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1 KENDRIYA VIDYALAYA SANGATHAN RAIPUR REGION MODULES for CLASS-XII MATHEMATICS Session -2015-16

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1

KENDRIYA VIDYALAYA SANGATHAN RAIPUR REGION

MODULES

for

CLASS-XII

MATHEMATICS

Session -2015-16XII Class Chapter wise weightage

S.No Chapter Weightage

2

1 Relations and Functions

(i) Equivalence Relation

(ii) Invertible functions

(iii) Binary Operations

6M

2 Inverse Trigonometric functions 4M

3 Matrices

(i) Practical Problems

(ii) Equality of Matrices

4M

4 Determinants

(Matrix equation & Practical Problems)

6M

5 Continuity

Logarithmic Differentiation

Parametric form with second order derivative

8M

6 Applications of Derivatives

(i) Increasing & Decreasing functions

(ii) Tangents and Normal

4M

7 Integration 4M

8 Application of Integrals 6M

9 Differential Equations

(i) Variable separable differential equation

(ii) Homogeneous Differential equation

(iii) Linear Differential equation

4M

10 Vectors 4M

11 Three Dimensional Geometry

(i) Shortest distance

(ii) Plane passing through intersection of two planes

6M

12 Linear Programming problems 6M

13 Probability

(i) Bayes Theorem

(ii) Probability distribution

(iii) Multiplication rule

6M

MODULES FOR CLASS XII (M.L.L.)

3

Relations and FunctionsConcept: - Types of relations

A relation R in a set A is called

(i) Reflexive, if (a ,a )∈R for every a∈ A

(ii) Symmetric, if (a1 , a2 )∈ R⇒ (a2 , a1 )∈ R , for all a1 , a2∈ A

(iii) Transitive, if (a1 , a2 )∈ R and (a2 , a3 )∈R implies that (a2 , a3 )∈R for all a1 , a2 , a3∈ A A relation R in a set A is said to be an equivalence relation if R is reflexive,Symmetric and Transitive.

PRACTICE PROBLEMSLEVEL –I

Let T be the set of all triangles in plane with R a relation in T given by R={(T1,T2):T1

is congruent to T2}. Shoe that R is an equivalence relation. Let L be the set of all lines in a plane and R be the relation in L defined as

R={(L1,L2):L1 is a perpendicular to L2}. Show that R is symmetric but neither

Show that the relation R in the set R of real numbers, defined as R={(a ,b) :a≤b2} is neither reflexive nor symmetric nor transitive.

LEVEL-II Show that the relation R in the set Z of integers given by R={(a, b):2 divides a-b} is

an equivalence relation

Show that the relation R in the set A={1,2,3,4,5}given by R = {(a,b):|a−b|is even }, is an equivalence relation. Show that all the element of {1,3,5} are related to each other and all the element of {2,4} are related to each other. But no element of {1,3,5} is related to any element of {2,4}.

Show that each of the relation R in the set A={x∈Z :0≤x≤12}, given by

(i) R=¿¿is a multiple of 4}

(ii)R={(a , b :a=b} is an equivalence relation. Find the set of all element related to 1 in each case

LEVEL-III Show that the relation R in the set A of points in a plane given by R = {(P,Q):

distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points P≠(0,0) is the circle passing through P with origin as centre.

Show that the relation R defined in the set A of all triangles as R = {(T1,T2):T1 is similar to T2}, is equivalence relation. Consider three right angle triangle T1 with sides 3,4,5,T2 with sides 5,12,13 and T3 with sides 6,8,10. Which triangle among T1,T2 and T3 are related?

Show that the relation R defined in the set A of all polygon as R = {(P1,P2):P1 and P2

have same number of sides}, is an equivalence relation. What is the set of all element in A related to the right angle triangle T with sides 3,4,and 5?

4

Let A={1,2,3,4…..,9} and R is the relation on AXA defined by (a,b)R(c,d) if a+d=b+c for (a,b),(c,d) in AXA. Prove that R is an equivalence relation and also

obtain the equivalent class [ (2,5 ) ] .Concept: One-one (injective),Onto (surjective) and bijective

Injective: - A function f : X→Y is define to be injective, if the image of distinct

element of X under f are distinct. For every x1 , x2∈ X , f (x1)= f ( x2 )⇒ x1=x2

Surjective:- A function f : X→Y is said to onto (surjective) if every element of Y is

the image of some element of X under f ,i.e for every y∈Y there exists an element x in X

such that f ( x )= y

Bijective: A function f : X→Y is said to be bijective if f is one –one and onto PRACTICE PROBLEMS LEVEL – I

Prove that the function f : R→R ,given by f ( x )=2 x is one - one and onto

Show that the function f : N→N given byf (1 )=f (2 )=1 and f ( x )=x−1for every x>2 is onto but not one-one

Find the number of all one –one function from set A = {1,2,3} to itself. LEVEL – II

Let A=R−{3}and B=R−{1}.consider the function f : A→B defined

by f ( x )=( x−2

x−3 ) . Is f one-one and onto? Justify your answer.

Let A = {-1,0,1,2}, B = {-4,-2,0,2} and f , g : A→B be a function defined by

f ( x )=x2−x , x∈ A and g( x )=2|x−1

2|−1 , x∈ A

. Are f and g equal? Justify your answer

LEVEL-III

Show that the function f : R→R given by f ( x )=x3is injective

Show that f : N→N given by f ( x )={x+1

x−1, x∈odd, x∈ even is both one - one and onto.

Concept :- Composition of function and Inverse of Function

Let f : A→B and g :B→C be two function .then the composition of f andg , denoted by

gof, is defined as the function gof: A→C given by gof ( x )=g( f (x ) ) ,∀ x∈ A

A function f : X→Y is defined to be invertible. If there exists a function g :Y →X Such

that gof=I X and fog=IY .The function g is called the inverse of f . If f is invertible, then f must be one one and onto.

5

PRACTICE PROBLEMS LEVEL – I

If f : R→R be given by f ( x )=(3−x3 )13

, then fof ( x ) is

Consider f : N→N , g: N →N and h : N→R defined asf ( x )=2 x , g( x )=3 y+4

and h( z )= sin z,∀ x , y and z in N. Show that ho (gof )=(hog)of .

Let f : R−{−4

3 }→R be a function defined as

f ( x )= 4 x3 x+4 . The show that

inverse of f is the map g :Range f →R−{−4

3 } is g( x )= 4 x

4−3 x . LEVEL – II

Consider f : R+→[4 ,∞)given byf ( x )=x2+3 . Show that f is invertible with the

inverse f−1

of given byf−1( y )=√ y−4 , where R+ is the set of all non-negative real number.

Let f : R→R be defined asf ( x )=10 x+7 .Find the function g : R→R such that gof= fog=1R .

Show that the function f : R→R defined by f ( x )= x

x2+1,∀ x∈R

is neither one-one nor onto. LEVEL – III

Consider f : R+→[−5 ,∞) given by f ( x )=9 x2+6 x−5 . Show that if f is

invertible with f−1( x )=( (√ y+6 )−1

3 ).

Let f : N→R be a function defined as f ( x )=4 x2+12 x+15 . Show that f : N→ S , where, S is the range of f , is invertible. Find the inverse of f .

Concept :- Binary Operations

A binary operation * on the set X is called commutative, if a∗b=b∗a , for every a ,b∈ X .

A binary operation *: A×A→A is said to be associative if

(a∗b )∗c=a∗(b∗c ) ,∀ a ,b , c ,∈ A .

Given a Binary operation *: A×A→A , an element e∈ A , if it exist, is called identity

for the operation *, if a∗e=a=e∗a ,∀a∈ A .

6

Given a binary operation *: A×A→A with the identity element e in A, an element a∈ A is said to be invertible with respect to the operation *,If there exist an element b in A such that a∗b=e=b∗a and b is called the inverse of a and is donated by a-1.

PRACTICE PROBLEMS LEVEL-I

Consider the binary operation ¿ on the set{1,2,3,4,5} defined by a∧b=min {a , b}. Write the operation table of the operation ¿ .

Let * be the binary operation on N given by a*b=L.C.M of a and b. Find (i)5*7,20*16

Let * be a binary operation on the set Q of rational numbers as follows:- (i)a*b=a-b (ii)a*b=a2+b2 (iii)a * b=a + ab

(iv)a * b = (a-b)2 (v)a * b =ab4 (vi)a * b = ab2

Find which of binary operation are commutative and which are associative? LEVEL – II

Determine which of the following binary operation on the set R are associative and which are commutative.

(a)a∗b=1∀a , b∈R (b)a∗b=(a+b)

2∀a , b∈R

Let A=N×N and * be the binary operation on A defined by (a , b )∗(c , d )=(a+b ,b+d ) .Show that * is commutative and associative. Find the identity element for *on A ,if any

Consider a binary operation *on N defined as a∗b=a3+b3.Choose the correct

answer. (A)Is * both associative and commutative? (B)Is * commutative but not associative? (C)Is * associative but not commutative? (D)Is * neither commutative nor associative

LEVEL – III

Consider the binary operation * : R×R→R and R×R→R defined as a∗b=|a−b|and a ob=a ,∀ a ,b∈ R .Show that * is commutative but not associative, o is

associative but not commutative. Further, show that ∀ a , b , c∈R , a∗( boc )=( a∗b) . [If it is so, we say that the operation * distributives over the operation o].Do o distributive over *? Justify your answer.

Given a non-empty set X, let *:P (X )×P(X )→P (X )be defined as A∗B=( A−B)∪(B−A ) ,∀ A , B∈P( X ) .Show that the empty setφ is the identify for

the operation * and all the element A of P(X) are invertible with A−1=A .

7

Define a binary operation * on the set {0,1,2,3,4,5,6} as

a∗b={a+ba+b−7

ifif

a+b<7a+b≥7

Write the operation table of the operation * and prove that zero is the identity for this

operation and each element a≠0 of the set of the invertible with 7-a being the inverse of a.

