g ( x ) ≠ 0 )
x→a
g lim g ( x ) x→a
x→a
sin x cos x
(ix) = lim
lim = 0
x →∞ x x →∞ x
x −a
n n
n n−m xn − an
(x) lim m = a and also lim = na n −1
x→a x − a m
m x→a x − a
x3 x5 x7
(xix) tan −1 x =x − + − + ......
3 5 7
3. Logrithmic limits : To evaluate the logarithmic limits we use following formulae.
x 2 x3
(i) log (1 + x ) =x − + − ......to ∞ where −1 < x ≤ 1
2 3
x 2 x3
(ii) log (1 − x ) =− x − − − ... to ∞ where −1 ≤ x < 1
2 3
log (1 + x )
(iii) lim =1 (iv) lim log e x = 1
x →0 x x →e
log (1 − x ) log a (1 + x )
(v) lim = −1 (vi) lim = log a e, a > 0, ≠ 1
x →0 x x →0 x
4. Exponential limits
(i) Based on series expansion
x 2 x3
We use (a) e x =1 + x + + + .....∞
2! 3!
( x log a ) ( x log a )
2 3
(b) a =
x
e x log a
=1 + ( x log a ) + + + ...
2! 3!
To evaluate the exponential limits we use the following results
ex −1 a x −1
(a) lim =1 (b) lim = log e a
x →0 x x →0 x
(ii) Based on the form 1∞ :
To evaluate the exponential form 1∞ we use the following results.
x
1
lim (1 + x )
1/ x
(a) =
e (b) lim 1 + =
e
x →0 x →∞
x
∞ if x >1
x =1
1 if
(c) lim x n = where n ∈ N
n →∞
0 if −1 < x < 1
does not exist if x ≤ −1
∞ if x >1 ∞ if x >1
1 if x =1 x =1
1 if
(d) lim x 2 n = 0 if −1 < x < 1 (e) lim x 2 n +1 = 0 if −1 < x < 1
n →∞ n →∞
1 if x = −1 −1 if x= −1
∞ if x < −1 −∞ if x < −1
(f) If lim f ( x ) = 1 and lim g ( x ) = ∞ .
x→a x→a
g( x) g( x) lim ( f ( x ) −1) g ( x )
Then lim { f ( x )}
= lim 1 + f ( x ) − 1 = e x →a
x→a x→a
g( x) g( x) lim g ( x ).log f ( x )
If lim ( f ( x ) ) is in the form of 00 or ∞ 0 then lim ( f ( x ) ) = e x →a
x→a x→a
JEEMAIN.GURU
Limit 115
5. L’ HOSPITAL’S RULE
Let f ( x ) and g ( x ) be two function differentiable in some neighbourhood of the point a, except may be
=
at the point ‘a’ itself. If lim f ( x ) lim
= g ( x) 0 .
x→a x→a
f ( x) f '( x)
or lim f ( x ) = lim g ( x ) = ∞ . Then lim = lim provided that the limit on the right exist or is
x→a x→a x→a g ( x) x→a g '( x)
±∞ .
d g ( x)
= f ( g ( x )).
dl
Newton Leibnitz’s formula say that,
dx dx
h( x )
and if l ( x ) = ∫ f ( t ) dt
g( x)
dh ( x ) dg ( x )
= f ( h ( x )). − f ( g ( x )).
dl
dx dx dx
JEEMAIN.GURU
Continuity Chapter 18
CONTINUITY OF A FUNCTION AT A POINT
A function ‘ f ’ is said to be continuous at a point a in the domain of f if the following conditions are
satisfied :
(i) f ( a ) exists (ii) lim f ( x ) exists finitely (iii) lim f ( x ) = f ( a ) .
x→a x→a
For the existence of lim f ( x ) it is necessary that lim f ( x ) and lim f ( x ) both exist finitely and both are
x→a x→a −0 x→a + 0
equal.
If any one or more of the above condition fail to be satisfied, the function f is said to be discontinuous at the
point a. Geometrically speaking, the graph of the function will exhibit a break at the point x = a
GEOMETRICAL MEANING OF CONTINUITY
(i) The function ‘f ’ will be continuous at x = a if there is no break in the graph of the function
y = f ( x ) at the point ( a, f ( a ) ) .
(ii) The function f ( x ) will be continuous in the closed interval [ a, b ] if the graph of y = f ( x ) is an
unbroken line (curved or straight) from the point ( a, f ( a ) ) to ( b, f ( b ) ) .
CONTINUITY OF A FUNCTION IN AN OPEN INTERVAL (a, b)
A function f is said to be continuous in ( a, b ) if f is continuous at each and every point ∈ ( a, b ) .
CONTINUITY OF A FUNCTION IN A CLOSED INTERVAL [a, b]
A function f is said to be continuous in a closed interval [ a, b ] if
(i) f is continuous in the open interval ( a, b ) , and
(ii) f is continuous at ‘a’ from the right
i.e., f ( a ) exists, lim f ( x ) exists finitely and lim f ( x ) = f ( a )
x→a + 0 x→a + 0
Continuity 117
For example, [ x ] and { x} both are discontinuous at all integral points, but their linear combination i.e.,
[ x ] + { x} is continuous for all values of x.
a +b a −b a +b a −b
Note : max ( a=
, b) + , max {a= , b} − .
2 2 2 2
*ACCUMULATION POINT (Cluster Point or Limit Point or Condensation Point)
Accumulation point : An accumulation point of a set of points is a point P such that there is at least one point
of the set distinct from P in any neighborhood of the given point; a point which is the limit of a sequence of
points of the set (for spaces ability). An accumulation point of a sequence is a point P such that there are an
infinite number of terms of the sequence in any neighborhood of P; e.g., the sequence 1, 12 , 1, 13 , 1, 14 , 1, 15 , ...
has two accumulation points the numbers 0 and 1
Note :
(a) Every constant function is everywhere continuous.
(b) The greatest integer function [ x ] is continuous at all points except at integeral points.
(c) The identity function I ( x ) = x, ∀x ∈ R is everywhere continuous.
(d) Modulus function x is everywhere continuous.
( x ) log a x, ( a > 0, a ≠ 1) is continuous on ( 0, ∞ ) .
(e) The exponential function f =
(f) A polynomial function a0 + a1 x + a2 x 2 + ... + an x n , an ≠ 0 and ai ∈ R, is everywhere
continuous.
(g) All trigonometrical functions, namely sin x, cos x, tan x, cot x, sec x, cosec x are
continuous at each point of their respective domains.
(h) Likewise, each inverse trigonometrical function (i.e. sin −1 x, cos −1 x, etc.) is continuous in
its domain.
CLASSIFICATION OF DISCONTINUITIES
Although discontinuity means failure of continuity it is interesting to note that all discontinuities are not
identical in character. Broadly speaking, discontinuities may be classified as
(i) Removable, where at a point x = a , lim f ( x ) exists but f ( a ) is either undefined or if defined, it
x→a
(ii) Irremovable where lim f ( x ) does not exist, even though f ( a ) exists or additionally f ( a ) also
x→a
does not exist.
In the first case the removal of discontinuity is achieved by modifying the definition of the function
suitably so that f ( a ) is equal to lim f ( x ) which is found existing.
x→a
In the second case when lim f ( x ) does not exist, no modification of definition at x = a can succeed
x→a
function will show an imperceptible break at x = a in as much as the point x = a is missing on it. It
just like a wire (of the shape of the graph) which is broken but whose ends are put together.
JEEMAIN.GURU
Appropriately, such a discontinuity is called missing point discontinuity This type of discontinuity
can, however, be removed by redefining the function such that f ( a ) = lim f ( x ) ; thus the
x→a
redefinition provides welding together the broken wire at the point of break.
(ii) In case a function f is defined such that lim f ( x ) ≠ f ( a ) , the graph of the function will show a
x→a
break at x = a together with a single point at x = a isolated from the main graph, thus justifying the
name isolated point discontinuity.
Irremovable type of discontinuity can be further classified as :
(i) Finite discontinuity
(ii) Infinite discontinuity
(iii) Oscillatory discontinuity.
In all these cases the value f ( a ) of the function at x = a (point of discontinuity) may exist or may not
exist but lim f ( x ) does not exist.
x→a
(i) Finite discontinuity: Here f ( a ) exists finitely, but it is either equal to lim− f ( x ) or lim+ f ( x ) and
x→a x→a
1 1 1
e.g. for f ( x ) = , f ( 4 ) is undefined. → ∞ for x → 4− and → ∞ for x → 4+
x−4 x−4 x−4
Showing x = 4, is a point of discontinuity occurring at ∞ , since both branches of the graph from the
left as well as from the right of the line x = 4 tend towards ∞,
(iii) Oscillatory discontinuity: For a function f, if f ( a ) is undefined as f ( x ) oscillates finitely as
x → a then x = a is called a point of finite oscillatory discontinuity, and incase f ( x ) oscillates
infinitely as x → a , then x = a is called a point of infinite oscillatory discontinuity.
PROPERTIES OF CONTINUOUS FUNCTIONS
Here we present two extremely useful properties of continuous functions ;
Let y = f ( x ) be a continuous function ∀ x ∈ [ a, b ] , then following results hold true.
(i) f is bounded between a and b. This simply means that we can find real numbers m1 and m2 such
m1 ≤ f ( x ) ≤ m2 ∀ x ∈ [ a, b ] .
(ii) Every value between f ( a ) and f ( b ) will be assumed by the function atleast once. This
property is called intermediate value theorem of continuous function.
In particular if f ( a ) . f ( b ) < 0 , then f ( x ) will become zero atleast once in ( a, b ) . It also means
that if f ( a ) and f ( b ) have opposite signs then the equation f ( x ) = 0 will have atleast one
real root in ( a, b ) .
JEEMAIN.GURU
119
Differentiation Chapter 19
DEFINITION
If f ( x ) is a function and a and a + h belongs to the domains of f , then the limit given by
f (a + h) − f (a)
lim , if it finitely exist, is called the derivative of f ( x ) with respect to ( or w.r.t ) x at x = a
h →0 h
and is denoted by f ' ( a ) ,
f (a + h) − f (a)
∴ f ′(a) = lim .
h →0 h
Note : f ′ ( a ) is the derivates of f ( x ) w.r.t x at x = a.
DIFFERENTIABILITY
(i) A function f is said to have left hand derivative at x = a iff f is defined in some (undeleted)
f (a + h) − f (a)
left neighbourhood of a and lim− exists finitely and its value is called the left
h →0 h
hand derivative at a and is denoted by f ′ ( a − ) .
(ii) A function f is said to have right-hand derivative at x = a iff f is defined in some (undeleted)
f (a + h) − f (a)
right neighbourhood of a and lim+ exists finitely and its value is called the
h →0 h
right hand derivative at a and is denoted by f ′ a + . ( )
(iii) A function f is said to have a derivative (or is differentiable) at a if f is defined in some
f (a + h) − f (a)
(undeleted) neighbourhood of a and lim exists finitely and its value is called
h →0 h
df ( x )
the derivative or differential coefficient of f at a and is denoted by f ′ ( a ) or
dx x = a
(iv) If a function f is differentiable at a point ‘ a ’ then it is also continuous at the point ‘ a ’. But,
converse may not be true. For example, f ( x ) = x is continuous at x = 0 but is not differentiable
at x = 0.
(v) A function f is differentiable at a point x = a and P ( a, f ( a ) ) is the corresponding point on
the graph of y = f ( x ) iff the curve does not have P as a corner point.
Note : From (iv) and (v) it is clear that if a function f is not differentiable at a point x = a then either the
function f is not continuous at x = a or the curve represented by y = f ( x ) has a corner at the point
( a, f ( a ) ) (i.e. the curve suddenly changes the direction)
(vi) A function f is differentiable (or derivable) on [ a, b ] if
(a) f is continuous at every point of ( a, b )
f (a + h) − f (a) f (b + h ) − f (b )
(b) lim+ and lim− both exist.
h →0
h h →0 h
A function f is said to be differentiable if it is differentiable at every point of the domain.
A function f is said to be everywhere differentiable if it is differentiable for each x ∈ R .
SOME STANDARD RESULTS ON DIFFERENTIABLITY
(i) Every polynomial function, every exponential function a x ( a > 0 ) and every constant function
are differentiable at each x ∈ R .
(ii)
The logarithmic functions, trigonometrical functions and inverse – trigonometrical functions are
always differentiable in their domains.
(iii) The sum, difference, product and quotient (under condition) of two differentiable functions is
differentiable.
(iv) The composition of differentiable functions (under condition) is a differentiable function.
DERIVATIVES OF SOME STANDARD FUNCTIONS
d
(i) ( c ) = 0 if c is a constant and conversely also.
dx
JEEMAIN.GURU
Differentiation 121
Test of constancy. If at all points of a certain interval f ′ ( x ) = 0, then the function f is constant in that
interval.
(ii)
d n
dx
( x ) = nx n −1 (iii)
d
dx
( ax + b ) = n ( ax + b ) .a
n n −1
d d
(iv) ( sin x ) = cos x (v) ( cos x ) = − sin x
dx dx
d d
(vi) ( tan x ) = sec2 x (vii) ( cot x ) = − cosec2 x
dx dx
d d
(viii) ( sec x ) = sec x tan x (ix) ( cosec x ) = − cosec x cot x
dx dx
−1
(x)
d
( )
sin −1 x =
1
(xi)
d
(
cos −1 x =)
dx 1− x 2 dx 1 − x2
−1
(xii)
d
dx
(
tan −1 x =) 1
1 + x2
(xiii)
d
dx
( )
cot −1 x =
1 + x2
−1
(xiv)
d
(
sec −1 x =) 1
(xv)
d
(
cosec −1 x = )
dx x x2 −1 dx x x2 −1
(xvi)
d x
dx
( )
a = a x log e a (xvii)
d x
dx
( )
e = ex
d 1
(xviii) ( log a x ) =
dx x log e a
d 1
( log e x ) =
dx x
(xix)
d
dx
=( x )
x
x
, x ≠ 0, y = x is not differentiable at x = 0
0 for x ∈ R − I 1 if x ∈ R − I
d
(xx) ( [ x ]) = does not exist for x ∈ I (xxi)
d
( { x}) =
dx dx does not exist if x ∈ I
SOME RULES FOR DIFFERENTIATION
d
1. The derivative of a constant function is zero , i.e. ( c ) = 0.
dx
2. The derivative of constant times a function is constant times the derivative of the function, i.e.
d
dx
{c. f ( x )}= c ⋅ { f ( x )} .
d
dx
3. The derivative of the sum or difference of two function is the sum or difference of their derivatives, i.e.,
d
dx
{ )}
f ( x ) ± g ( x=
d
dx
{ f ( x )} ± { g ( x )} .
d
dx
g ( x ) ⋅ { f ( x )} − f ( x ) ⋅ { g ( x )}
d d
d f ( x ) dx dx
i.e. =
dx g ( x ) { g ( x )}
2
Differentiation 123
⇒ y= f ( x) + y ⇒ y 2 =f ( x ) + y
dy
2=
y f ' ( x ) + dy / dx
dx
dy f ' ( x )
∴ =
dx 2 y − 1
f ( x ).... ∞
f ( x)
If y = f ( x ) then y = f ( x ) .
y
(ii)
y log f ( x )
∴ log y =
1 dy y ⋅ f ' ( x ) dy
= + log f ( x ) ⋅
y dx f ( x) dx
dy y2 f '( x)
∴ =
dx f ( x ) 1 − y log f ( x )
1
If y f ( x ) +
(iii) =
1
f ( x) +
1
f ( x) +
f ( x ) .....
JEEMAIN.GURU
dy y f '( x)
Then = .
dx 2 y − f ( x )
f ( a + 2h ) − 2 f ( a + h ) + f ( a )
12. f ′′ (α ) = lim and in general
h →0 h2
f (α + nh ) − nC1 f (α + ( n − 1) h ) + nC2 f (α + ( n − 2 ) h ) + .... + ( −1) f (α )
n
f ( n ) (α ) = lim
h →0 hn
d −1 1
13. f ( x ) =
dx x = f (α ) d f x
( )
dx x =α
(ii)
d
dx
( u )=
u du
u dx
(iii)
d
dx
( log f ( x ) ) =
1 d
f ( x ) dx
f ( x)
(iv)
d
dx
( )
a f ( x ) = a f ( x ) log a . f ′ ( x ) .
LEIBNITZ THEOREM AND nTH DERIVATIVES
Let f ( x ) and g ( x ) be functions both possessing derivatives up to n th order. Then,
dn
dx n
( f ( x= ) g ( x ) ) f n ( x ) g ( x ) + nC1 f n−1 ( x ) g1 ( x ) + nC2 f n−2 ( x ) g 2 ( x ) + ... +
n
Cr f n − r ( x ) g r ( x ) + ... + nCn f ( x ) g n ( x ) .
d n 1 ( −1) n ! d n
n
dn n π
dx n
= x ( )
n !; =
n
dx x
x n +1
; n ( sin
dx
= x ) sin x + n ,
2
dn π d n mx
dx n ( cos x ) = cos
x + n ; n e =
2 dx
m n e mx . ( )
SUCCESSIVE DIFFERENTIATION
( ax + b ) , m ∉ N , then y=n m ( m − 1)( m − 2 ) ..... ( m − n + 1)( ax + b ) .a n
m m−n
(i) If y =
( ax + b ) , m ∈ N , then
m
(ii) If y =
m ( m − 1)( m − 2 ) ..... ( m − n + 1)( ax + b )m − n .a n for n < m
=yn = m !a m , if n m
0, n>m
JEEMAIN.GURU
Differentiation 125
1
( −1) .n !( ax + b ) .a n
n − n −1
(iii) If y = , then yn =
ax + b
If y log ( ax + b ) , then yn = ( −1) ( n − 1)!a n ( ax + b )
n −1 −n
(iv) =
π
(v) =If y sin ( ax + b ) , then
= yn a n sin ax + b + n
2
π
If y cos ( ax + b ) , then
(vi) = = yn a n cos ax + b + n
2
(vii) If y = a x then yn = a x ( log e a ) .
n
PARTIAL DIFFERENTIATION
The partial differential coefficient of f ( x, y ) with respect to x is the ordinary differential coefficient of
∂f
f ( x, y ) when y is regarded as a constant. It is written as or Dx f or f x .
∂x
∂f f ( x + h, y ) − f ( x, y )
Thus, = lim
∂x h→0 h
∂f
Again, the partial differential coefficient of f ( x, y ) with respect to y is the ordinary differential
∂y
coefficient of f ( x, y ) when x is regarded as a constant.
∂f f ( x, y + k ) − f ( x, y )
Thus, = lim
∂y k →0 k
e.g., If z = f ( x, y ) = x + y 4 + 3xy 2 + x 2 y + x + 2 y
4
∂z ∂f
Then or or f x = 4 x 3 + 3 y 2 + 2 xy + 1 (Here y is regarded as constant)
∂x ∂x
∂z ∂f
or or f x = 4 y 3 + 6 xy + x 2 + 2 (Here x is regarded as constant)
∂y ∂y
∂2 f ∂2 f
= or f xy = f yx .
∂x ∂y ∂y ∂x
EULER’S THEOREM ON HOMOGENEOUS FUNCTIONS
If f ( x, y ) is a homogeneous function in x, y of degree n, then
∂f ∂f
x +y = nf
∂x ∂y
DEDUCTION OF EULER’S THEOREM
If f ( x, y ) is a homogeneous function in x, y of degree n, then
∂2 f ∂2 f ∂f
(i) x 2 +y ( n − 1)
=
∂x ∂x ∂y ∂x
∂2 f ∂2 f ∂f
(ii) x +y 2 = ( n − 1)
∂y ∂x ∂y ∂y
∂2 f ∂2 f 2 ∂ f
2
(iii) x 2 + 2 xy +y n ( n − 1) f ( x, y )
=
∂x ∂x ∂y ∂y 2
JEEMAIN.GURU
127
dy
2. Equation of tangent to the curve at P =
is y − y1 ( x − x1 )
dx ( x1 , y1 )
JEEMAIN.GURU
dy
3. If is zero, then the tangent to the curve y = f ( x ) at P is y = y1 which is parallel to x-axis.
dx ( x1 , y1 )
The equation of normal to the curve at P is given by x = x1 which is parallel to y-axis
dy 1
4. If ≠ 0, then slope of normal at P is − and equation of normal is
dx ( x1 , y1 ) dy
dx
( x1 , y1 )
1
y − y1 = − ( x − x1 )
dy
dx
( x1 , y1 )
dx
5. If = 0, then the tangent is perpendicular to x-axis and its equation is x = x1 or normal is parallel to x-
dy
axis and equation of normal is y = y1
ANGLE OF INTERSECTION OF TWO CURVES
Angle of intersection of two curves is the (acute) angle between the tangents to the two curves at their
point of intersection.
m − m2
If θ is the acute angle between the tangents, then tan θ = 1
1 + m1m2 Y
y = f ( x) y = g ( x)
dy
where m1 = value of at the common point for first curve. θ
dx
P ( x1 , y1 )
dy
and m2 = value of at the common point for the second curve. θ1 − θ 2 θ1
dx θ2
θ (θ1 − θ 2 ) ,
If θ is the required angle of intersection, then,= O X
where θ1 and θ 2 are the inclination of tangent to the curves y = f ( x ) and y = g ( x ) respectively at the
point P.
ORTHOGONAL AND TOUCHING CURVES
Two curves are said to be orthogonal (or intersect orthogonally) if the angle of intersection of two curves
is a right angle. i.e. if m1m2 = −1
Two curves touch each other if m1 = m2
1 1 1 1
Note : The curve ax 2 + by 2 = 1 and a′x 2 + b′y 2 =
1 cut each other orthogonally if − = −
a b a ′ b′
LENGTH OF TANGENT, NORMAL, SUB-TANGENT AND SUBNORMAL
Let the tangent and normal at the point P ( x, y ) on the curve meet the axis of x at the points T and N
respectively. Let M be the foot of the ordinates at P. Then,
y = f ( x)
= PT
(i) Length of the tangent = y cosec θ Y
tangent
2
dy P ( x, y )
y 1+ 2
normal
dx dx
=y 1 + cot 2 θ = =y 1 + θ
dy dy θ
X
dx O T M N
JEEMAIN.GURU
2
dy
(ii) = PN
Length of the normal = y sec θ =y 1 + tan θ =y 1 +
2
dx
y dx
= TM
(iii) Length of subtangent = θ
y cot= = y
dy dy
dx
dy
= MN
(iv) Length of subnormal = θ
y tan= y .
dx
y y
Geometrical Interpretation : If a function f ( x ) satisfies the above two conditions, then there exists at
least one point c between a and b at which tangent is parallel to the chord joining the point A ( a, f ( a ) )
and B ( b, f ( b ) ) .
y y
B ( b, f ( b ) ) B
( b, f ( b ) )
A
A ( a, f ( a ) )
( a, f ( a ))
(f) If a function f ( x ) is defined on ( a, b ) and f ′ ( x ) < 0 for all x ∈ ( a, b ) except for a finite
number of points where f ′ ( x ) = 0, then f ( x ) is strictly decreasing ( ↓ ) on ( a, b )
S
lim f ( x ) is greatest (or least) then we will say that point of Absolute Maxima (or Absolute Minima)
x→a+
to be concave upwards (or convex downwards) at P if in the immediate neighbourhood of P , the curve
lies above the tangent PT on both sides (as in figure I ).
y y T
T
P y = f(x)
y = f(x)
P
O x O x
Figure I Figure II
The curve is said to be concave downwards (or convex upwards) at P if in the immediate neighbourhood
of P, the curve lies below the tangent PT on both sides (as in figure II).
Criteria for concavity and convexity
At a point P on the curve y = f ( x ) , the curve is
(i) Convex downward if f ′′ ( x ) > 0
(ii) Concave downwards if f ′′ ( x ) < 0.
2. Point of inflexion
A point on a curve at which the curve changes from concavity to convexity or vice-versa is called a point
of inflexion. At a point of inflexion, the tangent to the curve crosses the curve.
y y
P1
P2
O x O x
dx 1 a+x dx 1 x−a
11.=∫ a 2 − x 2 2a log a − x + C. 12. = ∫ x 2 − a 2 2a log x + a + C.
dx x dx 1 x
13. ∫ = sin −1 + C. 14. ∫ = sec −1
a −x
2 2 a x x −a2 2 a a
dx
15. ∫ x +a2 2
= log x + x 2 + a 2 + C.
dx
16. ∫ x −a2 2
= log x + x 2 − a 2 + C.
1 a2 x
∫ a −x= x a − x + sin −1 + C.
2 2 2 2
17. dx
2 2 a
2
1 a
18. ∫ a 2 + x 2=
dx
2
x x 2 + a 2 + log x + x 2 + a 2 + C.
2
n
1 a2 n x x
19. ∫ x − a =
dx 2
x x 2 − a 2 − log x + x 2 − a 2 + C.
2
20. ∫ x=
dx + C , where n ≠ −1
2 2 n +1
INTEGRALS OF THE FORM ∫ e x {f ( x ) + f ' ( x )} dx :
If the integral is of the form ∫ e x { f ( x ) + f ' ( x )} dx, then by breaking this integral into two integrals, integrate
one integral by parts and keeping other integral as it is, by doing so, we get
∫ e f ( x ) + f ' ( x ) dx =e f ( x ) + c
x x
1.
JEEMAIN.GURU
∫e mf ( x ) + f ' ( x ) dx = e f ( x ) + c
mx mx
2.
f '( x) e mx f ( x )
3. ∫
e mx
f ( x ) + dx = +c
m m
INTEGRAL IS OF THE FORM ∫ xf ' ( x ) + f ( x ) dx :
If the integral is of the form ∫ xf ' ( x ) + f ( x ) dx then by breaking this integral into two integrals, integrate
one integral by parts and keeping other integral as it is , by doing so, we get,
∫ x f ' ( x ) + f ( x ) dx =x f ( x ) + c .
INTEGRALS OF THE FORM ∫ eaxsinbxdx, ∫ eaxcosbx dx :
e ax e ax b
∫=
e ax sin bx dx ( a sin bx −=
b cos bx ) + c sin bx − tan −1 + c
a 2 + b2 a 2 + b2 a
e ax e ax b
∫=
e ax .cos bx dx ( a cos bx +=
b sin bx ) + c cos bx − tan −1 + c
a 2 + b2 a 2 + b2 a
e ax e ax b
( bx + c ) dx
∫ e .sin=
ax
a sin ( bx + c ) − b cos=
( bx + c ) + k sin ( bx + c ) − tan −1 + k
a 2 + b2 a +b 2
2
a
e ax
e ax
b
∫ e .cos
ax
= ( bx + c ) dx 2 2 a cos ( bx + c ) + b sin= ( bx + c ) + k cos ( bx + c ) − tan −1 + k
a +b a +b
2 2
a
x2 + 1 x2 - 1 dx
INTEGRALS OF THE FORM ∫ 4 dx, ∫ 4 2
dx, ∫ 4 , where k ∈ R
2
x + kx + 1 x + kx + 1 x + kx 2 + 1
Working Method
(a) To evaluate these types of integrals divide the numerator and denominator by x 2
1 1
(b) Put x= + t or x= − t as required.
x x
x2 + a2 x2 − a2
∫ x 4 + kx 2 + a 4 dx , ∫ x 4 + kx 2 + a 4 dx, where k is a constant
These integrals can be obtained by dividing numerator and denominator by x 2 , then putting
a2 a2
x=
− t and x =
+ t respectively.
x x
STANDARD SUBSTITUTIONS
(a) For terms of the form x 2 + a 2 or x 2 + a 2 , put x = a tan θ or a cot θ θ ∈ ( 0, π / 2 )
(b) For terms of the form x 2 − a 2 or x 2 − a 2 , put x = a sec θ or a cosec θ θ ∈ ( 0, π / 2 )
(c) For terms of the form a 2 − x 2 or a 2 − x 2 , put x = a sin θ or a cos θ θ ∈ [ 0, π / 2]
(d) If both a + x , a − x are present, then put x = a cos θ .
x−a
(e) For the type ( x − a )( b − x ) , put x a cos 2 θ + b sin 2 θ
,= θ ∈ [ 0, π / 2]
b−x
x−a
(f) For the type ( x − a )( x − b ) , put x a sec 2 θ − b tan 2 θ
,= θ ∈ ( 0, π / 2 )
x −b
JEEMAIN.GURU
1
(i) For , n1 , n2 ∈ N (and > 1 ), again put ( x + a ) = t ( x + b )
( x + a) ( x + b)
n1 n2
Euler’s Substitution
The integral of the form ∫ R ( x, )
ax 2 + bx + c dx are calculated with the aid of one of the three Euler’s
substitution.
