MIND MAP CLASS-12 MATHEMATICS

Page 1

5

An element a X is said to be invertible with respect to the operation *, if there exists an element b X such that a*b = b*a = e, b X

If a*(b c) = (a*b) (a*c) a, b, c A (distribution of * from left) and (b c)*a = (b*a) (c*a) a, b, c A (distribution of * from right)

e X, if a*e = e*a = a, a X

If a*(b*c) = (a*b)* c, a,b, c X

For X is any set, if a*b = b*a, a, b, X

A binary operation '*' on a set A is a function * : AĂ—A A denoted by a* b i.e. a,b, A.

d

A function is said to be invertible iff it is both one-one and onto i.e. bijective Domain f B A {a, b, c, d} 1 a odomain 2 b {1, 2, 3, 4, 5,} C 3 c 4 Range {1, 3, 5}

A function f : x y is invertible, if a function g : y x such that gof = Ix and fog = Iy. Then, g is the inverse of f and it is denoted by f –1

It is a function that assigns a each member of its domain

f

B 1 2 3

If distinct elements of A have distinct images inB.

A a b c

Set B is the codomain of R e.g. B ={4, 5, 6, 7}

Relations and Functions

Set of second components of ordered The composition of functions f : A B and pairs e.g. {4, 7, 5} is the range. g : B C is denoted by gof, and is defined as gof : A C given by gof(x) = g(f(x)) x Set of first components of ordered pairs e.g. Let A = {4, 5, 6} and B = {4, 5, 6, 7} then {4, 5, 6} is the domain. R = {(4, 4), (5, 7), (6, 5)} is a relation in AĂ—B

A a b c d e

Two or more elements in A have the same image in B

If the set B is not entirely used up i.e. there exists at least one element in B that does not have pre image in A.

f

A a b c

A a b c d e

1 2 3 4

B

f

f

B 5 7 9

1 2 3 4

B

R is an equivalence relation, just in case R is reflexive, symmetric and transitive. Let A = {1, 2, 3}, R = {(1, 2), (1, 1), (2, 1), (2, 2), (3, 3)}. Here R is reflexive, symmetric and transitive, so R is an equivalence relation.

A relation R : A A is transitive if aRb, bRc aRc a,b,c A

A relation R : A A is symmetric if aRb bRa a, b A

A relation R : A A is reflexive if aRa a A

A relation R : A A is universal if a R b a, b A, R = AĂ—A.

A relation R : A A is empty if a R b a, b A, R = AĂ—A. For e.g. R ={(a,b) : a = b2}, A = {1, 5, 10}

If every element of A is related to itself only i.e. IA = {(a, a) : a = A}

If every element of B has pre-image in A, i.e. the set B is entirely used up i.e. if f(A) = B

A Relation R in a set A is a subset of AĂ—A. i.e, R AĂ—A.

Chapter - 1 Relations and Functions

–1

–1

x sin

x sin

x cos

–1

1 x

x

2

2

–1

–1

–1

cos

1 – x tan

1 – x tan

2

x x

1

x tan

–1

y tan

x

1

–1

x

1

x– y 1 xy

,if xy � –1

–1

2

–1

–1

x cos

–1

–1

2

–1

x tan

–1

y tan

–1

1 – xy

x y

X’

Y'

– � – �� – ��

–1  1

Y Â?Â? Â? Â?Â?

Y'

=

(–1,0) O (1,0) Y'

��•

– ,

Y

Y=cosec–1x X

Inverse Trigonometric Function

–1

c

x

Y=sec–1x X

se

2

1 – y ],

,if xy 1

2

2

The expressions sin–1x, cos–1x, tan–1x, cot–1x, sec–1x, cosec–1x are called inverse trigonometric function

(iii) tan

2

y cos [ xy – 1 – x

2

–1

y sin [ x 1 – y y 1 – x ],

if – 1 x , y 1and x y 0

(ii)cos

x ) = x , for all x – , –1

1, , i.e., R – (–1, 1)

–1

x sin

if – 1 x , y 1and x y 1

–1

–π Ď€ (cosecθ) = θ, for allθ , , 2 2

2

x

1

1 x

1– x

1

–1

cosec (cosec

cosec θ 0

–1

–1

cosec

sec

–1 x

2

c se

x

2

cosec

1– x

1 x

–1

–1

cot

–1 cos x

–1 sin x

–1

sec

–1

–1

1 – y ],

2

t Relations of Inverse ortan imp e etric Functions Som onom Trig

x – tan

sec

cosec

2

2

–1

–1

y cos [ xy 1 – x

2

(i)sin

X'

y=

–1 x 1

y=

y=

x

co se c

y = sec

x

x

–1

–1

y

y = cot

tan

–1

2

–π 2

Ď€

ge Dom ai n

Ra n

Domain

Range

m a in Do

e Rang

ain Dom

–π π , i.e., , 2 2

ng e

Y'

Â

tan

x

–1

X'

–1

O

Y

– �� Y'

Â?Â?

os–1 x

­Â? Â?Â? – Â? – Â?Â? y=cot–1x X

Y

y=

y=c

2

–π

y

2

Ď€

, i.e., 0,π –

y 0 and 0 y

Ď€

Ď€ 2

–1

–1

O – �� Y'

Â?Â?

Y

O

Y

– �� Y'

– � – ��

X'

– � – ��

X'

Â?Â?

y=sin–1x X 1

Â? Â?Â?

y=sin–1x X 1

Â? Â?Â?

2 –π π Notethat y 0, i.e., , – 0 2 2

2

–π

i.e., x –1 or x –1

– x –1 and 1 x

2

Ď€

π and y < π 2 Note that y

0 y<

€‚

–π π , i.e., , 2 2

1 x, and – x – 1 i.e. x – 1 or x 1

0 y , i.e., [0, ]

2

Ď€

All real numbers (– < x < )

0 y , i.e., [0, ]

–1 x 1

All real numbers (– < x )

y=sin–1x X 1

Â? Â?Â?