INVERSE TRIGONOMETRIC FUNCTIONSIntroduction

Principal Value Branch Table

Functions DomainPrincipal Value

Branches

 y = sin-1 x [-1, 1] [−π2

, π2 ]

 y = cos-1 x [-1, 1] [0 , π ]

 y = cosec-1 x  R – (-1, 1) [−π2

, π2 ]−{0 }

 y = sec-1 x  R – (-1, 1) [0 , π ]−{π

2 } y = tan-1 x  R (− π

2, π

2 ) y = cot-1 x  R (0 , π )

Properties of Inverse Trigonometric Functions:For suitable Values of domain, we have:1. (a) y = sin-1 x x = sin y (b) x = sin y y = sin-1 x2. (a) sin (sin-1 x) = x (b) sin-1 (sin x) = x

3. (a) sin-1 ( 1

x ) = cosec-1 x (b) cos-1 ( 1

x ) = sec-1 x(c) tan-1 ( 1

x ) = cot-1 x4. (a) cos-1 (-x) = - cos-1 x (b) cot-1 (-x) = - cot-1 x (c) sec-1 (-x) = - sec-1 x5. (a) sin-1 (-x) = - sin-1 x (b) tan-1 (-x) = - tan-1 x(c) cosec-1 (-x) = - cosec-1 x

6. (a) sin-1x + cos-1x =

π2 (b) tan-1x + cot-1x =

π2 (c) cosec-1x + sec-1x =

π2

7. (a) tan−1 x+ tan−1 y=tan−1 x+ y

1−xy (b) tan−1 x− tan−1 y=tan−1 x− y

1+xy

8.2 tan−1 x=sin−1 2x

1+x2=cos−1 1−x2

1+x2= tan−1 2 x1−x2

8

9. (a) sin-1x + sin-1y = sin-1 (x √1− y2+ y √1−x2)

(b) sin-1x - sin-1y = sin-1 (x √1− y2− y √1−x2 )

10. (a) cos-1x + cos -1y = cos -1 ( xy−√ (1−x2 ) (1− y2))

(b) cos -1x - cos -1y = cos -1 ( xy+√(1−x2) (1− y2 ))

Important Solved Problems

1. Write the principal value of tan−1√3−sec−1 (−2 ) .Solution:

tan−1√3−sec−1 (−2 )= tan−1 ( tan π3 )−sec−1 (−sec π

3 )¿ π

3−sec−1 [sec(π−π

3 )]=π3−(π−π

3 )=−π3

2. Using principal value , evaluate the following:

sin−1 (sin 3π5 )

Solution:

sin−1(sin 3 π5 )≠3 π

5as3 π

5∉[−π

2, π2 ]

sin−1(sin 3 π5 )=sin−1 [sin(π−2 π

5 )]=sin−1(sin 2 π5 )=2π

5

3. If tan−1√3+cot−1 x= π

2 , find xSolution:

tan−1√3+cot−1 x= π2

cot−1 x=π2−tan−1√3

tan-1x + cot-1x =

π2

= cot−1 √3⇒ x=√3

4. Prove that:cos−1(12

13 )+sin−1( 35 )=sin−1(56

65 )Solution:

9

cos−1(1213 )=x⇒ cos x=12

13

sin x=√1−cos2 x=√1−(1213 )

2=5

13⇒and sin−1(35 )= y⇒ sin y=3

5

cos y=√1−sin2 y=√1−(35 )2=4

5sin (x + y) = sin x cos y + cos x sin y

=

(513 )(45 )+(1213 )(35 )=56

65

⇒ x+ y=sin−1 (5665 )⇒cos−1(12

13 )+sin−1(35 )=sin−1 (5665 )

5. Prove that cot−1 [ √1+sin x+√1−sin x

√1+sin x−√1−sin x ]= x2

, x∈(0 , π4 )

Solution:

cot−1 [√1+sin x+√1−sin x√1+sin x−√1−sin x ]

¿cot−1 [√1+sin x+√1−sin x√1+sin x−√1−sin x

×√1+sin x+√1−sin x√1+sin x+√1−sin x ]⇒=cot−1[1+sin x+1−sin x+2√1+sin x√1−sin x

(1+sin x )−(1−sin x ) ]¿cot−1 [2+2√1−sin2 x

2sin x ]=cot−1[1+cos xsin x ]⇒=cot−1[2cos2 x

2

2sin x2 cos x

2 ]=cot−1 (cot x2 )=x

2

6. Prove the following: tan−1( 1

3 )+ tan−1(15 )+tan−1 (17 )+ tan−1( 1

8 )= π4 .

Solution:

LHS (tan−113+tan−11

5 )+( tan−1 17+ tan−1 1

8 )

=tan−1(13+1

5

1−13×1

5)+ tan−1(

17+1

8

1−17×1

8)

=tan−1(814 )+ tan−1 (1555 )=tan−1(47 )+ tan−1 (311 )

=tan−1(47 +

311

1−47 ×

311

)=tan−1(6565 )=tan−11=π

4 =RHS

10

7. If tan−1( x−1

x−2 )+ tan−1( x+1x+2 )=π

4 , then find the value of x.

Solution:

tan−1(x−1x−2 )+ tan−1(x+1

x+2 )=π4

tan−1(x−1x−2 )+ tan−1(x+1

x+2 )=tan−11

tan−1(x−1x−2 )=tan−1 1− tan−1(x+1

x+2 )= tan−1(1−x+1x+2

1+x+1x+2

)== tan−1(12 x+3 )⇒

x−1x−2=

12 x+3 ⇒2 x2−1=0⇒ x2=

12 ⇒ x=±

1√2

PRACTICE PROBLEMES:Level-1

1. Write the Principal value of the following:

(i) tan-1(−√3 ) (ii) sin-1(−12 ) (iii) cos-1(− 1

√2 )2. Evaluate:cot [ tan−1 a+cot−1 a ] .3. Prove: 3sin-1x = sin-1(3x-4x3)4. Find x if sec-1(2) + cosec-1x = /25. Solve tan-12x + tan-13x = /4

Level-21. Write the principal value of the following:

cos−1(cos 2 π3 )+sin−1 (sin 2 π

3 )

2. Write in the simplest form: tan−1[ √1+x2−1

x ] , x≠ 0

3. Prove that sin-1(8/17) + sin-1(3/5) = tan-1(77/36).4. Prove that 2 tan-1(1/2)+ tan-1(1/7) = tan-1(31/17)

5. Solve for x tan−1 2 x

1−x2+cot−1 1−x2

2 x=2 π

36. Find the value of x if sin[cot

-1(x+1)]=cos(tan-1x).

7. 2 sin−1 3

5−tan−117

31= π

4Level-3

1. Prove that : tan−1[ √1+x−√1−x

√1+x+√1−x ]= π4−1

2cos−1 x

2. Prove that: sin-1(12/13) + cos-1(4/5) + tan-1(63/16) = .

11

3. Prove that: tan-11+ tan-12 + tan-13 = .

4. Prove that: tan−1 x

y− tan−1( x− y

x+ y )=π4

5. Write in the simplest form:cos [2 tan−1(√ 1−x

1+x )] .6. If

tan−1 x+ tan−1 y+ tan−1 z=π2

, x, y, z>0 then find the value of xy + yz + zx.

7. Prove that cos−1[cos α+cos β

1+cosα . cos β ]=2 . tan−1 [ tan α2

. tan β2 ] .

8. Prove that tan−1[ √1+cos x+√1−cos x

√1+cos x−√1−cos x ]=π4− x

2; If π<x< 3 π

2 .

9.If cos−1 x+cos−1 y+cos−1 z=π . Then show that x2+y2+z2+2xyz=1.

12

MATRICES & DETERMINANTSSCHEMATIC DIAGRAM

INTRODUCTION: MATRIX: If mn elements can be arranged in the form of m row and n column in a rectangular array then this arrangement is called a matrix. Order of a matrix: A matrix having m row and n column is called a matrix of m× n order.Addition and subtraction of matrices: Two matrices A and B can be added or subtracted if they are of the same order i.e. if A and B are two matrices of order m× n then A± B is also a matrix of order m×n.Multiplication of matrices: The product of two matrices A and B can be defined if the number of rows of B is equal to the number of columns of A i.e. if A be an m× n matrix and B be an n × p matrix then the product of matrices A and B is another matrix of orderm× p.Transpose of a Matrix: If A = [aij] be an m × n matrix, then the matrix obtained by interchanging.the rows and columns of A is called the transpose of A. Transpose of the matrix A isdenoted byA¿∨AT .Properties of transpose of the Matrices: For any matrices A and B of suitable orders, we have

(i) ( ii ) ( KA )T=K AT ( iii ) ( A+B )T=AT+BT (iv )(AB)T=BT AT

Symmetric Matrix: A square matrix M is said to be symmetric if AT=A

e.g.[a bb c ] ,[ x y z

y u vz v w]

Note: there will be symmetry about the principal diagonal in Symmetric Matrix.Skew symmetric Matrix: A square matrix M is said to be skew symmetric if AT=−A

e.g.[ 0 e f−e 0 g−f −g 0 ]

Note: All the principal diagonal element of a skew symmetric Matrix are zero.Determinant: For every Square Matrix we can associate a number which is called the Determinant of the square Matrix.Determinant of a matrix of order oneLet A = [a ] be the matrix of order 1, then determinant of A is defined to be equal to a.Determinant of a matrix of order two

13

Let A¿ [a bx y ] be a Square Matrix of order2×2 then the determinant of A is denoted by |A|

and defined by |A|=|a bx y|= ay-bx

Determinant of a matrix of order 3×3: Let us consider the determinant of a square matrix of order

3×3 ,|A|= |a b cp q rx y z|

Expansion along first row |A|=a (qz− yr )−b ( pz−xr )+c (py−qx)We can expand the determinant with respect to any row or any column.Minors and cofactors:Minor of an elementa ijof a determinant is the determinant obtained bydeleting its ith row and jth column in which elementa ij lies. Minor of an elemena ijisdenoted by M ij.Cofactors: cofactors of an elementa ijdenoted by Aij and is defined by Aij=(−1)i+ j M ij where M ij is the minor of a ij.