(i) ax 2 + bx + c =t ± x a , If a > 0;
(ii) ax 2 + bx + c = tx ± c , If c > 0;
(iii) ax 2 + bx + c = ( x −α )t , If ax 2 + bx + c= a ( x − α )( x − β ) ,
i.e., If α is a real root of the trinomial ax 2 + bx + c =0.
Working Rule :
2 b c 2 b b2 c b2 b b 2 − 4ac
2
Write ax + bx +=
2
c a x + x + = a x + 2 x. + 2 + − 2 = a x + −
a a 2a 4a a 4a 2a 4a 2
Thus, ax 2 + bx + c will be reduced to the form A2 + X 2 or A2 − X 2 or X 2 − A2 , where X is a linear
expression in x and A is a constant.
X Y Substitution
Linear Linear z2 = Y
Quadratic Linear z2 = Y
1
Linear Quadratic z=
X
Y 1
Pure Quadratic Pure Quadratic z2 = or z =
X x
( )
n
x ± a2 + x2
∫(x ± ) dx
n
dx
(i) a2 + x2 (ii) ∫ (iii) ∫ dx
(x ± ) a +x
n 2 2
a +x
2 2
∫ R x, ( ax + b )
α /n
(ii) dx, where R means for a rational functional.
Working Rule : Put z= n
ax + b
(iii) ∫ R x, ( ax + b ) , ( ax + b ) dx
α /n β /m
Working Rule : Put z= ax + b, where p = L.C.M. for m and n
p
α
ax + b n
dx
(iv) ∫ R x,
cx + d
ax + b
Working Rule : Put z n = .
cx + d
∫ x ( a + bx )
m n p
6. Integrals of the form dx, Where m, n and p are rational numbers can be solved as follows :
(i) If p ∈ N , Expand the integral with the help of binomial theorem and then integrate.
(ii) If p is a negative integer, the integral reduces to the integral of a rational function by means of the
substitution x = t s , where s is the L.C.M. of denominators of the fractions m and n.
(iii) If
( m + 1) is an integer, the integral can be rationalized by the substitution a + bx n = t s where s
n
is the denominator of the fraction P.
(iv) If
( m + 1) + p is an integer, substitute ax − n + b =t s , where s is the denominator of the fraction P.
n
JEEMAIN.GURU
1
7. Integrals of the form ∫ (x − k) ax 2 + bx + c
dx , the substitution x − k =
1/ t reduces the integral
1 1
∫ (x − k) ax + bx + c
2
dx to the problem of integrating an expression of the form
At + Bt + C
2
.
dx
8. ∫ (x − k) r
ax 2 + bx + x
. Here we substitute, x − k =
1/ t.
Ax + B
9. Integrals of the form ∫
ax 2 + bx + c
dx are solved by isolating in the numerator, the derivative of the
quadratic appearing under the root sign and expanding the integral into the sum of two integrals.
Ax + B ( A / 2a )( 2ax + b ) + B − ( Ab / 2a ) dx
∫ ax 2 + bx + c dx = ∫ ax 2 + bx + c
A 2ax + b b dx
=
2a ∫ ax + bx + c
2
dx + B − A ∫
2a ax + bx + c
2
a0 x n + a1 x n −1 + a2 x n − 2 + ........ + an −1 x + an
10. Integrals of the form ∫ ax 2 + bx + c
dx are solved as follows
= (C x
0
n −1
+ C1 x n − 2 + ..... + Cn −1 ) ax 2 + bx + c + Cn ∫
dx
→ (I)
ax 2 + bx + c
Where C0 , C1 , C2 , ......., Cn are arbitrary constants
Now differentiating both of (I) w.r.t x and multiplying by ax 2 + bx + c we get.
a0 x n + a1 x n −1 + .......an −1 x + an
( 1
) ( )
= ( n − 1) C0 x n − 2 + ( n − 2 ) C1 x n −3 + .... + Cn − 2 ax 2 + bx + c + C0 x n −1 + C1 x n − 2 + ....... + Cn −1 ( 2ax + b ) + Cn
2
Where constant C0 , C1 , C2 , ........, Cn can be evaluate by comparity of like power of x four both sides.
dx
Substituting the values of C0 , C1 , C2 ......Cn in (I) and evaluating ∫ the given integral is
ax 2 + bx + c
determined completely.
(ax 2 + bx + c dx )
11. ∫ ( dx + e ) fx 2 + gx + h .
Here, we write, ax 2 + bx + = c A1 ( dx + e )( 2 fx + g ) + B1 ( dx + e ) + C1 where A1 , B1 and C1 are constants
which can be obtained by comparing the coefficient of like terms on both sides. And given integral will
reduce to the form
( 2 fx + g ) dx dx
A1 ∫ dx + B1 ∫ + C1 ∫
fx 2 + gx + h fx 2 + gx + h ( dx + e ) fx 2 + gx + h
JEEMAIN.GURU
dx dx dx
INTEGRALS OF THE FORM ∫ a + b cosx , ∫ a + b sinx and ∫ a + b sinx + c cosx
To evaluate such form proceed as follows :
x x
1 − tan 2 2 tan
=
1. Put cos x = 2 and sin x 2
x x
1 + tan 2 1 + tan 2
2 2
x x
2. Replace 1 + tan 2 in the numerator by sec 2 .
2 2
x 1 x
3. Put tan = t so that sec 2 dx = dt.
2 2 2
INTEGRATION OF TRIGONOMETRIC FUNCTIONS
1. Integral of the form ∫ sin m x cos n x dx :
(i) To evaluate the integrals of the form I = ∫ sin m x cos n x dx, where m and n are rational numbers.
(a) Substitute sin x = t ,if n is odd ;
(b) Substitute cos x = t , if m is odd ;
(c) Substitute tan x = t , if m + n is a negative even integer; and
The above substitution enables us to integrate any function of the form R ( sin x, cos x ) However, in
practice; it sometimes leads to very complex rational function. In some cases, the integral can be
simplified by :
(a) Substitute sin x = t , if the integral is of the form ∫ R ( sin x ) cos x dx.
(b) Substituting cos x = t , if the integral is of the form ∫ R ( cos x ) sin x dx.
dt
(c) Substituting=
tan x t ,=
i.e. dx , if the integral is dependent only on tan x.
1+ t2
(d) If the given function is R ( sin x + cos x ) cos 2 x
t ⇒ 1 + sin 2x =
Put sin x + cos x = t2 ⇒ 2 cos 2 x dx =
2 t dt
And if function is R ( sin x − cos x ) cos 2 x
Put sin x − cos x = t ⇒ 1 − sin 2x = t 2 ⇒ − 2 cos 2 x dx =2 t dt and proceed further.
INTEGRALS OF THE FORM :
p cos x + q sin x + r
(i) ∫ a cos x + b sin x + c dx
In this integral express numerator as l (Denominator) + m (d.c. of denominator) + n.
Find l , m, n by comparing the coefficients of sin x, cos x and constant term and split the integral into
sum of three integrals.
d.c. of (Denominator) dx
l ∫ dx + m ∫ dx + n ∫
Denominator a cos x + b sin x
p cos x + q sin x
(ii) ∫ dx
a cos x + b sin x
Express numerator as l (denominator) + m (d.c. of denominator) and find l and m by comparing the
coefficients of sin x, cos x.
JEEMAIN.GURU
∫ f ( x=
) dx F (=
x ) a F (b) − F ( a )
b b
a
The following formula due to Newton and Leibnitz therefore this is also called Newton Leibnitz formula.
PROPERTIES OF DEFINITE INTEGRALS
b a
1. ∫ f ( x ) dx = − ∫ f ( x ) dx
a b
b b
2. ∫ f ( x ) dx = ∫ f ( y ) dy
a a
b c b
3. ( x ) dx ∫ f ( x ) dx + ∫ f ( x ) dx, where a < c < b
∫ f=
a a c
2a a
4. ∫ f ( x ) dx= ∫ f ( x ) + f ( 2a − x ) dx
o o
0 if f ( 2a − x ) =
− f ( x)
2a
a
Corollary : ∫ f ( x ) dx =
o 2 ∫ f ( x ) dx if f ( 2a − x ) =f ( x)
0
b b
5. ∫ f ( x )=
a
dx ∫ f ( a + b − x ) dx
a
a a
Corollary : ∫ f (=
x ) dx ∫ f ( a − x ) dx
o o
a
∫ f ( x=
) dx ∫ f ( x ) + f ( − x ) dx
a
6.
0
−a
a
2 f ( x ) dx , if f ( − x ) =
f ( x ) i.e. f ( x ) is even
= ∫0
0 , if f ( − x ) =− f ( x ) i.e. f ( x ) is odd
a
f ( x) a
7. ∫ f ( x ) + f ( a − x ) dx = 2
o
b
f ( x) b−a
8. ∫ f ( x ) + f ( a + b − x ) dx =
a
2
9. If f ( x ) is a periodic function of period T i.e. f (T + x ) =f ( x ) , then
nT T nT T
(a) ∫ f ( x ) dx = n ∫ f ( x ) dx (b) ∫ f ( x ) dx= ( n − m ) ∫ f ( x ) dx
o o mT o
JEEMAIN.GURU
b b + nT a + nT
(c) ∫ f ( x ) dx = ∫ f ( x ) dx
a a + nT
(d) ∫ f ( x ) dx
a
is independent of a.
∫ f ( x ) dx =( b − a ) ∫ f ( ( b − a ) x + a ) dx
b 1
10.
a 0
b
11. If f ( x ) ≥ 0 on the interval [ a, b ] , then ∫ f ( x ) ≥ 0.
a
12. If a function f is integrable and non-negative on [ a, b ] and there exists a point c ∈ [ a, b ] of continuity of
f for which f ( c ) > 0, then ∫ f ( x ) dx > 0 ( a < b).
b
a
b b
13. If f ( x ) ≤ g ( x ) on the interval [ a, b ] , then ∫ f ( x ) dx ≤ ∫ g ( x ) dx
a a
b b
14. ∫ f ( x ) dx ≤ ∫ f ( x )
a a
dx
∫ f ( x ) ⋅ g ( x ) dx ≤ ∫ f ( x ) dx ⋅ ∫ g ( x ) dx
2 2
inequality holds i.e.
a a a
17. If a function f ( x ) is continuous on the interval [ a, b ] , then there exists a point c ∈ ( a, b ) such that
b
( x ) dx
∫ f= f ( c )( b − a ) , where a < c < b.
a
(a) Mean Value or Average Value of a function :
Mean Value or Average Value of a function on the interval [ a, b ] is given by
1
f ( x) > = f ( x ) dx
b
( b − a ) ∫a
(b) Root Mean Square Value (RMSV) :
Root Mean Square Value (RMSV) of a function y = f ( x ) on the interval ( a, b ) is
b y 2 dx
∫a
( b − a )
18. Holder’s Inequality for Integrals
1 1
b b p b q
1 1
∫ f ( x) g ( x)
dx ≤ ∫ f ( x ) dx ∫ g ( x ) dx where + = 1, p > 1, q > 1 .
p q
a a a p q
If p= q= 2 , this reduces to Cauchy-Schwarz Inequality for integrals.
JEEMAIN.GURU
p −1
f ( x)
The equality holds if and only if is a constant.
g ( x)
p p p
b b b
If p > 1, ∫ f ( x ) + g ( x ) dx ≤ ∫ f ( x ) dx + ∫ q ( x ) dx
p p p
a a a
f ( x)
The equality holds if and only if is a constant.
g ( x)
SECOND FUNDAMENTAL THEOREM
If f is continuous on [ a, b ] then
F ( x ) = ∫ f ( t ) dt
x
a
dF d x
is differentiable at every point x in [ a,= b ] and f ( t ) dt f ( x ) .
dx dx ∫a
=
Leibnitz’s rule
1. If f ( x ) is continuous on [ a, b ] , and u ( x ) and v ( x ) are differentiable functions of x whose values lie in
[ a, b] , then
d v( x )
f ( t )dt f {v ( x )} {v ( x )} − f {u ( x )} {u ( x )} .
d d
=
dx ∫u ( x ) dx dx
2. If the function φ ( x ) and ψ ( x ) are defined on [ a, b ] and differentiable at a point x ∈ ( a, b ) , and f ( x, t ) is
continuous then
d ψ ( x) = ψ ( x) ∂ dψ ( x ) dφ ( x )
dx ∫φ ( x )
f ( x , t ) dt
∫φ ( x ) ∂x
f ( x , t ) dt + f ( x ,ψ ( x ) ) − f ( x, φ ( x ) ) .
dx dx
SUMMATION OF SERIES WITH THE HELP OF DEFINITE INTEGRAL AS THE LIMIT OF A SUM :
If f ( x ) is a continuous and single valued function defined on the interval [ a, b ] , then the definite integral
b
∫ f ( x ) dx is defined as follows :
a
b
∫ f ( x )=
a
dx lim h f ( a + h ) + f ( a + 2h ) + ... + f ( a + nh ) where nh= b − a .
h →0
n b
or lim h∑ f ( a + rh ) =
∫ f ( x ) dx … (1)
n →∞
h →0 r =1 a
1
Put a = 0 and b =1 ⇒ nh =1 ⇒ h = .
n
r
1
1 1 n
Substitute=a 0,=b 1 and h = in (1), we get lim ∑ f = ∫ f ( x ) dx.
n n →∞ n
r =1 n 0
∫ f ( x ) dx = ( )
lim a ( r − 1) f ( a ) + rf ( ar ) + ...... r n −1 f ar n −1
b
Also
a r →1
JEEMAIN.GURU
IMPROPER INTEGRALS :
∞
∫ f ( x ) dx ∫ f ( x ) dx.
b
1. = lim
a b →+∞ a
∫ f ( x ) dt lim ∫ f ( x ) dx.
b b
2. =
−∞ a →−∞ a
+∞ +∞
f ( x ) dx f ( x ) dx + f ( x ) dx
c
3. ∫=
−∞ ∫ −∞ ∫ c
∴ Γ ( n + 1) = nΓ ( n ) .
when n is a positive integer, Γ ( n + 1) =
n!
∞ Γ (n)
(b) ∫ e − kx x n −1dx =, [ k > 0, n > 0]
0 kn
Γ (m) Γ ( n)
(c) B ( m, n ) = , [ m > 0, n > 0]
Γ (m + n)
π
( m ) Γ (1 − m )
(d) Γ= where ( 0 < m < 1) .
sin mπ
1
Corollary : putting m =
2
1 1 π 1
we get Γ Γ = = π ∴ Γ = π .
2 2 sin 1 π 2
2
m −1
∞ x dx ∞ x n −1dx
=
(e) B ( m, n ) ∫= ∫0 (1 + x )m+n .
(1 + x )
0 m+n
p +1 q +1
1 Γ Γ
π 2 2
(f) ∫ sin θ cos θ dθ
= p
2 q
, [ p > −1, q > −1]
0 p+q+2
2Γ
2
n −1 n − 3 n − 5 2
π π . . ... , when n is odd
n n−2 n−4 3
∫ sin xdx
= ∫ cos x dx
=
2 n 2 n
Walli’s Formula
n − 1 . n − 3 . n − 5 ... 3 . 1 . π ,
0 0
when n is even
n n − 2 n − 4 4 2 2
JEEMAIN.GURU
y = x3 ( x, y )
Symmetry about the origin :
The equation is unaltered when x and y are replaced
by –x and –y ; that is,
F ( x, y ) = F ( − x, − y ) . x
O
( − x, − y )
y
Symmetry about the line y = x
The equation is unaltered when x and y are interchanged, x+ y=
1
that is,
F ( x, y ) = F ( y , x ) . ( y, x )
( x, y )
x
O
x
O
2. ORIGIN
If there is no constant term in the equation of the given curve, then the curve passes through the origin.
In that case, the given equation is a polynomial equation in x and y then the tangents at the
origin are given by equating to zero the lowest degree terms in the equation of the given curve.
For example, the curve y 3 + x 3 + axy =0 passes through the origin and the tangents at the origin
are given by axy = 0 i.e. x = 0 and y = 0.
3. INTERSECTION WITH THE CO-ORDINATES AXES
(i) To find the points of intersection of the curve with X-axis, put y = 0 in the equation of the given
curve and get the corresponding values of x.
(ii) To find the points of intersection of the curve with Y-axis, put x = 0 in the equation of the given
curve and get the corresponding values of y.
4. ASYMPTOTES 5
x
As a point P on the graph of a function y = f ( x ) moves 4 y=
x −1
farther and farther away from the origin, it may happen 3
that the distance between P and some fixed line tends to 2
zero. In other words, the curve “approaches” the line 1
Asymptotes
as it gets further from the origin. In such a case, the line
x
is called an asymptote of the graph. For instance, the x- –4 –3 –2 –1 1 2 3 4 5
axis and y-axis are asymptotes of the curves y = 1/ x –1
and y = 1/ x 2 . –2
–3
–4
=
The graph of y x / ( x − 1) .
To find out the asymptotes of the curve if the equation is a polynomial equation in x and y .
(i) The vertical asymptotes or the asymptotes parallel to y-axis of the given curve are obtained by
equating to zero the coefficient of the highest power of y in the equation of the given curve.
(ii) The horizontal asymptotes or the asymptotes parallel to x-axis of the given curve are obtained by
equating to zero the coefficient of the highest power of x in the equation of the given curve.
Horizontal and vertical Asymptotes
A line y = b is a horizontal asymptote of the graph of a function y = f ( x ) if either
lim f ( x ) = b or lim f ( X ) = b
x →∞ x →∞
6. CRITICAL POINTS
These are the points at which either the derivative of the function is zero or it does not exist.
dy
Find out the values of x at which = 0.
dx
At such points y generally changes its character from an increasing function of x to a decreasing
function of x or vice-versa.
7. Trace the curve with the help of the above points.
x=a
x=b
O x- axis X
y- axis
d d
x = g (y)
=A x dy ∫ g ( y ) dy.
∫=
c c
y=c
O X
Y
3. The area bounded by the curve y = h ( x ) , x-axis and the y = h (x)
two ordinates x = a and x = b is given by
c b
=A ∫ y dx +
a
∫ y dx
c
O X
x=c
where c is a point in between a and b.
x=a x=b
Y
4. If we have two curve y = f ( x ) and y = g ( x ) , such that y = f ( x ) lies y = f (x)
above the curve y = g ( x ) then the area bounded between them and the
ordinates x = a and=x b ( b > a ) , is given by
b
= ∫ f ( x ) − g ( x ) dx
y = g (x)
A x=a x=b
a
i.e. upper curve area − lower curve area. O X
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5. =
The area bounded between the curve ( y), x φ ( y)
x f=
y=d
y = f (x) y = g (x)
6. The area bounded by the curves y = f ( x ) and y = g ( x ) between
the ordinates x = a and x = b is given by where x = c is the
point of intersection of the two curves is given by
f ( x ) dx + ∫ g ( x ) dx .
c b
=A ∫ a c
X
O x=a x=c x=b
PARAMETRIC FORM FOR THE AREA UNDER A CURVE
Let the curve be given by x = φ ( t ) ; y = Ψ ( t ) . If, on eliminating t , y can be expressed as a function of x (or x
as a function of y ), then the area in equation can be obtained by the formula.
If, however, the parameter can not be eliminated easily, then we use the formula :
b t2
Area = ∫ y dx= ∫ Ψ ( t ) ϕ ′ ( t ) dt
a t1
149
Hence, ∫ f ( x ) dx + ∫ g ( y ) dy =
C is the solution of (1).
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dy ∫ Pdx
+ Pye ∫ = Qe ∫
Pdx Pdx
∴ e
dx
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d ∫ P dx ∫ P dx
i.e., ye = Qe .
dx
ye ∫ ∫ P dx
∫ Qe dx + c
P dx
∴ integrating =
y e ∫ ∫ Qe ∫ dx + c
− P dx P dx
or= is the required solution.
If in the above equation if Q is zero, the general solution is y = Ce ∫ .
− P dx
Cor.1
Cor.2 If P be a constant and equal to −m , then the=
solution is y e mx ∫ e − mx Q dx + C .
dx
Linear differential equations of the form + Rx = s.
dy
Sometimes a linear differential equation can be put in the form
dx
+ Rx = s
dy
where R and S are functions of y alone or constants.
Note : y is independent variable and x is a dependent variable.
Bernoulli’s Equation
dy
An equation of the form + Py = Qy n ,
dx
Where P and Q are functions of x alone, is known as Bernoulli’s equation. It is easily reduced to the
linear form.
dy
Dividing both sides by y n , we get y − n + Py − n +1 = Q.
dx
dy dv dv
Putting y − n +1 = v, and hence ( −n + 1) y − n = the equation reduces to + (1 − n ) Pv = (1 − n ) Q.
dx dx dx
This being linear in v can be solved by the method of the previous article.
ORTHOGONAL TRAJECTORY
Any curve which cuts every member of a given family of curves at right angle is called an orthogonal
trajectory of the family. For example, each straight line y = mx passing through the origin, is an
orthogonal trajectory of the family of the circles x 2 + y 2 =. a2
Procedure for finding the orthogonal trajectory
(i) Let f ( x, y, c ) = 0 be the equation, where c is an arbitrary parameter.
(ii) Differentiate the given equation w.r.t. x and then eliminate c.
dy dx
(iii) Replace by − in the equation obtained in (ii).
dx dy
(iv) Solve the differential equation in (iii).
DIFFERENTIAL EQUATIONS OF FIRST ORDER BUT NOT OF FIRST DEGREE
The typical equation of the first order and the nth degree can be written as
P n + P1 p n −1 + P2 p n − 2 + ...... + Pn −1 p + Pn = 0 … (i)
dy
where p stands for and P1 , P2 ,....., Pn are function of x and y.
dx
The complete solution of such an equation would involve only one arbitrary constant.
The equations which are of first order but not of the first degree, the following types of equations are
discussed.
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155
3. The coordinates of the mid point of the line segment joining the two points A ( x1 , y1 ) and B ( x2 , y2 ) are
given by
x1 + x2 y1 + y2 B ( x2 , y2 )
, . P ( x, y )
2 2
A ( x1 , y1 )
AREA OF A TRIANGLE
Let ( x1 , y1 ) , ( x2 , y2 ) and ( x3 , y3 ) respectively be the coordinates of the vertices A, B, C of a triangle ABC.
Then the are of triangle ABC , is
1
= x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) … (1)
2
x1 y1 1
1
= x2 y2 1 … (2)
2 A
x3 y3 1
While using formula (1) or (2), order of the points ( x1 , y1 ) , ( x2 , y2 ) and
( x3 , y3 ) has not been taken into account. If we plot the points
A ( x1 , y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ) , then the area of the triangle as obtained
by using formula (1) or (2) will be positive or negative as the points A, B, C
are in anti-clockwise or clockwise directions,
B D C
A A
+ −
C B B C
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So, while finding the area of triangle ABC , we use the formula :
x1 y1 1
1 1
Area of ∆ ABC
= x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y=
2) | x2 y2 1 |
2 2
x3 y3 1
Note :
(a) If the three points A, B, C are collinear then area of ∆ ABC is zero.
(b) The area of a quadrilateral, whose vertices are A ( x1 , y1 ) , B ( x2 , y2 ) , C ( x3 , y3 ) and D ( x4 , y4 ) , is
1 x1 y1 x y2 x y3 y4
x
= | + 2 + 3 + 4 |
2 x2 y2 x3 y3 x4 y4 x1
y1
(c) The area of a polygon of n sides with vertices A1 ( x1 , y1 ) , A2 ( x2 , y2 ) ,...., An ( xn , yn ) is
1 x1 y1 x y2 x yn −1 xyn
= | + 2 + ... + n −1 + n |
2 x2 y2 x3 y3 xn yn x1
y1
(d) If a1 x + b1 y +=
c1 0, a2 x + b2 y + =
c2 0 and a3 x + b3 y + c3 =0 are the equations of the sides of a
2
a1 b1 c1
1
triangle, then the area of the triangle is = a2 b2 c2
2 C1C2C3
a3 b3 c3
where C1 , C2 , C3 are the cofactors of c1 , c2 , c3 in the determinant
i.e. C1 = a2b3 − a3b2 , C2 =a3b1 − a1b3 and = C3 a1b2 − a2b1.
LOCUS
When a point moves in a plane under certain geometrical conditions, the point traces out a path. This path of a
moving point is called its locus.
EQUATION OF LOCUS
The equation to a locus is the relation which exists between the coordinates of any point on the path and which
holds for no other point except those lying on the path.
PROCEDURE FOR FINDING THE EQUATION OF THE LOCUS OF A POINT
(i) If we are finding the equation of the locus of the locus of a point P, assign coordinates ( h, k ) to P.
(ii) Express the given conditions as equations in terms of the known quantities to facilitate calculations.
We sometimes include some unknown quantities known as parameters.
(iii) Eliminate the parameters, so that the eliminate contains only h, k and known quantities.
(iv) Replace h by x, and k by y, in the eliminate. The resulting equation would be the equation of the
locus of P.
(v) If x and y coordinates of the moving point are obtained in terms of a third variable t (called the
parameter), eliminate t to obtain the relation in x and y and simplify this relation. This will give
the required equation of locus.
TRANSLATION OF AXES Y Y′
The translation of axes involves the shifting of the origin to a new point, the new
axes remaining parallel to the original axes. P
Let OX , OY be the original axes and O′ be the new origin. Let coordinates Y y
of O′ referred to original axes i.e. OX , OY be ( h, k ) . ( h, k )
X′
O′ X M′
X
O N M
x
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Let O′X ′ and O′Y ′ be drawn parallel to and in the same direction as OX and OY respectively. Let P
be any point in the plane having coordinates ( x, y ) referred to old axes and ( X , Y ) referred to new axes.
Then, x =OM =ON + NM =ON + O′M ′ and y = MP = MM ′ + M ′P = NO′ + M ′P
=h + X =X + h = k + Y = Y + k.
Thus, we get x =X + h, y =Y + k ⇒ X =x − h, Y =y − k .
Thus, the point whose coordinates were ( x, y ) has new coordinates ( x − h, y − k ) .
ROTATION OF AXES
Y
ROTATION OF AXES WITHOUT CHANGING THE ORIGIN Y′
P ( x, y )
Let OX , OY be the original axes and OX ′, OY ′ be the new axes obtained
( x′, y′)
by rotating OX and OY through an angle θ in the anticlockwise sense.
θ y′
Let P be any point in the plane having coordinates ( x, y ) w.r.t. axes y X′
OX and OY and ( x′, y′ ) w.r.t. axes OX ′ and OY ′. θ
x′
Then, θ
X
x
= x x′ cos θ − y′ sin θ = x′ x cos θ + y sin θ O
and .
= y x′ sin θ + y′ cos θ y′ =
− x sin θ + y cos θ
Y Y′
CHANGE OF ORIGIN AND ROTATION OF AXES P ( x, y )
If origin is changed to O′ ( h, k ) and axes are rotated about the new origin O′ by θ ( x′, y′)
X′
angle θ in the anticlockwise sense such that the new coordinates of P ( x, y ) θ
become ( x′, y′ ) then the equations of transformation will be O′
x= h + x′ cos θ − y′ sin θ and y= k + x′ sin θ + y′ cos θ . X
O
GENERAL EQUATION OF A STRAIGHT LINE
An equation of the form ax + by + c = 0, where a, b, c are constants and a, b are not simultaneously zero,
always represents a straight line.
SLOPE OF A LINE
π
If a line makes an angle θ θ ≠ with the positive direction of x-axis, then tan θ is the slope or gradient of
2
that line. It is usually denoted by m . i.e. m = tan θ .
y1 − y2 y2 − y1
Slope of the line joining two points ( x1 , y1 ) and ( x2= , y2 ) = .
x1 − x2 x2 − x1
INTERCEPT OF A LINE ON THE AXES
(i) Intercept of a line on x-axis
If a line cuts x-axis at ( a, 0 ) , then a is called the intercept of the line on x-axis. a is called the
length of the intercept of the line on x-axis. Intercept of a line on x-axis may be positive or negative.
(ii) Intercept of a line on y-axis
If a line cuts y-axis at ( 0, b ) , then b is called the intercept of the line on y-axis and b is called the
length of the intercept of the line on y-axis. Intercept of a line on y-axis may be positive or negative.
EQUATIONS OF LINES PARALLEL TO AXES
1. Equation of x-axis : The equation of x-axis is y = 0.
2. Equation of y-axis : The equation of y-axis is x = 0.
3. Equation of a line parallel to y-axis : The equation of the straight line parallel to y-axis at a distance
a from it on the positive side of x-axis is x = a.
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4. Equation of a line parallel to x-axis : The equation of the straight line parallel to x-axis at a distance b
from it on the positive side of y-axis is y = b.