The range of an inverse trigonometric function is the principle value branch and those values which lie in the principal value branch is called the principal value of inverse trigonometric function –1 y = sin x

x

•sec (sec ) = , for all [0, ], –– 2 –1 •sec (sec x) = x, for all x (– , –1) [1, ], i.e., R – (–1, 1)

–1

1– x

x

2

–1

2

–

•cot–1 (cot ) = , for all (0, ) •cos (cot–1x) = x, for all x R

(

cot

–1

1– x

(iii) tan

x

x – cos

–1

if – 1 x , y 1and x y

(ii)cos

–1

2

–

(

2

cot

cot

2

–1

y sin [ x 1 – y – y 1 – x ],

–1

– •tan–1 (tan ) = , for all –– , –– 2 2 –1 •tan (tan x) = x, for all x R

1 x

x

1– x

2

2

–1

–1

t y = co

•cos–1 (cos ) = , for all [0, ] •cos (cos–1x) = x, for all x [–1, 1]

–1

x

1– x

–π Ď€ • sin (sinθ) = θ, for all θ , 2 2 –1 •sin (sin x ) = x , for all x [–1, 1]

–1

–1

(vi) tan

(v) cos

(iv) sin

–1

x – sin

if – 1 x , y 1and x y 1

–1

ng Ra

(i)sin

x

1 (iii) tan–1x = cot–1 –– , x > 0 x

1 (i) sin–1x = cosec–1 –– , x [–1, 1] x 1 (ii) cos–1x = sec–1 –– , x [–1, 1] x

(i) sin–1(–x) = – sin–1 x, x [–1, 1] (ii) cos–1(–x) = – cos–1 x, x [–1, 1] (iii) tan–1(–x) = – tan–1 x, x R (iv) cot–1(–x) = – cot–1 x, x R (v) cosec–1(–x) = – cosec–1 x, x R– (–1, 1) (vi) sec–1(–x) = – sec–1 x, x R– (–1, 1) (vii) sin–1x + cos–1 x = –– , x [–1, 1] 2 (viii) sec–1x + cos–1 x = –– , x R– (–1, 1) 2 (ix) tan–1x + cot–1 x = –– , x R 2 –1 1 –1 –1 2 (x) 2 sin x sin [2 x 1 – x ], x 2 2 –1 –1 2 (xi) cos x = cos [2x – 1], 0 x 1 –1 2 = 2 – cos (2x –1), –1 x 0 –1 2 x , x –1, 1

sin 1 x 2 1– x 2 –1 x 0 (xii) 2 tan x cos –1 , 1 x 2 2x tan –1 , –1 x 1 1 x 2 – –1 –1 (xiii) sin x = sin [3x – 4x3], –– x –– 2 2 – (xiv) 3cos–1x = cos–1 [4x3 – 3x], –– x 2 3 1 1 –1 –1 3 x – x (xv) tan x = tan

x

, if – 1–3 x 2  3 3

e

Chapter - 2 Inverse Trigonometric Functions

y

–1

Ra

Addition to the elements of any row or column, the corresponding elements of any other row or column multiplied by any non-zero number r so ro w Interchange of any two

Multiplication of the elements of any row or column by a non -zero number

0 = –1

–1 =A 0

–2 10

0 1 –2 ,B= 0 4 6

a c

then, AB =

e.g. A =

ae ce

bg dg

f h

af + bh cf + dh

b e and B = g d

when Cik = aij . b jx f 1

n

Let A = [aij]mĂ—n and B = [bjk]nĂ—p , then C = AB = [Cik]mĂ—p

1 e.g. A = 2 2 A+B 2

e.g. A =

2 5

–4 6 –12 , 3A = 15 18 6

1 2 3 T then A –1 3 5 5 4 7

1 –1 5 2 3 4 , 3 5 7

Product of two matrices is zero matrix, as it is not necessary that one of the matrices is a zero matrix 0 –1 3 5 0 0 e. g . A , B= , AB = 0 2 4 0 0 0

Existence of non - zero matrices whose product is the zero matrix

A new matrix whose rows are columns of the original, it is denoted by AT or A' e. g . A

Matrices

e.g. 2 5 –4 1 3 , 2 4 1 2

ij 1 n

0 4 e.g. 1 , –2 3 1 5 2 1

atrix

ij

=0 i j

but

e. g .

ij

Allentries are zero 0 0 0 0 0 , e. g . 0 0 0 0 1 2 2 0 0 0 3 3

1 0 0 1 0 , e. g . 0 1 0 0 1 2 2 0 0 1 3 3

aij 0 when i j and a 1 when i j

2 0 0 e. g . 0 2 0 0 0 2

aij 0, i j and aij k (scalar), i j

2 0 0 e. g . 0 5 0 0 0 4 3 3

a

All non-diagonalentries are zero i.e.

1 2 5 e. g . 3 7 –4 0 –1 –2 3 3

A aij n n

Here m = n (no. of rows = no.of columns)

1 4 / 2 2 – 1 0 9 0 1 2 3 1 0 0

2 3 2 3

A = [aij]mĂ—n and B = [bij]mĂ—n are said to be equal, if aij = bij i, j

rm ala Sc

ij m 1

It is of theform A [ a ]

It is of theform A [ a ]

A matrix of order mĂ—n is an ordered rectangular array of numbers or functions having 'm' rows and 'n' columns. The matrix A = [aij]mĂ—n is given by a11 a12 ....a1n A a21 a22 ....a2n a m1 am 2 ....amn m n

If A = [aij]mĂ—n, then kA = [kaij]mĂ—n

–1 2 1 = A T , A = 1 2 0

1 2

Let A = [aij]mĂ—n and B = [bij]mĂ—n , than A+B = [aij + bij]

A

T

If A T

If A T = A, e.g. A =

2 1 0 = – A, A = 1

A square matrix [A]nĂ—n is said to be an invertible matrix if there exists another square matrix [B]nĂ—n such that AB = BA = I

Ci Cj

Ri Rj or

Ci Ci + kCj

Ri Ri + kRj or

Ci kCj

Ri kRj or

Chapter - 3 MAtrices

1

1

1

2

2

2

3

3

3

b

2

b

1

b

2

3

b

d1 If A 0, then solution of system of c x 1 linear equations is given by X =A–1B. c , x y and B d This method of solving a system of 2 2 c d linear equations is also known as matrix z 3 3 method.

2 a3

3

x d1 y d or AX = B 2 2 c z d3 3

1

c

c

1

b

b

3

21

32

a

22

12

a

a

33

a

23

13

a

a

ij

13

A

12

11

A

, then adj A A

where A ' s are coafactors of A.

31

a

If A a

11

a

Let A = [aij]nĂ—n be a square matrix of order n and Cij (or Aij) be the cofactor of aij in the determinant A . Then, the adjoint of A is defined as the transpose of the cofactor matrix and it is denoted by adj (A).

then A is known as singular matrix.

If A is a square matrix, and A = 0,

then A is known as non-singular matrix.