Adjoint of a Matrix: Let A = [α βγ δ ] be a Matrix of order 2 ×2

Then adj(A) = [ δ −β−γ α ]

Again letA = [ x y zp q ra b c] be a Matrix of order3 ×3

Then adj(A) =

[ |q rb c| −|p r

a c| |p qa b|

−|y zb c| |x z

a c| −|x ya b|

|y zq r| −|x z

p r| |x yp q| ]

T

¿ [ qc−br −( pc−ar ) pb−aq−( yc−bz) xc−az −(bx−ay )

yr−qz −( xr−pz) xq−py ]T

=[qc−br bz−ay yr−qzar−pc xc−az pz−xrpb−aq ay−bx xq−py ]

Inverse of a Matrix: Inverse of a Square Matrix A is defined asA−1=adj(A )|A|

Note: If A be a given Square Matrix of order n then(i) A(adj(A) = adj(A)A=|A|Iwhere I is the Identity Matrix of order n.(ii) A square Matrix A is said to be singular and non-singular according as

|A|=0∧|A|≠ 0(iii) |adj(A)|= |A|n−1¿)

IMPORTANT SOLVED PROBLEMS

14

Q1. If A=[−245 ] , B=[1 3 −6 ] , Verify that (AB)1=B1A1

Solution: - We have

If A=[−245 ] , B=[1 3 −6 ]

Then AB =[−245 ] [1 3 −6 ] = [−2 −6 12

4 12 −245 15 −30 ]

Now A1 = [−2 4 5 ] , B1= [ 13−6]

B1A1 = [ 13−6] [−2 4 5 ] = [−2 4 5

−6 12 1512 −24 −30] = (AB)1

Clearly (AB)1=B1A1

Q2. If x [23]+ y [−11 ]=[10

5 ] then find the value of x and y.

Sol. Given x [23]+ y [−11 ]=[10

5 ][2 x3 x ]+[− y

y ]=[105 ]or[2 x− y

3 x+ y ]=[105 ]

So 2x – y = 10 and 3x + y = 5 On solving we get x = 3 and y = -4

Q3. If F(x) = [cosx −sinx 0sinx cosx 0

0 0 1] prove that F(x) F(y) = F(x+y)

Sol. Given F(x) = [cosx −sin x 0sinx cosx 0

0 0 1] so F(y) = [cosy −siny 0siny cosy 0

0 0 1] Hence F(x) .F(y) =

[cosx −sinx 0sinx cosx 0

0 0 1] [cosy −siny 0siny cosy 0

0 0 1]=[cosxcosy−sinxsiny −cosxsiny−sinxcosy 0sinxcosy+cosxsiny −sinxsiny+cosxcosy 0

0 0 1] = [cos (x+ y) −sin (x+ y ) 0

sin (x+ y ) cos (x+ y ) 00 0 1]

Hence F(x) F(y) = F(x+y)

Q4. Express the given Matrix as the sum of a symmetric and skew symmetric matrix

15

A= [ 6 −2 2−2 3 −12 −1 3 ]

Sol. Here AT=[ 6 −2 2−2 3 −12 −1 3 ]

P = 12

( A+AT )=12 [ 12 −4 4

−4 6 −24 −2 6 ]=[ 6 −2 2

−2 3 −12 −1 3 ]

Now PT=P so P = 12

( A+AT ) is a symmetric Matrix.

Also let Q = = 12

( A−AT )=12 [0 0 0

0 0 00 0 0]=[0 0 0

0 0 00 0 0 ]

QT=−Q Hence Q is an Skew Symmetric Matrix.

Now P + Q= [ 6 −2 2−2 3 −12 −1 3 ]+[0 0 0

0 0 00 0 0]=[

6 −2 2−2 3 −12 −1 3 ]=A

Thus A is represented as the sum of a symmetric and skew symmetric matrix.

Q5. Using the property of determinant prove that|a−b−c 2 a 2 a2b b−c−a 2b2c 2c c−a−b|=(a+b+c)3

Sol.ApplyingR1 → R1+R2+R3 we get

L.H.S = |a+b+c a+b+c a+b+c2b b−c−a 2b2 c 2c c−a−b|

Taking common a + b + c from first Row we get

L.H.S =(a + b+ c) | 1 1 12 b b−c−a 2 b2c 2 c c−a−b|

Now applying C2→ C2−C1 ,C3→C3−C1 we get

L.H.S =(a + b+ c) | 1 0 02 b −(a+b+c) 02c 0 −(a+b+c )|

Expanding along first Row L.H.S = (a+b+c)[(a+b+c)2−0 ]=(a+b+c)3 = R.H.S

Hence proved

Q6. Solve the system of equations x + 2y – 3z = - 4, 2x + 3y + 2z = 2, 3x – 3y – 4z = 11 Sol. The given system of equation can be written as A X = B where

16

A = [1 2 −32 3 23 −3 −4]X=[ x

yz ]B=[−4

211 ]

Now |A|=|1 2 −32 3 23 −3 −4|=−6+28+45=67

adj(A) = [−6 14 −1517 5 913 −8 −1 ]

T

=[ −6 17 1314 5 −8−15 9 −1]Now A−1=

adj(A )|A|

= 167 [ −6 17 13

14 5 −8−15 9 −1]

Hence X=A−1 B=167 [−6 17 13

14 5 −8−15 9 −1] [

−42

11 ]= 167 [ 24+34+143

−56+10−8860+18−11 ]

So[ xyz ]= 167 [ 201

−13467 ]=[ 3

−21 ]

Hence x = 3 , y = -2 , z = 1Flow chart:

Step 1.Write the given system of equation in the form of A X = B Step2. Find |A| Step3. Find adj(A)

Step4. FindA−1=adj(A )|A|

Step5. Find X=A−1 B Step6 Find the value of x , y and z

ASSIGNMENTS(i). Order, Addition, Multiplication and transpose of matrices:

LEVEL I1. If a matrix has 6 elements, what are the possible orders it can have?

2. Construct a 3 × 2 matrix whose elements are given by aij = |i – 3j |

3. If A = , B = , then find A –2 B.

4. If A = and B = , write the order of AB and BA.

LEVEL II

1. For the following matrices A and B, verify (AB)T = BTAT, where A = , B = .2. Give example of matrices A & B such that AB = O, but BA ≠ O, where O is a zero matrix and A, B are both non zero matrices.3. If B is skew symmetric matrix, write whether the matrix (ABAT) is symmetric or skew symmetric.

4. If A = and I = , find a and b so that A2 + aI = bA

17

LEVEL III

1. If A = , then find the value of A2 –3A + 2I2. Express the matrix A as the sum of a symmetric and a skew symmetric matrix, where:

A =

3. If A = [ cosθ sinθ−sinθ cosθ ]then prove that= An=[ cosnθ sinnθ

−sinnθ cosnθ ] , nϵN

(ii) Cofactors & Adjoint of a matrix LEVEL I

1. Find the co-factor of a12 in A =

2. Find the adjoint of the matrix A = LEVEL II Verify A(adjA) = (adjA) A = I if

1. A =

2. A = (iii)Inverse of a Matrix & Applications

LEVEL I

1. If A = , write A-1 in terms of A 2. If A is square matrix satisfying A2 = I, then what is the inverse of A ?

3. For what value of k , the matrix A = is not invertible ? LEVEL II

1. If A = , show that A2 –5A – 14I = 0. Hence find A-1

2. If A, B, C are three non-zero square matrices of same order, find the condition on A such that AB = AC B = C.

3. Find the number of all possible matrices A of order 3 × 3 with each entry 0 or 1 and

for which

A[ xyz ]=[100 ] has exactly two distinct solutions.

LEVEL III

1. If A = , find A-1 and hence solve the following system of equations: 2x – 3y + 5z = 11, 3x + 2y – 4z = - 5, x + y – 2z = - 3.2. Using matrices, solve the following system of equations:

18

(i) x + 2y - 3z = - 4 , 2x + 3y + 2z = 2 , 3x - 3y – 4z = 11 (ii) 4x + 3y + 2z = 60 , x + 2y + 3z = 45 , 6x + 2y + 3z = 70

3. Find the product AB, where A = , B = and use it to Solve the equations x – y = 3, 2x + 3y + 4z = 17, y + 2z = 7

4. Using matrices solve the following system of equations:

1x−1

y+1

z=4

2x+1

y−3

z=0

1x+1

y+1

z=2

5. A trust caring for indicate children gets rupees 30,000/- every month from its donors. The trust spends half of the funds received for medical and educational care of the children and for that it charges 2% of the spent amount from them and deposits the balance note in a private bank to get the money multiplied so that in future the trust goes on functioning regularly. What % of interest should the trust get from the bank to get a total of rupees 1800/- every month? Use matrix method to find the rate of interest? Do u think people should donate to such trusts?

6. Using elementary transformations, find the inverse of the matrix

(iv)To Find The Difference Between LEVEL I

1. Evaluate |Cos 15 Sin15

Sin75 Cos 75|

2. What is the value of , where I is identity matrix of order 3? 3. If A is non singular matrix of order 3 and = 3, then find

4. For what valve of a, is a singular matrix? LEVEL II

1. If A is a square matrix of order 3 such that = 64, find 2. If A is a nonsingular matrix of order 3 and = 7, then find

LEVEL III

1. If A = and 3 = 125, then find a. 2. A square matrix A, of order 3, has = 5, find

(v).Properties of DeterminantsLEVEL I

1. Find positive valve of x if =

19

2. Evaluate LEVEL II

Using properties of determinants, prove the following :

1.

abc4bacc

bacbaacb

2.