Y
EQUATION OF A STRAIGHT LINE IN VARIOUS FORMS
1. Slope-intercept form
The equation of a straight line whose slope is m and which cuts an
intercept c on the y-axis is given by =
y mx + c.
c
θ X
O
Y
2. Point-slope form P ( x, y )
The equation of a straight line passing through the point ( x1 , y1 ) and
having slope m is given by A ( x1 , y1 )
y − y1= m ( x − x1 ) . θ
X' O X
Y
3. Two-point form P ( x, y )
The equation of a straight line passing through two points ( x1 , y1 )
B ( x2 , y2 )
and ( x2 , y2 ) is given by A ( x1 , y1 )
y2 − y1
=
y − y1 ( x − x1 ) . θ
x2 − x1 X' O X
Y
4. Intercept form
The equation of a straight line which cuts off intercepts a and b on
b
x y
x-axis and y-axis respectively is given by + = 1.
a b
X
O a
Y
5. Normal form (or perpendicular form)
The equation of a straight line upon which the length of the perpendicular
from the origin is p and the perpendicular makes an angle α with the
positive direction of x-axis is given by L
p
x cos α + y sin α =p.
α
Note : In normal form of equation of a straight line p is always taken as X
O
positive and α is measured from positive direction of x-axis in
anticlockwise direction between 0 and 2π . Y
6. Parametric form or symmetric form
P ( x, y )
The equation of a straight line passing through the point ( x1 , y1 ) and
making an angle θ with the positive direction of x-axis is A ( x1 , y1 )
x − x1 y − y1
= = r where 0 ≤ θ < π θ
cos θ sin θ X' O X
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m1 − m2
provided no line is ⊥ to x-axis and the acute angle θ is given by tan θ = .
1 + m1m2
Note :
(a) If both the lines are ⊥ to x-axis then the angle between them is 0°.
(b) If any of the two lines is perpendicular to x-axis, then the slope of that line is infinite.
m
1− 2
m1 − m2 m1 1
Let m1 = ∞. Then= tan θ = =
1 + m1m2 1
+ m2 m2
m1
or θ= 90° − α where tan α = m2
(c) The two lines are parallel if and only if m1 = m2 .
(d) The two lines are ⊥ if and only if m1 × m2 = −1.
CONDITION FOR TWO LINES TO BE COINCIDENT, PARALLEL, PERPENDICULAR OR
INTERSECTING
Two lines a1 x + b1 y + c1 =0 and a2 x + b2 y + c2 = 0 are
a1 b1 c1 a1 b1 c1
(i) Coincident, if = = ; (ii) Parallel, if = ≠ ;
a2 b2 c2 a2 b2 c2
(iii) Perpendicular, if a1a2 + b1b2 = 0;
a b
(iv) Intersecting, if 1 ≠ 1 i.e. if they are neither co-incident nor parallel.
a2 b2
EQUATION OF A LINE PARALLEL TO A GIVEN LINE
The equation of a line parallel to a given line ax + by + c = 0 is ax + by + k = 0, where k is a constant.
Thus to write the equation of any line parallel to a given line, do not change the coefficient of x and y
and change the constant term only.
EQUATION OF A LINE PERPENDICULAR TO A GIVEN LINE
The equation of a line perpendicular to a given line ax + by + c = 0 is bx − ay + k = 0, where k is a constant.
Thus to write the equation of any line perpendicular to a given line interchange the coefficients of x and y
then change the sign of any one of them and finally change the constant term.
POINT OF INTERSECTION OF TWO GIVEN LINES
Let the two given lines be a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0.
Solving these two equations, the point of intersection of the given two lines is given by
b1c2 − b2 c1 c1a2 − c2 a1
, .
a1b2 − a2b1 a1b2 − a2b1
INTERIOR ANGLES OF A TRIANGLE : To find the interior angles of a triangle arrange the slopes of the
sides in decreasing order i.e., m1 > m2 > m3 . Then apply
m1 − m2 m2 − m3 m3 − m1
= tan α = , tan β = , tan γ
1 + m1m2 1 + m2 m3 1 + m3 m1
LINES THROUGH THE INTERSECTION OF TWO GIVEN LINES
The equation of any line passing through the point of intersection of the lines a1 x + b1 y + c1 = 0
0 is ( a1 x + b1 y + c1 ) + k ( a2 x + b2 y + c2 ) =
and a2 x + b2 y + c2 = 0,
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where k is a parameter. The value of k can be obtained by using one more conditions which the required line
satisfies.
CONDITIONS OF CONCURRENCE
The family of given lines are said to be concurrent if they meet in a point.
WORKING RULE TO PROVE THAT THREE GIVEN LINES ARE CONCURRENT
1. The three lines a1 x + b1 y=
+ c1 0, a2 x + b2 y=+ c2 0, a3 x + b3 y + c3 =
0
a1 b1 c1 l1 l2
are concurrent if a2 b2 c2 = 0
l3
a3 b3 c3
2. The three lines= P 0,= Q 0 and R = 0 are concurrent if there exist constants l , m and n, not all zero at
the same time, such that lP + mQ + nR = 0.
This method is particularly useful in theoretical results.
POSITION OF TWO POINTS RELATIVE TO A LINE
Two points ( x1 , y1 ) and ( x2 , y2 ) are on the same side or on opposite sides of the line ax + by + c =0 according
as the expressions ax1 + by1 + c and ax2 + by2 + c have same sign or opposite signs.
LENGTH OF PERPENDICULAR FROM A POINT ON A LINE
The length of the perpendicular from the point (α , β ) to the line ax + by + c =0 is given by
(α , β )
aα + bβ + c
p= . p
a 2 + b2
ax + by + c =0
DISTANCE BETWEEN TWO PARALLEL LINES
The distance between two parallel lines ax + by + c1 =
0 and ax + by + c2 =
0 is given by
c1 − c2
d= .
a 2 + b2 d
Note : (a) The distance between two parallel lines can also be obtained by taking a suitable point (take y = 0
and find x or take x = 0 and find y ) on one straight line and then finding the length of the
perpendicular from this point to the second line.
(b) Area of a parallelogram or a rhombus, equations of whose sides are given, can be obtained by
using the following formula
D C
p1 p2
= =
Area p1 p2 cosec θ
sin θ p1
where p= 1 =
DL distance between lines AB and CD , M
p2
=
p2 = distance between lines AD and BC ,
BM
θ
θ = angle between adjacent sides AB and AD. A
B
L
In the case of a rhombus, p1 = p2
p12
∴ Area of rhombus = .
sin θ
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1
Also, area of rhombus = d1d 2
2
where d1 and d 2 are the lengths of two ⊥ diagonals of a rhombus.
a1 x + b1 y + d1 =
0
(c) If the equation of sides of a parallelogram are as shown
then its area is given by a2 x + b2 y + d 2 =
0 a2 x + b2 y + c2 =
0
( c1 − d1 )( c2 − d 2 )
a1b2 − a2b1 a1 x + b1 y + c1 =
0
D L2 = 0 C
(d) If equation of sides of a parallelogram are
= L1 0,= L2 0,= L3 0,= L4 0 then L2 L3 − L1 L4 = 0 represents the
L1 = 0 L3 = 0
diagonal BD. Again L1 L2 − L3 L4 = 0 represents the diagonal AC .
A L4 = 0 B
A ( x1 , y1 )
STANDARD POINTS OF A TRIANGLE
Centroid of a Triangle
E
The point of intersection of the medians of the triangle is called the centroid of
the triangle. The centroid divides the medians in the ratio 2 : 1. F G
C ( x3 , y3 )
The coordinates of the centroid of a triangle with vertices ( x1 , y1 ) , ( x2 , y2 ) and
( x3 , y3 ) are D
B ( x2 , y2 )
x1 + x2 + x3 y1 + y2 + y3
, .
3 3
(i) If P is any point in the plane of the triangle ABC and G is the centroid then
PA2 + PB 2 + PC 2 = GA2 + GB 2 + GC 2 +3PG 2
(ii) If G is the centroid of the triangle ABC , then
AB 2 + BC 2 + CA2= 3 GA2 + GB 2 + GC 2
A ( x1 , y1 )
Incentre of a Triangle
The point of intersection of the internal bisectors of the angles of a triangle is
b
called the incentre of the triangle.
The coordinates of the incetre of a triangle with vertices ( x1 , y1 ) , ( x2 , y2 ) and c
I
C ( x3 , y3 )
( x3 , y3 ) are
a
ax1 + bx2 + cx3 ay1 + by2 + cy3
a+b+c , a + b + c
. B ( x2 , y2 )
Ex-centres of a Triangle
A circle touches one side outside the triangle and the other two extended sides then circle
is known as excircle. A
I2
Let ABC be a triangle then there are three excircles, with three excentres I1 , I 2 , I 3 I3
A ( x1 , y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ) then I1
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CIRCUMCENTRE A ( x1 , y1 )
The circumcentre of a triangle is the point of intersection of the perpendicular
bisectors of the sides of a triangle. It is the centre of the circle which passes
through the vertices of the triangle and so its distance from the vertices of the F E
triangle is same and this distance is known as the circum-radius of the triangle.
If angles of triangle ∆ ABC i.e. A, B, C and vertices of triangle O
C ( x3 , y3 )
A ( x1 , y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ) are given, then circumcentre of the triangle B
( x2 , y2 ) D
ABC is
x1 sin 2 A + x2 sin 2 B + x3 sin 2C y1 sin 2 A + y2 sin 2 B + y3 sin 2C
sin 2 A + sin 2 B + sin 2C
,
sin 2 A + sin 2 B + sin 2C
ORTHOCENTRE
The orthocentre of a triangle is the point of intersection of altitudes. A
If angles of a ∆ ABC , i.e. A, B and C and vertices of triangle
A ( x1 , y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ) are given, then orthocentre of ∆ ABC is
F H E
165
Case II : If ∆ ≠ 0, then
(i) a = b and h = 0, then (1) is circle
(ii) h2 = ab, then (1) is parabola
(iii) h 2 < ab, then (1) is ellipse
(iv) h 2 > ab, then (1) is hyperbola.
3. INFORMATION ABOUT PAIR OF STRAIGHT LINES
General equation of second degree is
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 if abc + 2 fgh − af 2 − bg 2 − ch 2 = 0
and h > ab, then (1) represents pair of intersecting lines. Formula for the slopes of the lines, sum and
2
product of the slopes, angles between the lines and condition for the perpendicularity of lines will
remain same as it was in case of second degree homogeneous equation.
hf − bg hg − af
Let point of intersection of lines given by (1) be ( x1= , y1 ) , then x1 = and y1 .
ab − h 2
ab − h 2
4. Equation of the angle bisectors of the angles formed by (1) are given by
( x − x1 ) − ( y − y1 ) ( x − x1 )( y − y1 ) .
2 2
=
a −b h
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ax + 2hxy + by + ( 2 gx + 2 fy ) −
2 2
+ c− =
0. lx + my + n =0
n n
O
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167
Circle Chapter 26
CIRCLE
A circle is the locus of point which moves in a plane such that its distance from a fixed point is constant. The
fixed point is called the centre and the constant distance is called the radius of the circle.
STANDARD EQUATION OF A CIRCLE
The equation of a circle with the centre at (α , β ) and radius a, is ( x − α ) + ( y − β ) =
2 2
1. a2
2. If the centre of the circle is at the origin and the radius is a, then the equation of circle is x 2 + y 2 =
a2.
GENERAL EQUATION OF A CIRCLE
The general equation of a circle is of the form x 2 + y 2 + 2 gx + 2 fy + c =0, … (1)
Where g , f and c are constants.
The coordinates of the centre are ( − g , − f ) and radius = g 2 + f 2 − c.
CONDITIONS FOR GENERAL EQUATION OF SECOND DEGREE TO REPRESENT A CIRCLE
A general equation of second degree ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c =0 in x, y represents a circle if
1. Coefficient of x = coefficient of y i.e. a = b,
2 2
O X
or x 2 + y 2 − 2hx − 2ky + h 2 =
0
Y
2. Circle with centre at the point (h, k) and which touches the axis of y
Since the circle touches the y-axis, the radius of the circle = h h
( h, k )
∴ Equation of the circle is ( x − h ) + ( y − k ) =
2 2
h2
or x 2 + y 2 − 2hx − 2ky + k 2 =
0. O
X
3. Circle with radius a and which touches both the coordinate axes
Since the centre of the circle may be in any of the four quadrants, ( − a, a ) ( a, a )
therefore it will be any one of the four points ( ± a, ± a ) . Thus, there are
X
four circles of radius a touching both the coordinate axes, and their
equations are ( x ± a ) + ( y ± a ) =
2 2
a 2 or x 2 + y 2 ± 2ax ± 2ay + a 2 =
0. ( − a, − a ) ( a, − a )
C C C
Note : Let S be a circle and P ( x1 , y1 ) and Q ( x2 , y2 ) be two points in the plane of S , then they lie
(a) on the same side of S iff S1 and S 2 have same sign,
(b) on the opposite sides of S iff S1 and S 2 have opposite signs.
Circle 169
(b) l intersects S in one and only one point iff d = a i.e. the line touches C
the circle iff perpendicular distance from the centre to the line is equal
to the radius of the circle. a=d
l
am a
The coordinates of the point of contact are ± , .
1+ m 1 + m2
2
CONDITION OF TANGENCY
y mx + c will be a tangent to the circle x 2 + y 2 =
The straight line = a 2 if c = ±a 1 + m2 .
Note : A line will touch a circle if and only if the length of the ⊥ from the centre of the circle to the line
is equal to the radius of the circle.
NORMAL TO A CIRCLE AT A GIVEN POINT
The normal to a circle, at any point on the circle, is a straight line which is ⊥ to the tangent to the circle at that
point and always passes through the centre of the circle.
1. Equation of the normal to the circle x 2 + y 2 =a 2 at the point ( x1 , y1 ) on it is
x y
= .
x1 y1
2. Equation of the normal to the circle x 2 + y 2 + 2 gx + 2 fy + c =0 at the point ( x1 , y1 ) on it is
x − x1 y − y1
= .
x1 + g y1 + f
LENGTH OF TANGENTS
Let PQ and PR be two tangents drawn from P ( x1 , y1 ) to the circle S1
Q
x 2 + y 2 + 2 gx + 2 fy + c =.
0 Then PQ = PR and the length of tangent drawn from
point P is given by P
PQ = PR = x + y + 2 gx1 + 2 fy1 + c =S1 .
2 2 ( x1 , y1 ) R
1 1
PAIR OF TANGENTS
Form a given point P ( x1 , y1 ) two tangents PQ and PR can be drawn to the circle Q
director circle is a concentric circle whose radius is 2 times the radius of the given
circle.
Director circle of circle x 2 + y 2 + 2 gx + 2 fy + c =0 is
x 2 + y 2 + 2 gx + 2 fy + 2c − g 2 − f 2 =
0
POWER OF A POINT WITH RESPECT TO A CIRCLE
Let P ( x1 , y1 ) be point and secant PAB, drawn. T
The power of P ( x1 , y1 ) w.r.t. A
B
S = x 2 + y 2 + 2 gx + 2 fy + c = 0 P
S=0
C
is equal to PA . PB, which is x12 + y12 + 2 gx1 + 2 fy1 + c D
Circle 171
[Power of a point P is positive, negative or zero according to position of P outside, inside or on the
circle respectively]
DIAMETER OF A CIRCLE
Diameter
The locus of the middle points of a system of parallel chords of a circle is called a x + my =
0
diameter of the circle. B
O
P ( h, k )
The equation of the diameter bisecting parallel chords =
y mx + c ( c is a parameter) of
=
y mx + c
the circle x 2 + y 2 =
a 2 is x + my =
0.
A
P
Q Q
C
C T1 T1
P
R R
Direct common
tangents
r1
r2
C1 P
T C2
Transverse common
tangents
In this case four common tangents can be drawn to the two circles, in which two are direct
common tangents and the other two are transverse common tangents.
The points P, T of intersection of direct common tangents and transverse common tangents
respectively, always lie on the line joining the centres of the two circles and divide it externally and
internally respectively in the ratio of their radii.
JEEMAIN.GURU
Circle 173
C1 P r1 C T r1
= = ( externally ) and 1 ( internally ) Hence, the ordinates of P and T are
C2 P r2 C2T r2
rx −r x r y −r y rx +r x ry +r y
P ≡ 1 2 2 1 , 1 2 2 1 and T ≡ 1 2 2 1 , 1 2 2 1 .
r1 − r2 r1 − r2 r1 + r2 r1 + r2
Direct common
Case II : When C1C2= r1 + r2 i.e., the distance between the centres is tangents
equal to the sum of radii.
In this case two direct common tangents are real and distinct P
C1 T C2
while transverse tangents are coincident.
Transverse common
Tangent
Direct common
Case III : When C1C2 < r1 + r2 i.e., the distance between the centres is less than
Tangents
sum of radii.
In this case two direct common tangents are real and distinct while
the transverse tangents are imaginary. C1 C2 P
Case IV : When C1C2= r1 − r2 , i.e., the distance between the centres is Tangent at the
equal to the difference of the radii. Point of
r2 contact
In this case two tangents are real and coincident while the
C1 C2 P
other two tangents are imaginary.
r1
Case V : When C1C2 < r1 − r2 , i.e., the distance between the centres is less than the
r2
difference of the radii. C1
In this case, all the four common tangents are imaginary. C2
r1
Note :
(a) The direct common tangents to two circles meet on the line joining centres C1 and C2 , and divide it
externally in the ratio of the radii.
(b) The transverse common tangents also meet on the line of centres C1 and C2 , and divide it internally
in the ratio of the radii.
WORKING RULE TO FIND TRANSVERSE COMMON TANGENTS
All the steps except the 2nd step are the same as above. Here in the second step the point R ( h, k ) will divide
C1C2 internally in the ratio r1 : r2 .
r1
r2
C1 C2 P
R
Note :
(a) When two circles are real and non-intersecting, 4 common tangents can be drawn.
(b) When two circles touch each other externally, 3 common tangents can be drawn to the circles.
(c) When two circles intersect each other at two real and distinct points, two common tangents can
be drawn to the circles.
(d) When two circle touch each other internally one common tangent can be drawn to the circles.
IMAGE OF THE CIRCLE BY THE LINE MIRROR
Let the circle be S ≡ x 2 + y 2 + 2 gx + 2 fy + c = 0 and the line be L = lx + my + n = 0.
The radius of the image circle will be the same as that of the given circle.
lx + my + n =0
Let the centre of the image circle be ( x1 , y1 ) .
Slope of C1C2 × slope of line L = −1 … (1)
and midpoint of C1C2 lies on lx + my + n =0
x −g y1 − f r r
⇒ l 1 + m +n = 0 … (2)
2 2 C2 ( − g , − f ) C1 ( x1 , y1 )
Solving (1) and (2), we get ( x1 , y1 ) .
⇒ Required image circle will be Given circle Image circle
( )
2
( x − x1 ) + ( y − y1 )=
2 2
g2 + f 2 − c .
A′ B′
ANGLE OF INTERSECTION OF TWO CIRCLES S=0 S′ = 0
The angle of intersection between two circles S = 0 and S ′ = 0 is defined as the angle π −θ
r1
P
r2
between their tangents at their point of intersection. θ
C1 C2
If S ≡ x 2 + y 2 + 2 g1 x + 2 f1 y + c1 =
0 B A
Q
S ′ ≡ x 2 + y 2 + 2 g 2 x + 2 f 2 y + c2 =
0
are two circles with radii r1 , r2 and d be the distance between their centres then the angle of intersection
θ between them is given by
JEEMAIN.GURU
Circle 175
r12 + r22 − d 2
cos θ =
2r1r2
2 ( g1 g 2 + f1 f 2 ) − ( c1 + c2 )
or cos θ = .
2 g12 + f12 − c1 g 22 + f 22 − c2
P
Condition of Orthogonality : If the angle of intersection of the two circles is a right
angle (θ= 90° ) , then such circles are called orthogonal circles and condition for C1
90°
C2
( − g1 , − f1 ) ( − g2 , − f2 )
orthogonality is 2 g1 g 2 + 2 f1 f 2 =
c1 + c2 .
FAMILY OF CIRCLES
1. The equation of the family of circles passing through the point of intersection
of circle S = 0 and a line L = 0 is given as S + λ L = 0, (where λ is a
parameter)
S =0 S + λL =
0
L=0
2. The equation of the family of circles passing through the point of intersection of
two given circles S = 0 and S ′ = 0 is given as S + λ S ′ = 0, (where λ is a
parameter, λ ≠ −1 ). But it is better to find first the equation of common
S =0 S′ = 0
0 and then use S + λ ( S − S ′ ) =
S − S′ = 0
S + λS′ =
0
3. The equation of the family of circles touching the circle S = 0 and the line
L = 0 at their point of contact P is
S + λL = 0, (where λ is a parameter) S =0
L=0 S + λL =
0
RADICAL AXIS
The radical axis of two circles is the locus of a point which moves such that the lengths of the tangents drawn
from it to the two circles are equal.
P ( x1 , y1 ) P ( x1 , y1 )
B
B
A A
C1 C2 C1 C2
S1 = 0 S2 = 0
Circle 177
2. The equation of a co-axial system of circles, where the equation of any two circles of the system are
S1 ≡ x 2 + y 2 + 2 g1 x + 2 f1 y + c1 =
0 and S 2 ≡ x 2 + y 2 + 2 g 2 x + 2 f 2 y + c2 =
0
respectively, is S1 + λ ( S1 − S 2 ) =
0 or S 2 + λ1 ( S1 − S2 ) =
0
λ S 2 0, ( λ ≠ −1)
Other form S1 + =
S 2 + λ ( S1 − S 2 ) =
0
S1 + λ ( S1 + S 2 ) =
0
S1 + λ ( S1 − S 2 ) =
0
S1 + λ S 2 =
0 S2 = 0
S1 = 0 S2 = 0 S1 = 0
S1–S2 = 0
3. The equation of a system of co-axial circles in the simplest form is x 2 + y 2 + 2 gx + c =0, where g is a
variable and c is a constant.
LIMITING POINTS
Limiting points of a system of co-axial circles are the centres of the point circles belonging to the family
(Circles whose radii are zero are called point circles).
Let the circle is x 2 + y 2 + 2 gx + c =0 … (i)
where g is a variable and c is a constant.
∴ Centre and the radius of (i) are ( − g , 0 ) and (g 2
)
− c respectively. Let g 2 − c =0 ⇒ g =± c
Thus we get the two limiting points of the given co-axial system as ( )
c , 0 and − c , 0 ( )
Clearly the above limiting points are real and distinct, real and coincident or imaginary according as
c > , = , < 0.
A
Lex
( r1 − r2 ) B
d B r2
r r2
C
r1 Lin
r1 + r2
r2 Lin
A
Nine-point circle : The circle through the midpoints of the sides of a triangle passes through the feet of the
altitudes and the midpoints of the lines joining the orthocentre to the vertices. This circle is called the nine-point
circle of the triangle.
Note :
(a) The radius of the nine point circle is half the radius of the circumcircle of the triangle ABC
(b) Centre is midpoint of the line segment joining orthocenter and circumcentre.
A A
F E P
P
B′ B′
C′ E C′ E
H H
Q R Q R
B B
A′ D C A′ D C
L
A
Simson’s line : The feet L, M , N of the perpendicular on the sides BC , CA, AB of any X
∆ ABC from any point X on the circumcircle of the triangle are collinear. The line M
LMN is called the Simson’s line or the pedal line of the point X with respect to ∆ ABC.
N
B C
N′
A
If H is the orthocentre of ∆ ABC and AH produced meets BC at D and the
H E
circumcircle of ∆ ABC at P, then HD = DP.
B D C
P
JEEMAIN.GURU
179
x − x1 y − y1
=
ax1 + hy1 + g hx1 + by1 + f
Note :
To find the equation of tangent at P ( x1 , y1 ) to the conic ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c =0 … (*)
We use the following steps :
We replace the terms containing
(i) x 2 and y 2 by x x1 and y y1 respectively
xy +x y
(ii) xy by 1 1
2
x + x1 y + y1
(iii) x and y by and respectively in equation (*)
2 2
JEEMAIN.GURU
181
Parabola Chapter 28
PARABOLA
A parabola is the locus of a point which moves in a plane in such a way that its distance from a fixed point is
always equal to its distance from a fixed straight line.
This fixed point is called focus and the fixed straight line is called directrix.
Let S be the focus, ZZ ′ be the directrix and P be any point on the parabola. Then by definition,
PS = PM
where PM is the length of the perpendicular from P on the directrix ZM.
STANDARD EQUATION OF THE PARABOLA
Let S be the focus, zz ′ be the directrix of the parabola and (x, y) be any point on parabola, then standard form
of the parabola is y 2 = 4ax .
Some terms related to parabola
Z′ Y ( h, 2 ah )
Directrix F
( a, 2a ) Q
Focal chord
L
P x=a
M
Focal distance Double ordinate
Vertex Focus
S ( a, 0 )
X
Z A Axis
x+a=0
Latus rectum
F′
L′
( a, − 2a ) Q′
Y′
( h, − 2 ah )
Some other standard forms of parabola are
(i) Parabola opening to left i.e. y 2 = −4ax (ii) Parabola opening upwards i.e. x 2 = 4ay ,
y Y
L
S ( 0, a )
L′ L
X
( − a, 0 ) S O
X
O
L′
x=a y 2 = −4ax y = −a
x = 4ay
2
(iii) Parabola opening downwards i.e., x 2 = −4ay Y
y=a
O
X
L′ L
S ( 0, − a )
x 2 = −4ay
JEEMAIN.GURU
if t1t2 = −1 . S
R O
X′
R ( at1t2 , a ( t1 + t2 ) ) . Q ( at , 2at )
2
2 2
Y′
JEEMAIN.GURU
Parabola 183
Note :
(a) m1 + m2 + m3 = 0 i.e., the sum of the slopes of the normals at co- m3
normals points is zero.
(b) The sum of the ordinates of the co-normal points
(i.e., −2am1 − 2am2 − 2am3 = −2a ( m1 + m2 + m3 ) =
0 ) is zero.
(c) The centroid of the triangle formed by the co-normal points lies on the axis of the parabola [the
vertices of the triangle formed by the co-normals points are ( am12 , − 2am1 ) , ( am22 , − 2am2 ) and
−2a ( m1 + m2 + m3 ) −2a
( am , − 2am )
2
3 3 ∴ y-coordinate of the centroid=
3
=
3
× 0= 0 . Hence, the
centroid lies on the x-axis i.e. axis of the parabola].
JEEMAIN.GURU
(d) If three normals drawn to any parabola y 2 = 4ax from a given point ( h, 0 ) be real, then
h > 2a .
POSITION OF A POINT WITH RESPECT TO A PARABOLA
The point ( x1 , y1 ) lies outside, on or inside the parabola y 2 = 4ax according as y12 − 4ax1 >, = or < 0 .
( x3 , y3 ) R
DIAMETERS OF A PARABOLA
Y ( x1 , y1 )
The locus of the middle points of a system of parallel chords is called a P
diameter of the parabola. The diameter is a straight line parallel to the
=
y mx + c
axis of the parabola.
R ( h, k )
The equation of the diameter bisecting chords of the parabola X′ O X
y = 4ax of slope m is y =
2 2 a . Diameter
m
Tangents drawn at the ends of any of these chords meet on the Q
diameter of these chords. Y′ ( x2 , y2 )
Y′ A
P
JEEMAIN.GURU
Parabola 185
CONJUGATE POINTS
If the polar of P ( x1 , y1 ) passes through Q ( x2 , y2 ) , then the polar of Q will pass through P and such
points are said to be conjugate points.
CONJUGATE LINES
If the pole of line ax + by + c = 0 lies on the another line a1 x + b1 y + c1 =0 then the pole of the second line
will lie on the first and such lines are said to be conjugate lines.
Note :
(a) Polar of the focus is the directrix.
(b) Any tangent is the polar of its point of contact.
(c) The point of intersection of the polars of two points Q and R is the pole of QR.
SOME MORE IMPORTANT FACTS ABOUT PARABOLA
1. The parametric equations of the parabola or the coordinates of any point on it are = x at= 2
, y 2at.
2. The tangents at the extremities of any focal chord intersect at right angles on the directrix.
3. The locus of the point of intersection of perpendicular tangents to the parabola is its directrix.
4. The area of the triangle formed by any three points on the parabola is twice the area of the triangle formed
by the tangents at these points.