If A is a square matrix and A 0,

If A is a square matrix of order n×n, then (i) A(adj A) = (adj A)A = A ln (ii) adj (AT ) = (adj A)T (iii) adj A = A n–1, provided A 0

23

A

22

21

A

A

33

A

32

31

A

A ,

Let A and B be square matrices of same order, then (i) A square matrix is invertible if and only if it is non-singular (ii) (A–1)–1 = A (iii) (A')–1 = (A–1)' (iv) (AB)–1 = B–1 A–1 (v) A–1 = A –1 (vi) AA–1 = A–1 A = I (vii) (kA)–1 = –1 A–1, if k 0 k

2

x1 x2 x 3

y

y

3

2

1

y

1 1

1

x ( y – y3 ) x2 ( y3 – y1 ) x3 ( y1 – y2 ) 2 1 2

1

area of C

1

i.e.

x1 x2 x 3

3

2

1

y

y

y

1

1 0

1

Three points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear, if and only if the area of triangle formed by these three points is zero.

e. g .

a1 a2 p1 p1 u u 1 1

If each element of a row (or column of determinant is the sum of the two more terms, then the determinants can be expressed as the sum of the two or more determinants

A unique number (real or complex) can be associated, to every square matrix A =[aij] of order m, is called determinant of matrix A. It is denoted by A or det (A)

Determinants

A system of equations is consistent if it has one or more solutions while it is inconsistence if it has no solutions

Let the vertices of a ABC are A(x1, y1) B(x2, y2) and C(x3, y3). Then in the form of determinant,

Let A be a non-singular (i.e. A 0) square matrix. Then, a square matrix B, such that AB = BA = I is called of A inverse of A and it is denoted by A–1 A –1 i.e. A adj ( A). A

a1 Where A = a 2 a3

2

a1

It can be written in the matrix form as a

1

a x + b y + c z d , a x b y c z d and a x b y c z d

Let system of linear equation in three variables be

Chapter - 4 Determinants

v

q

b

21

23

33

33

a

23

13

a a

a1

1

k

v

q

c a 2 r p 2 w u2

v

q

b

c r w

2

a3

a1 e. g . a

2 a3

e. g . A a

a1 If each element of a row or a column of a determinant e. g . a2 is multiplied by a constant a3 k, then the value of new determinant are k times the value of the original determinant

If any two rows or columns are interchanged then the determina nt retains its absolute value, but its sign is changed.

b

31

23

11

23

12

and so on.

=M

a33

a

11

22

–a . a )

2

3

c

a1 1 c1 b

c

2

1

3

b

2

1

b

b

2

c 3

1

3

b

2

kb

1

b

2

3

b

b

b

3

3

b

a

c 1

1

2

c 3 c

kc

2

3

c

c

1 c k 2 1 c 3 c

ka1 , a 1 2 c a3 3 c

2 2

b

a

23 33

a

a

a3 , a 1 2 c a3 3 2

1

c

c

a1 k a 2 a3

3

b

2

b

1

b

b

3

b

2

1

c 3 c

0

c 2

a22 a 32

0 A 0

1

1

b

b

3

b

2

0

b

b

1

b

b

Mij, cofactor of a11 = (–1)1+1 M11 A11

The value of a determinant remains unchanged if its rows and columns are interchanged.

c a 1 r p 1 w u1

32

32

–a . a ) – a

a22 a23 , Minor of a = M = 12 12 a33 a32

32

a

22

12

a a

21

(a .a

33

, then minor of a

13

22

22

(a .a

21

– a .a

11

=a

22

If the minor Mij is multiplied by (–1)i+j, then we obtain the factor of aij. It is denoted by Aij or Cij, i.e., Aij = (–1)i+j

a22 a32

32

a

23

13

a

a

11

= a .a

a1 If each element of a e g . . A a row or column is 2 zero, then determinant a3 becomes zero

=

21 31

a

31

a

22

12

a

a

22

12

a a

–a . a ) a

11

a

33

31

If A = a

(a .a

a

21

11

a

21

11

a a

A = a

A =

(cot

–1

x) =

x) =

2

2

1+ x –1

1– x 2 1

1– x 2 1

1

; x ĂŽR

; x ĂŽR

; x <1

; x <1

d

1

xÂŽa

(v)

xÂŽa

(ii)

(i)

d

dx

dx d x

x

x

e

( a ) = a log a , a > 0

x

(e ) = e

X' 0 Y'

a

[a, f(a)]

c

dx

d

( x n ) = nx n –1

(iii)

dx

d

x =

2 x

1

d ĂŚ 1Ăś – n –1 (ii) ç á = – nx dx è x n ø

(i)

b–a

f(b) – f(a)

b

X

(vi)

(v)

(iv)

dx

dx d

dx d

2

x

(cosec x) = –cosec x cot x

(sec x) = sec x tan x

(cot x) = –cosec

(tan x) = sec 2 x

(cos x) = – sin x

X' Y'

a

c

f(c)

b

X

X' Y'

0

Y

a

xÂŽa+

b

X

x–5 x – 4 –1

when x š 5

d d d {f ( x ) . g ( x )}= f ( x ) {g ( x )}+ g ( x ) {f ( x )} dx dx dx (v) Quotient Rule : – d d g ( x) {f ( x )}– f ( x ) {g ( x )} d Ï f ( x) ß dx dx í ý= dx Î g ( x ) Þ {g ( x )}2

(iv) Product Rule : –

A function f(x) is said to be differentiable at a point x=a, if left hand derivative at (x=a) equals to Right hand derivative at (x=a) i.e. LHD at (x=a) = RHD (at x=a). f(a+h) – f(a) , h>0 where Right hand derivative, R f'(a) = lim– hŽ0 h f(a–h) – f(a) Left hand derivative, L f'(a) = lim+ , h>0 d hŽ0 –h (i) (k ) = 0 dx where, ' k ' is constant d d (ii) {k . f ( x )}= k {f ( x )} dx dx where, ' k ' is a constant d d d (iii) {f ( x ) ¹ g ( x )}= {f ( x )}¹ {g ( x )} dx dx dx

Let g ( x ) =

If the limit of the function exists but does not equal the value of the function at that point

Let f and g are two real functions, and also continuous at real number k, then, (i) f+g is continuous at x=k (ii) f–g is continuous at x=k (iii) f.g is continuous at x=k (iv) mf is continuous, where m is any constant f (v) – is continuous at x=k, where g(k) š 0 g

A function f(x) is continuous in an internal (a, b) if f(x) is continuous at every point in that internal and f is continuous in [a, b] if f is continuous in the interval (a, b) and also at the point a from the right and at the point b than the right x, x –4 Let f ( x ) x 2, 4 x 4 Is f continuous on the interval (2, 5) ?