322

22

22

22

ba1ba1a2b2

a2ba1ab2b2ab2ba1

3. = (1 + pxyz)(x - y)(y - z) (z - x)

LEVEL III1. Using properties of determinants, solve the following for x :

a. = 0

b. = 0

c. = 0 2. If a, b, c, are positive and unequal, show that the following determinant is negative:

=

3.

222

2

2

2

cba11ccbca

bc1babacab1a

4.

abc3cbabaaccbaccbba

cba333

5.

0baabbaaccaaccbbccb

22

22

22

6.

3

22

22

22

)cabcab(ababbaba

accacacabccbcbbc

20

7. = 2abc( a + b + c)3

8. If a, b, c are real numbers, and

0accbbacbbaacbaaccb

Show that either a + b +c = 0 or a = b = c.ANSWERS

1. Order, Addition, Multiplication and transpose of matrices:LEVEL I

1. 16, 61 , 23 , 32 2. 3. 4. 22, 33 LEVEL II

3.skew symmetric 4. a = 8, b = 8

LEVEL III.

1. 2. + (ii). Cofactors &Adjoint of a matrix

LEVEL I

1. 46 2.

(iii)Inverse of a Matrix & ApplicationsLEVEL I

1. A-1 = -A 2. A-1 = A 3. k = 17

LEVEL II

1. 3. 512LEVEL III

1.x = 1, y = 2, z = 3. 2. (i) x = 3, ,y = -2, z = 1. (ii) x=7,y=4,z=10 3. AB = 6I, x = , y = - 1, z =

21

4.x = ½, y = -1, z = 1. 6.

(iv). To Find The Difference Between

LEVEL I

1. 0 2. 27 3.24 4.LEVEL II

1. 8 2. 49

LEVEL III 1.a = 3 2. 125(v). Properties of Determinants

LEVEL I1. x = 4 2. + + +

LEVEL II 2. [Hint: Apply C1 C1–bC3 and C2 C2+aC3]

CONTINUITY AND DIFFERENTIABILITYConcept :- Continuity Suppose f is a real function on a subset of the real numbers and let c be a point in the domain of f.

then f is continuous at c if limx→ c

f ( x )= f (c ) or

limh→0

f ( c−h )= f (c )= limh→0

f ( c+h )

PRACTICE PROBLEMS

LEVEL – I

Examine whether the function f given by f ( x )=x2 is continuous at x = 0

Discuss the continuity of the function f given by f ( x )=|x| at x = 0 Show that every polynomial function is continuous

LEVEL- II

Show that the function f defined by f ( x )=|1−x|+|x|, where x is any real number, is a continuous function .

22

Find the relationship between a and b so that the function f defined by

f ( x )={ax+1 ,bx+3

ifif

x≤3x>3 is continuous at x=3 .

For what value of λ is the function defined by f ( x )={λ ( x2−2 x ) ,

4 x+1ifif

x≤0x>0

continuous at x=0 ? What about continuity at x=1?

LEVEL-III

Let

f ( x )={1−sin3 x3cos2 x

if x< π2

a if x=π2

b (1−sin x )(π−2 x )2

if x> π2

If f ( x ) be a continuous function at x=π

2 , find a and b.

If the function

f ( x )={ 3ax+b if x>111 if x=15ax−2b if x<1 is continuous at x = 1,find the value of a

and b.

Find the value of k, so that a function f ( x )={2x+2−164 x−16

,if x ≠2

k , if x=2is continuous at x=2.

Concept :- Differentiation of Implicit function, logarithmic functions, functions in parametric forms, second order derivatives, Rolle’s theorem and

lagrange’s Mean value Theorem

log (m×n )=logm+log n

log (mn )=log m−logn

log mn=n log m

log ab=logc blogc a

log aa=1

23

PRACTICE PROBLEMSLEVEL – I

Find dydx : 2 x+3 y=siny

Find dydx : y=sin2¿

Find dydx :y=tan (x+ y )

Find dydx :x3+ x2 y+x y2+ y3=81

Find dydx :sin2 x+co s2 y=1

Verify Rolle’s theorem for the function y=x2+2 , a=−2∧b=2 If f : [−5 , 5 ]→ R is differentiable function and if f ' (x) does not vanish anywhere, then

prove that f (−5)≠ f (5) LEVEL-II

Differentiate √ ( x−3 ) (x2+4 )3 x2+4 x+5 w.r.t x.

Find

dydx of the function:- x

y+ yx=1 .

Find

d2 ydx2

, if y=x3+tan x .

If x=√asin−1 t , y=√acos−1 t, show that

dydx

=− yx .

Differentiate the following w.r.t x (i)tan−1(sin x

1+cos x ) (ii) sin−1 ( 2x+1

1+4x ) If y=3 e2 x+2e3 x

, prove that

d2 ydx2 −5 dy

dx+6 y=0

.

Differentiate the function :- (x+1x )

x+x

(1+ 1x )

.

Differentiate the function :- ( x cos x )x+( x sin x )1x .

Differentiate the function :- xsin x+ (sin x )cos x

.

Differentiate the function :- xx cos x+ x2+1

x2−1 .

Find dydx of the function:- x

y+ yx=1 .

24

Find

dydx of the function:- (cos x )y=(cos y )x .

x=a (θ−sin θ ) , y=a (1+cosθ ) .

x=a (cos t+logtan t

2 ) , y=a sin t.

Verify Mean Value Theorem if f ( x )=x2−4 x−3in the interval [a, b], where a= 1 and b = 4. LEVEL –III

Finddydx , if y

x+x y+x x=ab.

If y=( tan−1 x )2 ,show that (x2+1 )2 y2+2 x (x2+1 ) y1=2

If ( x−a )2+( y−b )2=c2, for some

c>0

prove that

[1+( dydx )

2]32

d2 ydx 2

is a constant independent of a and b

If cos y=xcos (a+ y ) with cos a≠±1 ,prove that dydx

=cos2 (a+ y )sin a

If x √1+ y+ y √1+x=0 ,for, -1< x <1 , Prove that

dydx

=− 1(1+x )2

If x=cos ( log y ) , −1≤x≤1 , Show that (1−x2 ) d

2 ydx 2 −x dy

dx− y=0

Ify=sin−1 x , show that (1−x2 ) d

2 ydx 2 −x dy

dx=0

.

Find

dydx of the function:- y

x=x y.

If x=sin3 t

√cos2t, y=cos3 t

√cos2 t . Find dydx

Differentiate √ ( x−3 ) (x2+4 )3 x2+4 x+5 w.r.t x.

For a positive constant a find dydx ,where

y=at+

1t , x=(t+ 1

t )a

If x=a (cos t+t sin t ) and y=a (sin t−t cos t ) , find

d2 ydx2

Ify=500 e7 x+600 e−7 x,show that

d2 ydx2 =49 y

25

Examine the validity and conclusion of the Lagrange’s mean value theorem for the function f ( x )=√x2−4 in the interval [2, 4]

Applications of DerivativesConcept :- Increasing or Decreasing function:- Let I be an open interval contained in the domain of a real valued function f . Then f is said to be

(i) Increasing on I if x1<x2 in I⇒ f (x1)≤f (x2 ) ,∀ x1 , x2∈ I

(ii) Strictly increasing on I if x1<x2 in I⇒ f (x1)< f (x2) ,∀ x1 , x2∈ I

(iii) Decreasing on I if x1>x2 in I⇒ f (x1)≥f (x2 ) ,∀ x1 , x2∈ I

(iv) Strictly decreasing on I if x1>x2 in I⇒ f (x1)> f (x2) ,∀ x1 , x2∈ I

Theorem ;- Let f be continuous on [a ,b ] and differentiable on the open interval(a , b ) then

(i) f is increasing in [a ,b ] if f ' ( x )>0 for each x∈ (a ,b )

(ii) f is decreasing in [a ,b ] if f ' ( x )<0 for each x∈ (a ,b )

(iii) f is a constant function in [a , b ] if f ' ( x )=0 for each x∈ (a ,b )

Theorem ;- Let f be continuous on [a ,b ] and differentiable on the open interval(a , b ) then

26

(i) f is strictly increasing in (a , b ) if f ' ( x )>0 for each x∈ (a ,b )

(ii) f is decreasing in (a , b ) if f ' ( x )<0 for each x∈ (a ,b )

(iii) f is a constant function in (a , b ) if f ' ( x )=0 for each x∈ (a ,b )

PRACTICE PROBLEMSLEVEL-I

Show that f(x)=2x+3 is strictly increasing. Prove that f(x)=cosx is

(i) Strictly decreasing in (0,π )(ii) Strictly increasing in (π ,2π )(iii) Neither increasing nor decreasing in (0,2π ).

Find the least value of a such that the function given by f(x)=x2+ax+1 is strictly increasing on (1,2)

LEVEL-II Find the intervals in which the function f is given by f ( x )=4 x3−6 x2−72 x+30 is (a)

strictly increasing (b) strictly decreasing.

Find the interval in which the function given by f ( x )=sin3 x , x∈[0 , π

2 ] is (a) increasing (b)decreasing

Find the interval in which the function f given byf ( x )=sin x+cos x , 0≤x≤2π is strictly increasing or strictly decreasing.

Find the interval in whichy=x2 e−x is increasing. The length of rectangle is decreasing at the rate of 5cm/min and the width y is

increasing at the rate of 4cm/min .When x = 8cm and y = 6cm Find the rate of change of (a)the perimeter and (b) the area of rectangle.

A balloon, which always remains spherical, has a variable diameter32

(2 x+1 ) find the

rate of change of its volume with respect to x.