5. The circle described on any focal chord of a parabola as diameter touches the directrix.
6. If the normal at the point ( at12 , 2at1 ) meets the parabola again at ( at22 , 2at2 ) then t2 =−t1 − 2 / t1.
7. Three normals can be drawn from a point ( x1 , y1 ) to the parabola. The points where these normals meet
the parabola are called feet of the normals or conormal points. The sum of the slopes of these normals is
zero and the sum of the ordinates of the feet of these normals is also zero.
8. If the normals ‘ t1 ’ and ‘ t2 ’ meet on the parabola then t1 t2 = 2.
9. The pole of any focal chord of the parabola lies on its directrix.
10. A diameter of the parabola is parallel to its axis and the tangent at the point where it meets the parabola is
parallel to the system of chords bisected by the diameter and tangent at the ends of any of the parallel
chords of this diameter meet on the diameter.
11. The harmonic mean between the focal radii of any focal chord of a parabola is equal to semi-latus rectum.
12. If the tangent and normal at any point P on the
parabola meet the axis of the parabola in T and G Y
(x , y )
1 1
respectively, the P
(i) ST = SG = SP, S being the focus M θ
Z θ
(ii) PSK = π / 2, where K is the point where the K
X′ θ
tangent at P meets the directrix. X
T ( − x , 0) O S
1 N G
(iii) The tangent at P is equally inclined to the axis
of the parabola and the focal distance of P.
(iv) length of the subtangent is twice the abscissa at
the point of the tangency for the parabola Y′
y = 4ax
2
(v) length of the sub-normal is always of constant length and is equal to semi latus rectum of the
parabola i.e. 2a
13. If we draw a circle taking any focal radius as diameter will touch the tangent at the vertex.
14. The foot of ⊥ from focus on any tangent to the parabola is the point where the tangent meets the
tangent at vertex.
i.e. here Z lies on y-axis and SZ = OS .SP
JEEMAIN.GURU
Ellipse Chapter 29
DEFINITION
An ellipse is the locus of a point which moves in such a way that its distance from a fixed point is in constant
ratio (<1) to its distance from a fixed line. The fixed point is called the focus and fixed line is called the
directrix and the constant ratio is called the eccentricity of the ellipse, denoted by e.
Directrix
Directrix
STANDARD EQUATION OF THE ELLIPSE
Let S be the focus, ZM be the directrix of the ellipse and P ( x, y ) is any
M′ (0, b) B p(x, y) M
point on the ellipse, then by definition
Z′ C Z
x2 y 2
= 1 where b 2 = a 2 (1 − e 2 )
X′ X
+ A′ S ′ (–ae, 0) S (ae, 0) A
a 2 b2 (–a,0) (a, 0)
(0,–b) B ′
Y x = –a/e
Y′ x = a/e
Y= a/e Z K
A(0 , a)
(0 ,ae)
S
X' X
The other form of equation of ellipse is B' C 0,0 B
(0 ,–ae)
(–b, 0) (b, 0)
x2 y 2 S'
+ =
1
b2 a 2 A ' (0 , – a)
where= b 2 a 2 (1 − e 2 ) Y= –a/e Z' K'
Y'
PARAMETRIC FORM OF THE ELLIPSE
x2 y 2
Let the equation of ellipse in standard form will be given by + =1.
a 2 b2
Then the equation of ellipse in the parametric form will be given= by x a= cos θ , y b sin θ , where θ is the
eccentric angle which is such that 0 ≤ θ < 2π . Therefore, coordinate of any point P on the ellipse will be given
by ( a cos θ , b sin θ ) .
x2 y 2 P (outside)
Let P ( x1 , y1 ) be any point and let 2 + 2 = 1 is the equation of an ellipse. P (on)
a b
P (inside)
x2 y 2 X
The point lies outside, on or inside the ellipse if S1 = 12 + 12 − 1 >, =, < 0 . C
a b
Ellipse 187
x y
cos θ + sin θ =1. ( x1 , y1 ) P
a b B
Equation of normal in different forms
x2 y2
1. Point form: The equation of the normal at ( x1 , y1 ) to the ellipse 2 + 2 =
1 is
a b
a 2 x b2 y
− =a 2 − b 2 .
x1 y1
x2 y 2
2. Parametric form : The equation of the normal to the ellipse + 1 at ( a cos θ , b sin θ ) is
=
a 2 b2
ax sec θ − by cosec θ =
a 2 − b2 .
x2 y 2
3. Slope form : If m is the slope of the normal to the ellipse 2 + 2 =1 , then the equation of normal is
a b
=y mx ±
(
m a 2 − b2 ) .
a 2 + b2 m2
Y
AUXILIARY CIRCLE
The circle described on the major axis of an ellipse as diameter is called an Q P(x, y)
x2 y 2
auxiliary circle of the ellipse. If 2 + 2 =1 is an ellipse then its auxiliary θ
a b X′ C M X
circle is x 2 + y 2 =.
a2
x2 y2
2
+ 2 =
1
a b
ECCENTRIC ANGLE OF A POINT Y′
x2 y 2
Let P be any point on the ellipse 2 + 2 = 1 . Draw PM perpendicular from P on the major axis of the ellipse
a b
and produce MP to meet the auxiliary circle in Q. Join CQ. The angle ∠XCQ = θ is called the eccentric angle
of the point P on the ellipse.
Note that the angle ∠XCP is not the eccentric angle of point P.
EQUATION OF PAIR OF TANGENTS
The combined equation of pair of tangents PA and PB is given by SS 1 =T2
x2 y 2 x2 y 2 xx yy
where S ≡ 2 + 2 − 1, S1 ≡ 12 + 12 − 1, T ≡ 21 + 21 − 1
a b a b a b
JEEMAIN.GURU
DIRECTOR CIRCLE
The director circle is the locus of point from which perpendicular tangents are drawn to the ellipse. Hence locus
of P ( x1, y1 ) i.e., equation of director circle is x 2 + y 2 = a 2 + b 2 .
CHORD OF CONTACT Y
If PQ and PR be the tangents through point P ( x1 , y1 ) to the ellipse (x1, y1) Q
P
x2 y 2 X'
+ =1 then QR is called chord of contact and the equation of the C X
a 2 b2 R
xx yy
chord of contact QR is 21 + 21 = 1 for ( x1 , y1 ) .
a b Y'
EQUATION OF CHORD WITH MID POINT (x 1, y 1 )
x2 y 2
The equation of the chord of the ellipse 2 + 2 = 1 whose mid point be ( x1 , y1 ) is given by T = S1
a b
Q
xx yy x2 y 2
where T = 21 + 21 − 1, S1 = 12 + 12 − 1. P ( x1 , y1 )
a b a b
R
Ellipse 189
x2 y 2
Equation of polar : Equation of polar of the point ( x1 , y1 ) with respect to ellipse + =
1 is given by
a 2 b2
xx1 yy1
+ = 1 , i.e., T = 0 .
a 2 b2
PROPERTIES OF POLE AND POLAR
1. If the polar of P ( x1 , y1 ) passes through Q ( x2 , y2 ) , then the polar of Q ( x2 , y2 ) passes through P ( x1 , y1 )
and such points are said to be conjugate points.
2. If the pole of a line l1 x + m1 y + n1 = 0 lies on the another line l2 x + m2 y + n2 = 0 , then the pole of the
second line will be on the first and such lines are said to be conjugate lines.
3. Pole of a given line is same as point of intersection of tangents at its extremities.
Y
DIAMETER OF THE ELLIPSE
Definition : A line through the center of an ellipse is called a diameter of =
y mx + c
the ellipse.
The equation of the diameter bisecting the chords = y mx + c of slope m of X' X
−b 2
x2 y 2 b2 x2 y 2
+ = y= 2 x
the ellipse + =
1 is y =
− x, which is passing through (0, 0) a 2 b2
1 am
a 2 b2 a2m
Y'
Y
Conjugate diameter : Two diameter of an ellipse are said to be
conjugate diameter if each bisects all chords parallel to the other. The
coordinates of the four extremities of two conjugate diameters are B A
Q
P ( a cos θ , b sin θ ) ; P ' ( −a cos θ , −b sin θ ) , P
90°
θ
Q ( −a sin θ , b cos θ ) ; Q ' ( a sin θ , −b cos θ ) X′ X
C
=If y m=
1 x and y m2 x be two conjugate diameters of an ellipse, then
P´ Q´
−b 2 B´
m1m2 = 2 . A´
a
Y′
1. Properties of diameters :
(i) The tangents at the extremity of any diameter is parallel to the chords it bisects or parallel to the
conjugate diameter ,
(ii) The tangents at the ends of any chord meet on the diameter which bisects the chord.
D P ( a cos φ , b sin φ )
(ii) The sum of the squares of any two conjugate semidiameters of an
ellipse is constant and equal to the sum of the squares of the semi
axes of the ellipse i.e., CP2+CD2 = a2 + b2. S′ C S
P′ D′
JEEMAIN.GURU
(iii) The product of the focal distances of a point on an ellipse is equal to the square of the semi diameter
which is conjugate to the diameter through the point
i.e. SP.S ′P = CD 2 .
Q
(iv) The tangents at the extremities of a pair of conjugate diameters M
D P
form a parallelogram whose area is constant and equal to
R' R
product of the axes. i.e. C
Area of parallelogram = (2a) (2b)
P' D'
= Area of rectangle contained under major and minor axes. Q'
3. Equi-conjugate diameters: Two conjugate diameters are called equi-conjugate, if their lengths are equal
i.e., (CP)2 = (CD)2.
∴ = CD
CP =
(a 2
+ b2 )
for equi-conjugate diameters.
2
SUBTANGENT AND SUBNORMAL
Let the tangent and normal at P ( x1 , y1 ) meet the x-axis at A and B respectively. Length of subtangent at
x2 y 2
P ( x1 , y1 ) to the ellipse+ = 1 is Y
a 2 b2
A′
a2 P ( x1 , y1 )
DA =CA − CD = − x1.
x1 X′ X
C B D A
2 2
x y
Length of sub-normal at P ( x1 , y1 ) to the ellipse2
+ 2 =
1 is
a b
b2 b2 Y′
( )
BD =CD − CB =x1 − x1 − 2 x1 = 2 x1 =1 − e 2 x1.
a a
m Q
PROPERTIES OF THE ELLIPSE t
B P
PN 2 b2 M
1. = 2 F
AN ⋅ A′N a A′ S′ C G N
(i) AS ⋅ A′S = D S A T
b2
g
(ii) CN ⋅ CT = a2 B′
(iii) Cn ⋅ Ct =b2
2. The locus of the feet of the perpendiculars M and m from the foci on any tangent to ellipse is its
auxiliary circle
The product of the two perpendicular distances from the foci on any tangent of a ellipse is equals to b 2
i.e. SM ⋅ S ′m′ = b2 .
3. CG = e 2 . CN = e 2 x1 .
4. The tangent and normal at a point on an ellipse bisect the angles between the focal radii to the point
= SG e= . SP, S ′G e. S ′P.
5. If the tangent on an ellipse meets the directrix in F , then PF subtends a right angle at the corresponding
focus i.e. ∠ PSF = 90°
6. Tangents at the ends of a focal chord intersect on the directrix.
JEEMAIN.GURU
Ellipse 191
6. If the normal at any point P meets the major and minor axes in G and g and CD is the perpendicular
upon the normal, then PD × PG = b 2 and PD × Pg = a2.
7. Tangents at the ends of any chord meet on the diameter which bisects the chord.
8. The sum of distances of any point P on the ellipse from the focus S and S ′ is 2a i.e. PS + PS ′ =
2a
9. The ratio of y-coordinates of corresponding points on ellipse and Auxiliary circle = b : a
b2
10. The Harmonic mean of focal radii of any focal chord is equal to semi-latus rectum = .
a
11. If α , β , γ , δ be the eccentric angles of the four concyclic points on an ellipse then
α + β +γ =+ δ 2nπ , n ∈ I .
12. If eccentric angles of feet P, Q, R, S of these normals be α , β , γ , δ then
α + β + γ + δ= ( 2n + 1) π , n ∈ I
13. The necessary and sufficient condition for the normals at three α , β , γ points on the ellipse to be
concurrent if sin ( β + γ ) + sin ( γ + α ) + sin (α + β ) =
0.
JEEMAIN.GURU
Hyperbola Chapter 30
DEFINITION
A hyperbola is the locus of a point which moves in the plane in such a way that the ratio of its distance from a
fixed point in the same plane to its distance from a fixed line is always constant which is greater than unity.
X′ X
( −ae, 0 ) S′ ( −a, 0 ) A′ Z′ C Z A ( a, 0 ) S
( ae, 0 )
L1′ L′
Q′
x = − a / e Y′ x = a / e
CONJUGATE HYPERBOLA
This hyperbola whose transverse and conjugate axis are respectively the conjugate and transverse axis of a
given hyperbola is called conjugate hyperbola of the given hyperbola.
Y
( 0, be′)
S′
Z B ( 0, b ) y = b / e′
X′ C X
Z′
B ( 0, − b ) y = −b / e′
S′
( 0, − be′)
Y′
x2 y 2 y 2 x2
Equation of conjugate hyperbola of a given hyperbola − =
1 is − =
1
a 2 b2 b2 a 2
JEEMAIN.GURU
Hyperbola 193
Difference between both hyperbola will be clear from the following table :
x2 y 2
Hyperbola − + = 1 or
x2 y 2 a 2 b2
− =
1
a 2 b2 x2 y 2
Imp. terms − = −1
a 2 b2
Centre ( 0, 0 ) ( 0, 0 )
Length of transverse axis 2a 2b
Length of conjugate axis 2b 2a
Foci ( ± ae, 0 ) ( 0, ± be′)
Equation of directrices x = ± a/e y = ± b / e′
a 2 + b2 a 2 + b2
Eccentricity e= 2 e′ = 2
a b
Length of latus rectum 2b 2 / a 2a 2 / b
( a sec φ , b tan φ ) ( a tan φ , b sec φ )
Parametric co-ordinates
0 ≤ φ < 2π 0 ≤ φ < 2π
(a) If P lies on right branch (a) If P lies on upper branch
= ex1 − a , S ′=
SP P ex1 + a SP = e′y1 − b, S ′P = e′y1 + b
Focal radii
(b) If P lies on left branch (b) If P lies on lower branch
SP =−ex1 + a, S ′P =−ex1 − a SP = −e′y1 + b, S ′P = −e′y1 − b
Difference of focal radii
S ′P − SP 2a 2b
Note :
1 1
(a) If e, e′ are eccentricity of hyperbola & conjugate hyperbola then + 2 =
1.
e e′
2
Hyperbola 195
x2 y 2 x12 y12 xx yy
S = 2 − 2 − 1, S1 = 2 − 2 − 1, T = 21 − 21 − 1 (x 1, y1) R
a b a b a b
The equation of the chord joining the points P ( a sec φ1 , b tan φ1 ) and Q ( a sec φ2 , b tan φ2 ) is
x φ −φ y φ +φ φ +φ
cos 1 2 − sin 1 2 = cos 1 2
a 2 b 2 2
Point of intersection of tangents
Point of intersection of tangents at P ( a sec φ1 , b tan φ1 ) and Q ( a sec φ2 , b tan φ2 ) is
φ1 − φ2 φ +φ
cos 2 sin 1 2
a , b 2
φ +
cos 1 2 φ φ +φ
cos 1 2
2 2
Point of intersection of normals
Coordinates of point of intersection of normals at P ( a sec φ1 , b tan φ1 ) and Q ( a sec φ2 , b tan φ2 ) is
φ −φ φ +φ
2 cos 1 2 sin 1 2
a + b sec φ1 sec φ2 2 , − a + b tan φ tan φ 2 .
2 2 2
a φ +φ φ +φ
1 2
b
cos 1 2 cos 1 2
2 2
Q´ B´
A
Polar
Q A′ A´ Polar
Pole
X´ P (x1, y1)
X X´ X
Q′ Pole P
B B′ ( x1 , y1 ) A
B
Q
PROPERTIES OF POLE AND POLAR
1. If the polar of P ( x1 , y1 ) passes through Q ( x2 , y2 ) , then the polar of Q ( x2 , y2 ) goes through P ( x1 , y1 )
and such points are said to be conjugate points.
JEEMAIN.GURU
Hyperbola 197
2. If the pole of a line lx + my + n = 0 lies on the another line l1 x + m1 y + n1 = 0 then pole of the second line
will lie on the first and such lines are said to be conjugate lines.
3. Pole of a given line is same as point of intersection of tangents as its extremities.
DIAMETER OF THE HYPERBOLA
The locus of the middle point of a system of parallel chords of a Y
hyperbola is called a diameter and the point where the diameter
(x1,y1) P
intersects the hyperbola is called the vertex of the diameter.
R(h, k)
x2 y 2
Let = y mx + c system of parallel chords to 2 − 2 = 1 for X' X
a b C
different chords then the equation of diameter of the hyperbola (x2,y2) Q
b2 x
is y = 2 , which is passing through (0, 0). Y'
am
Conjugate diameter : Two diameter are said to be conjugate when each bisects all chords parallel to the other.
b2
=
If y m= 1 x , y m 2 x be conjugates, diameters, then m 1 m 2 = .
a2
Y
( a tan θ , b sec θ )
Note : If the two extremities of a diameter lie in the first and
P ( a sec θ , b tan θ )
third quadrants, the extremities of the conjugate d
diameter also lie in the first and third quadrants.
The coordinates of the four extremities of two conjugate
diameters are shown in the adjoining figure. C
Caution : The extremities d and d ′ of the conjugate P′
diameter do not lie on the hyperbola. d′
( − a sec θ , ( − a tan θ ,
− b tan θ ) − b sec θ )
X
a2
1 1
C A N B
b2
= 2=
a
( )
x1 e 2 − 1 x1 .
ASYMPTOTES OF A HYPERBOLA Y´
If the length of the perpendicular let fall from a point on a hyperbola to a straight line tends to zero as the point
on the hyperbola moves to infinity along the hyperbola, then the straight line is called an asymptote of the
hyperbola.
x2 y 2 b x y
The equations of two asymptotes of the hyperbola 2 − 2 = 1 and y = ± x or ± = 0.
a b a a b
SOME IMPORTANT POINTS ABOUT ASYMPTOTES
x2 y 2 x2 y 2
1. The combined equation of the asymptotes of the hyperbola 2 − 2 = 1 is 2 − 2 = 0.
a b a b
JEEMAIN.GURU
Asymptotes
B
A
X´ C X
A´
x2 y 2
− =
1
a 2 b2
x2 y 2
Y´ − =
−1
a 2 b2
4. The equation of the pair of asymptotes differ the hyperbola and the conjugate hyperbola by the same
constant only i.e., Hyperbola – Asymptotes = Asymptotes – Conjugate hyperbola or
x2 y 2 x2 y 2 x2 y 2 x2 y 2
2 − 2 − 1 − 2 − 2 = 2 − 2 − 2 − 2 + 1
a b a b a b a b
5. The asymptotes pass through the centre of the hyperbola.
6. The bisector of the angles between the asymptotes are the coordinate axes.
x2 y 2 b
7. The angle between the asymptotes of the hyperbola S = 0 i.e., 2 − 2 = 1 is 2 tan −1 or 2sec −1 e .
a b a
8. Asymptotes are equally inclined to the axes of the hyperbola.
(c ) ( )
2, c 2 and −c 2, − c 2 directrices are x + y =±c 2. F2
4. Equation of the chord joining points t 1 and t 2 : The equation of the chord joining two
c c
points ct1 , and ct2 , on the hyperbola xy = c 2 ⇒ x + y t1t2 = c ( t1 + t2 ) .
t1 t2
JEEMAIN.GURU
Hyperbola 199
{ } { }
c t1t2 ( t12 + t1t2 + t22 ) + 1 c t13t23 + ( t12 + t1t2 + t22 )
, .
t1t2 ( t1 + t2 ) t1t2 ( t1 + t2 )
c c 1
(iii) If the normal at P ct , cuts the rectangular hyperbola xy = c 2 at Q ct ′, then t ′ = − 3 .
t t′ t
7. Equation of diameter of rectangular hyperbola xy = c 2 is y + mx = 0 ( m is the slope of the chord joining
two points lies on the rectangular hyperbola)
Two diameters y + m1 x = 0 and y + m2 x = 0 are conjugate diameter if m1 + m2 = 0.
2 2 2 2
PROPERTIES OF HYPERBOLA x / a – y / b = 1
1. If PN be the ordinate of a point P on the hyperbola and the tangent at P meets the transverse axis in T,
then ON .OT = a 2 , O being the origin.
2. If PM be drawn perpendiculars to the conjugate axis from a point p on the hyperbola and the tangent at P
meets the conjugate axis in T, then OM .OT = − b 2 ; O , being the origin.
3. If the normal at P on the hyperbola meets the transverse axis in G, then SG = eSP ; S being a foci and e
the eccentricity of the hyperbola.
4. The tangent and normal at any point of a hyperbola bisect the angle between the focal radii to that point.
5. The locus of the feet of the perpendiculars from the foci on a tangent to a hyperbola is the auxiliary circle.
6. The product of the length of the perpendicular drawn from foci on any tangent to hyperbola is b2 .
JEEMAIN.GURU
x2 y 2
7. From any point on the hyperbola − =1 perpendicular are drawn to the asymptotes then product is
a 2 b2
a 2b 2 a2
and for rectangular hyperbola =
a 2 + b2 2
8. If a circle cuts the rectangular hyperbola xy = 1 in ( xr , yr ) (four points) r = 1, 2, 3, 4 then
1
= y= 1 y2 y3 y4 1.
x1 x2 x3 x4
9. A rectangular hyperbola with centre at C is cut by any circle of radius R in four points L, M, N, P then the
value of CL2 + CM 2 + CN 2 + CP 2 = 4R2 .
10. If a triangle is inscribed in a rectangular hyperbola then the orthocenter of triangle lies on the rectangular
hyperbola.
11. The portion of tangent intercepted between the asymptotes at any point of the hyperbola is bisected by the
point of contact.
12. Whenever any circle and any hyperbola cut each other at four points the mean position of these four
points is the mid point of the line segment joining centre of hyperbola and centre of circle.
b2
13. The harmonic mean of focal radi for any focal chord =
a
14. Tangent drawn at the ends of any focal chord meet on the directrix.
x2 y 2
15. Local of point of intersection two perpendicular tangents to the hyperbola 2 − 2 = 1 is a circle called
a b
director circle whose equation is x 2 + y 2 = a 2 − b 2
and if a < b then there is no real point from where we can draw two perpendicular tangents to the
hyperbola.
16. The portion of tangent between point of contact and the point where it cuts the directrix subtend 90°
angle at the focus.
JEEMAIN.GURU
201
Vectors Chapter 31
Physical Quantity : The quantity by means of which we describe the laws of physics are called physical
quantities.
Scalars : The physical quantities which have magnitude only are called scalars. These quantities are specified
by number and a unit.
Examples : Length, mass, volume, area, temperature, density, work etc.
Vectors : The physical quantities which have magnitude and direction both and also obey the triangle law of
addition are called vectors.
Examples : Displacement, velocity, acceleration, force, momentum and couple.
ADDITION OF VECTOR C
(i) Triangle law of addition: If two vectors are represented by two
consecutive sides of a triangle then their sum is represented by the
third side of the triangle from tail of the first vector to the head of the c= a + b
b
second vector. This is known as the triangle law of vector addition. Thus,
=
If AB a= , BC b, and = AC c
A B
then AB + BC = AC i.e a + b = c a
(ii) Parallelogram Law of Addition : If two vector are represented by two
adjacent of a parallelogram, then their sum is represented by the diagonal of the parallelogram.
Thus, if= O A a= , O B b, and = O C c then O A + O B = OC
i.e. a + b = c where OC is a diagonal of the parallelogram OABC.
JEEMAIN.GURU
Vectors 203
B C
(iii) Addition in component from:
If a = a1 i + a 2 j + a 3 k and b = b1 i + b 2 j + b3 k their sum is defined b
as a + b = ( a + b ) i + ( a + b ) j + ( a + b ) k
1 1 2 2 3 3
A ( r1 )
y´
MORE ABOUT THE POSITION VECTORS
OA + AB = OB
r1
∴ = OB − OA
AB
r2 B ( r2 )
= p.v. of B − p.v. of A= r2 − r1
⇒ AB= r2 − r1 O
JEEMAIN.GURU
Vectors 205
n m
= λa + µ b, where λ = and µ = .
m+n m+n
r λa + µ b, where λ + µ =
Thus, p.v. of any point P on AB can always be taken as = 1.
A
2. EXTERNAL DIVISION
The p.v. of the point Q, which divides externally the join of two given points
B
A and B whose position vectors are a and b in the given ratio m : n, is a
b
mb − ma
OQ = Q
m−n O
SYSTEM OF VECTORS
An ordered set of three non-coplaner vectors is called system of vectors.
( a, b, c ) are said to be forming right handed system if [a b c] > 0 and said to be forming left handed system if
[a b c] < 0 .
LINEAR COMBINATION
A vector r is said to be a linear combination of the given vectors a, b, c ...etc., if there exist a system of scalars
x, y, z , ...etc. such that
r = xa + yb + zc + ...
LINEARLY DEPENDENT AND INDEPENDENT SYSTEM OF VECTORS
The system of n vectors a1 , a 2 , ..., a n is said to be linearly dependent, if there exist scalars x1 , x2 ,... xn not all
zero such that
x1a1 + x2a 2 + ... + xna n =0 … (1)
The same system of vectors is said to be linearly independent, if x1a1 + x2a 2 + ... + xna n = 0 implies that
x1= x2= ...= xn= 0 is the only solution.
COLLINEARITY OF THREE POINTS
The necessary and sufficient condition for three points with position vectors a, b and c to be collinear is that
there exist three scalars x, y, z not all zero such that
xa + yb + zc = 0 where x + y + z =0.
TEST OF COLLINEARITY OF TWO VECTORS
To prove that two vectors a and b are collinear, find a scalar m such that one of the vectors is m times the
other. In case no such scalar m exists, then the two vectors will be non-collinear vectors.
TEST OF COLLINEARITY OF THREE POINTS
Method I
To prove that three points A, B, C are collinear, find the vectors AB and AC and show that there exists
a scalar m such that AB = mAC.
If no such scalar m exists, then the points are not collinear.
Method II
To prove that three points A, B, C with position vectors a, b, c respectively are collinear, find three
scalars x, y, z (not all zero) such that
xa + yb + zc = 0, where x + y + z =0.
If no such scalars x, y, z exist then the points are not collinear.
JEEMAIN.GURU
Vectors 207
5. Scalar product of two perpendicular vectors is zero i.e. if a and b are two perpendicular vectors, then
a ⋅b = 0.
However, if a ⋅ b = 0 ⇒ Either a = 0 or b = 0 or a ⊥ b.
6. Scalar product of mutually orthogonal unit vectors i, j, k :
i ⋅ i = j ⋅ j = k ⋅ k = 1
and i ⋅ j = j ⋅ k = k ⋅ i = 0.
7. Scalar product of two vectors in terms of components : If
a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k ,
then a ⋅ b= a1b1 + a2b2 + a3b3
Thus, the scalar product of two vectors is equal to the sum of the product of their corresponding
components.
8. Angle between two vectors in terms of the components of the given vectors.
If θ is the angle between two vectors a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k , then
a ⋅b
cos θ =
a b
a1b1 + a2b2 + a3b3
=
a12 + a22 + a32 b12 + b22 + b32
9. Components of a vector b along and perpendicular to vector a
a ⋅b
Component of b along a = 2 a
a
a ⋅b
Component of b perpendicular to a = b − 2 a
a
10. ( ) ( ) ( )
Any vector r can be expressed as r = r ⋅ i i + r ⋅ j j + r ⋅ k k.
(a − b )
2
2. = a 2 + b 2 − 2a ⋅ b
JEEMAIN.GURU
3. ( a + b ) ⋅ ( a − b ) = a2 − b2 .
( a ⋅ b ) ≤ a2b 2 this relation is known as Cauchy – Schwarz is inequality
2
4.
WORK DONE BY A FORCE
Work done by a force F in displacing a particle from A to B is defined by
W= F ⋅ AB.
Note : If a number of forces are acting on a particle, then the sum of the works done by the separate
forces is equal to the work done by the resultant force
CROSS PRODUCT OR VECTOR PRODUCT OF TWO VECTORS
The vector product or cross product of two vectors a and b is defined as a
a×b
vector, written as a × b, whose
(i) modulus is a b sin θ , θ being the angle between the directions of a B
and b and 0 ≤ θ ≤ π .
n b
(ii) direction is that of the unit vector n̂ which is perpendicular to both a
and b such that a, b and n̂ form a right handed system. θ
Thus, a × b =a b sin θ nˆ . O a
A
Vectors 209
N
(iii) The sense of A is such that the direction of description of the boundary of the curve and the sense of A
corresponded to the rotation of a right hand screw. The sense of A will be reversed, if we reverse the
direction of description of the boundary of the area.