xŽa–

0

Y

A function f(x) is said to be continuous at a point x=a, if (LHL)x=a, if (RHL)x=a = f(a) or lim f(x) = f(a) where, xÂŽa (LHL)x=a = lim im f(x) and (RHL)x=a = lim f(x)

If a function f(x) (i) is continuous in the closed interval [a, b] (ii) is differentiable in an open interval (a, b) (ii) f(a) = f(b), then there will be atleast one point c in the interval (a, b), such that f'(c)=0

(sin x) = cos x

dx d

dx d

dx d

(iii)

(ii)

(i )

d

[c, f(c)]

[b, f(b)]

Continuity Differentiability and Differentiation

f'(c)

Y

If a function f(x) is (i) continuous in the closed interval [a, b] (ii) differentiable in an open interval (a, b) Then, there will be atleast one point c. where a<c<b, such that

Differentiation (The process of finding derivative of a function)

xÂŽa

(ii) both are continuous at x=a (iii) both are differentiable at x=a (iv) f'(x) and g'(x) are defined at f(x) the point x=a, then lim = xŽa g(x) f'(x) lim provided that g'(a) š 0 xŽa g'(x) Above rule is also applicable, if lim xŽa f(x) = ¼ and lim g(x) = ¼

If f(x) and g(x) are two function of x such that (i) lim f(x) = lim g(x) = 0

dx (i) (log x ) = , x > 0 1+ x e d 1 dx x –1 (sec x ) = ; x >1 d 1 2 dx x x –1 (ii) (log x ) = d –1 a –1 dx x log a (vi) (cosec x ) = ; x >1 e 2 dx x x –1

(iv)

(tan

–1

x) =

x) =

–1

–1

(cos

(sin

dx d

dx d

dx d

(iii)

(ii)

(i)

d

y = f(t) and x = g(t), dy dy /dt f '(t ) \ = = dx dx /dt g '(t ) If y = f(x) and z = g(x) are two functions dy dy / dx f '( x ) = = dz dz / dx g '( x ) dy In such a case, to find ,differentiate dx both sides of the given relation w.r.t. x keeping in mind that the derivative of d f dy f(y) w.r.t. x is . dx dx d dy e.g. (i) (tan y ) = sec 2 y dx dx d 4 3 dy (ii) (y ) = 4y dx dx

Let y be a real valued function which is a composite of two function say y = f(u) , where u = g(x) dy dy du Then, = . = f '( a ) . g '( x ) dx du dx d ĂŠ Ăš i.e. ĂŤ f {g ( x )}Ăť = f ' {g ( x )}. g '( x ) dx The derivative of the composite function is the product of their derivatives.

(i) If y = u(x) v(x) w(x), then log y = log u(x)+ log v(x) + log w(x) u(x) (ii)If y = ,then log y=log u(x) – log v(x) v(x) (iii)If y = [f(x)]g(x), then log y = g(x) log [f(x)]

y = [f(x)]g(x) or y= f1(x) f2(x) f3(x).....or f1(x) f2(x) f3(x)..... y= g1(x) g2(x) g3(x).....

The functions, which can be evaluated by using logarithmic differentiation, are of type

can be differentiated with respect to x. dx 2 d Ì dy Ü d y It is writeen as of f "( x ). ç øá = è 2 dx dx dx 3 d Ì d2yÜ d y or f "'( x ) ç á = Again 3 dx è dx 2 ø dx

dy

Chapter - 5 Continuity Differentiability and Differentiation

1

(x – x )

dy dx ( x1 , y1 )

m

–1

where, m

1

y–y =

where, m

dy dx ( x1 , y1 )

y–y1= m (x–x1)

–1 dy dx ( x1 , y1 )

dy or f dx ( x1 , y1 )

An absolute maximum point is a point where the function obtains its greatest possible value

An absolute minimum point is a point where the function obtains its least possible value

The points at which a function changes its nature from decreasing to increasing or viceversa are called turning points.

1

'( x )

a Maxim and Minima m

Application of Derivatives

A line is perpendicular to the tangent at the point of contact is called the normal

A line which touches the curve at a single point is called tangent at a point.

Maxima and M i ni

f ( x x) f ( x) f '( x) x

If f '(x) < 0, then f is decreasing on the interval

If f ' (x) > 0, then f is increasing on the interval

A function which is either increasing or decreasing is a given interval I.

Let y = f (x) be any function of x. Let x be the small change in x and y be the corresponding change in y, i.e., y = f (x+ x) –f (x) then, dy = f'(x). dx dy or dy = x is a good approximation of y, when dx dx = x is relatively small, we denote it by dy y.

x 100 x is called percentage error in x.

x If x is an error in x, then x is called relative error in x.

The error x in x, is called the absolute error in x.

If x is an error in x, then

Local

A point C is the interior of the domain of f, is called local minima, if there exists >0 such that f(x) f (a), x (a– , a+ ).

The derivative of a function may be used to determine whether the function is increasing or decreasing on any intervals in its domain

A point C is the interior of the domain of f, is called local maxima, if there exists a >0 such that f(x) f (a), x (a– , a+ ).

dx dx , where 0 dt dt

If f' (x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima, such a point is called a point of inflection.

dy dy dx dt

If x = f (t) and y=g(t), then

the rate of change of y with respect to x at x = x0

( (

dy —– represents the rate of change of y dx dy with respect to x and —– ofx = x represents dx 0

Let y = f (x) be a function of x. Then,

Let f be a function on an open interval I and f be continuous at a critical point c in I. Then, (i) If f' (x) changes sign from positive to negative as a x increasing through c, then c is a point of local maxima. (ii) If f' (x) changes sign from positive to negative as x increasing through c, then c is a point of local minima. (iii) If f' (x) does not change sign as x increasing through c, then c is neither a point of local maxima nor a point of local minima. Such a point is called a point of inflection.

Let f (x) be a function defined on an interval I and c I. Let f be twice differentiable at c. Then, (i) x = c is a point of local maxima, if f '(c) = 0 and f"(c)<0 (ii) x = c is a point of local minima, if f '(c) = 0 and f "(c)>0 (iii) the test fails, if f '(c) = 0 and f"(c) = 0

Chapter - 6 Applications of Derivatives

a

(iv)

B

px q

2

A

B

2

A

B

C

A

B

2

C

x – a x

2

bx c

px 2 qx r

2

A

x – a

x bx c

2

Bx C ,

x – a x – b x – a x – a x – b

px 2 qx r

x – a x – b x – c x – a x – b x – c

px 2 qx r

x – a x – a x – a

, a b

f ( x)

ax

x

x

n 1

n 1

C, n – 1

x

dx log x C

C

C

log a

a

x

ax

a

e

C, n – 1

11. cosec x dx log cosec x – cot x C log tan

x C 2

Ï€ x 10. sec x dx log sec x tan x C log tan C 4 2

9. cot x dx log sin x C – log cosec x C

8. tan x dx – log cos x C log sec x C

7. cos x dx sin x C

6. sin x dx – cos x C

5.