A particle moves along the curve6 y=x3+2 .Find the point on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

LEVEL – III

Find the interval in which the function f given by f ( x )= 4 sin x−2 x−xcos x

2+cos x is (i) increasing (ii) decreasing

A water tank has the shape of an inverted right circular cone with its axis vertical and

vertex lower most. Its semi vertical angle istan−1 (0 .5 ) .Water is poured into it at a

27

constant rate of 5 cubic meters per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 4 m.

Find the interval in which the function f given by f ( x )=x3+ 1

x3, x≠0

is (i) increasing (ii) decreasing

Concept :Tangents and Normal , Approximation

Tangents and Normal:- The slope of the tangent to the curve y=f ( x ) at the

point (x0 , y0 ) is given by

dydx

]( x0 , y0 )

The slope of the normal to the curve y=f ( x ) at the point (x0 , y0 ) is given by −1

The equation of a tangent at (x0 , y0 ) to the curve y=f ( x ) is given by

y− y0=f ' (x0) ( x−x0)

The equation of a normal at (x0 , y0 ) to the curve y=f ( x ) is given by

y− y0=−1

f ' ( x0)(x−x0)

Tangent line parallel to x-axis then equation of the tangent y= y0

Tangent line parallel to y-axis then equation of the tangent x=x0

Approximations:

(i) The differential of x, denoted bydx , is defined bydx=Δx .

(ii) The differential of y, denoted bydy , is defined by dy=f ' (x0 ) dx or

dy=( dydx )Δx

.

PRACTICE PROBLEMS

LEVEL -I

Find the slope of the tangent to the curve y=3 x2−4 x at x = 4 Find the slope of the tangent to the curve y=x2−3 x+2 at the point whose x- coordinate is 3

Find the slope of the normal to the curve x=a cos3θ , y=a sin3θ , at , θ=π

4LEVEL –II

Find the point at which the tangent to the curve y=√4 x−3−1has its slope 23 .

Find the equation of the tangent line to the curve y=x2−2 x+7 which is (i)

28

parallel to the line 2 x− y+9=0 (ii) Perpendicular to the line 5 y−15 x=13

Find the point on the curve x2+ y2−2 x−3=0at which the tangents are the

parallel to the x-axis.LEVEL-III

Find the equation of the normal to the curve y=x3+2x+6 which are parallel to the line x + 14y + 4 = 0.

Prove that the curves x= y2and xy=k cut at right angle if 8 k3=1

Find the equation of the tangent to the curve y=√3 x−2 which is parallel to the line

4 x−2 y+5=0

INTEGRATIONINTRODUCTION

IF f(x) is derivative of function g(x), then g(x) is known as antiderivative or integral of f(x)

i.e., ddx (g(x)) = f(x) ⇔∫ f ( x )dx=g ( x )

STANDARD SET OF FORMULAS* Where c is an arbitrary constant.

1. ∫ xn dx= xn+1

n+1+c(n¿ -1)

2. ∫ dx = x + c

3. ∫ 1x dx = log |x| + c

4. ∫cos xdx=sin x+c

5. ∫sin x dx=¿−cos x+c ¿

6. ∫ sec2 x dx=tan x+c

7. ∫ cosec2 x dx=−cot x+c

8. ∫ sec x tan x dx=sec x+c

9. ∫ cosec x cot xdx=−cosec x+c

10. ∫ ex dx=ex+c

11. ∫ tan xdx=log|sec x|+c

12. ∫cot x dx=log|sin x|+c

13. ∫ sec x dx=log|sec x+ tan x|+c

14. ∫ cosec x dx=log|cosec x−cot x|+c

29

15. ∫ 1√1−x2 dx = sin -1 x + c

16. ∫ 11+x2 dx=tan−1 x+c

17. ∫ 1x √x2−1

dx=sec−1 x+c

18. ∫ ax dx= ax

log a+c

19. ∫ 1√ x

dx=2√x+c

INTEGRALS OF LINEAR FUNCTIONS

1. ∫ ( ax+b )n dx= (ax+b )n+1

(n+1 ) a+c

2. ∫ 1ax+b

dx= log (ax+b)a

+c

3. ∫sin ( ax+b ) dx=−cos(ax+b)a

+c

In the same way if ax +b comes in the place of x, in the standard set of formulas, then divide the integral by a

SPECIAL INTEGRALS

1. ∫ 1x2−a2 dx= 1

2alog|x−a

x+a | + c

2. ∫ 1a2−x2 dx= 1

2alog|a+x

a−x|+c

3. ∫ 1

x2+a2dx=1

atan−1 x

a+c

4. ∫ 1√ x2−a2

dx=log|x+√x2−a2|+c

5. ∫ 1√ x2+a2

dx=log∨x+√x2+a2∨¿¿ + c

6. ∫ 1√a2−x2

dx=sin−1( xa )+c

7. ∫√x2+a2 dx= x2 √x2+a2+ a2

2log|x+√x2+a2|+c

8. ∫√x2−a2 dx = x2 √x2−a2−a2

2log|x+√ x2−a2|+c

9. ∫√a2−x2 dx = x2 √a2−x2 +a2

2sin−1( x

a )+c

INTEGRATION BY PARTS1. ∫u . v dx=u .∫v dx−∫¿¿¿¿

OR

30

The integral of product of two functions = (first function) x integral of the second function – integral of [(differential coefficient of the first function ) × (integral of the second function)] We can choose first and second function according to I L A T E where I → inverse trigonometric function → logarithmic function , A →algebraic function T →trigonometric function E → exponential function

2. ∫ ex [f (x )+ f 1 (x )] dx=ex f ( x )+c .Working Rule for different types of integrals

1. Integration of trigonometric functionWorking Rule (a) Express the given integrand as the algebraic sum of the functions of the following forms (i) Sin kα , (ii) cos kα ,(iii) tan kα , (iv) cot kα , (v) sec kα ,(vi) cosec kα ,(vii) sec2 kα , (viii) cosec2 kα , (ix) sec kα tan kα (x) cosec kα cot kα For this use the following formulae whichever applicable

(i) sin2 x=1−cos2 x2 (ii) cos2 x=1+cos 2 x

2

(iii) sin3 x=3 sin x−sin3 x4 (iv)cos3 x=3 cos x+cos3 x

4 (v) tan2 x = sec2 x – 1 (vi) cot2 x = cosec2x – 1 (Vii) 2sin x sin y = cos (x – y ) – cos ( x + y) (vii) 2 cos x cos y = cos (x + y) + cos (x – y ) (ix) 2 sin x cos y = sin (x + y ) + sin (x – y ) (x) 2cos x sin y = sin (x + y ) – sin ( x – y )

2. Integration by substitution

(a) Consider I=∫ f ( x )dx

Put x=g(t) so that dxdt

=g1( t ) .

We write dx=g1(t). Thus I=∫ f ( x )dx=∫ f (g ( t )) g1( t )dt(b) When the integrand is the product of two functions and one of them is a function g (x) and the other is k g’ (x), where k is a constant then Put g (x) = t

3. Integration of the types ∫ dxa x2+bx+c

,∫ dx√a x2+bx+c

, and ∫ (a x2+bx+c )dx

In this three forms change a x2+bx+c in the form A2 + X2 , X2 – A2, or A2 – X2

Where X is of the form x +k and A is a constant ( by completing square method) Then integral can be find by using any of the special integral formulae

31

4. Integration of the types ∫ px+qa x2+bx+c

dx ,∫ px+q√a x2+bx+c

dx and

∫ ( px+q )√a x2+bx+c dx

In this three forms split the linear px +q = λddx

(a x2+bx+c )+μ

Then divide the integral into two integrals The first integral can be find out by method of substitution and the second integral by completing square method as explained in 3

I,e., to evaluate ∫ px+q√a x2+bx+c

dx = ∫ λ (2 ax+b )+μa x2+bx+c

dx

= λ∫ 2ax+ba x2+bx+c

dx+μ∫ dxa x2+bx+c

↓ ↓ Find by substitution method + by completing square method

5. Integration of rational functions In the case of rational function, if the degree of the numerator is equal or greater than degree of the denominator , then first divide the numerator by denominator and write it as

NumeratorDenominator

=Quotient+ RemainderDenominatior , then integrate

6. Integration by partial fractionsIntegration by partial fraction is applicable for rational functions. There first we must check thatdegree of the numerator is less than degree of the denominator, if not, divide

the numerator by denominator and write as Numerator

Denominator=Quotient+ Remainder

Denominatior and

proceed for partial fraction of Remainder

Denominatior

Sl. No.