MOMENT OF A FORCE ABOUT A POINT
The vector moment or torque m of a force F acting at a point A about the point O is given F
by O
M = r × F = OA × F
where r = OA is the p.v. of the point A w.r.t. the point O. r
Note : The algebraic sum of the moments of a system of forces about any point is A
equal to the moment of their resultant about the same point.
TRIPLE PRODUCTS
SCALAR TRIPLE PRODUCT
If a, b, c be three vectors, the scalar product a ⋅ ( b × c ) of the two vectors a and b × c is a scalar quantity,
called the scalar triple product of these vectors a, b and c and is denoted by [a b c ] .
GEOMETRICAL INTERPRETATION OF SCALAR TRIPLE PRODUCT
JEEMAIN.GURU
Vectors 211
O
1 1 [ a, b , c ]
= AB × AC
3 2
AB × AC c
a
1
= [a, b, c]
6
1 A C
= Volume of parallelopiped whose coterminous edges are a, b, c b
6
1
= [OA, OB, OC]
6
VECTOR TRIPLE PRODUCT B
Let a, b, c be any three vectors, then the vectors a × ( b × c ) and ( a × b ) × c are called vector triple product of
a, b, c .
a × (b × c) = (a ⋅ c) b − (a ⋅ b ) c i.e., (First . Third) Second – (First . Second) Third.
Note
1. Vector triple product is not associative
i.e. (a × b ) × c ≠ a × (b × c).
2. (b × c) × a =− a × ( b × c )
=− ( a ⋅ c ) ⋅ b − ( a ⋅ b ) c
= ( a ⋅ b ) c − ( a ⋅ c ) b.
3. ( ) ( ) (
2r =i × r × i + j × r × j + k × r × k ) ( ) ( ) ( )
and 2r = i × r × i + j + r × j + k × r × k
SCALAR PRODUCT OF FOUR VECTORS
If a, b, c, d be four vectors, then ( a × b ) ⋅ ( c × d ) is the scalar product of four vectors.
a ⋅c a ⋅d
(a × b ) ⋅ (c × d ) = = ( a ⋅ c )( b ⋅ d ) − ( a ⋅ d )( b ⋅ c )
b ⋅c b ⋅d
VECTOR PRODUCT OF FOUR VECTORS
If a, b, c, d be four vectors, then ( a × b ) × ( c × d ) is the vector product of four vectors.
EXPANSION OF VECTOR PRODUCT OF FOUR VECTORS
( c × d ) [a b d ] c − [a b c ] d
( a × b ) ×=
= [a c d ] b − [b c d ] a.
Note :
1. [b c d ] a − [a c d ] b + [a b d ] c − [a b c ] d =
O.
2. Any vector r can be expressed in terms of three non-coplanar vectors a, b, c in the form
r=
[r b c ] a + [r c a ] b + [r a b ] c .
[a b c ]
RECIPROCAL SYSTEM OF VECTORS
If a, b, c be three non-coplanar vectors so that [a b c ] ≠ 0, then the three vectors a′, b′, c′ defined by the
equations
b×c c×a a×b
= a′ = , b′ = , c′
[a b c ] [a b c ] [a b c ]
JEEMAIN.GURU
Vectors 213
8. Orthocenter formula
The position vector of the orthocenter of ∆ ABC is
a tan A + b tan B + c tan C
.
tan A + tan B + tan C A
C A
B
10. If ABCD is a parallelogram whose diagonals intersect at P if O be the origin then
P
OA + OB + OC + OD = 4OP
D C
O
11. If a, b, c are position vector of three points respectively then a × b + b × c + c × a is perpendicular to the
plane of ABC.
12. (a) [a + b, b + c, c + a ] =
2 [ a, b , c ]
(b) [a − b, b − c, c − a] =0
[a × b, b × c, c × a] =
[a, b, c]
2
(c)
a ⋅p a ⋅q a ⋅r
(d) [a b c ] [p q r ] = b ⋅ p b ⋅ q b ⋅ r
c⋅p c ⋅q c ⋅r
(e) r [a b c ] = [r b c ] a + [r c a ] b + [r a b ] c
(f) r [a b=
c] ( r. a )( b × c ) + ( r. b )( c × a ) + ( r. c )( a × b )
l m n
(g) ( a × b ) [ l m n ] =
a .l a .m a .n
b .l b .m b .n P (α )
13. Perpendicular distance of a point from a line
Let L is the foot of perpendicular drawn from P ( α ) on the line
r = a λ+b . Since r denotes the position vector of any point on the line
r = a λ+b.
So, let the position vector of L beλ a + b.
A B
( aα− b) . r= a + λb L
Position vector of L = a − b
b
2
( aα− b) .
and PL =α( a − ) − b2
b
The length PL, is the magnitude of PL , and is the required length of perpendicular.
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= =
( b1 × b 2 ) ⋅ ( a2 − a1 ) [b1 , b 2 , a 2 − a1 ]
b1 × b 2 b1 × b 2
Shortest distance between two parallel lines
r a2 + µb
The shortest distance between the parallel lines r= a1 + λb an d =
( a 2 − a1 ) × b
is given by d= .
b
16. Vector equation of a plane through the point A ( a ) and perpendicular to the vector n is (r − a) ⋅ n = 0 or
r ⋅ n = a ⋅ n or r . n = d, where d = a . n . This is known as the scalar product form of a plane.
17. Vector equation of a plane normal to unit vector n̂ and at a distance d from the origin is r.nˆ = d . If n is
not a unit vector, then to reduce the equation r . n = d to normal form we divide both sides by n to
n d d
obtain =
r⋅ =
or r ⋅ nˆ
n n n
18. The equation of the plane passing through a point having position vector a and parallel to b and c is
r =a + λb + µ c or [r b c] =[a b c] where λ and µ are scalars.
19. Vector equation of a plane passing through a point a, b, c is
r = (1 − s − t ) a + sb + tc or r ⋅ ( b × c + c × a + a × b ) = [a b c ] .
20. The equation of any plane through the intersection of planes
r ⋅ n1 = d1 and r ⋅ n 2 = d 2 is r ⋅ ( n1 + λn 2 ) = d1 + λ d 2 where λ is an arbitrary constant.
21. The perpendicular distance of a point having position vector a from the plane r ⋅ n =d is given by
a ⋅n − d
p= .
n
22. Perpendicular distance of a point P( r ) from a plane passing through the points a, b and c is given by
PM =
(r − a ) .(b × c + c × a + a × b ) .
b×c + c×a + a×b
23. Angle between line and plane: If θ is the angle between a line r= a + λb and the plane r ⋅ n = d , then
b.n
sin θ = .
b n
24. The equation of sphere with centre at C (c) and radius ‘a’ is r − c =
a.
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215
Spherical polar co-ordinates : The measures of quantities r , θ , φ are known as spherical or three
dimensional polar co-ordinates of the point P. If the rectangular cartesian co-ordinates of P are ( x, y, z )
= cos θ , u r sin θ .
then z r=
∴= x cos φ
u= r sin θ cos φ=
, y sin φ
u= r sin θ sin φ and
= z r cos θ .
u x2 + y 2 y
Also, r 2 = x 2 + y 2 + z 2 and tan θ=
= ; tan φ= .
z z x
2. Distance Formula : Distance between two points P ( x1 , y1 , z1 ) and Q ( x2 , y2 , z2 ) is given by
d= ( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 )
2 2 2
3. Section Formula
(i) Co-ordinates of the point which divides the join of P ( x1 , y1 , z1 ) and Q ( x2 , y2 , z2 ) internally in
mx + nx1 my2 + ny1 mz2 + nz1
the ratio m : n are 2 , , .
m+n m+n m+n
(ii) Co-ordinates of the point which divides the join of P ( x1 , y1 , z1 ) and Q ( x2 , y2 , z2 ) externally in
mx − nx1 my2 − ny1 mz2 − nz1
the ratio m : n are 2 , , .
m−n m−n m−n
4. Triangle and tetrahedron
(i) Co- ordinates of the centroid
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a b c
l= ± ; m= ± ; n=± .
a 2 + b2 + c2 a 2 + b2 + c2 a 2 + b2 + c2
where + ve or – ve sign is to be taken in all the three.
(b) A line can have many sets of direction ratio which are proportional to each other.
(c) If P ( x1 , y1 , z1 ) and Q ( x2 , y2 , z2 ) are any two points, then direction-ratios of PQ (or line) PQ are
( x2 − x1 , y2 − y1 , z2 − z1 ) .
(d) Angle between two vectors (or line) whose direction-ratios are ( a1 , b1 , c1 ) and ( a2 , b2 , c2 ) is given
by
a1a2 + b1b2 + c1c2
∴ cos θ = .
a12 + b12 + c12 . a2 2 + b2 2 + c2 2
a1 b1 c1
(i) If these are ' ⊥ ' ; a1a2 + b1b2 + c1c2 = 0. (ii) If these are ' ' ; = = .
a2 b2 c2
7. Projection
Projection of a line joining the points P ( x1 , y1 , z1 ) and Q ( x2 , y2 , z2 ) on another line AB whose direction
cosines are l, m and n. If the line segment PQ makes angle θ with the line AB then Q
θ
P
A B
P´ Q´
Or
(r − a)⋅ n =0; n is a vector normal to plane.
Vector form
where a is the position vector of a point in the plane.
(v) Equation of a plane through three given points
The equation of the plane passing through three non-collinear points ( x1 , y1 , z1 ) , ( z2 , y2 , z2 )
and ( x3 , y3 , z3 ) is
( x − x1 ) ( y − y1 ) ( z − z1 )
( x2 − x1 ) ( y2 − y1 ) ( z2 − z1 ) =
0.
( x3 − x1 ) ( y3 − y1 ) ( z3 − z1 )
2. Equation of Systems of Planes
(i) The equation ax + by + cz + k = 0 represents a system of plane parallel to the plane
ax + by + cz + d =
0 , k being parameter.
Or
Vector form r .n = k
(ii) The equation ax + by + cz + k = 0 , represents a system of planes perpendicular to the
x y z
line = = .
a b c
(iii) The equation ( a1 x + b1 y + c1 z + d1 ) + k ( a2 x + b2 y + c2 z + d 2 ) =
0 .represents a system of planes
passing through the intersection of the planes a1 x + b1 y + c1 z= + d1 0, and a2 x + b2 y + c2 z=
+ d 2 0.
k being a parameter.
(iv) The equation A ( x − x1 ) + B ( y − y1 ) + C ( z − z1 ) =
0 represents a system of planes passing through
the point ( x1 , y1 , z1 ) where A, B, C are parameters.
3. Angle Between Two Planes
The angle θ between the planes a1 x + b1 y + c1 z + d= 1 0, a2 x + b2 y + c2 z + d=
2 0, is given by
a1a2 + b1b2 + c1c2
cos θ =
( a12 + b12 + c12 ) ( a22 + b22 + c22 )
4. Parallelism and Perpendicularity of Two Planes
a1 b1 c1
The planes a1 x + b1 y + c1 z + d= 1 0, a2 x + b2 y + c2 z + d=
2 0 are parallel if and only if = = ; and
a2 b2 c2
perpendicular if and only if a1a2 + b1b2 + c1c2 = 0
5. Two Sides of a Plane
Two points P ( x1 , y1 , z1 ) and Q ( x2 , y2 , z2 ) lie on the same or different sides of the plane ax + by + cz + d = 0
according as ax1 + by1 + cz1 + d and ax2 + by2 + cz2 + d are of the same or of different signs.
6. Length of the Perpendicular from a Point to a Plane
The perpendicular distance of the point ( x1 , y1 , z1 ) from the plane lx + my + nz =
p
is p − lx1 − my1 − nz1 , where l, m, n are direction cosines of the normal to the plane and p is the length of
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the perpendicular on the plane from the origin. The perpendicular distance of the point ( x1 , y1 , z1 ) from
ax1 + by1 + cz1 + d
the plane ax + by + cz + d =
0 is
(a 2
+ b2 + c2 )
STRAIGHT LINE
1. (i) Vector equation of a line passing through a point ‘A’ whose position vector is a w.r.t. some
origin O and is parallel to a given vector m is given by :
Z
r= a + λ m ; λ is a constant.
(ii) Cartesian equation
m
If a straight line passes through a given point ( x1 , y1 , z1 ) and has
A
direction cosines l, m, n, then the coordinates of any point on it P
r
satisfy the equations a
O X
x − x1 y − y1 z − z1
= = =( r )
l m n
These equations enable us to write down the coordinates of any Y
point (x, y, z) on the line in terms of its distance r is
x =+x1 lr , y =+ y1 mr , z =+ z1 nr. This form is called the distance form of the equation of a
line.
A line passing through a point A ( x1 , y1 , z1 ) and having direction-ratios a, b, c is given by
x − x1 y − y1 z − z1
= = = λ , say.
a b c
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a1 b1 c1
If these are (i) parallel = = (ii) perpendicular a1a2 + b1b2 + c1c2 = 0.
a2 b2 c2
5. To find equation of a line parallel to given line
Vector form : Given equation is r= a1 + λb1 .
Equation of a line parallel to it through point B ( a2 ) is = r a2 + µ b1 .
x − x1 y − y1 z − z1
Cartesian form : Let the given line be = = then a line parallel to it and passing
a b c
x − x2 y − y2 z − z2
through point B ( x2 , y2 , z2 ) has equation = = .
a b c
6. To find equation of line perpendicular to two given lines
Vector form
Let given lines be: r1 = a1 + λb1 and r2 = a2 + µ b2 .
A line perpendicular to them is r= a3 + α b3 where b1.b3 = 0 and b2 .b3= 0 or b3= b1 × b2 .
Cartesian form
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x − x1 y − y1 z − z1 x − x2 y − y2 z − z2
Let given lines be : = = and = =
a1 b1 c1 a2 b2 c2
x − x3 y − y3 z − z3
Then a line through ( x3 , y3 , z3 ) and ⊥ to given lines is: = =
a b c
where aa1 + bb1 + cc1 = 0 and aa2 + bb2 + cc2 = 0.
Find a, b, c by rule of cross multiplication.
7. To find whether two given lines cut or not
Vector form. Let given lines be r1 = a1 + λb1 and r2 = a2 + µ b2 .
Since r is the position vector of an arbitrary point on the lines, if these intersect, for some values
of λ and µ these points must coincide i.e., a1 + λb1 =a2 + µ b2 . Equate components of iˆ, ˆj , kˆ,
Solve any two equations for ‘ λ ’ and ‘ µ ’. Put this value in third equation, if it is satisfied lines
intersect, otherwise not. To find point of intersection put λ (or µ ) in given equation.
x − x1 y − y1 z − z1 x − x2 y − y2 z − z2
Let given lines be L1 : = = = λ , say and L2 : = = = µ , say.
a1 b1 c1 a2 b2 c2
Any point on line L 1 is ( x1 + λ a1 , y1 + λb1 , z1 + µ c1 ) and on line L2 is ( x2 + µ a2 , y2 + µ b2 , z2 + µ c2 ) .
If lines intersect these two points must coincide for some value of λ and µ . Equate the corresponding
co-ordinates then proceed as in vector form.
If l 1 and l 2 are two skew lines, then the straight line which is perpendicular to each of these two non-
intersecting lines is called the “Line of shortest distance “There is one and only one line perpendicular to
each of lines l1 and l2 .”
10. Shortest distance between two skew lines
Vector form
Let two lines be r1 =
a1 + λb1 and r2 =
a2 + µ b2 then Shortest Distance =
(
( a2 − a1 ) . b1 × b2 )
.
b1 × b2
Cartesian form
x − x1 y − y1 z − z1 x − x2 y − y2 z − z2
Let two skew lines be, = = and = =
l1 m1 n1 l2 m2 n2
Therefore, the shortest distance between the lines is given by
x2 − x1 y2 − y1 z2 − z1
l1 m1 n1
l2 m2 n2
d=
( m1n2 − m2 n1 ) + ( n1l2 − l1n2 ) + ( l1m2 − m1l2 )
2 2 2
Q ( x1 + ar , y1 + br , z1 + cr )
SPHERE
1. Equation of a sphere having C ( c ) as it’s centre and a as it’s radius:
r −c = a (Vector form) a P
C
Cartesian form: Let center be C ( x1 , y1 , z1 ) and radius = a.
Then equation is: ( x − x1 ) + ( y − y1 ) + ( z − z1 ) = r
2 2 2
a2. …(1)
c
If the centre is at the origin, then equation (1) takes the form
x2 + y 2 + z 2 =a2 O
which is known as the standard form of the equation of the sphere.
2. General equation of sphere
0 with centre ( −u , −v, − w ) i.e.,
The general equation of a sphere is x 2 + y 2 + z 2 + 2ux + 2vy + 2 wz + d =
( x − x1 )( x − x2 ) + ( y − y1 )( y − y2 ) + ( z − z1 )( z − z2 ) =
0.
4. Section of a sphere by a plane
Consider a sphere intersected by a plane. The set of points common to both
sphere is always a circle. The equation of the sphere and the plane taken C
together represent the plane section.
Let C be the centre of the sphere and M be the foot of the perpendicular from C
on the plane. Then M is the centre of the circle and radius of the circle is given M
P Q
by =
PM CP 2 − CM 2 .
The centre M of the circle is the point of intersection of the plane and line CM
which passes through C and is perpendicular to the given plane.
6. Great circle: The section of a sphere by a plane through the centre of the sphere is a great circle. Its
centre and radius are the same as those of the given sphere.
7. Condition of tangency of a plane to a sphere
A plane touches a given sphere if the perpendicular distance from the centre of the sphere to the plane is
equal to the radius of the sphere. The plane lx + my + nz =p touches the sphere
0 If ( ul + vm + wn − p ) = ( l 2 + m 2 + n 2 )( u 2 + v 2 + w2 − d ) .
x 2 + y 2 + z 2 + 2ux + 2vy + 2 wz + d =
2
225
Probability Chapter 33
SOME BASIC DEFINITIONS
EXPERIMENT
An operation which results in some well defined outcome is called an experiment.
RANDOM EXPERIMENT
An experiment whose outcome cannot be predicted with certainty is called a random experiment.
For example, tossing of a fair coin or throwing an unbiased die or drawing a card from a well shuffled
pack of 52 cards is a random experiment.
SAMPLE SPACE
The set of all possible outcomes of a random experiment is called the sample space.
It is usually denoted by S .
For example, if we toss a coin, there are two possible outcomes, a head ( H ) or a tail (T ) .
So, the sample space in this experiment is given by S = { H , T }.
EVENT
A subset of the sample space S is called an Event.
Note :
(a) Sample space S plays the same role as the universal set for all problems related to the particular
experiment.
(b) φ is also a subset of S which is called an impossible event.
(c) S is also a subset of S which called a sure event.
SIMPLE EVENT
An event having only a single sample point is called a simple event.
For example, when a coin is tossed, the sample space S = { H , T }.
Let=
E1 {=
H } the event of occurrence of head and
=
E2 {=
T } the event of occurrence of tail.
Then, E1 and E2 are simple events.
MIXED EVENT
A subset of the sample space S which contains more than one element is called a mixed event.
For example, when a coin is tossed, the sample space S = { H , T }.
Let E {=
= H , T } the event of occurrence of a head or a tail.
Then, E is a mixed event.
EQUALLY LIKELY EVENTS
A set of events is said to be equally likely if none of them is expected to occur in preference to the other.
For example, when a fair coin is tossed, then occurrence of head or tail are equally likely cases and there
is no reason to expect a ‘head’ or a ‘tail’ in preference to the other.
EXHAUSTIVE EVENTS
A set of events is said to be exhaustive if the performance of the experiment always results in the occurrence of
atleast one of them.
For example, when a die is thrown, then the events
A1 = {1, 2} and A2 = {2, 3, 4} are not exhaustive as we can get 5 as outcome of the experiment
which is not the member of any of the events A1 and A2 .
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But, if we consider the events E1 = {1, 2, 3} and E2 = {2, 4, 5, 6} then the set of events E1 , E2 is exhaustive.
MUTUALLY EXCLUSIVE EVENTS
A set of events is said to be mutually exclusive if occurrence of one of them precludes the occurrence of any of
the remaining events.
φ for i ≠ j .
Thus, E1 , E2 , ...., En are mutually exclusive if and only if Ei ∩ E j =
For example, when a coin is tossed, the event of occurrence of a head and the event of occurrence of a tail
are mutually exclusive events.
INDEPENDENT EVENTS
Two events are said to be independent, if the occurrence of one does not depend on the occurrence of the other.
For example, when a coin is tossed twice, the event of occurrence of head in the first throw and the event
of occurrence of head in the second throw are independent events.
COMPLEMENT OF AN EVENT
The complement of an event E , denoted by E or E ′ or E c , is the set of all sample points of the space other
than the sample points in E.
For example, when a die is thrown, sample space
S = {1, 2, 3, 4, 5, 6}.
If E = {1, 2, 3, 4} , then E = {5, 6}.
Note that E ∪ E = S.
MUTUALLY EXCLUSIVE AND EXHAUSTIVE EVENTS
A set of events E1 , E2 ,..., En of a sample space S form a mutually exclusive and exhaustive system of events,
if
(i) Ei ∩ E j =φ for i ≠ j and
(ii) E1 ∪ E2 ∪ ... ∪ En =S.
For example, when a die is thrown, sample space S = {1, 2, 3, 4, 5, 6}.
=
Let E1 {1,=
3, 5} the event of occurrence of an odd number and
=E2 {=
2, 4, 6} the event of occurrence of an even number.
Then, E1 ∪ E2 = S and E1 ∩ E2 =φ.
PROBABILITY OF OCCURRENCE OF AN EVENT
Let S be a sample space, then the probability of occurrence of an event E is denoted by P ( E ) and is
defined as
n ( E ) number of elements in E
P=(E) =
n ( S ) number of elements in S
number of cases favourable to event E
=
total number of cases
Note :
(a) 0 ≤ P ( E ) ≤ 1, i.e. the probability of occurrence of an event is a number lying between 0 and 1.
(b) P (φ ) = 0, i.e. probability of occurrence of an impossible event is 0.
(c) P ( S ) = 1, i.e. probability of occurrence of a sure event is 1.
ODDS IN FAVOUR OF AN EVENT AND ODDS AGAINST AN EVENT
If the number of ways in which an event can occur be m and the number of ways in which it does not occur be
n, then
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Probability 227
m
(i) odds in favour of the event = and
n
n
(ii) odds against the event = .
m
a
If odds in favour of an event are a : b then the probability of the occurrence of that event is and the
a+b
b
probability of the non-occurrence of that event is
.
a+b
SOME IMPORTANT RESULTS ON PROBABILITY
1. P ( A ) = 1 − P ( A)
2. If A and B are any two events, then P ( A ∪ B=
) P ( A) + P ( B ) − P ( A ∩ B ) .
3. φ and hence P ( A ∩ B ) =
If A and B are mutually exclusive events, then A ∩ B = 0.
∴ P ( A ∪ B=
) P ( A) + P ( B ) .
4. If A, B, C are any three events, then P ( A ∪ B ∪ C=) P ( A) + P ( B ) + P ( C ) − P ( A ∩ B )
− P ( B ∩ C ) − P ( C ∩ A) + P ( A ∩ B ∩ C ) .
5. If A, B, C are mutually exclusive events, then A ∩ B= φ , B ∩ C= φ , C ∩ A= φ , A ∩ B ∩ C= φ and
hence P ( A ∩ =
B ) 0, P ( B ∩ C
= ) 0, P ( C ∩ =
A ) 0, P ( A ∩ B ∩ C ) =0.
∴ P ( A ∪ B ∪ C=) P ( A) + P ( B ) + P ( C ) .
6. P ( A ∩ B ) =−
1 P ( A ∪ B ).
7. P ( A ∪ B ) =−
1 P ( A ∩ B ).
8. P ( A )= P ( A ∩ B ) + P ( A ∩ B )
9. P ( B )= P ( B ∩ A ) + P ( B ∩ A )
10. If A1 , A2 , ...., An are independent events, then P ( A1 ∩ A2 ∩ ... ∩ An=
) P ( A1 ) ⋅ P ( A2 ) ,..., P ( An ) .
11. If A1 , A2 ,..., An are mutually exclusive events, then
P ( A1 ∪ A2 ∪ ... ∪ A=
n) P ( A1 ) + P ( A2 ) + ... + P ( An ) .
12. If A1 , A2 ,..., An are exhaustive events, then P ( A1 ∪ A2 ∪ ... ∪ An ) =
1.
13. If A1 , A2 ,..., An are mutually exclusive and exhaustive events, then
P ( A1 ∪ A2 ∪ ... ∪ A=
n) P ( A1 ) + P ( A2 ) + ... + P ( A=
n) 1.
14. If A1 , A2 ,..., An are n events, then
(a) P ( A1 ∪ A2 ∪ ... ∪ An ) ≤ P ( A1 ) + P ( A2 ) + ... + P ( An )
(b) P ( A1 ∩ A2 ∩ ... ∩ An ) ≥ 1 − P ( A1 ) − P ( A2 ) ... − P ( An )
CONDITIONAL PROBABILITY
Let A and B be two events associated with a random experiment. Then, the probability of occurrence of A
under the condition that B has already occurred and P ( B ) ≠ 0, is called the conditional probability and it is
denoted by P ( A / B ) .
Thus P A ( B) = Probability of occurrence of A, given that B has already happened.
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P ( A ∩ B) n ( A ∩ B)
= =
P ( B) n ( B)
Similarly, P B( A) = Probability of occurrence of B, given that A has already happened.
P ( A ∩ B) n ( A ∩ B)
= = .
P ( A) n ( A)
Sometimes, P A ( B ) is also used to denote the probability of occurrence of A when B occurs.
Probability 229
for i = 1, 2,......, n. We can think of Ei ’s as the causes that lead to the outcome of an experiment. The
probabilities P ( Ei ) , i = 1, 2,...., n are called prior probabilities. Suppose the experiment results in an outcome of
event A, where P ( A ) > 0. We have to find the probability that the observed event A was due to cause Ei , that
is, we seek the conditional probability P ( Ei / A ) . These probabilities are called posterior probabilities, given by
P ( Ei ) ⋅ P ( A / Ei )
Baye’s rule as P ( Ei / A ) = n
.
∑ P(E ) P( A/ E )
k =1
k k
RANDOM VARIABLE
A random variable is a real valued function whose domain is the sample space of a random experiment.
A random variable is usually denoted by the capital letters X , Y , Z , ...., etc.
DISCRETE RANDOM VARIABLE
A random variable which can take only finite or countably infinite number of values is called a discrete random
variable.
CONTINUOUS RANDOM VARIABLE
A random variable which can take any value between two given limits is called a continuous random variable.
Geometrical method for probability: When the number of points in the sample space is infinite, it becomes
difficult to apply classical definition of probability. For instance if we are interested to find the probabilities that
a point selected at random from the interval [1,6] lies either in the interval [1,2] or [5,6], we cannot apply the
classical definition of probability. In this case we define the probability as follows:
Measure of region A
P { x ∈ A} =
Measure of the sample spaceS
where measure stands for length, area or volume depending upon whether S is a one – dimensional,
two – dimensional or three dimensional region.
Probability distribution : Let S be a sample space. A random variable X is a function from the set S to R ,
the set of real numbers.
For example, the sample space for a throw of a pair of dice is
{11, 12, , 16
21, 22, , 26
S=
61, 62, , 6 6}
X (12 ) 3,=
Let X be the sum of numbers on the dice. Then= X ( 43) 7, etc. Also, { X = 7} is the event
{61, 52, 43, 34, 25, 16}. In general, if X is a random variable defined on the sample space S and r is a real
number, then { X = r} is an event.
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=
If the random variable X takes n distinct values x1 , x2 ,...., xn , then { X x=
1} , { X x2 } ,.....,
= { X xn } are
mutually exclusive and exhaustive events.
Now, since ( X = xi ) is an event, we can talk of P ( X = xi ) . X = x1 X = x3
X = x2
If P ( X = xi )= Pi ( where 1 ≤ i ≤ n ) , then the system of numbers.
x1 x2 xn X = x4 X = xn
is said to be the probability distribution of the random
p1 p2 pn
variable X .
n
The expectation (mean) of the random variable X is defined as E ( X ) = ∑ pi xi and the variance of X is
i =1
n n
defined as var ( X ) =∑ pi ( xi − E ( X ) ) =∑ pi xi2 − ( E ( X ) ) .