1

4. a dx

x

3. e dx

n 1

f ( x)

n 1

. dx log f ( x ) C , f ( x ) 0

2. e dx e C

x

1. x dx

n

(ii)

f '( x )

(i) {f (x )} . f '(x ). dx =

n

such that dx = g'(t) dt.

f (x). dx = f{g (t)}. g' (t), if we substitute x=g(t),

where x 2 + bx + c cannot be factorised further

(v)

A

x – a x – b x – a x – b

(iii)

(ii)

(i)

px q

p(x) p(x) –––– can be integrated by expressing –––– q(x) q(x) as the sum of partial fractions of the following types :-

1

–1

x C

x C

x C

–1

–1

dx cos

dx sin

dx tan

1 – x2

–1

1 – x2

1

1 x 2 1 –1 dx cot x C 19. – 1 x 2

18.

17.

16.

2

15. cosec x dx – cot x C

14. sec x dx tan x C

2

13. cosec x cot x dx – cosec x C

12. sec x tan x dx sec x C

E = Exponential function

T = Trigonometric function

A = Algebraic function

L = Logarithmic function

I = Inverse trigonometric function

to its position in ILATE, where

Here, we can choose first function according

25.

24.

23.

22.

21.

20.

a x

2

1 2

x a

2

dx

x –a

2

dx

–1

–1

a

x

x C

C

–1

x C

a

1

tan

–1

x C a

log x x 2 a 2 C

log x x 2 – a 2 C

sin

dx cosec

dx sec

dx

2

2

a2 – x2

dx

x x –1

2

–1

x x –1

2

1

(iii) Again further use the integration by parts and simplify it

sec x 1 cosec x

dx

d

A

B ( x – a)2

C ( x – b)

x

1

dx log x C

[F( x )] f ( x ), then f ( x ). dx F( x ) C

Method:- To solve this problem, we use the concept of method of substitution. Here first of all put x = t2, then apply the formula i.e.

1 dx x x

( x – a)

x2 – a2

dx

a2 – x2

1

n

31. ax b dx

30. x 2 a 2 dx

29. a 2 – x 2 dx

1

2

n 1

a

n 1

2

2

2

2

2

2

a

2

a

2

a

–1

x a

C

log x x 2 a 2 C

sin

log x x 2 – a 2 C

C, n – 1

x a

2

a –x

2

1 ax b

2

x

2

x

2

2

x –a –

x a

C

C

a–x

a x

x–a

log

log

2a

x

2a

1

dx

(i) [f (x) g (x)] dx = f (x). dx g (x). dx (ii) For any real number k, k. f (x). dx k f (x) dx (iii) If functions f1, f2, f3,......fn, are functions and k1, k2,.......kn are real numbers, then [k1 f1 (x) + k2 f2 (x) +.........+ kn fn (x)] dx = k1 f1(x) dx + k2 f2 (x). dx +.........+ kn fn (x). dx

dx

Evalute

( x – a ) 2 ( x – b)

px 2 qx r

Method:- To solve this problem first we simplify the trigonometric function, then we use the method of substitution by putting sin x = t, then apply the property of partial fraction :

Evalute

28. x 2 – a 2 dx

27.

26.

d (ii) Use theformula: u.v dx u v. dx – dx (u ). v dx dx I II

Evaluate e–2x sinx dx Method:(i) Use integration by part method i.e. consider sin x as 1st function and e–2x as 2nd function constant

Indefinite Integrals

[f(x). g(x)] dx = f(x). g(x). dx – {f '(x). g(x). dx}. dx

For functions f(x) and g(x), we have

Chapter - 7 IntegralS (Topic 1)

b

b

b

a

a

a

f ( x ) dx f (t ) dt

n

or nh b – a

h 0

b–a

c

b

b

a

a

b

b

c

a

a

a

0

0

– a

(ix )

a

a

2

a+b

if f ( a x ) – f (b – x )

f ( x ) dx , if f ( a x ) f (b – x )

a

x

Then, A'( x ) f ( x ), for all x [ a , b ]

a

A ( x ) f ( x ) . dx

f (t ). dt .

Definite Integrals

Evaluate the integral, so obtained by usual method

g(a)

becomes

g (b)

Find the limits of integration in new system of variable i.e. the lower limit g (a) and the upper limit g (b) and the given integral

a 2 f ( x ) dx , if f ( x )is even i.e.f (– x ) f ( x ) f ( x ) dx [ f ( x ) f (– x )] dx 0 0 if f ( x ) is odd i.e. f (– x ) – f ( x ) 0,

b 0, or f ( x ) dx a 2

a 2a 2 f ( x ) dx , if f (2 a – x ) f ( x ) (viii ) f ( x ) dx 0 0 if f (2 a – x ) – f ( x ) 0,

0

(vii) f ( x ) dx f ( x ) dx f (2 a – x ) dx

2a

a

(vi) f ( x ) dx f ( a b – x ) dt

0

0

(v) f ( x ) dx f ( a – x ) dx

a

a

(iv) f ( x ) dx 0

a

a

(iii) f ( x ) dx f ( x ) dx f ( x ) dx , where a c b

b

a

(ii) f ( x ) dx – f ( x ) dx

(i)

where, h

a

f ( x ) . dx lim h f ( a ) f ( a h ) f ( a 2 h ) ....... f {a ( n – 1) h} ,

b

Chapter - 7 Integrals (Topic 2)

b

a

a

b

F(b ) – F( a )

[F(b ) C] – [F( a ) C]

F(b ) – F( a )

b

Calculate F(b ) – F( a )

a

Evaluate f ( x ) . dx [F( x )]a

a

There is no need to keep integration constant C b b f ( x ) . dx [F( x ) C]

f ( x ) . dx [F( x )]a F(b ) – F( a )

b

sin –1 x dx (1– x 2 )3/ 2

Method:- (i) Firstly, put sin–1 (x) = t and transform the limit of x in t. (ii) Further integrate and substitute the limits to get the required result

0

Evaluate

1/ 2

Consider a definite integral: b a f. [g (x)] . g' (x) dx Substitute g (x) = t g' (x) dx = dt