Form of the rational functions Form of the rational functions

1 px+q( x−a )(x−b)

Ax−a

+ Bx−b

32

2 px+q(x−a)2

Ax−a

+ B(x−a)2

3. p x2+qx+r( x−a ) ( x−b ) x−c¿

¿A

x−a+ B

x−b+ C

x−c

4 p x2+qx+r( x−a )2 ( x−b )

Ax−a

+ B(x−a)2+

Cx−b

5 p x2+qx+r( x−a )(x2+bx+c )

Where x2 + bx + c cannot be factorized further

Ax−a

+ Bx+Cx2+bx+c

DEFINITE INTEGRATIONWorking Rule for different types of definite integrals

1. Problems in which integral can be found by direct use of standard formula or by transformation method Working Rule (i). Find the indefinite integral without constant c (ii). Then put the upper limit b in the place of x and lower limit a in the place of x and subtract the second value from the first. This will be the required definite integral.2. Problems in which definite integral can be found by substitution method Working Rule When definite integral is to be found by substitution then change the lower and upper limits of integration. If substitution is z = φ(x) and lower limit integration is a and upper limit is b Then new lower and upper limits will be φ(a) and φ(b) respectively.Properties of Definite integrals

1. ∫a

b

f ( x )dx=¿∫a

b

f (t ) dt ¿

2. ∫a

b

f ( x )dx=¿∫b

a

f ( x )dx ¿. In particular, ∫a

a

f ( x )dx=¿¿ 0

3. ∫a

b

f ( x )dx=¿∫a

c

f ( x )dx ¿ + ∫c

b

f ( x )dx, a < c < b

4. ∫a

b

f ( x )dx=¿∫a

b

f (a+b−x ) dx¿

33

5. ∫0

a

f ( x )dx=¿∫0

a

f (a−x ) dx ¿

6. ∫0

2 a

f ( x )dx=¿∫0

a

f ( x )dx ¿ + ∫0

a

f (2 a−x ) dx

7. ∫0

2 a

f ( x )dx=¿2∫0

a

f ( x ) dx¿ , if f (2a−x ) = f ( x ) and

=0 , if f (2a−x ) = −f ( x )

8. (i) ∫−a

a

f (x ) dx=¿2∫0

a

f ( x ) dx¿, if f is an even function, i.e., if f (-x) = f (x)

(ii) ∫−a

a

f (x ) dx=¿0¿, if f is an odd function, i.e., if f (−¿x) = −f (x)

Problem based on property

∫a

b

f ( x )dx=¿∫a

c

f ( x )dx ¿ + ∫c

b

f ( x )dx, a < c < b

Working RuleThis property should be used if the integrand is different in different parts of the

interval [a,b] in which function is to be integrand. This property should also be used when the integrand (function which is to be integrated) is under modulus sign or is discontinuous at some points in interval [a,b]. In case integrand contains modulus then equate the functions whose modulus occur to zero and from this find those values of x which lie between lower and upper limits of definite integration and then use the property.Problem based on property

∫0

a

f ( x )dx=¿∫0

a

f (a−x ) dx ¿

Working Rule

Let I = ∫0

a

f ( x )dx

Then I = ∫0

a

f (a− x ) dx

(1) + (2) => 2I = ∫0

a

f ( x )dx + ∫0

a

f (a− x ) dx

I = 12∫0

a

{f (x )+ f (a−x ) }dx

This property should be used when f ( x ) + f ( a−x ) becomes an integral function of x.Problem based on property

34

∫−a

a

f (x ) dx=¿¿ 0, if f ( x ) is an odd function and ∫−a

a

f (x ) dx=¿¿ 2∫0

a

f ( x )dx, if f ( x ) is an even

function.Working Rule

This property should be used only when limits are equal and opposite and the function which is to be integrated is either odd/ even.

PROBLEM BAESD ON LIMIT OF SUM Working rule

∫a

b

f ( x )dx = limh→ 0

h {f (a )+ f (a+h )+….+f (a+(n−1 )h )}

Where nh = b – a The following results are used for evaluating questions based on limit of sum.

(i) 1 + 2 + 3 + ….. + (n-1) = ∑k=1

n−1

k=n(n−1)2

(ii) 12 + 22 + 32 + …. + (n−1)2 = ∑ (n−1 ) n(2 n−1)6

(iii) 13 + 23 + 33 + ….. + (n−1)3 = [ (n−1 ) n2 ]

2

(iv) a + ar + …… + ar n−1 = a[rn−1]r−1

(r≠1)

IMPORTANT SOLVED PROBLEMSEvaluate the following integrals

1. ∫ (1+ log X)2

Xdx

Solution

put 1+log x = t 1x

dx=dt

∫ (1+ log X)2

Xdx = ∫ t 2 dt

= t3

3+c

= (1+ logx )3

3+c

2. ∫ ex

√5−4 ex−e2 x dx

Solution Put ex=t thenex dx=dt

∫ ex

√5−4 ex−e2 x dx=∫ dt√5−4 t−t2

= ∫ dt

√−(t 2+4 t−5)

35

= ∫ dt

√−(t 2+4 t+4−4−5)

= ∫ dt

√−{(t+2 )2−9 }

= ∫ dt√32−¿¿¿

¿

= sin−1 t+23 + C = sin−1( ex+2

3 )+C

3. ∫√tanx dx Solution

Put tan x = t 2then sec2x dx = 2t dt => dx = 2t dt1+t 4

∫ √tanx dx = ∫ t 2 t dt1+ t4 = ∫ 2 t2

1+t 4 dt

= ∫2

1t2+ t2

dt (by dividing nr and dr by t2 )

= ∫(1+ 1

t2 )+(1− 1t 2 )

t 2+ 1t 2

dt

= ∫1+ 1

t2

t 2+ 1t 2

dt+∫1− 1

t2

t2+ 1t2

dt

= ∫1+ 1

t2

( t−1t )

2

+2dt+∫

1− 1t 2

(t+ 1t )

2

−2dt

= ∫ duu2+2

+∫ dvv2−2

( 1st integral put t−1t = u

then (1+ 1t2 )dt=du

2nd integral putt+1t=v then(1− 1

t 2 )dt=dv

= 1√2

tan−1( u√2 )+ 1

2√2log|v−√2

v+√2|+C

= 1√2

tan−1( t−1t

√2 )+ 12√2

log|t+ 1t−√2

t+ 1t+√2 |+C

36

= 1√2

tan−1( t 2−1√2 t )+ 1

2√2log|t 2+1−√2t

t 2+1+√2t |+C

= 1√2

tan−1( tanx−1√2 tanx )+ 1

2√2log|tanx+1−√2 tanx

tanx+1−√2 tanx|+C

4. ∫ 5 x+3√ x2+4 x+10

dx

Solution

5x + 3 = A (2x + 4 ) + B = > A = 52∧B=−7

∫ 5 x+3√ x2+4 x+10

dx=∫52

(2 x+4 )−7

√ x2+4 x+10dx

= ∫52

(2x+4 )

√ x2+4 x+10dx+∫ 7

√x2+4 x+10dx

= 52∫

(2 x+4 )

√x2+4 x+10dx+7∫ 1

√ x2+4 x+10dx

= 52∫

dt√t

+7∫ 1√ x2+4 x+4−4+10

dx

= 52

×2√ t+7∫ 1

√ ( x+2 )2+6 dx

= 5 √ x2+4 x+10+7 log|x+2+√x2+4 x+10|+C

5. ∫0

π x tan xsec x+tan x

dx

Solution

I = ∫0

π x tan xsec x+tan x

dx -------------------(1)

Again I = ∫0

π ( π−x ) tan ( π−x )sec ( π−x )+ tan (π−x )

dx Using the property ∫0

a

f ( x )dx=∫0

a

f (a−x ) dx

I = ∫0

π (π−x ) tan xsec x+tan x

dx -------------------- (2)

Adding (1) and (2) we get

2 I = ∫0

π x tan xsec x+tan x

dx+∫0

π ( π−x ) tan xsec x+ tan x

dx

= π∫0

π tanxsec x+ tan x

dx = π∫0

π tanx(sec x−tan x )sec2 x−tan2 x

dx

= π∫0

π

¿¿) dx = π∫0

π

( tan x sec x−sec2 x+1 )dx

¿ π ⌈ sec x− tan x+x ⌉ π0

37

I = π ( π2−1)

6. Evaluate ∫1

4

(x¿¿2¿−x)dx¿¿ using limit of sum

Solution

Comparing the given integral with ∫a

b

f ( x )dx

a = 1, b = 4 f (x ) = x2 – x ∴nh=4−1=3 and f ( a+ (n-1)h )= f ( 1 + (n – 1 ) h ) = ( 1 + (n – 1 ) h ) 2 - [1 + (n – 1 ) h]

= 1 + 2 (n-1 )h + (n – 1 ) 2 h2 -1 – (n – 1 ) h= (n – 1 ) h + (n – 1 ) 2 h2

∫a

b

f ( x )dx = limh→ 0

h∑0

n−1

f (a+ (n−1 ) h )

∫1

4

(x¿¿2¿−x)dx¿¿ = limh→ 0

h {∑0

n−1

(n−1 )h+ (n−1 )2 h2 }

= limh→ 0

h {∑ (n−1 )h+∑ (n−1 )2 h2 }

= limh→ 0

h {h∑ (n−1 )+h2∑ (n−1 )2 }

= limh→ 0

h2 n(n−1)2 +h3 n (n−1 )(2n−1)

6

= limh→ 0

nh (nh−h)2 +

nh (nh−h )(2nh−h)6

= limh→ 0

3(3−h)2 +

3 (3−h )(6−h)6

= 92 +

546

= 272

PRACTICE PROBLEMSLEVEL I

Evaluate the following integrals1. ∫ (2 x−3 cosx+ex )dx

2. ∫ x4+5 x2+3√x

dx

3. ∫sin 3 x dx

4. ∫sin 3 x cos5 x dx

5. ∫sin 3 x cos5 xdx

38

6. ∫ dxx2+4 x+1

7. ∫ dx√ x2+5 x+8

8. ∫0

π4

tan2 x dx

9. ∫ x2 ex dx

10. ∫0

1 xx2+1

dx

39

LEVEL IIEvaluate the following integrals

1. ∫ sin xsin ( x−a )

dx

2. ∫ 2(1−x ) (1+x2)

dx

3. ∫ x+2√ ( x−5 )(x−3)

dx

4. ∫ ex( sin 4 x−41−cos4 x )dx

5. ∫ x2 tan−1 x dx

6. ∫ dxcos2 x+sin 2 x

7. ∫ 2 x(1+x2 )(3+x2)

dx

9. ∫ tan−1√ 1−x1+x

dx

10. ∫0

π xtan xsec x cosec x

dx

11. ∫0

π x1+sinx

dx

12. ∫−1

3

¿ x3−x∨¿dx ¿

13. ∫1

3 √x√ x+√5−x

dx

14. ∫0

π4

log (1+ tan x ) dx

15. ∫0

2

(x¿¿2+2)dx ¿ as a limit of sum

16. ∫0

π2

( 5Sinx+3 CosxSinx+Cosx )dx

LEVEL IIIEvaluate the following integrals

1. ∫ sin xsin 4 x

dx

2. ∫(√ tan x¿+√cot x )dx¿

39

40

3. ∫ sin xcos xsin4 x+cos4 x

dx

4. ∫ √ 1−√ x1+√ x

dx

5. ∫sin ( log x ) dx

6. ∫ log ( x+√x2+a2 )dx

7 ∫0

π

x log (sin x )dx

8. ∫1

2

(2x2+x+7 )dx as a limit of sum

9.∫ xdx

1+x tan x

10∫ x4

( x−1 )( x2+1)dx

11. ∫ [ log ( log x )+ 1

( log x )2 ]dx

40

41

APPLICATION OF INTEGRATIONINTRODUCTION

Area under Simple Curves(i)

Area bounded by the curve y = f(x), the x-axis and between the ordinates at x = a and x = b is given by

Area = ∫a

b

y dx = ∫a

b

f (x )dx

(ii)

41

42

Area bounded by the curve y = f(x), the y axis and between abscissas at y = c and y = d is given by

Area = ∫c

d

x dx = ∫c

d

g ( y ) dy

Where y = f ( x )=¿x=g ( y )Note: If area lies below x-axis or to left side of y-axis, then it is negative and in such a case we like its absolute value. (numerical value)

4.