2 2
=i 1 =i 1
Bernoullian Trials, set of n trials is said to be Bernoullian if
(a) The value of n is finite i.e. number of trials are finite
(b) Each and every trial/experiment is independent
(c) Trial (Experiment) consist only two out comes namely success and failure.
(d) Probability of success & failure for each trial is fixed (same)
3. Binomial probability distribution : A random variable X which takes values 0, 1, 2, ..., n is said to
follow binomial distribution if its probability distribution function is given by
P ( X= r=
) nCr p r q n − r , =
r 0, 1, 2, ...., n
where p, q > 0 such that p + q =
1.
The notation X ~ B ( n, p ) is generally used to denote that the random variable X follows binomial
distribution with parameters n and p.
We have P ( X =+ 0) P ( X = 1) + ... + P ( X =
n).
= nC0 p 0 q n −0 + nC1 p1q n −1 + ... + nCn p n q n − n = ( q + p ) = 1n = 1
n
Now probability of
(a) Occurrence of the event exactly r times P ( X= r= ) nCr q n − r p r .
n
(b) Occurrence of the event at least r times P ( =
X ≥ r) n
Cr q n − r p r + ...
= + pn ∑
X =r
n
C X p X q n− X .
Note :
1. If the probability of happening of an event in one trial be p, then the probability of successive
happening of that event in r trials is p r .
2. If n trials constitute an experiment and the experiment is repeated N times, then the frequencies of
0, 1, 2, ...., n successes are given by
N .P. ( X 0 )=
= , N .P. ( X 1=
) , N .P. ( X 2 ) , ....,
= N .P. ( X n ) .
(i) Mean and variance of the binomial distribution : The binomial probability distribution is
JEEMAIN.GURU
Probability 231
X 0 1 2 .... n
P( X ) n
C0 q n p 0 n
C1 q n −1 p n
C2 q n − 2 p 2 .... n
Cn q 0 p n
n n
=
The binomial probability distribution is ∑ X i pi ∑
= X. C q pn
X
n− X X
np,
=i 1 =X 1
The variance of the Binomial distribution is σ 2 = npq and the standard deviation is σ = ( npq ).
(ii) Use of multinomial expansion : If a die has m faces marked with the numbers 1, 2, 3, ...m and
if such n dice are thrown, then the probability that the sum of the number exhibited on the upper
(x + x )
n
p
2
+ x 3 + ... + x m
faces equal to p is given by the coefficient of x in the expansion of .
mn
4. Poission distribution : It is the limiting case of B.D. under the following condition
(i) Number of trials are very-very large i.e. n → ∞
(ii) p → 0 (Here p is not exactly 0 but nearly approaches to zero)
(iii) np = λ , a finite quantity ( λ is called parameter)
→ Probability of r success for poission distribution is given by
e−λ λ r
P ( X= r=
) ,=
r 0, 1, 2, ....
r!
→ For poission distribution recurrence formula is given by
λ
P ( r + 1) = P ( r )
r +1
Note :
(a) For Poission Distribution mean = variance= λ= np.
(b) If X and Y are independent poission variates with parameter λ1 and λ2 then X + Y also has
poission distribution with parameters λ1 + λ2 .
5. Normal Distribution
For a normal distribution, number of trials are infinite.
The Normal probability function or distribution is given by
1 x−µ
2
1 − x−µ
P ( X= x=) e 2 σ where = z known as standard variate, −∞ < x < ∞
σ 2π σ
Facts About Normal Distribution P(z)
(i) It is limiting case of B.D. i.e. B ( n, p ) when n → ∞
(ii) mean = mode = Median
(iii) Total area under a standard normal (curve) distribution is 1.
(iv) Normal curve is bell-shaped and uni-model
(v) Q3 − Median = Median − Q1 z
1 µ − 3σ o µ + 3σ
(vi) ( X r=
∫ P= ) dx 1 =µ mean
= mode
= median
−1
X −µ
If X ~ N ( µ , σ 2 ) and Z = then P ( µ − σ ≤ x ≤ µ + σ ) = P ( −1 ≤ z ≤ 1) = .6827
σ
P ( µ − 2σ ≤ x ≤ µ + 2 σ ) = P ( −2 ≤ z ≤ 2 ) = .9544
P ( µ − 3σ ≤ x ≤ µ + 3σ )= P ( −3 ≤ z ≤ 3)= .9973
JEEMAIN.GURU
Note :
The probabilities P ( z1 ≤ z ≤ z2 ) , P ( z1 < z < z2 )
P ( z1 < z ≤ z2 ) , P ( z1 < z < z2 ) are all treated to be same.
JEEMAIN.GURU
233
INTRODUCTION
For given data, a single value of the variable representing the entire data is selected which described the
characteristics of the data. Averages are, generally, the central part of the distribution and therefore, they are
also called the measures of central tendency.
The following are the five measures of central tendency
(1) Arithmetic mean (2) Geometric mean (3) Harmonic mean
(4) Median (5) Mode
I. ARITHMETIC MEAN
Arithmetic mean is the most important among the mathematical mean.
According to Horace Secrist.
“The arithmetic mean is the amount secured by dividing the sum of values of the items in a series by
their number.”
1. Simple arithmetic mean in individual series (Ungrouped data)
(i) Direct method
If the series in this case be x1 , x2 , x3 ,......, xn ; then the arithmetic mean x is given by
Sum of the seires
x=
Number of terms
x1 + x2 + x3 + ... + xn 1 n
= i.e., x =
n
∑ xi
n i =1
f1 x1 + f 2 x2 + ... + f n xn ∑fx i i
=x = i =1
f1 + f 2 + ... + f n n
∑f
i =1
i
3. Mean of the composite series : If xi , ( i = 1, 2,....., k ) are the means of k-component series of sizes
ni , ( i = 1, 2, ...., k ) respectively, then the mean x of the composite series obtained on combining the
component series is given by the formula
n
n1 x1 + n2 x2 + ... + nk xk ∑n x i i
=x = i =1
.
n1 + n2 + ... + nk n
∑n
i =1
i
∑ f ( x − x) =
n
i.e., i i 0 , where fi is the frequency of xi (1 ≤ i ≤ n ) .
i =1
(ii) In a statistical data, the sum of squares of the deviations of items A.M. is least
n
∑ f (x − x)
2
i.e., i i is least.
i =1
(iii) If each of the n given observations be doubled, then their mean is doubled.
(iv) If x is the mean of x1 , x2 ,....., xn then the mean of ax1 , ax2 , ....., axn where a is any
number different from zero, is a x .
II. GEOMETRIC MEAN
1. If x1 , x2 , x3 ,..., xn are n values of a variate x, none of them being zero, then geometric mean (G.M.) is
given by
G.M. = ( x1 ⋅ x2 ⋅ x3 ..... xn )
1/ n
1
or log ( G.M.
= ) ( log x1 + log x2 + .... + log xn ) .
n
log x1 + log x2 + .... + log xn
or G.M. = antilog .
n
2. In case of frequency distribution, G.M. of n values x1 , x2 ,..... xn of a variate x occurring with frequency
f1 , f 2 , ...., f n is given by
(x )
1/ N
=
G.M. 1
f1
⋅ x2f2 ... xnfn , where N = f1 + f 2 + .... + f n .
n
∑ fi log xi
G.M . = antilog
i =1
or
N
III. HARMONIC MEAN
1. The harmonic mean of n items x1 , x2 ,....., xn is defined as
n
H.M. = .
1 1 1
+ + .... +
x1 x2 xn
2. If the frequency distribution is f1 , f 2 , f 3 ,....., f n respectively,
JEEMAIN.GURU
f1 + f 2 + f3 + ... + f n
then H.M. = .
f1 f 2 fn
+ + ... +
x1 x2 xn
RELATION BETWEEN A.M. AND H.M.
The arithmetic mean (A.M.), geometric mean (G.M.) and harmonic mean (H.M.) for a given set of observation
of a series are related as under :
A.M. ≥ G.M. ≥ H.M.
Equality sign holds only when all the observation in the series are same.
IV. MEDIAN
Median is the middle most or the central value of the variate in a set of observations, when the observations are
arranged either in ascending or in descending order of their magnitudes. It divides the arranged series in two
equal parts.
1. CALCULATION OF MEDIAN
(i) Individual series : If the data is raw, arrange in ascending or descending order. Let n be the number
of observations.
n +1
If n is odd, Median = value of th item.
2
1 n n
If n is even, = Median value of th item + value of + 1 th item
2 2 2
(ii) Discrete series : In this case, we first find the cumulative frequencies of the variables arranged in
ascending or descending order and the median is given by
n +1
Median = th observation, where n is the cumulative frequency.
2
(iii) For grouped or continuous series : In this case, following formula can be used.
N
−C
l+ ×i
2
(a) For series in ascending order, Median =
f
Where l = Lower limit of the median class
f = Frequency of the median class
N = The sum of all frequencies
i = The width of the median class
C = The cumulative frequency of the class preceding to median class.
(b) For series in descending order
N
2 −C n
Median = u − × i , where u = upper limit of the median class, N = ∑ fi .
f i =1
As median divides a distribution into two equal parts, similarly the quartiles, quantiles, deciles and
percentiles divide the distribution respectively into 4, 5, 10 and 100 equal parts.
The j th quartile is given by
JEEMAIN.GURU
N
j 4 −C
Qj = l + i; j= 1, 2, 3.
f
Q1 is the lower quartile,
Q2 is the median and
Q3 is called the upper quartile.
2. LOWER QUARTILE
(i) Discrete series
n +1
Q1 = size of th item
4
(ii) Continuous series
N
−C
l+ ×i
4
Q1 =
f
3. UPPER QUARTILE
(i) Discrete series
3 ( n + 1)
Q3 = size of th item
4
(ii) Continuous series
3N
−C
l+ ×i
4
Q3 =
f
4. DECILE
Decile divide total frequencies N into ten equal parts.
N× j
−C
Dj = l + 10 × i, [where j = 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
f
5. PERCENTILE
Percentile divide total frequencies N into hundred equal parts and
N ×k
−C
Pk = l+ 100 × i, where k = 1, 2, 3, 4, 5, ......, 99.
f
V. MODE
Mode is that value in a series which occurs most frequently. In a frequency distribution, mode is that variate
which has the maximum frequency .
For continuous series, mode is calculated as,
f1 − f 0
Mode = l1 + ×i
2 f1 − f 0 − f 2
Where, l1 = The lower limit of the model class
f1 = The frequency of the model class
f 0 = The frequency of the class preceding the model class
JEEMAIN.GURU
SYMMETRIC DISTRIBUTION
A distribution is a symmetric distribution if the values of mean, mode and median coincide. In a symmetric
distribution frequencies are symmetrically distributed on both sides of the centre point of the frequency curve.
Q3 − Q1
i.e. Q.D. =
2
Q3 − Q1
and coefficient of quartile deviation = ,
Q3 + Q1
where Q3 is the third or upper quartile and Q1 is the lowest or first quartile.
2. Mean deviation
The arithmetic average of the deviations (all taking positive) from the mean, median or mode is known as
mean deviation.
(i) Mean deviation from ungrouped data (or individual series)
Mean deviation =
∑ x−M ,
n
where x − M means the modulus of the deviation of the variate from the mean (mean, median or
mode) and n is the number of terms.
(ii) Mean deviation from continuous series :
Mean= deviation
∑=
f x−M ∑ f dM ,
n n
where n = ∑ f . and dM= x − M the deviation of each variate from the mean M .
3. Standard deviation
Standard deviation (or S.D.) is the square root of the arithmetic mean of the square of deviations of
various values from their arithmetic mean and is generally denoted by σ read as sigma.
(i) Coefficient of standard deviation : To compare the dispersion of two frequency distributions the
relative measure of standard deviation is computed which is known as coefficient
of standard deviation and is given by
σ
Coefficient of S.D. = , where x is the A.M.
x
(ii) Standard deviation from individual series
∑(x − x )
2
σ=
N
where, x = The arithmetic mean of series
N = The total frequency.
(iii) Standard deviations from continuous series
∑ f (x − x)
2
σ= i i
N
where, x = Arithmetic mean of series
xi = Mid value of the class
fi = Frequency of the corresponding xi
N = ∑f = The total frequency
Short cut method :
∑ fd ∑ fd ∑d ∑d
2 2 2 2
(i) σ
= − =(ii) σ −
N N N N
where, d = x − A = Deviation from the assumed mean A
JEEMAIN.GURU
then σ 2
deviation of two series,=
1
n1 + n2 ( ) ( )
n1 σ 12 + d12 + n2 σ 22 + d 22 ,
n x +n x
where d1 =− x1 x , d 2 =−
x2 x and x = 1 1 2 2 .
n1 + n2
SKEWNESS
∑ ( xi − µ )
3
= Sk
M − Mo
= 3
(M − Md ) , − 3 ≤ S ≤ 3 where σ is standard deviation.
σ σ k
241
Cov ( x, y )=
1 n
(
∑ xi − x
n i =1
)( y − y )i
1 n 1 n ∑ xi ∑ yi
= ∑
n i =1
x y
i i −=x y ∑
n i =1
xi yi − .
n n
METHOD OF FINDING CORRELATION
(i) Scatter diagram
With the help of scatter diagram we can develop the relationship .i.e., correlation between the
variable quantities by taking one of them on x − axis and other on y − axis , using all distinct order
pair of these two variable quantities we get a picture in the form of dots, due to this reason. We
called it dot diagram also. If these dots lies in the same straight line then with the help of the
direction it is said to +ve correlation and for obtuse angle it is said to be negative correlation.
(ii) Karl Pearson’s co-efficient of correlation
We denote coefficient of correlation by rxy .
Cov ( x,y ) Cov ( x, y ) Cov ( x, y )
∴ rxy = or = or =
σ xσ y Var x Var y SD x .SD y
where σ x , σ y are standard deviation (SD) of variables x and y respectively.
1 n
(
∑ xi − x )
2
σ x2
= …(i)
n i =1
1 n 2 ∑ xi
2
=
n
∑ xi − n
…(ii)
∑ dx
2
1
=
n
∑ d 2 x − …(iii)
n
JEEMAIN.GURU
Cov ( x, y )
Now rxy = …(i)
σ xσ y
=
∑ ( x − x )( y − y )
i i
…(ii)
∑(x − x ) ∑( y − y )
2 2
i i
=
∑ ( x − x )( y − y )
i i
=
( ∑ x )( ∑ y )
n∑ xi yi − i i
∑(x − x ) ∑( y − y ) n x −
∑
( ∑ x ) × n∑ y − ( ∑ y )
2 2 2 2
2 2
i i i i
i i
Shortcut method
n∑ ( d x d y ) − ( ∑ d x ) ( ∑ d y )
rxy = …(iv)
n∑ d 2 x − ( ∑ d x ) × n∑ d 2 y − ( ∑ d y )
2 2
ρ = 1− 6 ∑d 2
where d= R1 − R2
(
n n2 − 1 )
PROPERTIES OF CORRELATION
(i) The values of r always lies between −1 and 1, including itself the numbers −1, and 1.
i.e., −1 ≤ rxy ≤ 1 or −1 ≤ r ≤ 1.
(ii) If 0 < r < 1 known as positive correlation.
(iii) If 0 < r < .5, known as low positive correlation.
(iv) If 0.5 < r < 1, known as high positive correlation.
(v) If r = +1, known as perfect positive correlation.
(vi) If r = −1, known as perfect negative correlation.
(vii) If −0.5 < r < 0, known as low negative correlation.
(viii) If −1 < r < −0.5, known as high negative correlation.
(ix) If r = 0 then x, y are said to uncorrelated but it does not indicate that x, y are independent.
(x) If x, y are two independent variable then r ( x, y ) = 0.
(xi) The value of rxy .i.e., coefficient to correlation is independent of the change in origin and scale.
(xii) Coefficient of correlation is purely a number which has no unit of measurement.
(xiii) If the variables are connected by the equation Ax + By + k = 0 then,
rxy = −1 if AB > 0 and
γ xy = 1 if AB < 0
(xiv) For three variables x, y, z if = z Ax + By then
σ z 2 = A2σ x 2 + B 2σ y 2 + ( 2 ABσ xσ y ) .r , where r = coefficient of correlation between x and y .
(xv) For variable x and y
JEEMAIN.GURU
AB
r ( Ax + k1 , By + k2 ) = r ( x, y ) such that A, B ≠ 0
AB
DEDUCTIONS
(i) Standard error
1− r2
( SE ( r ) ) = n r = rxy ,
n = Number of observation.
PE ( r ) Probable error
=
(ii) = .6745
SE ( r ) Standerd error
(iii) r < PE ( r ) , No evidence of r.
(iv) r > 6 PE ( r ) , existence of correlation is definite
(v) r 2 is known as coefficient of determination.
REGRESSION ANALYSIS
According to British scientist “Sir Francis Galton” Regression means stepping back towards the average.
Regression may be linear or curvilinear.
Equation of Regression lines
Line of regression y on x
σy
y=
−y r ( x −=
x) byx ( x − x )
σx
Line of regression x on y
σx
x=
−x r ( y −=
y) bxy ( y − y )
σy
Where byx and bxy are regression coefficient y on x and y respectively and defined as
σy σx
byx = r and bxy = r
σx σy
Facts of regression lines:
(i) Two regression equations intersects at mean values of the data’s ( xi , yi ) .i.e., regression lines
intersect at x , y .
(ii) To determine x , y , solve the two regression lines.
(iii) The sign. of byx , bxy and r are always same.
(iv) The regression line y on x is used to estimate the value of y for the given value of the variables
x Here y is dependent and x is independent variable.
(v) The regression line x on y is used to determined the value of x for the given value of y. In this
case x is dependent on y.
=
(1 − r ) 2
σ x .σ y
…(ii)
r σ 2x + σ 2 y
byx .bxy − 1
= …(iii)
byx + bxy
where θ be the angle between the lines of regression.
rσ y
b= = m= slope of regression line y on x
yx
σx 1
σx
b= r = m= slope of regression line x on y .
xy
σy 2
(1 − r )
2
2
( )
2
(i) sin 2 θ = 2
∴ sin 2 θ =
1− r2
σ 2x + σ 2 y
(1 − r )
2
2
+r
2
σ x .σ y
∴ sin θ ≤ 1 − r 2
(ii) If m1m2 = −1 .i.e., regression lines are ⊥ to each other.
π
Then= θ
r 0,= In this line y on x parallel to x − axis and line x on y parallel to y − axis
2
(iii) If θ = 0, π The lines of regression are coincident, overlapping symmetrical or identical
1
∴ 1 , ∴ byx =
byx bxy =
bxy
(iv) If a, b, c, d ,∈ R ( a, c having opposite sign ) , r is coefficient of correlation between x and y then
coefficient of correlation between ax + b and cx + d is given by. r ( ax + b, cx + d ) =
−rxy
1 1 2
(i) +
If r , byx , bxy are positive then ≥
byx bxy r
(ii) If the two variates x and y are having perfect correlation between them, then they will be
x y σ x y −σ y x
connected by + = 1 where a =
a b σy
σ x y −σ y x
b= S n σ x 1 − r 2 and Standard error of
, Standard error estimate of x is given by=
σx
Sy σ y 1− r2
estimate of y is given by=
JEEMAIN.GURU
245
Statics Chapter 36
DEFINITIONS
Matter
Anything that occupies space and is perceived by our senses is matter. Table, cup, water, etc. are example
of matter.
Body
A body is a portion of matter occupying finite space. It has, therefore, a definite volume and a definite
mass.
Particle
A particle is a body indefinitely small in size, so that the distance between its different parts are
negligible. It may be regarded as a mathematical point associated with mass
Rigid Body
A body is said to be rigid when it does not change its shape and size when subjected to external forces i.e.
a rigid body is a body the distance between any two points of which always remains the same .
Force
Force is an agent which changes or tends to change the state of rest or uniform motion of a body.
Note :
Force is a vector quantity
REPRESENTATION OF A FORCE
A force is completely known if we know the following data about it :
(i) its magnitude
(ii) its direction
(iii) its point of application.
Thus, we can completely represent a force by a straight line AB drawn through the point of application
along the line of action of the force, the length of the line AB representing the magnitude of the force and
the order of the letters A, B specifying the direction.
MECHANICS
It is the science which deals with moving bodies or bodies at rest under the action of some forces.
Mechanics
Dynamics Statics
DYNAMICS
It is that branch of mechanics which deals with the action of forces on bodies in motion.
STATICS
It is that branch of mechanics which deals with the action of forces on bodies, the forces being so
arranged that the bodies are at rest.
FORCES IN STATICS
1. Action and Reaction
Whenever one body is in contact with another body, they apply equal and opposite
F2 F1
forces at the point of contact. Such pair forces are called action and reaction pairs
F 2 = −F1
JEEMAIN.GURU
(i) When a rod AB rests with one end B upon a smooth plane, the reaction is R
along the normal to the plane at the point of contact.
(ii) When a rod rests over a smooth peg, the reaction at the point of contact is ⊥ to the rod.
R A R
S
A C B B
R B
(iii) If one point of a body is in contact with the surface of another body, the
reaction at the point of contact is ⊥ to the surface, e.g . the equilibrium of a S
ladder in contact with the ground and a wall [both being smooth].
A C
R
(iv) when a rod rests completely within a hollow sphere, the reactions at the S
extremities of the rod are along the normals at those points and will pass
through centre of the hollow sphere . B
A
2. Weight
Everybody is attracted towards the centre of the earth with a force proportional to its mass (the quantity of
matter in the body).This force is called the weight of the body. If m is the mass of the body and g, is the
acceleration due to gravity, then its weight W = mg .
3. Tension or Thrust
Whenever a string is used to support a weight or drag a body, there is a force of pull along the string. This
force is called tension. Similarly, if some rod be compressed, a force will be exerted. This type of force is
called thrust.
Note :
(i) The tension in a string is the same throughout. When two string are knotted together, the tensions
in the two portions are different.
(ii) When a weight W hangs by a string, the tension in the string must be equal to the weight suspended.
i.e. T = W .
(iii) The tension of a string always acts in a direction diverging away from the body under consideration
and acts along the string.
Statics 247
AB and AD of a parallelogram ABCD, then their resultant R is represented in magnitude and direction by
the diagonal AC.
RESULTANT OF TWO FORCES B C
If two concurrent forces P and Q are inclined at an angle α to each
Q R
other, then the magnitude R of their resultant is given by
α
R = P 2 + Q 2 + 2 PQ cos α . If R makes an angle θ with the direction of θ
O
Q sin α P A
P, then tan θ =
P + Q cos α
PARTICULAR CASES
1. When P and Q are at right angles to each other i.e. α= 90°
Q
In this case=R P 2 + Q 2 and tan θ = .
P
α α
2. When P = Q. = In this case, R 2= P cos and θ .
2 2
3. When P and Q are in the same direction i.e.α = 0
In this case, R is in the same direction as P and Q and R= P + Q.
This is called the greatest resultant of the two forces.
4. When P and Q are in the opposite direction ( i.e.= α 180° ) and P > Q.
In this case, R is in the direction of P and R= P − Q.
This is called the least resultant of the two forces.
COMPONENT OF A FORCE IN TWO DIRECTIONS :
The component of a force R in two directions making angles α and β with the line of action of R on
and opposite sides of it are B C
OC.sin β R sin β R
= F1 =
sin (α + β ) sin (α + β )
F2 β
OC.sin α R sin α
and= F2 = α
sin (α + β ) sin (α + β ) O F1 A
A
TRIANGLE LAW OF FORCES
If three forces, acting at a point, be represented in magnitude and direction by
the three sides of triangle, taken in order, they are in equilibrium. F3 F2
F1 + F2 + F3 =0
B C
F1
POLYGON LAW OF FORCES
If any number of forces acting on a particle be represented in magnitude and A3
P4 P3
direction by the sides of a polygon taken in order, the forces shall be in
equilibrium. A4 A2
P1 + P2 + ...... + Pn =
0 P2
Ai
A A1
P1
JEEMAIN.GURU
λ – μ THEOREM B
The resultant of two forces, acting at a point O along OA and
OB and represented in magnitude λ . OA and µ . OB, is
µ OB
represented by a force ( λ + µ ) .OC , where C is a point on C
AB such that λ CA = µ CB i.e.C divides AB in the ratio
( λ + µ ) OC
µ : λ. In vector notation the above statement can be written
as : λ.OA + µ .OB= ( λ + µ ) .OC , where C is a point on AB
O
dividing it in the ratio µ : λ. λ OA A
Note :
In the above theorem, if λ= µ= 1, then OA + OB = 2OC , where C is the mid point of AB , i.e. the
resultant of two forces OA and OB is 2OC , where C is the mid point of AB.
EQUILIBRIUM OF FORCES
A system of forces acting on a body is said to be in equilibrium if it produces no change in the motion of
the body i.e.
(i) Vector sum of all forces is equal to zero and
(ii) Vector sum of all the moments of these forces about any point is zero.
EQUILIBRIUM OF TWO FORCES
Two forces acting at a point are in equilibrium if and only if they,
(i) are equal in magnitude
(ii) act along the same line
(iii) have opposite directions.
CONDITION OF EQUILIBRIUM OF A NUMBER OF COPLANAR CONCURRENT FORCES
A given number of forces acting at a point are in equilibrium if and only if the algebraic sum of their
resolved parts in each of the two perpendicular directions OX and OY vanish separately.
PARALLEL FORCES
1. Like parallel forces : Two parallel forces are said to be like parallel forces when they act in the same
direction.
P R Q
A C B
The resultant R of two like parallel forces P and Q is equal in magnitude of the sum of the magnitudes
of forces and R acts in the same direction as the forces P and Q and at the point on the line segment
joining the point of action P and Q, which divides it in the ratio Q : P internally.
JEEMAIN.GURU
Statics 249
A p
B
P
2. Moment of couple
The moment of a couple is obtained in magnitude by multiplying the magnitude of one of the forces
forming the couple and perpendicular distance between the lines of action of the force. The perpendicular
distance between the forces is called the arm of the couple. The moment of the couple is regarded as
positive or negative according as it has a tendency to turn the body in the anticlockwise or clockwise
direction.
Moment of a couple = Force × Arm of the couple = P. p
3. Sign of the moment of a couple : The moment of a couple is taken with positive or negative sign
according as it has a tendency to turn the body in the anticlockwise or clockwise direction.
P P
B A
A B
Negative couple Positive couple
P P
JEEMAIN.GURU
Note : A couple can not be balanced by a single force, but can be balanced by a couple of opposite sign.
TRIANGLE THEOREM OF COUPLES
If three forces acting on a body be represented in magnitude, direction
and line of action by the sides of triangle taken in order, then they are A P
equivalent to a couple whose moment is represented by twice the area of D E
P
triangle.
Q
Consider the force P along AE , Q along CA and R along AB. R
These forces are three concurrent forces acting at A and represented in
magnitude and direction by the sides BC , CA and AB of ∆ ABC . So, by
B L P C
the triangle law of forces, they are in equilibrium.
The remaining two forces P along AD and P along BC form a couple, whose moment is,
=m P= . AL BC. AL
1
Since, = ( BC. AL ) Area of the ∆ ABC
2
∴ Moment = BC = . AL 2 ( Area of ∆ ABC ) .
EQUILIBRIUM OF COPLANAR FORCES A
1. If three forces keep a body in equilibrium, they must be coplanar.
2. If three forces acting in one plane upon a rigid body keep it in α β
equilibrium, they must either meet in a point or be parallel.
3. When more than three forces acting on a rigid body, keep it in θ
equilibrium, then it is not necessary that they meet at a point. A system B C
m P n
of coplanar forces acting upon a rigid body will be in equilibrium if the
algebraic sum of their resolved parts in any two mutually perpendicular directions vanish separately, and
if the algebraic sum of their moments about any point in their plane is zero.
i.e., =X 0,= Y 0,= G 0 or= R 0,= G 0.
4. A system of coplanar forces acting upon a rigid body will be in equilibrium if the algebraic sum of the
moments of the forces about each of three non-collinear points is zero.
5. Trigonometrical theorem : If P is any point on the base BC of ∆ ABC such that BP : CP = m : n.
Then, (i) ( m + n ) cot θ = m cot α − n cot β ,
where ∠ BAP = α , ∠ CAP = β.
(ii) ( n + m ) cot θ =n cot B − m cot C.
VARIGNON’S THEOREM OF MOMENTS
The algebraic sum of moments of any number of coplanar forces about any point in their plane is equal to
the moment of their resultant about the same point.
FRICTION AND FORCE OF FRICTION
The property by virtue of which a resisting force is created between two rough bodies which prevents the
sliding of one body over the other is called the friction and this force which always acts in the direction
opposite to that in which the body has a tendency to slide or move is called forces of friction.
LIMITING FRICTION
When one body is just on the point of sliding on another body, the force of friction called into play attains
its maximum value and is called limiting friction and the equilibrium then is said to be limiting
equilibrium.