Va ria

De fin itio n

A Differential Equation of the form

dy +Py =Q, dx where P, Q are constants or functions or ‘x’ only is called a first order linear Differential Equation. ∫ P . dx = Q.e ∫ P . dx dx + c Its solution is ye ∫ dy eg: + 3y = 2 x has solution dx 3. dx ∫ 3. dx ∫ ye = ∫ 2 x.e dx + c ⇒ ye3 x = 2 ∫ xe3 x + c

- 9

Differential Equations

E

• A function which satisfies the given Differential Equation is called its solution. • The solution which contains as many arbitrary constants as the order of the Differential equations is called a general solution. • The solution free from arbitrary constants is called particular solution. • E.g. y = ex + 1 is a solution of y”–y’ = 0. Since y’= e x and y”= e x ⇒ y”–y’ = ex– ex=0

The highest exponent of the highest order derivative.  2 3 d y dy e.g. the degree of  2  + = 0 is three.  dx  dx Order and degree (if defined) of a Differential Equation are always positive integers.

e of Differential Equation Degre s

It is the order of the highest order derivative of the al dependent variable with respect to the independent i t en variable involved in the given differential equation. dy x d2y = e is one and order of 2 + x = 0 e.g. the order of dx dx is two.

To form a Differential Equation from a given function we differentiate the function successively as many times as the no. of arbitrary constants are given in the function. e.g. Let the function be y = ax + b, then we have to differentiate it two times, since there are 2 arbitrary constants a and b. ∴ y’= a ⇒ y”= 0. Thus, y”= 0 is the required Differential Equation.

A Differential Equation which can dy f ( x, y) = be expressed in the form dx g( x, y ) Homogenous dx g( x, y ) = or where, f(x,y) and Differential Equations dy f ( x, y ) ns g(x, y) are homogeneous functions of atio u same degree is called a homogenous Eq al differential equations. eg: (x2+xy)dy=(x2+y2)dx y To solve this, we substitute =v⇒y=vx. x

x ⇒ = c⇒x=cy is the solution. y

Integrating both sides, logx = logy + logc

dx dy = x y

Formation of Differential Equations

eg., ydx = xdy can be solved as

Li n e a rD iff e r en ti

An equation involving derivatives of the dependent variable with respect to independent dy d2y variable (variables) and constant. e.g.: x + xy 2 + k = 0. If there is only one independent dx dx dy d 2 y variable, then we call it as an ordinary differential equation. e.g.: – 2=0 + dx dx 2

Ordi nary Differentia l E q u Ord atio er ns of Eq Diff er ua t i on s

It is used to solve such an equation in which variables can be separated completely.

- 8 DIFFERENTIAL EQUATIONS Chapter Mind mapChapter : learning made simple

ble ara p Se ble od eth M tial ren e f if fD n o o ion uati t u q l So

i

n

2

i

n

i

n

.......

2

xn

pn

1

i

j

j =1

1

1

2

P(E ). P(A/E )

2

i j

n

n

j

1

j 1

j

= P ( E ). P(A/E )

1

n

2

n

where P(F) 0

or P (E F) = P(F). P(E/F)

where P(E) 0

P (E F) = P(E). P(F/E)

2

r......

n

n

P(X=r) =

Ďƒ npq

n

(i) P [(E F)/G] = P(E/G) + P(F/G) – P(E F)/G) (ii) P [(E F)/G] = P(E/G) + P(F/G), if E and F are mutually exclusive (iii) P [E'/G] = 1 – P(E/G) (iv) P(S/G) = P(G/G) = 1

P(F/E) =

Possible outcomes of a trial or experiment

= np

P(E F)

P(E)

P(F E)

where P(E) 0

P(F) where P(F) 0

P(E/F) =

Probability of Occurence of event E, when event F has already Occured

Probability

Cr p r.q n–r

2 = npq

If E and F are two events of sample space S and G is an event of S

P(E ). P(A/E ) + P(E ). P(A/E ) +.......+ P(E ).P(A/E )

1

n

P(E ). P(A/E )

P (A) = P(E ). P(A/E ) + P(E ). P(A/E ) + .......+ P(E ). P(A/E )

1

or P(E /A) =

i

P (E /A) =

P(E ). P(A/E )

S = {(H, T), (T, H), (H, H), (T, T)}. If X denotes the number of heads in two tosses, then X is a random variable X(H, T) = 1, X(H,H) = 2, X(T, H) = 1, variable X(T, T) = 0 the values of X are 0, 1 and 2

.......

x3

p3

x2

p2

(or X) x p i 1 i i

x1

2

p1

i 1 i

x 2 p – ( )2

X P(X)

Ďƒ = Variance

i 1 i

2

i 1 i i

x p – ( )

n

i 1 i

x 2 p – x p

n

Ďƒ = Var ( x )

2

1

C0p0qn nC1p1qn–1 nC2p2qn–2 ....nCrprqn–r nCnpn

n

0

Independent trails in which have only two outcomes, i.e., success or failure e.g., tossing an unbiased coin

P(X)

X

Chapter - 9 Probability

Number of favourablecases Totalnumber of equally likely cases

n

m

=

n

m

P(happening of atleast one of the events) = 1–P (happening of none of the events) = 1– [(1–p1). (1–p2). (1–p3)......(1–pn)]

P(A B C) = P(A)Ă—P(B)Ă—P(C)

P(A B) P(A)Ă—P(B)

(vii) P(B) = P(A B) + P(B A)

(vi) P(A) = P(A B) + P(A B)

(v) P(A B) = 1– P (A B)

(iv) P(A B) = 1– P(A B)

(iii) P[(A B) (A B)] = P(A) + P(B) – 2P(A B)

(ii) P(A B) = P(A) – P(A B)

(i) P(A B) = P(B) –P(A B)

P(A or Bor C) = P(A B C) = P(A) + P(B) + P(C)

P(A or B) = P(A B) = P(A) + P(B)

P(A B) – P(B C) – P(A C) – (A B C)

P(Either A or Bor C) = P(A) + P(B) + P(C) –

P(A or Bor both) = P(A) + P(B) – P(A B)

S im ilarly, P (A ) = 1 – P (A ) = 1 –

P (A ) + P (A ) = 1

If A is th e co m p lem en t even t o f A , th en

P (A) = p =

If the occurrence of one event affects the occurrence of the other events

If the occurrence of one event does not affect the occurrence of the other events

R(OR)

n

n

B( b )

where0

x2 – x1

( y 2 – y1 ) ( z 2 – z1 ) ,

thedistance between two points.