Area bounded by two curves y=f (x ) and y=¿ g(x ), such that 0 ≤ g (x)≤ f (x ) for all x∈[a ,b] and between the ordinates at x=a , x=b is given by

Area ¿∫a

b

{f ( x )−g (x ) }dx

Finding the area enclosed between a curve, X- axis and two ordinates or a curve , Y- axis and two abscissa

WORKING RULE

42

43

1. Draw the rough sketch of the given curve2. Find whether the required area is included between two ordinate or two abscissa3. (a) If the required area is included between two ordinates x = a and x= b then use

the formula ∫a

b

y dx

(b) If the required area is included between two abscissa y = c and = d then use the

Formula ∫c

d

x dy

Finding the area included between two curvesWORKING RULE

1. Draw the graph of the given curves.2. Mark the region whose area is to be determined.3. Find whether the area is bounded between two given curves and two ordinates or between the two given curves and two abscissa. (a) If the required area is bounded between 2 ordinates , use the formula

∫a

b

[ f ( x )−g ( x )¿ ]dx¿

(b) If the required area is included between two abscissa use the formula

∫a

b

[ f ( y )−g ( y ) ] dy

SOME IMPORTANT POINTS TO BE KEPT IN MIND FOR SKETCHING THE GRAPH1. y2 = 4ax is a parabola with vertex at origin,

symmetric to X axis and right of origin

2. y2 = - 4ax is a parabola with vertex at origin, symmetric to X axis and left of origin

3. x2 = 4ay is a parabola with vertex at origin, symmetric to y axis and above origin

4. x2 = - 4ay is a parabola with vertex at origin, symmetric to y axis and below origin

5. x2

a2 + y2

b2 = 1 is an ellipse symmetric to both axis,

Cut x axis at (±a,0) and y axis at (0,± b ¿6. x2+ y2=r2 is a circle symmetric to both the axes

with centre at origin and radius r

43

44

7. (x−h)2+( y−k )2=r2 is a circle with centre at (h,k) and radius r.8. ax +by + c = 0 representing a straight line

9. Graph of

y=|x|

10. Graph of y=|x−3|

IMPORTANT SOLVED PROBLEMS1. Calculate the area of the region bounded by the parabola y =x2 and x = y2 Solution

44

45

Parabola y 2 = x is symmetrical to X- axis and x2 = y is symmetrical to y axis Solving both the equations we get the point of intersection of the two curves as (0,0) and (1,1)

Required area = ∫0

1

(√ x¿−x2)dx¿

= [ x32

32

− x3

3 ]10 =

23−1

3 = 13 sq. units

2. Find the area of the region ; { (x, y ) : y2 ≤ 4 x , 4 x2+4 y2≤ 9} Solution:

Curves y2 =4x, parabola symmetric to x axis and the circle x2+ y2=94 circle with centre at

(0,0 ) and radius 32

Sketch both the curves and shaded the area

The point of intersection of y2 =4x and x2+ y2=94 are the points (0, 0 ) ( 1

2,√2)

Required area = 2 × Area of OBALO = 2×[ area of OBLO + area of BLAB]

= 2 (∫012

2√ x+∫12

32

√ 94−x2) dx

45

46

= 2 {(2 x32

32 ) 1

20+[ x

2 √ 94−x2+ 9

8sin−1 2 x

3 ]3212 }

= 9 π8− 9

4sin−1( 1

3 )+ √26 sq. units

3. Find the area bounded between the lines y = 2x + 1 , y = 3x + 1 , x= 4 using integration Solution: Draw the rough sketch and shaded the area

Area enclosed = ∫0

4

(f ( x )−g ( x ))dx

= ∫0

4

¿¿) dx

= ∫0

4

x dx

= [ x2

2 ]40 = 8 sq.units

4. Find the area of the region { (x,y): 0≤ y≤ x2+1 ,0 ≤ y≤ x+1 , 0 ≤ x ≤ 2}

Solution Sketch the region whose area is to be found out.

46

47

The point of intersection of y = x2 +1 and y = x+1 are he points (0,1 ) and (1,2 ) The required area = area of the region OPQRSTO = area of the region OTQPO + area of the region TSRQT

= ∫0

1

(x¿¿2+1)dx+∫1

2

( x+1 ) dx ¿

= [ x3

3+x ]10 + [ x2

2+x ]21

= 236 sq.units

5. Find the area cut off from the parabola 4y = 3 x2 by the line 2y = 3x + 12 Solution Given 4y = 3x2 and 3x – 2y +12 = 0 Solve both the equation we get the point of intersecction of both the curves (-2,3) and (4,12) Required area = area of AOBA

= ∫−2

4

[ 3 x+122

¿−3 x2

4]¿ dx

= 27 sq.unts

47

48

PRACTICE PROBLEMSLEVEL I

1. Find the area of the region bounded by the parabola y2 = 4ax , its axis and two ordinates x = 4 and x = 92. Find the area bounded by the parabola x2 = y , y axis and the line y =13. Find the area bounded by the curve y = 4x – x2 , x axis and the ordinates x = 1 and x = 34. Find the area of the region bounded by the curve y2 = 4x and the line x = 3

LEVEL II1. Find the area of the region { (x,y) : x2 + y2 ≤ 4 , x+ y ≥2 }2. Find the of the circle x2 + y2 = a2

3. Find the area of the region { (x , y) : y2 ≤6 x , x2+ y2≤ 16 }4. Sketch the region common to the circle x2 + y2 = 4 and the parabola y2 = 4 x. Also find the area of the region by integration.

5. Find the area of the ellipse x2

a2 +y2

b2 = 1

6. Find the area of the smaller region bounded by the ellipse x2

9+ y2

4 = 1 and the straight line

x3+ y

2=1

7. Find the area enclosed by the curve x = 3 cos t, y = 2 sin t.8. Find the area bounded by the lines x +2y = 2, y – x = 1 and 2x + y = 7

9. Find the area of the region bounded by the parabola y=34

x2 and the line 3x – 2y +12 =0

10. Using integration, find the area of the triangle ABC with vertices A (-1, 0 ), B (1 , 3 ) and C (3, 2)

LEVEL III1. Using integration find the area of the following region { (x , y ) :|x+2|≤ y≤√20−x2}2. Sketch the region enclosed between the circles x2 + y2 = 1 and x2  + (y-1)2 = 1. Also find the area of the region using integration3. Find the area of the region lying above the x axis and included between the circle x2 + y2 = 8x and the parabola y2 = 4x 4. Using the method of integration , find the area bounded by the curve |x| + |y| = 1

48

49

DIFFERENTIAL EQUATIONSINTRODUCTION

1. Problems based on the order and degree of the differential equations Working rule

(a) In order to find the order of a differential equations, see the highest derivative in the given differential equation. Write down the order of this highest order derivatives.(b) In order to find the degree of a differential equation write down the power of the highest order derivative after making the derivatives occurring in the given differential equation free from radicals and fractions.

2. Problems based on formation of differential equation. Working rule(a) Write the given equation.(b) Differentiate the given equation w.r.t. independent variable of x as many times as the number of arbitrary constants.(c) Eliminate the arbitrary constants from given equation and the equations obtained by differentiation.

3. Problems based on solution of differential equation in which variables are separable. Working ruleThis differential equation can be solved by the variable separable method which can be put in the form

dydx = f ( x ) . g ( y ) .

i.e., in which dydx can be expressed as the product of two functions, one of which is

a function of x only and the other a function of y only.

In order to solve the equation dydx

=f ( x ) . g ( y ) .Write down this equation in the

form dy

g ( y )=f ( x ) . dx, then the solution will be ∫ dy

g( y )=∫ f ( x ) dx+c , where C is an

arbitrary constant.4. Problems based on solution of differential equations which are homogeneous.

Working rule

(a) Write down the given differential equation in the form dydx = f ( x , y )

(b) If f ( kx , ky )=f ( x , y ) . then differential equation is homogeneous.

(c) In order to solve, put y=vx ,so that dydx

=v+x dvdx and then separate the

variables x and v.

49

50

(d) Now solve the obtained differential equation by the variable separable

method. At the end put yx in place of v.

5. Working Rule for Linear Differential Equation of first degree:

(a) Type 1. dydx

+Py=Q , where P and Q are constants or function of x only

General solution is y . IF=∫ (Q. IF ) dx+C , where IF=e∫ Pdx.

(b) Type 2. dxdy

+Px=Q , where P and Q are constants or function of y only

General solution is x . IF=∫ (Q . IF ) d y+C , where IF=e∫ Pdy. Note: Particular solution can be obtained after getting the value of parameter C by substituting the given initial values of the variables.