STATIC FRICTION
When a body in contact with another body is in any position of equilibrium but not limiting equilibrium ,
then the friction exerted is called static friction. Thus, static friction is less than the limiting friction.
JEEMAIN.GURU
Statics 251
DYNAMIC FRICTION
When motion ensues by one body sliding on the other, the friction exerted between the bodies is called
dynamic friction.
Dynamic Friction
we have R cos λ =N and Rsin λ =µ N ⇒ tan λ =µ Hence, the coefficient of friction is equal to the
tangent of the angle of friction.
CONE OF FRICTION N
The right cone described with its vertex at the point of contact of two rough bodies
and having the common normal at the point of contact as axis and the angle of λ R
friction as the semi–vertical angle , is called the cone of fraction.
O F
LEAST FORCE ON HORIZONTAL PLANE
The least force required to pull a body of weight W on the rough horizontal plane is W sin λ .
LEAST FORCE ON INCLINED PLANE
Let α be the inclination of rough inclined plane to the horizontal and λ , the angle of friction.
1. If α = λ , then the body is in limiting equilibrium and is just on the point of moving downwards.
2. If α < λ , then the least force required to pull a body of weight W down the plane is
W sin ( λ − α ) .
3. If α > λ , then the body cannot rest on the plane under its own weight and reaction of the plane.
So, the question of finding the least force does not arise.
Note : The least force required to pull a body of weight W up an inclined rough plane is W sin (α + λ ) .
CENTRE OF GRAVITY
The centre of gravity of a body or system of particles rigidly connected together, is that point through
which the line of action of the weight of the body always passes.
CENTRE OF GRAVITY OF A NUMBER OF PARTICLES ARRANGED IN A STRAIGHT LINE
If n particles of weights w1 , w2 , w3 ,..., wn be placed at points A1 , A2 , A3 ,..., An on the straight line OAn such
that the distance of these points from O are x1 , x2 ,..., xn respectively, Then, the distance of their centre of
gravity G (say) from O is given by
A1 A2 G An
O
x1
x2
xn
w1 w2
x wn
∑w x . ( w1 + w2 + ... + wn )
OG= x= i i
∑w i
Statics 253
Dynamics Chapter 37
DYNAMICS
Dynamics is that branch of mechanics which deals with the motion of bodies under the action of forces
Dynamics
Kinematics Kinetics
KINEMATICS
Kinematics is the study of geometry of motion without reference to the cause of motion. It deals with
displacement, velocity, acceleration etc. and we establish relation between these and time without reference to
the cause of motion.
KINETICS
Kinetics is the study of relationship between the forces and the resulting motion of bodies on which they act.
REST AND MOTION
A particle is said to be at rest if it does not change its position with respect to its surroundings and is said to be
in motion if it changes its position with respect to its surroundings.
PATH
The straight line or the curve along which an object moves is called its path. If the path is a straight line, the
object is said to have rectilinear motion and if the path is a curve the object is said to have
curvilinear motion.
SPEED
The speed of a moving point is the rate at which it describes its path. Speed is a scalar quantity.
AVERAGE SPEED
Average speed of a moving object in a time interval is defined as the distance travelled by the object during that
time interval divided by the time interval.
Thus, average speed in a time interval
= distance travelled in the given time interval
time interval
Y
DISPLACEMENT
The displacement of a moving point is the distance covered by it in a definite B
direction.
This is the shortest distance between initial and fixed point and its direction is Q
along the line from initial to final point.
A
X P
O P Q O X
A displacement has two fundamental characteristics-magnitude and direction. So it is a vector quantity.
VELOCITY
The rate of change of displacement of a moving particle is called its velocity.
Velocity has magnitude as well as direction. So it is a vector quantity.
In M.K.S. system, its magnitude is measured in m/sec and in C.G.S. system, its magnitude is measured in
cm/sec.
1×100 5
Velocity of 1 Km/hr = m/sec. = m/sec
60 × 60 18
JEEMAIN.GURU
Dynamics 255
22
Velocity of 1 mile/hr = ft/sec.
15
VELOCITY AT A POINT
Consider the motion of a particle along the straight line OX and let O be a point on it. Let P be the position
of the particle at a time t and let OP = x.
X
O x P δx Q
Let Q be the position of the particle at time ( t + δ t ) and let OQ= x + δ x. The displacement of the particle in
time ( t + δ t ) − t .i.e., δ t is PQ and PQ = OQ − OP = ( x + δ x) − x = δ x.
∴ Displacement of the particle in time δ t in δ x.
δx
Average velocity during the interval δ t = .
δt
δ x dx
∴ Velocity at time t is given= by, v lim = .
δ t →0 δ t dt
This is known as instantaneous velocity at t.
dx
Thus, v = .
dt
UNIFORM VELOCITY
A particle is said to move with uniform velocity if it moves in a constant direction and covers equal distances in
equal intervals of time, however small these intervals may be.
CONSTANT VELOCITY
A particle is said to move with constant velocity if it covers equal distances in equal intervals of time. Even if it
is not moving in a constant direction.
AVERAGE VELOCITY
The average velocity of a particle in a given interval of time is given by the ratio of the total displacement
undergone to the total time taken.
Total displacement
Thus, average velocity =
Total time
ACCLERATION AND RETARDATION
The acceleration of a particle is the rate of increase in its velocity while retardation is the rate of decrease in
velocity.
Thus, retardation is negative acceleration.
Unit of acceleration in C.G.S. system it is cm / sec 2 and in M.K.S. system it is m / sec 2 .
Acceleration has magnitude as well as direction. So, it is a vector quantity.
EXPRESSION FOR ACCLERATION
dv
If v is the velocity of a moving particle at a time t , then the acceleration at time t is given by a = .
dt
OTHER EXPRESSIONS FOR ACCLERATION
dv d dx d 2 x dx
1. = a = = 2
= v
dt dt dt dt dt
dv dv dx dv dx
2. = a == . v= v
dt dx dt dx dt
JEEMAIN.GURU
dv d 2 x dV
Thus, =
a = = 2
v .
dt dt dx
Note :
(a) If a = 0, the particle is said to be moving with uniform velocity.
(b) If a > 0 then v increases with time.
(c) If a < 0 then v decreases with time.
(d) If v and a have the same sign then the speed of the particle is increasing.
(e) If v and a have the opposite signs then the speed of the particle is decreasing.
UNIFORM ACCLERATION
A particle is said to be moving with uniform acceleration, if equal changes in velocity take place in equal
intervals of time, however small these interval may be.
EQUATIONS OF MOTION
If a particle moves along a straight line with initial velocity u and constant acceleration f , then the following
relations are known as equations of motion :
1
(i) v= u + ft (ii) =
s ut + ft 2 (iii) v 2 − u 2 = 2 fs
2
where ' v ' is the velocity after time t and s, the distance travelled in this time.
DISPLACEMENT IN THE nth SECOND
The distance travelled by a particle in the nth second of its motion in a straight line is given by
1
Sn = u + f ( 2n − 1)
2
MOTION UNDER GRAVITY
When a body is allowed to fall towards the earth, it will move vertically downwards with an acceleration which
is always the same at the same place on the earth but varies slightly from place to place. This acceleration is
called acceleration due to gravity. Its value in F.P.S. system is 32 ft / sec 2 , in C.G.S. system is 981
cms / sec 2 and in M.K.S. system is 9.8 m / sec 2 . It is always denoted by g.
The acceleration due to gravity always acts vertically downwards. If the body moves downwards, then the
effect of acceleration due to gravity is to increase its velocity. If the body moves upwards, then the effect of
acceleration due to gravity is that of retardation .i.e., the velocity of the body decreases. Hence, g is taken
positive for the downwards motion and negative for the upward motion of a body.
DOWNWARD MOTION
If a body is projected vertically downwards from a point at a height h above the earth’s surface with velocity
u , the equation of its motion are
1
(i) v= u + gt (ii) h= ut + gt 2
2
1
(iii) v= 2
u 2 + 2 gh (iv) hn = u + g ( 2n − 1)
2
Where hn denotes the distance covered in the nth second
UPWARD MOTION
When a body is projected vertically upwards with initial velocity u , then it moves in a straight line with
constant retardation g. So, the equations of motion in this case are :
1
(i) v= u − gt (ii) h= ut − gt 2
2
1
(iii) v=2
u 2 − 2 gh and (iv) hn = u − g ( 2n − 1) .
2
JEEMAIN.GURU
Dynamics 257
But v = u corresponds to the starting position, so we get v = −u Thus, the magnitude of the velocity
on reaching the ground is equal to the magnitude of velocity of projection and its direction is vertically
downwards.
PROJECTILE MOTION
A particle projected in any direction is called a projectile. Y
TRAJECTORY u
The path described by the particle is called its trajectory. A
In figure, the curve OAB is the trajectory of the projectile.
VELOCITY OF PROJECTION
The velocity with which the particle is projected is called the velocity of
projection.
In figure, u is the velocity of projection. α
ANGLE OF PROJECTION X
O N B
It is the angle which the direction of projection makes with the horizontal.
In figure, α is the angle of projection.
RANGE
The distance between the point of projection and the point where the projectile hits a given plane through the
point of projection is called its range.
When the given plane is horizontal, it is called the horizontal range.
In figure, OB is the horizontal range of the projectile.
TIME OF FLIGHT
The time taken between the instant of projection and the instant when the projectile meets a fixed plane through
the point of projection is called the time of flight. In figure, the time taken by the projectile in moving from O to
B is the time of flight.
GREATEST HEIGHT
The maximum height reached by the particle above the point of projection during its motion is called the
greatest height.
In figure, the distance AN is the greatest height.
SOME IMPORTANT RESULTS
Let a particle be projected from a point O with velocity u . Let the horizontal Y
and vertical lines OX and OY in the plane of motion be taken as axes of
v A
reference. Let α be the angle of projection Let P ( x, y ) be the position of the
particle at any time t and v be the velocity of the particle at P. Let the direction p ( x, y )
Dynamics 259
x = u cos α t
(i) parametric form 1 2
= y u sin α t − 2 gt
gx 2
(ii) General form = y x tan α − 2 .
2u cos 2 α
It is a parabola.
u 2 sin α .cos α u 2 sin 2 α
Its vertex is A , .
g 2g
u 2 sin 2 2α −u 2 cos 2α
Its focus is S ,
2 g 2g
2
u
Its directrix is y = .
2g
2u 2 cos 2 α
Latus rectum =
g
2
= [Horizontal component of velocity ] 2
g
It is the reciprocal of the numerical coefficient of x 2 in the equation of the trajectory.
2u sin α 2
3. Time of= flight = [Vertical component of velocity]
g g
u 2 sin 2α 2u cos α .u sin α
4. =
Horizontal range, R =
g g
2
= [horizontal component of velocity] × [vertical component of velocity]
g
u 2 sin 2 α
5. Greatest height attained =
2g
1
= [vertical component of velocity ] 2
2g
u2
6. Maximum horizontal range = .
g
π
7. Angle of projection for maximum horizontal range = .
4
8. For a given range, there are two directions of projection which are complements of each other .i.e., α and
900 − α .
u2
9. Locus of the focus of trajectory is x 2 + y 2 = 2 .
4g
2u 2 y
10. Locus of the vertex of trajectory is x 2 + 4 y 2 = .
g
u 2 sin 2 α − 2 gh
11. Velocity of the projectile at a height=
h u − 2 gh , Its direction is , θ = tan
2 −1
.
4 cos α
JEEMAIN.GURU
Dynamics 261
= mass × acceleration
This is called the fundamental equation of Dynamics.
Note : The equation P = mf is valid only when the mass of the body in motion remains constant where
as the force may be constant or variable. If the force is constant, the acceleration is uniform.
UNITS OF FORCE
1. Dyne: In C.G.S system, the unit of force is dyne. 1 dyne is the force which produces an acceleration of 1
cm / sec 2 in a mass of 1g.
2. Newton: In MKS system, the unit of force is Newton. 1N is the force which produces an acceleration of 1
m / sec 2 in a mass of 1 kg.
3. Poundal: In FPS system, the unit of force is poundal. 1 poundal is the force which produces an
acceleration of 1 ft / sec 2 in a mass of one pound.
Note : These units of force are called absolute units because their values are the same everywhere and do
not depend upon the value of ‘g’ which varies from place to place on the Earth’s surface.
GRAVITATIONAL UNITS
The units of force which depend on the value of ‘g’ are called gravitational units of force.
In CGS system, 1 gm. Wt. = g dynes =981 dynes.
In MKS system, 1 kg. wt.= g Newtons = 9.8 N.
In FPS system, 1 lb. wt. = g Poundal =32 poundals.
Note : While using the formula P = mf , P is always measured in absolute units .i.e., in poundals or dynes
or Newtons.
THIRD LAW OF MOTION
To every action, there is an equal and opposite reaction. This law asserts that forces occur in pairs. When a
book is placed on the table, the book presses the table with a certain force (which is action) and the table in turn
presses the book with an equal but opposite force (which is reaction)
MOTION OF A LIFT MOVING UPWARDS R
Suppose a body of mass m is carried by a lift upwards with an acceleration f .
The forces acting upon the body are:
(i) The normal reaction R of the plane of the lift acting vertically upwards. f
(ii) The weight mg of the body acting vertically downwards.
Since the lift is moving upwards, we have R > mg . Lift
∴ Total upward forces acting on the body= R − mg .
∴ Equation of motion is R − mg =
mf
mg
R m ( g + f ).
∴ =
Acceleration f = 1
( m − m2 ) g
m1 + m2
Tension in the string T
f T T f
2m m g
T= 1 2 T
m1 + m2
4m1m2 g m2 g
Force on the pulley= 2= T .
m1 + m2 m1 g
Dynamics 263
Conservation of Linear momentum : If the external force acting on a system of particles ( a body) is zero,
then the net linear momentum of the system ( a body) is conserved.
Thus in the absence of external forces.
Momentum before collision = Momentum after collision
Mathematically m1.u1 + m2 .u2 =m1.v1 + m2 .v2 .
where m1 = mass of the first body, m2 = mass of the second body,
u1 = initial velocity of the first body, u2 = initial velocity of the second body,
v1 = final velocity of the first body, v2 = final velocity of the second body.
Coefficient of Restitution : When the two bodies collide directly, then the ratio of relative velocity after
collision to the relative velocity before collision is a fixed quantity. The quantity
v −v
e= − 1 2
u1 − u2
is called the coefficient of restitution. Here u1 , u2 are initial velocity and v1 , v2 are final velocity of two bodies.
The value of e lies between 0 and 1. For a perfectly elastic collision e = 1 and for a perfectly inelastic collision
e =0.
Impact : When two bodies strike against each other, they are said to have an impact. It is of two kinds : Direct
and Oblique.
Newton’s experimental law of impact : It states that when two elastic bodies collide, their relative velocity
along the common before impact and is in opposite direction.
If u1 and u2 be the velocities of the two bodies before impact along the common normal at their point of
contact and v1 and v2 be their velocities after impact in the same direction, v1 − v2 =−e ( u1 − u2 )
Case I : If the two bodies move in direction shown in diagram given below, then
(a) v1 − v2 = −e ( u1 − u2 )
and (b) m1v1 + m2 v2 = m1u1 + m2u2
u1 m1 u2 m2 m1 v1 m2 v2
A B A′ B′
Case II : If the direction of motion of two bodies before and after the impact are as shown below, then by the
laws of direct impact, we have
(a) v1 − v2 =−e u1 − ( −u2 ) or v1 − v2 =−e ( u1 + u2 )
and (b) m1v1 + m2 v2 = m1u1 + m2 ( −u2 ) or m1v1 + m2 v2 = m1u1 − m2u2
u1 m1 u2 m2 m1 v1 m2 v2
A′ B′
Case III : If the two bodies move in directions as shown below, then by the law of direct impact, we have
(a) v1 − ( −v2 ) =−e −u1 − ( −u2 ) or v1 + v2 =−e ( u2 − u1 )
and (b) m1v1 + m2 ( −v2 ) = m1 ( −u1 ) + m2 ( −u2 ) or m1v1 − m2 v2 =
− ( m1u1 + m2u2 )
u1 m1 v1 u2 m2 v2 v1 v2
A B A′ B′
JEEMAIN.GURU
Direct impact of two smooth spheres : Two smooth spheres of masses m1 and
u1
m2 moving along their line of centres with velocities u1 and u2 (measured in the u2
same sense) impinge directly. To find their velocities immediately after impact, e
being the coefficient of restitution between them. m1 m2
Let v1 and v2 be the velocities of the two spheres immediately after impact,
measured along their line of centres in the same direction in which u1 and u2 are
measured. As the spheres are smooth, the impulsive action and reaction between v2
them will be along the common normal at the point of contact. From the principle v1
of conservation of momentum,
m1v1 + m2 v2 = m1u1 + m2u2 … (i)
Also from Newton’ experimental law of impact of two bodies, v2 − v1= e ( u1 − u2 ) … (ii)
Multiplying (ii) by m2 and subtracting from (i), we get
⇒ ( m1 + m2 ) v1 =
( m1 − em2 ) u1 + m2u2 (1 + e )
v = 1
( m − em2 ) u1 + m2u2 (1 + e )
∴ … (iii)
m1 + m2
1
Again, the only force acting on the spheres during the impact is along the line of centres. Hence the total
momentum in that direction is unaltered.
∴ m1v1 cos θ + m2 v2 cos φ = m1u1 cos α + m2u2 cos β … (iv)
The equations (i), (ii), (iii) and (iv) determine the unknown quantities.
Impact of a smooth sphere on a fixed smooth plane : Let a smooth sphere of mass m moving with velocity
u in a direction making an angle α with the vertical strike a fixed smooth horizontal plane and let v be the
velocity of the sphere at an angle θ to the vertical after impact.
Since, both the sphere and the plane are smooth, so there is no charge in velocity parallel to the horizontal
plane.
∴ v sin θ = u sin α … (i)
JEEMAIN.GURU
Dynamics 265
And by Newton’s law, along the normal CN , velocity of separation = e. (velocity of approach)
∴ v cos θ − 0 = eu cos α
v cos θ = eu cos α … (ii)
Dividing (i) by (ii), we get cot θ = e cot α
Particular case : If α = 0 then form (i), v sin θ = 0 ⇒ sin θ = 0
∴ θ =0; v ≠ 0 and from (ii) v = eu
Thus if a smooth sphere strikes a smooth horizontal plane normally, then it will rebound along the normal
with velocity, e times the velocity of impact i.e. velocity of rebound = e. (velocity before impact).
Rebounds of a particle on a smooth plane : If a smooth ball falls from a height h upon a fixed smooth
horizontal plane, and if e is the coefficient of restitution, then whole time before the rebounding ends
2h 1 + e
= ⋅ .
g 1− e
1 + e2
And the total distance described before finishing rebounding = h.
1 − e2
WORK, POWER AND ENERGY
1. Work : Work is said to be done by a force when its point of application undergoes a displacement. In
other word, when a body is displaced due to the action of a force, then the forces are said to do work.
Work is a scalar quantity.
Work done = force × displacement of body in the direction of force d
W = F.d
θ
W = F d cos θ , where θ is the angle between F and d . A
F
2. Power : The rate of doing work is called power. It is the amount of work that an agent is capable of doing
in a unit time. 1 watt = 107 ergs per sec = 1 joule per sec., 1 H .P. = 746 watt.
3. Energy : Energy of a body is its capacity to do work it is of two kinds :
(i) Kinetic energy : Kinetic energy is the capacity to do work by virtue of its motion and is
measured by the work which the body can do against any force applied to stop it, before the
velocity is destroyed.
1
K .E. = mv 2
2
(ii) Potential energy : The potential energy of a body is the capacity to do work by virtue of its
position of configuration. Potential energy of a particle having mass ‘ m ’ at height ‘ h ’ above the
surface of the earth.
P.E. = mgh. Also, K .E. + P.E. = Constant.
JEEMAIN.GURU
INTRODUCTION
Logic was extensively developed in Greece. In the middle ages the treatises of Aristotle concerning logic were
re-discovered. The axiomatic approach to logic was first proposed by George Boole. On this account logic
relative to mathematics is sometimes called Boolean logic. It is also called Mathematical logic or more recently
symbolic logic.
The word ‘Logic’ means “the science of reasoning”. It is the study and analysis of the nature of valid
arguments.
STATEMENTS OR PROPOSITIONS
Propositions : A statement or a proposition is an assertive (or declarative) sentence which is either true or false
but not both a true statement is called valid statement. If a statement is false, then it is called invalid statement.
Open statement : A declarative sentence containing variable ( s ) is an open statement if it becomes a statement
when the variable ( s ) is (are) replaced by some definite value ( s ).
Truth Set : The set of all those values of the variable ( s ) in an open statement for which it becomes a true
statement is called the truth set of the open statement.
Truth Value : The truth or falsity of a statement is called its truth value.
If a statement is true, then we say that its truth value is ‘True’ or ‘ T ’. On the other hand the truth value of
a false statement is ‘False’ or ‘ F ’.
Logical variables : In the study of logic, statements are represented by lower case letters such as p, q, r , s.
These letters are called logical variables.
For example, the statement ‘The sun is a star’ may be represented or denoted by p and we write
p : The sun is a star
Similarly, we may denote the statement
14 − 5 =−2.
Quantifiers : The symbol ∀ (stands for ‘for all’) and ∃ (stands for ‘there exists’) are known as quantifiers.
In other word, quantifiers are symbols used to denote a group of words or a phrase.
The symbols ∀ and ∃ are known as existential quantifiers.
An open sentence used with quantifiers always becomes a statement.
Quantifies statements : The statements containing quantifiers are known as quantified statements.
x 2 > 0.∀x ∈ R is a quantified statement. Its truth value is T .
USE OF VENN DIAGRAMS IN CHECKING TRUTH AND FALSITY OF STATEMENTS
Venn diagrams are used to represent truth and falsity of statements or propositions. For this, let us consider the
statement : “All teachers are scholars”. Let us assume that this statement is true. To represent the truth of the
above statement, we define the following sets
U
U = the set of all human beings S
T
S = the set of all scholars
And T = the set of all teachers x
Clearly, S ⊂ U and T ⊂ U
According to the above statement, if follows that T ⊂ S . Thus, the truth of the above statement can be
represented by the Venn diagram shown in
Now, if we consider the statement : “All the scholars are teachers”, It is evident from the Venn diagram
that there is a scholar x who is not a teacher. Therefore, the above statement is false and its truth value is ‘ F ’.
Thus, we can also check the truth and falsity of other statements which are connected to a given statement.
JEEMAIN.GURU
TYPES OF STATEMENTS
In Mathematical logic there are two types of statements.
1. Simple statements : Any statement or proposition whose truth value does not explicity depend on
another statement is said to be a simple statement.
In other words, a statement is said to be simple if it cannot be broken down into simpler statements, that
is, if it is not composed of simpler statements.
2. Compound statements : If a statement is combination of two or more simple statements, then it is said to
be a compound statement or a compound proposition.
TRUTH TABLES
Definition : A table that shows the relationship between the truth value of a compound statement S ( p, q, r ,...)
and the truth values of its sub-statement p, q, r ,... etc, is called the truth table of statement S .
BASIC LOGICAL CONNECTIVES OR LOGICAL OPERATORS
Definition : The phrases or words which connect simple statements are called logical connectives or sentential
connectives or simply connectives or logical operators.
List of some possible connectives, their symbols and the nature of the compound statement formed by them.
(i) Conjunction : Any two simple statements can be connected by the word “and” to form a compound
statement called the conjunction of the original statements.
Symbolically if p and q are two simple statements, then p ∧ q denotes the conjunction of p and q is
read as “ p and q ”.
(ii) Disjunction or alternation : Any two statements can be connected by the word “or” to form a compound
statement called the disjunction of the original statements.
Symbolically, if p and q are two simple statements, then p ∨ q denotes the disjunction, of p and q and
is read as “ p or q ”.
(iii) Negation : The denial of a statement p is called its negation, written as ~ p.
Note :
Negation is called a connective although it does not combine two or more statements. In fact, it only
modifies a statement.
(iv) Implication or conditional statements : Any two statements connected by the connective phrase “if ….
then” give rise to a compound statement which is known as an implication or a conditional statement.
If p and q are two statements forming the implication ‘if p then q ’, then we denote this implication by
" p ⇒ q " or " p → q ".
In the implication " p ⇒ q ", p is the antecedent and q is the consequent.
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p q p⇒q
T T T
T F F
F T T
F F T
p q p⇒q q⇒ p p⇔q= ( p ⇒ q) ∧ (q ⇒ p)
T T T T T
T F F T F
F T T F F
F F T T T
LOGICAL EQUIVALENCE
Logically equivalent statement : Two compound statements S1 ( p, q, r , ...) and S 2 ( p, q, r...) are said to be
logically equivalent, or simply equivalent if they have the same truth values for all logically possibilities.
If statements S1 ( p, q, r ...) and S 2 ( p, q, r ...) are logically equivalent, then we write
S1 ( p, q, r ,...) ≡ S 2 ( p, q, r ...)
It follows from the above definition that two statements S1 and S 2 are logically equivalent if they have
identical truth tables i.e., the entries in the last column of the truth tables are same.
NEGATION OF COMPOUND STATEMENTS
Writing the negation of compound statements having conjunction, disjunctions, implication, equivalence, etc, is
not very simple.
1. Negation of conjuntion : If p and q are two statements, then ~ ( p ∧ q ) ≡ ( ~ p ∨ ~ q )
2. Negation of disjuntion : If p and q are two statements, then ~ ( p ∨ q ) ≡ ( ~ p ∧ ~ q )
3. Negation of implication : If p and q are two statements, then ~ ( p ⇒ q ) ≡ ( p ∧ ~ q )
4. Negation of biconditional statement or equivalence :
If p and q are two statements, then ~ ( p ⇔ q ) ≡ ( p ∧ q ) ∨ ( q ∧ ~ p ) .
TAUTOLOGIES AND CONTRADICTIONS
Let p, q, r , ... be statements, then any statement involving p, q, r , ... and the logical connectives
∧, ∨, ~, ⇒, ⇔ is called a statement pattern or a Well Formed Formula (WFF).
For example
(i) p ∨ q
(ii) p ⇒ q
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(iii) (( p ∧ q ) ∨ r ) ⇒ ( s∧ ~ s )
(iv) ( p ⇒ q ) ⇔ ( ~ q ⇒~ p ) etc.
are statement patterns. A statement is also a statement pattern.
Thus, we can define statement pattern as follows.
Statement pattern : A compound statement with the repetitive use of the logical connectives is called a
statement pattern or a well-formed formula.
Tautology : A statement pattern is called a tautology, if it is always true, whatever may be the truth values of
constitute statements.
A tautology is called a theorem or a logically valid statement pattern. A tautology, contains only T in the
last column of its truth table.
Contradiction : A statement pattern is called a contradiction, if it is always false, whatever may the truth
values of its constitute statements.
In the last column of the truth table of contradiction there is always F .
The negation of a tautology is a contradiction and vice versa.
Arguments : An argument is the assertion that statement S, follows from the other statement, S1 , S 2 ....., S n .
We denote the argument by ( S1 , S 2 ,...., S n ; S ) . The statement S is called the conclusion and the statement
S1 , S 2 ,...., S n are called hypotheses (or premises).
Validity of an argument. An argument consisting of the hypotheses S1 , S 2 ...., S n and conclusion S is said to
be valid if S is true whenever all S1 , S 2 ,...., S n are true.
Working rule to test an argument for validity:
1. Construct a truth table showing the truth values of all the hypotheses and the conclusion.
2. Find the rows (called critical rows) in which all the hypotheses are true. In case there exists no critical
row, the argument is considered as invalid.
3. If in each critical row the conclusion is also true, then that argument is said to be valid otherwise the
argument is invalid. (i.e., if there is at least one critical row in which the conclusion is false, then the
argument is invalid).
ALGEBRA OF STATEMENTS
The following are some laws of algebra of statements.