P1 P2

P1 P2 x 2 – x1 i ( y 2 – y1 ) j ( z 2 – z1 ) k 2 2 2

B( b )

m

A( a )

OR

mb – na m–n

R( OR )

A( a )

m

mb na OR m n

a . b a b cos

form aright handedsystem.

perpendicular to both a and b , such that a b and n

where, n isa unit vector

such that 0< <

aĂ—b a b sin n ,

k

a2

b3

1 a b 2

c1

c2

c3

A

D

AB AC AD

• xi , y j and z k arecalled vector components of r

a AB

B

C

B

The vector sum of two coinitials vector is given by diagonal of the parallelogram whose adjacent sides are given vectors, if AB , A C are given vectors, then

A

• Quantities which have only magnitude and no direction • Quantities which have magnitude as well as direction.

• x, y and z are called scalar components of r

< <

• i, j and k are unit vectors along the X-axis, Y-axis and Z-axis

Vectors

a b 1 1 a b c a2 b2 a3 b3

In OP ( or r )=xi y j z k

Scalar product of a and b c , ie., a. b c iscalled the scalar triple product of a b c

Chapter - 10 Vectors

If is multiplied to vector AB , then the magnitude is | times that of vector AB and direction remains same if opposite AB if

• Areaof triangle =

• Areaof parallelogram = a b

i j then a b a1 a2 b b 1 2

•If a and b aretwo vectors

<

x

r

, m=

r

y

, and n=

r

z

2

2

2

AB+BC+CA=0.

B

A

C

• The vector sum of the three sides of a triangle taken in order is 0 i.e.,

• In  ABC, we have AB BC CA, where CA is resultant vector.

(i) Zero vector (initial and terminal points coincide) (ii) Unit vector (magnitude is unity) (iii) Coinitial vectors (same initial points) (iv) Collinear vectors (parallel to the same line) (v) Equal vectors (same magnitude and direction) (vi) Negative of a vector (same magnitude and opposite direction)

1 2 3 , , and direction cosines are 14 14 14

Direction ratios are (1, 2, 3)

• Numbers l r ,m r and n r are proportional to the direction cosines are called direction ratios of vector r • For e.g: If AB = i+2 j +3 k , then r 1 4 9 14

r x y z

x l r , y m r and z n r

l=

• Direction cosines are generally donated by letters l, m and n i.e., l = cos , m = cos , n = cos . x y z • cos = , cos = and cos = r r r

is 2i 3 j 5 k and its magnitude is 2 2 3 2 5 2 38

Position vector of a point P(x,y,z) is xi yj zk and its magnitude is OP(r)= x 2 y 2 z 2 , where O(0,0,0) be the origin. Position vector of P (2, 3, 5)

• Magnitude of vector AB is AB • Any line which can be described by a vector called its direction vector.

z2 – z1 c1 c2

l

2

1

2

m

2

1

m

2

b

1

b

n

2

1

2

1

n

c

c

1 2

1 2

1 2

1

2

1 2

cos

or l l m m n n 0

1 2

If a a b b c c 0

or

a

1

a

l

b1 λ b 2

2 1

2 1

2

a b c

1

1

2 2

2 2

a b c

2 1

2

2 2 2

Two straight lines are coplanar, if they are either parallel or intersecting cosθ

Lines which are neither parallel nor intersecting.

cos l1 l2 m1 m2 n1 n2

Three dimensional Geometry Topic-1 : Line

2 – a . b 2 d – a – b

d ( a – x1 )2 (b – y1 )2 ( c – z1 )2 – ( a – x1 )l (b – y1 ) m ( c – z1 ) n 2

b ( a 2 – a1 ) d b

1

a . a b .b c . c

If b 1 . b 2 0

PQ ( x2 – x1 )2 ( y2 – y1 )2 ( z2 – z1 )2

(b1c2 – b2c1 )2 ( c1a2 – c2a1 )2 ( a1 b2 – a2b1 )2

y2 – y1 b1 b2

(b1 b 2 ) . ( a 2 – a1 ) d b1 b 2

d

x2 – x1 a1 a2

1

2

b .b

1 2

b .b

2

1

2

2

1

2

2

1

2

n

l

a 2

c 2

2

,

for direction ratios a , b and c. x–x y–y z–z 1 1 1 a b c

r = a +λ b

2

x–x y–y z–z 1 1 1 x2 – x y2 – y z2 – z 1 1 1

r a λ (b – a ), λ ε R

form

b a b c

2

for direction cosines l , m and n. x–x y–y z–z 1 1 1 n m l

,m

Car te si a n

2

2

a b c

2

a b c

2

• Direction Ratios of a line are the numbers which are proportional to the Direction Cosines of the line if l, m, n are the Direction Cosines and a, b, c are Direction Ratios of a line, then

where PQ ( x – x ) ( y – y ) ( z – z )

• Direction Cosines of a line are the cosines of the angles made by the line with the positive direction of the co-ordinate axes. • If l, m, n are the Direction Cosines of a line, 2 2 2 then l +m + n = 1. • Direction Cosines of a line joining P (x1, y1, z1) and Q (x2, y2, z2) x –x y –y z –z 2 1 2 1 2 1 are , , , PQ PQ PQ

Chapter - 11 Three Dimensional Geometry (Topic 1)

Lines

2

1

1

2

1

2

1

2

2

2

2

2

a b c

2

2

n1 n2

n1 . n2

2

2

1

2

y – y1

y 2 – y1

y3 – y1

x – x1

x2 – x1

x3 – x1

z3 – z1

z 2 – z1 0

z – z1

r – a . b – a × c – a = 0

1

2

2

1

a b c

2

2

a1 . a b1 .b c1 . c

cosθ =

a b c

cosθ =

2

aa1 bb1 cc1

a b c

2

2

1

a b c

2

b n

b.n

b n

a

x

b

y

c

z

PL

1

where l 2 m 2 n 2 1

lx my nz p

Three dimensional Geometry Topic-2 : Plane

– ax1 +by1 +cz1 + d where, r = al +bm+cn

x1 +lr , y1 +mr , z1 +nr d

1

1

1

2

1

2

1

2

1

2

1

2

2

1

2

a x b y c z d λ a x b y c z d 0

1

1

1

where, d a . n

b y – y c z – z 0

or r . n d

ax + by + cz + d = 0 where (a, b, c) are the direction ratios and d is any constant

or r . n a . n

a x – x1

r . n – d r . n – d 0

•Equation of xy – plane is z = 0 •Equation of yz – plane is x = 0 •Equation of zx – plane is y = 0

2

2

r – a . n 0

a b c 2

ax by cz d

of Planes ation Equ

r . n d

n

a.n – d

b.n

1

aa1 bb1 cc1

a2 b2 c2

or cosθ =

sinθ =

or cosθ

sin θ =

Chapter - 11 Three Dimensional Geometry (Topic 2)

X'

O

y=c

y=d

Y'

Y

X

x=f(y)

where f (y) g (y) in [d, c]

c

A = [ f ( y ) – g ( y )] dy

d

O Y'

)

a

x=b

X

b

a

b c

X'

O

Y

Y'

x=a

y=f(x)

x=c

Application of the Integrals

A = [ f x – g x ] dx , where f x g x in [ a , b ]

A = [ f ( x ) – g ( x )] dx [ g ( x ) – f ( x )] dx

c

x=a

y = g (x)

y = f (x)

The areas of the region enclosed between two curves y = f (x), y = g (x) and the lines x = a, x = b is given by

X'

Y

x=b

y=g(x)

X

Chapter - 12 Applications of the Integrals

If f (x) g (x) in [a, c] and f (x) g (x) in [c, b] where a < c < b, then the area of the region bounded by the curves is

The area of the region enclosed between curves x = f (y), x = g (y) and the lines y = c, y = d is given by

x=g(y

b

O

Y

Y'

x=a

a

dx

y

y = f(x)

x=b

d

X'

c

O

Y

Y'

y=c

y=d

c

x=f(y)

A = x dy or f ( y ) dy.

d

X

The area of the region bounded by the curve x = f (y) right to the Y – axis and between the lines y=c and y=d (d > c) is given by

X'

a

A = y dx or f ( x ) dx.

b

X

The area of the region bounded by the curve y = f (x) above X – axis and between the lines x=a and x=b (b > a) is given by

Total Revenue Number of units sold

p.x R —=——=p x x

P(x) = R(x) – C(x)

Difference between revenue function and cost function.

R(x) = C(x)

When total revenue equals to total cost incurred

d dc MC = — (C) = — dx dx

It is the rate of change of total cost with respect to x (output).

1 d (MC – AC) — (AC) = — 2 dx

If C is the total cost of producing and marketing x units of commodity, then

AR =

It is the revenue generated per unit of output sold

Applications of Calculus

R or R(x) = p.x

Demand (x)

• It represents the functional relationship between demand and price of a commodity • If p denotes the price per unit and x is the number of units demanded by a consumer at that price, then (i) demand function in explicit form, x = f (p) (ii)demand function in implicit form, f (x, p) = 0

Where, C = total cost, x = number of commodities

C AC = — x

It represents the cost per unit,

TC = TFC + TVC

Total cost = total fixed cost + total variable cost

TFC = TC, when x = 0

Total fixed cost,

The total cost function is expressed as C = c(x), explicitly and f (c, x) = 0 is the implicit form where x denotes the quantity produced of a certain commodity at total cost c.

Total revenue = selling price per unit of the commodity × quantity sold.

dR d MR = — =— (px) dx dx

It is the rate of change of total revenue with respect to quality sold.

Chapter - 13 Applications of Calculus

Price (p)

,y

n

y

,y

n

y

x2 –

xy –

n

x

2

x. y

n

ve

e Positiv

Negat i

b yx

2

b xy

r

tanθ

d1

Y (x , y ) 1 1

d2

dn

(xn, yn)

1– r 2 bxy byx

Linear Regression

Properties of Regression Coefficient and Lines of Regression

(iv) Intersection points, (X, Y)

(iii) P ( X, Y ) 1

(ii)

(i ) r bxy byx

(i)

O

Y

O

y = f(x)

Y (iii)

O

The best fitting line of y on x

X

Y (ii)

X

X

The main purpose of curve fitting is to estimate one of the variable from the other

•It is a basic and commonly used The straight line type of predictive analysis which gives the •These regression estimates are best fit in the least used to explain the relationship square sense to between one dependent variable the given sets of and one or more independent date variables To construct a scatter Y (1, 9) diagram, x variable is 9 (2, 8) taken on X-axis and y 8 (3, 7) 7 variable is taken on 6 (4, 6) (5, 5) Y-axis 5 (6, 4) 4 e.g. Let us draw a (7, 3) 3 scatter diagram for (8, 2) 2 observations 1 (2, 8), (3, 7), (4, 6), (5, 5), X O 1 2 3 4 5 6 7 8 (6, 4), (7, 3), (8, 2)

The line drawn •It is a form of mathematical by least squares regression analysis that finds x. y (x2, y2) method is called Line of best fit for a data set xy – X x line of best fit • Provides a visual n O r. 2 demonstration of y y 2 2 2 relationship between y2 – where, d d ..... d is minimum and it is best fitting curve. n n 1 2 the data points

x

y

Perfect Positive Relation X

X

Perfect Negative Relation

where r is correlation coefficient

b yx r .

bxy

O

Y

O

Y

n

x

n

x

bxy Ă— byx = r2, 0 bxy Ă— byx

where, x

y – y b yx ( x – x )

where, x

x – x bxy ( y – y )

Chapter - 14 Linear Regression

Any point outside the feasible region

Feasible region of system of linear inequalities is said to be bounded, if it can be enclosed within a circle.

Set of all points available in the feasible region.

When a feasible solution optimises the objective function

Find the optimal solution

Find the solutions and feasible region of inequation by mathematical method

Maximise or minimises Z = ax +by subject to the constraints a1x+b1y (< – or >) – C1 a2x+b2y (< or >) – – C2 x >0, y > – 0

Write the given LPP in mathematical form

Consider all constraints as linear equations

Draw the graph

Shade the common region i.e., find the feasible region If region is bounded, then the obtained value is required maximum or minimum value of the objective function

Otherwise, it is said to be unbounded.

The region other than the feasible region is called infeasible region.

Linear Programming

Check the feasible region is bounded or unbounded

Chapter - 15 Linear Programming

The common region determined by all the constraints including nonnegative constraints x > 0, y > 0 of a linear programming problem.

•Diet problems •Manufacturing problem •Transportation problems

• Problem which seeks to maximise or minimise a linear function • Linear programming problems are spacial type of optimisation problems.

Inequations or equations in the variables of a linear programming problem which describe the conditions under which the optimisation is to be accomplish

The linear function z = k1 x1 + k2 x2 +........+kn xn which is to be maximised or minimised

Maximum value of an objective function

A linear programming problem is a simple technique for finding the optimal value (maximum or minimum) value of a linear function of several variables.

If region is unbounded, then no need to go to next step

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