1. Idempotent laws : For any statement p, we have
(a) p ∨ p ≡ p (b) p ∧ p ≡ p
2. Commutative laws : For any two statements p and q, we have
(a) p ∨ q ≡ q ∨ p (b) p ∧ p ≡ q ∧ p
3. Association laws : For any three statements p, q, r , we have
(a) ( p ∨ q) ∨ r ≡ p ∨ (q ∨ r ) (b) ( p ∧ q) ∧ r ≡ p ∧ (q ∧ r )
4. Distributive laws : For any three statements p, q, r we have
(a) p ∧ ( p ∨ q ) ≡ ( p ∧ q ) ∨ ( q ∧ r ) (b) p ∨ ( p ∧ q ) ≡ ( p ∨ q ) ∧ ( q ∨ r )
5. Demorgan’s laws : If p and q are two statements, then
(a) ~ ( p ∧ q ) ≡~ p∨ ~ q (b) ~ ( p ∨ q ) ≡~ p ∧ ~ q
6. Identity laws : If t and c denote a tautology and a contradiction respectively, then for any statement p,
we have
(a) p ∧ t ≡ p (b) p ∨ c ≡ p (c) p ∨ t ≡ t (d) p∧c ≡ c
7. Complement laws : For any statements p, we have
(a) p ∨ ~ p = t (b) p ∧ ~ p =
c (c) ~ t = c (d) ~c=t
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271
BOOLEAN ALGEBRA
Definition : Boolean algebra is a tool for studying and applying mathematical logic which was originated by
the English mathematician George Boole.
A non empty set B together with two operations denoted by ‘ ∨ ’ and ‘ ∧ ’ is said to be a boolean algebra
if the following axioms hold :
(i) For all x, y ∈ B
(a) x ∨ y ∈ B (Closure property for ∨ )
(b) x ∧ y ∈ B (Closure property for ∧ )
(ii) For all x, y ∈ B
(a) x ∨ y = y ∨ x (Commutative law for ∨ )
(b) x ∧ y = y ∧ x (Commutative law for ∧ )
(iii) For all x, y and z in B,
(a) ( x ∨ y ) ∨ z =x ∨ ( y ∨ z ) (Associative law of ∨ )
(b) ( x ∧ y ) ∧ z =x ∧ ( y ∧ z ) (Associative law of ∧ )
(iv) For all x, y and z in B,
(a) x ∨ ( y ∧ z ) = ( x ∨ y ) ∧ ( x ∨ z ) (Distributive law of ∨ over ∧ )
(b) x ∧ ( y ∨ z ) = ( x ∧ y ) ∨ ( x ∧ z ) (Distributive law of ∧ over ∨ )
(v) There exist elements denoted by 0 and 1 in B such that for all x ∈ B,
(a) x ∨ 0 = x ( 0 is identity for ∨ )
(b) x ∧ 1 =x (1 is identity for ∧ )
(vi) For each x ∈ B, there exists an element denoted by x′, called the complement or negation of x in
B such that
(a) x ∨ x′ = 1
(b) x ∧ x′ = 0 (Complement laws)
PRINCIPLE OF DUALITY
The dual of any statement in a boolean algebra B is the statement obtained by interchanging the operation
∨ and ∧, and simultaneously interchanging the elements 0 and 1 in the original statement.
In a boolean algebra, the zero element 0 and the unit element 1 are unique.
Let B be a boolean algebra. Then, for any x and y in B, we have
(a) x= ∨ x x ( a′ ) x=
∧x x
(b) ∨ 1 1 ( b′ ) x =
x= ∧0 0
(c) x ∨ ( x ∧ y) =
x ( c′ ) x ∧ ( x ∨ y ) =
x
(d) 0′ = 1 ( d ′) 1′ = 0
(e) ( x′)′ = x
(f) ( x ∨ y )′ =x′ ∧ y′ ( f )′ ( x ∧ y )′ =x′ ∨ y′
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IMPORTANT POINTS
We sometimes designate a Boolean algebra by ( B, '∨ ', '∧ ', '′', 0, 1) in order to emphasise its six parts; namely
the set B, the two binary operations ‘ ∨ ’ and ‘ ∧ ’, the complement operation ‘ ' ’ and the two special elements 0
and 1. These special elements are called the zero element and the unit element. However, it may be noted that
the symbols 0 and 1 do not necessarily represent the number zero and one.
Note :
For P ( A ) , the set of all subsets of a set A, the operations ∪ and ∩ play the roles of ‘ ∨ ’ and ‘ ∧ ’,
A and φ play the role of 1 and 0, and complementation plays the role of ‘ ' ’.
BOOLEAN EXPRESSION : Let ( B, '∨ ', '∧ ', '′', 0, 1) be a Boolean algebra and x1 , x2 ,..., xn are in B. Then,
Boolean expressions in x1 , x2 ,..., xn are defined recursively as follows:
(i) 0,1, x1 , x2 ,..., xn are all Boolean expressions.
(ii) If x and y are Boolean expression, then
(a) x′ (b) x ∨ y ( or x + y ) (c) x ∧ y ( or x. y ) (d) x′ ∨ ( y ∧ z )
are also Boolean expressions. Here x′ , x ∨ y , and x ∧ y are called monomials and the expression
x′ ∨ ( y ∧ z ) is called a polynomial.
We denote a Boolean expression X in x1 , x2 ,..., xn by X ( x1 , x2 ,..., xn ) .
BOOLEAN FUNCTIONS
Definition : Any expression like x ∧ x′, a ∧ b′, a ∧ ( b ∨ c′ ) ∨ ( a′ ∧ b′ ∧ c ) consisting of combinations by ∨
and ∧ of finite number of elements of a Boolean Algebra B is called a boolean function.
Let B = {a, b, c, ....} be a boolean algebra by a constant we mean any symbol as 0 and 1, which represents
a specified element of B.
By a variable we mean a symbol which represents a arbitrary element of B
If in the expression x′ ∨ ( y ∧ z ) we replace ∨ by ‘ + ’ and ∧ by ‘.’, we get x′ + y.z.
SWITCHING CIRCUITS
One of the major practical application of Boolean algebra is to the switching systems (an electrical network
consisting of switches) that involves two state devices. The simplest possible example of such a device is an
ordinary ON-OFF switch.
By a switch we mean a contact or a device in an electric circuit which lets (or does not let) the current to
flow through the circuit. The switch can assume two states ‘closed’ or ‘open’ (ON or OFF). In the first case the
current flows and in the second the current does not flow.
Symbols a, b, c, p, q, r , x, y, z , ..... etc. will denote switches in a circuit.
There are two basic ways in which switches are generally interconnected.
(i) Series (ii) Parallel
(i) Series : Two switches a, b are said to be connected ‘in series’ if the current can pass only when both
are in closed state and the current does not flow if any one or both are open. The following diagram will
show this circuit given by a ∧ b .
a b
a
(ii) Parallel : Two switches a, b are said to be connected ‘in parallel’
if current flows when any one or both are closed, and current does
not pass when both are open. The following diagram will represent
this circuit given by a ∨ b. b
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If two switches in a circuit be such that both are open (closed) simultaneously, we shall represent them by
the same letter. Again if two switches be such that one is open and the other is closed, we represent them
by a and a′.
The value of a close switch or when it is on is equal to 1 and when it is open or off is equal to 0.
An open switch r is indicated in the diagram as follows :
S1 r S2
A closed switch r is indicated in the diagram as follows :
S1 r S2
Boolean operations on switching circuits
(i) Boolean Multiplication : The two switches r and s in the series will perform the operation of Boolean
multiplication.
S1 r s
S2
Clearly, the current will not pass from point S1 to S 2 when either or both r , s are open . It will pass only
when both are closed.
r s r∧s
1 1 1
1 0 0
0 1 0
0 0 0
The operation is true only in one of the four cases i.e. when both the switches are closed.
(ii) Boolean Addition : In the case of an operation of addition the two switches will be in the parallel series
as shown below.
S1 S2
The circuit shows that the current will pass when either or both the switches are closed. It will not pass
only when both are open.
r s r∨s
1 1 1
1 0 1
0 1 1
0 0 0
The operation is not true only in one of the four cases i.e., when both r and s are open.
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r
S1 S2
q
S1 S2
s q
r r
S1 S2
s q
r u
q v
S1 S2
s w
SIMPLIFICATION OF CIRCUITS
Simplification of a circuit would normally mean the least complicated circuit with minimum cost and best
results. This would be governed by various factors like the cost of equipment, positioning and number of
switches, types of material used etc. For us, simplification of circuits would mean lesser number of switches
which we achieve by using different properties of Boolean algebra.
e.g., Consider the circuits given by ( a ∧ b ) ∨ ( a ∧ c )
This is represented by
a b
b
a c
a
Since ( a ∧ b ) ∨ ( a ∧ c ) =a ∧ ( b ∨ c ) c
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x1
AND x1 ∧ x2
x2
Input Output
x1 x2 x1 ∧ x2
1 1 1
1 0 0
0 1 0
0 0 0
Input Output
x1 x2 x1 ∨ x2
1 1 1
1 0 1
0 1 1
0 0 0
Input Output
x x′
1 0
0 1
COMBINATIONAL CIRCUIT
x1
AND
x2 s
OR NOT
x3
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In the above figure, output s is uniquely defined for each combination of inputs x1 , x2 and x3 . Such a circuit is
called a combinatorial circuit or combinational circuit.
x1
AND NOT s
x2
In the above figure, if= x1 1,= x2 0, then the inputs to the AND gate are 1 and 0 and so the output of the AND
gate is ‘0’ (Minimum of 1 and 0). This is the input of NOT gate which gives the output s = 1. But the diagram
states that x2 = s i.e. 0 = 1 , a contradiction.
∴ The output s is not uniquely defined. This type of circuit is not a combinatorial circuit.
Two combinatorial circuits : Circuit having inputs x1 , x2 , .... xn and a single output are said to be
combinatorial circuit if, the circuits receive the same input, they produce the same output i.e., if the input/output
tables are identical.
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277
a>0
X′ X
O
ax + b < 0 ax + b > 0
Y′
The graph of ax + b > 0 and ax + b < 0 are obtained by dividing xy - plane in two semi-planes by the
b
line x = − (which is parallel to y-axis). Similarly for cy + d > 0 and cy + d < 0.
a
Y
cy + d > 0 d
c>0 y= −
c
X′ X
cy + d < 0
Y′
(ii) Linear Inequation in two variables : General form of these inequations are
ax + by > c, ax + by < c. If any ordered pair ( x1 , y1 ) satisfies an inequation, then it is said to be a
solution of the inequation.
The graph of these inequations is given below ( for c > 0 ) :
Y
ax + by =
c
ax + by < c
X′ X
O
ax + by > c
Y′
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9. Convex set : In linear programming problems feasible solution is generally a polygon in first quadrant.
This polygon is convex. It means if two points of polygon are connected by a line, then the line must be
inside the polygon. For example,
A A
B B
(i) (ii) (iii) (iv)
Fig. (i) and (ii) are convex set while (iii) and (iv) are not convex set.
MATHEMATICAL FORMULATION OF A LINEAR PROGRAMMING PROBLEM
There are mainly four steps in the mathematical formulation of a linear programming problem, as mathematical
model. We will discuss formulation of those problems which involve only two variables.
1. Identify the decision variables and assign symbols x and y to them. These decision variables are those
quantities whose values we wish to determine.
2. Identify the set of constraints and express them as linear equations/inequations in terms of the decision
variables. These constraints are the given conditions.
3. Identify the objective function and express it as a linear function of decision variables. It may take the
form of maximizing profit or production or minimizing cost.
4. Add the non-negativity restrictions on the decision variables, as in the physical problems, negative values
of decision variables have no valid interpretation.
GRAPHICAL SOLUTION OF TWO VARIABLE LINEAR PROGRAMMING PROBLEM
There are two techniques of solving an L.P.P. by graphical method. These are :
(1) Corner point method (2) Iso-profit or Iso-cost method
1. Corner point method
Working Rule :
(i) Formulate mathematically the L.P.P.
(ii) Draw graph for every constraint.
(iii) Find the feasible solution region.
(iv) Find the coordinates of the vertices of feasible solution region.
(v) Calculate the value of objective function at these vertices.
(vi) Optimal value (minimum or maximum) is the required solution.
(vii) If there is no possibility to determine the point at which the suitable solution found, then the
solution of problem is unbounded.
(viii) If feasible region is empty, then there is no solution for the problem.
(ix) Nearer to the origin, the objective function is minimum and that of further from the origin, the
objective function is maximum.
2. Iso-profit or Iso-cost method :
Working Rule :
(i) Find the feasible region of the L.P.P.
(ii) Assign a constant value Z1 to Z and draw the corresponding line of the objective function.
(iii) Assign another value Z 2 to Z and draw the corresponding line of the objective function.
(iv) If Z1 < Z 2 , ( Z1 > Z 2 ) , then in case of maximization (minimization) move the line
1 1 corresponding to Z1 to the line P2 Q2 corresponding to Z 2 parallel to itself as far as possible,
PQ
until the farthest point within the feasible region is touched by this line. The coordinates of the
point give maximum (minimum) value of the objective function.
(v) The problem with more equations/inequations can be handled easily by this method.
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(vi) In case of unbounded region, it either finds an optimal solution or declares an unbounded solution.
Unbounded solutions are not considered optimal solution.
TO FIND THE VERTICES OF SIMPLE FEASIBLE REGION WITHOUT DRAWING A GRAPH
1. Bounded region : The region surrounded by the inequations ax + by ≤ m and cx + dy ≤ n in first quadrant
is called bounded region. It is of the form of triangle, quadrilateral or polygon. Change these inequations
into equations, then by putting x = 0 and y = 0, we get the solution. Also by solving the equations we get
the vertices of bounded region.
The maximum value of objective function lies at one of the vertex in bounded region.
2. Unbounded region : The region surrounded by the inequations ax + by ≥ m and cx + dy ≥ n in first
quadrant, is called unbounded region.
Change the inequation in equations and solve for x = 0 and y = 0. Thus we get the vertices of
feasible region.
The minimum value of objective function lies at one of the vertex in unbounded region
but there is no existence of maximum value.
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281
X X X
O O O
y = –1
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(0, 1)
y=1
X X X
O O O
y = –1
2 tanh x
= =
(i) sinh 2 x 2sinh x cosh x
1 − tanh 2 x
1 + tanh 2 x
(ii) cosh 2 x = cosh 2 x + sinh 2 x = 2 cosh 2 x − 1 =1 + 2sinh 2 x = 2
1 − tanh x
=
(iii) 2 cosh 2
x cosh 2 x + 1 (iv) =
2sinh 2
x cosh 2 x − 1
2 tanh x
(v) tanh 2 x = (vi) =
sinh 3 x 3sinh x + 4sinh 3 x
1 + tanh 2 x
3 tanh x + tanh 3 x
(vii) =
cosh 3 x 4 cosh 3 x − 3cosh x (viii) tanh 3 x =
1 + 3 tanh 2 x
6. (i) cosh x + sinh x = ex (ii) cosh x − sinh x = e− x
(iii) ( cosh x + sinh x ) =cosh nx + sinh nx
n
1 1
(i) Let sinh −1 x = y ⇒ x =
sinh y ⇒ cosech y = cosec −1
⇒ y=
x x
(ii) cosh y =1 + sinh 2 y =1 + x2
=y cosh −1 1 + x 2 ⇒ sinh −1 x = cosh −1 1 + x 2
sinh y sinh y x
(iii) =
tanh y = =
cosh y 1 + sinh 2 y 1 + x2
x x
∴ y= tanh −1 ⇒ sinh −1 x =tanh −1
1+ x 2
1 + x2
1 + sinh 2 y 1 + x2
(iv) =
coth y =
sinh y x
1 + x2 1 + x2
∴ coth −1
y= ⇒ sinh −1 x =
coth −1
x x
1 1 1
(v) =
sech y = =
cosh y 1 + sinh 2 y 1 + x2
1 1
y = sech −1 ⇒ sinh −1 x = sech −1
1+ x 2
1 + x2
1
(vi) Also, sinh −1 x = cosech −1
x
1
From the above, it is clear that coth −1 x = tanh −1
x
1 1
sech −1 x = cosh −1 ; cosech −1 = sinh −1
x x
Note : If x is real then all above six inverse functions are single valued.
1 1+ x 1 x +1
(iii) tanh −1 x
= log , x <1 =
(iv) coth −1 x
log , x >1
2 1− x 2 x −1
1 + 1 − x2 1 + 1 + x2
(v) sech −1 x log
= , 0 < x ≤1 =
(vi) cosec −1 x log , ( x ≠ 0)
x x
1
Note : Formulae for values of cosech −1 x, sech −1 x and coth −1 x may be obtained by replacing x by in the
x
values of sinh −1 x, cosh −1 x and tanh −1 x respectively.
Separation of inverse trigonometric and inverse hyperbolic functions
If sin (α + i β ) =+
x iy then (α + i β ) , is called the inverse sine of ( x + iy ) .
We can write it as, sin −1 ( x + iy )= α + i β .
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( x + iy )
1
( ) (1 − x )
2
1. cos −1 = cos −1 x 2 + y 2 − 2
+ y2 + 4x2 y 2
2
1
( ) (1 − x )
2
+ cosh −1 x 2 + y 2 + 2
+ y2 + 4x2 y 2
2
π π 1
( ) (
sin −1 ( x + iy ) = − cos −1 ( x + iy ) = − cos −1 x 2 + y 2 − 1 − x 2 + y 2 + 4 x 2 y 2 )
2
2.
2 2 2
i
( ) ( )
2
− cosh −1 x 2 + y 2 + 1 − x 2 + y 2 + 4x2 y 2
2
−1 i x 2 + (1 + y )2
1 −1 2x i −1 2y 1 2x
tan ( x + iy )
3. = −1
tan 2
=
+ tanh 2
tan 2
+ log 2
− − + + − − x + (1 − y )
2 2 2 2
2 1 x y 2 1 x y 2 1 x y 4
4. sin −1 ( cos θ + i sin
= θ ) cos −1 ( )
sin θ + i sinh −1( sin θ ) or cos ( sin θ ) + i log ( sin θ + 1 + sin θ )
−1
π i 1 + sin θ
6. tan −1 ( cos θ + i sin θ ) = + log , ( cos θ ) > 0
4 4 1 − sin θ
π 1 1 + sin θ
and tan −1 ( cos θ + i sin θ ) =−
+ log , ( cos θ ) < 0
4 4 1 − sin θ
Since, each inverse hyperbolic function can be expressed in terms of logarithmic function, therefore for
separation into real and imaginary parts of inverse hyperbolic function of complex quantities use the
appropriate method.
Note : Both inverse circular and inverse hyperbolic functions are many valued.
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INTRODUCTION
The limitations of analytical methods have led the engineers and scientists to evolve graphical and numerical
methods. The graphical methods, though simple, give results to a low degree of accuracy. Numerical methods
can, however, be derived which are more accurate.
SIGNIFICANT DIGITS AND ROUNDING OFF OF NUMBERS
1. Significant digits : The significant digits in a number are determined by the following rules :
(i) All non-zero digits in a number are significant.
(ii) All zeros between two non-zero digits are significant.
(iii) If a number having embedded decimal point ends with a non-zero or a sequences of zeros, then all
these zeros are significant digits.
(iv) All zeros preceding a non-zero digit are non-significant.
2. Rounding off of numbers : If a number is to be rounded off to n significant digits, then we follow the
following rules :
(i) Discard all digits to the right of the nth digit.
(ii) If the ( n + 1) digit is greater than 5 or it is 5 followed by a non-zero digit, then nth digit is
th
(iii) If the ( n + 1) digit is 5 and is followed by zero or zeros, then nth digit is increased by 1 if it is
th
y = f2 ( x ).
The abscissae of the points of intersection of these two graphs are X′ X
O
the real roots of f ( x ) = 0.
A real root of
Y′ f1 ( x ) = f 2 ( x )
Y
y = fx
2. Location Theorem : Let y = f ( x ) be a real-valued, continuous
function defined on [ a, b ].
f (a) A real root of
If f ( a ) and f ( b ) have opposite signs i.e., f ( a ) f ( b ) < 0, then f ( x) = 0
(iii) If f ( x1 ) = 0, then x1 is the required root or otherwise if f ( x1 ) is negative then root will be in
( x1 , b )
and if f ( x1 ) is positive then root will be in ( a, x1 ) .
(iv) Repeat it until you get the root nearest to the actual root.
2. Method of false position or Regula-Falsi method : This is the oldest method of finding the real root of
an equation f ( x ) = 0 and closely resembles the bisection method. Here we choose two points x0 and x1
such that f ( x0 ) and f ( x1 ) are of opposite signs i.e., the graph of y = f ( x ) crosses the x-axis between
these points. This indicates that a root lies between x0 and x1 , consequently f ( x0 ) f ( x1 ) < 0.
Equation of the chord joining the points
f ( x1 ) − f ( x0 )
− f ( x0 )
A x0 , f ( x0 ) and B x1 , f ( x1 ) is y= ( x − x0 ) … (i)
1 x −x
0
The method consists in replacing the curve AB by means of the chord AB and taking the point of
intersection of the chord with the x-axis as an approximation to the root. So the abscissa of the point
x1 − x0
where the chord cuts the x-axis ( y = 0 ) is given by x= x0 − f ( x0 ) … (ii)
f ( x1 ) − f ( x0 )
2
f ( xn )
In general, xn += xn − ,
f ′ ( xn )
1
y − f ( x0=
) f ′ ( x0 )( x − x0 ) . A1
f ( x0 ) A2
It cuts the x-axis at x= x0 − , which is a first
f ′ ( x0 )
1 X
O x2 x1
approximation to the root α . α
If A1 is the point corresponding to x1 on the curve, then the tangent at A1 will cut the x-axis of x2
which is nearer to α and is, therefore, a second approximation to the root. Repeating this process, we
approach to the root α quite rapidly. Hence the method consists in replacing the part of the curve
between the point A0 and the x-axis by means of the tangent to the curve at A0 .
NUMERICAL INTEGRATION
It is the process of computing the value of a definite integral when we are given a set of numerical values
of the integrand f ( x ) corresponding to some values of the independent variable x.
If I = ∫ y.dx. Then I represents the area of the region R under the curve y = f ( x ) between the
b
a
ordinates=x a=
, x b and the x-axis.
1. Trapezoidal rule : Let y = f ( x ) be a function defined on [ a, b ] which is divided into n equal sub-
intervals each of width h so that b − a = nh.
Let the values of f ( x ) for ( n + 1) equidistant arguments
x0 = a, x1 = x0 + h, x2 = x0 + 2h, ......, xn = x0 + nh = b be y0 , y1 , y2 , ....... yn respectively.
x0 + nh
∫ f ( x ) dx = ∫
b
Then y dx
a x0
1
= h ( y0 + yn ) + ( y1 + y2 + ..... + yn −1 )
2
h
= ( y0 + yn ) + 2 ( y1 + y2 + .... + yn −1 )
2
This rule is known as Trapezoidal rule.
The geometrical significance of this rule is that the curve y = f ( x ) is replaced by n straight lines
joining the points ( x0 , y0 ) and ( x1 , y1 ) ; ( x1 , y1 ) and ( x2 , y2 ) ;....... ; ( xn −1 , yn −1 ) and ( xn , yn ) . The area
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bounded by the curve y = f ( x ) . The ordinate x = x0 and x = xn and the x-axis, is then approximately
equivalent to the sum of the areas of the n trapeziums obtained.
2. Simpson’s one third rule : Let y = f ( x ) be a function defined on [ a, b ] which is divided into n (an
even number) equal parts each of width h, so that b − a =nh.
Suppose the function y = f ( x ) attains values y0 , y1 , y2 ,..... yn at n + 1 equidistant points
x=
0 a, x=1 x0 + h, x2 = x0 + 2h,....... xn = x0 + nh = b respectively. Then
x0 + nh h
∫a f ( x )= ( y0 + yn ) + 4 ( y1 + y3 + y5 + ... + yn −1 ) + 2 ( y2 + y4 + ... + yn − 2 )
b
dx ∫ y=
dx
x0 3
= (one-third of the distance between two consecutive ordinates)
× [(sum of the extreme ordinates) +4 (sum of odd ordinates) +2 (sum of even ordinates)]
This formula is known as Simpson’s one-third rule. Its geometric significance is that we replace the
n
graph of the given function by arcs of second degree polynomials, or parabolas with vertical axes. It is
2
to be note here that the interval [ a, b ] is divided into an even number of subinterval of equal width.
Simpson’s rule yield more accurate results than the trapezoidal rule. Small size of
interval gives more accuracy.
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2. Here I have proved the wrong statement that π = 2 2. Find the mistake in the following proof.
π
Proof : Consider the integral I = ∫ xf ( sin x ) dx, where f ( sin x ) is any function of sin x. In accordance with
0
a standard treatment, make the substitution x= π − x ', and then drop dashes.
I =∫ (π − x ) f {sin (π − x )} d ( −=
π
x) ∫ (π − x ) f ( sin x ) dx.
0
Thus
π 0
π π
Hence 2 ∫ xf ( sin x ) dx = π ∫ f ( sin x ) dx.
0 0
Take, in particular,
f ( u ) = u sin −1 ( u ) , so that f ( sin
= x ) sin
= x.x x sin x.
Then the relation is
π π
2 ∫ x 2 sin xdx = π ∫ x sin xdx.
0 0
π π
∫ = π − 4, ∫ x sin xdx = π .
2 2
But x sin xdx
0 0
Hence 2 (π 2=
− 4 ) π 2 , or
= π 2 8.
so that π = 2 2.
3. Here I have proved the wrong statement that every angle is multiple of two right angles. Find the mistake
in following proof.
Proof : Let θ be an angle (complex) satisfying the relation tan θ = i.
Then, if A is any angle,
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5. Here I have proved the wrong statement that 2 = 1. Find the mistake in the following proof.
f ( x ) dx f ( x ) dx − ∫ f ( x ) dx.
2 2 1
Then ∫=
1 ∫0 0
If we write x = 2 y in the first integral on the right, then
∫ f ( x ) dx 2=
∫ f ( 2 y ) dy 2∫ f ( 2 x ) dx,
2 1 1
=
0 0 0
on renaming the variable. Suppose, in particular, that the function f ( x ) is such that
1
f ( 2x) = f ( x)
2
for all values of x. Then
1
f ( x ) dx ( ) ∫0 f ( x ) dx= 0.
2 1 1
∫1 = 2∫
0 2
f x dx −
1 1
Now the relation f ( 2 x ) = f ( x ) is satisfied by the function f ( x ) =
2 x
2 dx
Hence ∫ = 0, so that log 2 = 0 or 2 = 1.
1 x
π
6. Here I have proved the wrong statement that, if f (θ ) is any function of θ , then ∫ f (θ ) cos θ dθ = 0.
0
Find the mistake in the following proof.
Proof : Substitute sin θ = t so that cos θ dθ = dt , and write f {sin −1 t} = g ( t )
∫ g ( t ) dt = 0.
0
The limits of integration are 0, 0 = and sin π 0. Hence the integral is
sin ce sin 0 0=
0
Corollary : The special case when f (θ ) = cos θ is of interest. Then the integral is
π
π 1 π 1 1 1 1
∫ θ dθ
cos=
2
∫ (1 + cos 2θ ) d=
θ θ + sin 2θ = π , hence π = 0.
0 2 0
2 4 0 2 2
7. Here I have proved the wrong statement that, 1 = 0. Find the mistake in the following proof.
Proof : S = 1 − 1 + 1 − 1 + 1 − 1 + ...
Then, grouping in pairs,
S = (1 − 1) + (1 − 1) + (1 − 1) + ... = 0 + 0 + 0 + ... = 0.
Also, grouping alternatively in pairs,
S =1 − (1 − 1) − (1 − 1) − (1 − 1) − ... =1 − 0 − 0 − 0... =1.
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Hence 1 = 0.
8. Here I have proved the wrong statement that, –1 is positive. Find the mistake in the following proof.
Proof : Let S =1 + 2 + 4 + 8 + 16 + 32 + ...
Then S is positive. Also, multiplying each side by 2,
2 S = 2 + 4 + 8 + 16 + 32 + ... = S − 1
Hence S = −1, so that – 1 is positive.
9. Here I have proved the wrong statement that, 0 is positive ( i.e., greater than zero).
Find the mistake in the following proof.
1 1 1 1 1 1 1
Proof : Write u =1 + + + + ... and v = + + + + ...
3 5 7 2 4 6 8
1 1 1
Then 2v =1 + + + + ... =u + v so that u − v = 0.
2 3 4
But, on subtracting corresponding terms,
1 1 1 1 1 1 1
u − v =1 − + − + − + − + ...
2 3 4 5 6 7 8
where each bracketed terms is greater than zero.
Thus u − v is greater that zero or 0 is greater that zero.
10. Here I have proved the wrong statement that −2π = 0. Find the mistake in the following proof.
2π i
Proof : As we know that e =cos 2π + i sin 2π =1 it follows that for any x
ei( x + 2 x ) ⇒ ( eix ) = ( )
i
e i ( x + 2π ) e −( x + 2 x )
i
=
eix e=
ix 2 xi
.e ⇒ e− x =
e − x .e −2 x ⇒ e −2 x = 1 ⇒ −2π = 0
⇒ e− x =
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