IMPORTANT SOLVED PROBLEMS1. Solve the differential equation (x + y ) dy + ( x – y ) dx = 0 Solution

(x + y ) dy + ( x – y ) dx = 0dydx

= y−xx+ y

Put y = vxdydx

=v+x dvdx

∴ v+x dvdx

= vx−xx+vx

v+x dvdx

= v−1v+1

x dvdx

= v−1v+1

−v

x dvdx

=−(1+v2)

1+v

1+v1+v2 dv=−dx

xIntegrating both sides we get

∫ 1+v1+v2 dv = - ∫ dx

x

∫ dv1+v2 +

12∫

2v1+v2 dv=−log x+c

tan−1 v+12

log|1+v2|=−log x+c

50

51

tan−1 yx+ 1

2log(1+ y2

x2 )=−log x+c

tan−1 yx+ 1

2log ( x2+ y2 )=c

2. Solve x √1− y2dx+ y √1−x2dy=0 Solution

x √1− y2dx = -y √1−x2dyx

√1−x2dx= − y

√1− y2dy

Integrating both sides

∫ x√1−x2

dx = - ∫ y√1− y2

dy

−12 ∫ dt

√t =

12∫

du√u

( put t = 1 – x2 and put u = 1 – y2 )

−√ t=√u + C √1−x2+√1− y2=C

3. Solve the differential equation dydx

−1x

. y=2 x2

Solution

The diff.eqn is in the form dydx

+Py=Q

Where P = −1x and Q = 2x2

I.F = e∫ Pdx=e∫−1

x dx=e−log x=1

x Multiplying both sides of diff.eqn by I.F we get

1x

dydx

− 1x2 y=2 x

ddx ( y . 1

x )=2 x

Integrating both sides w.r.t.x we get

y . 1x=x2+c

y = x3 + Cx

4. Solve the diff. eqn xdydx

= y−x tan yx

Solution

51

52

dydx

=y−x tan y

xx

Put y = vx => dydx

=v+x dvdx

∴ v+x dvdx

= vx−x tanvx

v+x dvdx

= v – tan v

x dvdx

=−tan v

cot v dv = - dxx

Integrating both sides ,we get

∫cot v dv=−∫ dxx

log sin v = - log x + log C

log [x. sin ( yx )¿=log c

x sin ( yx ) = C

5. xy dydx

=( x+2 )( y+2) , find the equation of the curve passing through the points (1, -1)

Solution

xy dydx

=( x+2 )( y+2)

yy+2

dy= x+2x

dx

(1− 2y+2 )dy=(1+ 2

x )dx

Integrating both sides weget Y –2 log (y + 2 ) = x +2 log x + C

The curve is pasing through the the point (1,-1) - 1 –2 log 1 = 1 + 2 log 1 + C => C = -2 The equation of the line is y−x=2 log[x(y+2)¿ –2

6. Solve the differential equation (x2−1 ) dydx

+2 xy= 2x2−1

Solution

(x2−1 ) dydx

+2 xy= 2x2−1

52

53

dydx

+( 2 xx2−1 ). y= 2

(x2−1 )2

I.F. = e∫ 2 x

x2−1dx=e log (x2−1 ) = (x2−1 )

ddx ( y (x2−1))= 2

x2−1Integrating both sides w.r.t x we get

y (x2−1 )=∫ 2x2−1

dx

y (x2−1 )= log( x−1x+1 )+C

-

PRACTICE PROBLEMSLEVEL I

1. Find the order and degree of the following differential equation.

( d3 yd x3 )

2

−x ( dydx )

3

=0

2. Show that y=Ax+ Bx is a solution of the differential equation.

x2 d2 ydx2 +x dy

dx− y=0

3. Form the differential equation corresponding to y2−2 ay+x2=a2 by eliminating a.4. Form the differential equation representing the family of curves y=a sin(x+b), where

a ,b are arbitrary constants.5. Solve the differential equation.sec2 x tan y dx+sec2 y tan x dy=06. Solve the differential equation.

dydx

=1+ y2

1+x2

LEVEL IISolve the following differential equations.

1.dydx

+ 1x

. y=2 x2

2. (x¿¿2+xy)dy= (x2+ y2 )dx ¿

53

54

3.dydx

+ 4 xx2+1

y+ 1(x2+1 )2 = 0

4. sin x dydx

+cos x . y=cos x . sin2 x

5. 3ex tan y dx+ (1−e2 ) sec2 y dy=0 , giventhat y=π4

, when x=1.

6. (x3+ y3 )dy−x2 y dx=0

7. cos2 x dydx

+ y= tan x

8.dydx

+ y=cos x−sin x

9. x log x dydx

+ y= 2x

log x

LEVEL III Solve the following differential equations

1. y – x dydx = a ( y2+x2 dy

dx )2. ( 1 + sin2x ) dy + ( 1 + y2 ) cos x dx = 0, given that x =

π2 , y = 0

3. ( x3 + x2 + x +1) dydx = 2x2 + x

VECTORS

54

55

SUMMARY1. Position vector of a point P(x,y,z) is given as OP(r) ¿ x i+ y j+z k, and its magnitude by

√ x2+ y2+z2.

2. The scalar components of a vector are its direction ratios, and represent its projections along the respective axes.

3. The magnitude(r),direction ratios (a , b , c )and direction cosines ( l , m, n ) of any vector are

related as: l=ar ,m=b

r ,n=cr

4. The vector sum of the three sides of a triangle taken in order is 0 .5. The vector sum of two coinitial vectors is given by the diagonal of the parallelogram

whose adjacents sides are the given vectors.6. The multiplication of a given vector by a scalar α, changes the magnitude of the given

vector by the multiple │α│, and keeps the direction same (or makes it opposite) according as the value of α is positive (or negative).

7. For a given vector a , the vector a=¿ a|a| gives the unit vector in the direction of a.

8. The position vector of a point R dividing a line segment joining the points P and Q whose position vectors are a and b respectively , in the ratio m : n

(i) internally , is given by R=m b+n am+n

.

(ii) externally, is given byR=m b−n am−n

.

(iii) if R is the mid point of PQ, then R= b+ a2

.

9. The scalar product of two given vectors a and b having angle θ between them is defined as a . b= ǀ aǀ ǀbǀ cosθ. Also, when a . b is given, the angle ‘θ ' between the vectorsa andb may be determined by

cosθ= a. bǀ a ǀ ǀ b ǀ

.

10. Product is given as a× b=|a||b|sinθ n ,where n is aunit vector perpendicular to the plane containing a and b such that a , b , n form right handed system of co-ordinate axes.

11. If we have two vectors a and b , given in component form as

a=a1 i+b1 j+c1 k and b=a2 i+b2 j+c2 k and λ any scalar, then

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a+ b=(a1+a2 ) i+(b1+b2 ) j+ (c1+c2 ) kλa=λa1 i + λb1 j+ λc1 ka . b= a1a2+b1b2+c1c2

And a × b=| i j ka1 b1 c1

a2 b2 c2|

Practice problems

LEVEL-11) Find the projection of i - j on i + j .2) If |a| =2,|b| =√3 and a .b =√3.Find the angle between a and b .3) Find the value of λ when the projection of a =λ i + j +4k on b =2i +6 j+3k is 4 units.4) Find a . ( b x c ) ,if a=2 i+ j+3 k, b=−i+2 j+k and c=3 i+ j+2 k Ans:−10 5) Show that the four points A, B, C and D with position vectors 4 i+5 j+ k, − j−k ,

3 i+9 j+4 k and 4 ( −i+ j+k ) respectively are coplanar.

LEVEL-21. Find the angle between the vectors i -2 j+3k and 3 i -2 j+k .

2. Find |a−b| if |a| =2,|b| =3 and a .b=4.

3. If a is a unit vector and (x - a¿ . (x + a¿ =15. Find |x| .

4. If a and b are two vectors such that |a+b| =|a| then prove that the vector 2a+ b is perpendicular to b .

5. If vectors a =2 i +2 j+3 k ,b =i + j+k and c =3i + j are such that a+ λ b is perpendicular toc. Find the value of λ.

6. If A and B be two points with position vectors 2 i− j+2 k and i+2 j respectively. Find the position vector of the point which divides AB in 1 : 2 internally.

LEVEL-3

1. Find the value of p so that a =2i + p j +k and b =i -2 j+3k are perpendicular to each other.

2. Find |a| if |a|=2|b| and (a +b ) . (a -b ) = 12.

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3. If |a+b| =60,|a−b| =40 and |b| =46. Find |a| .

4. If a= (3 i+2 j−3 k ) andb = (4 i+7 j−3 k ) Find vector projection of ain the direction of b.

5. The two adjacent sides of a parallelogram are (2 i−4 j+5 k )∧ ( i−2 j−3 k ).Find the unit

vectors parallel to its diagonals. Also find its area.

6. If ( i+ j+k ), (2 i+5 j−3 k ) , (3 i+2 j−2k )∧ ( i−6 j−k )are the position vectors of points A, B, C & D respectively, then find the angle between AB & CD. Deduce that AB & CD are parallel.

7. Find the value of λ such that the vectors (3 i+λ j+5 k ) , ( i+2 j−3 k )∧ (2 i− j+k )

are coplanar.8. The scalar product of the vector a=i+ j+k with a unit vector along the sum of the vectors

b=2 i+4 j−5 k and c=μ i+2 j+3 k is equal to one .Find the value of μ and hence find the

unit vector along b+ c Ans:μ=¿1,17

(3 i+6 j−2 k )

9. Find the value of λ if the points A( -1,4,-3 ),B(3,𝛌,-5),C(-3,8,-5) and D(-3,2,1) are coplanar. Ans: λ=2

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3D GEOMETRY

INTRODUCTIONSummary

1. Distance formula: Distance between two points A (x1 , y1 , z1 ) and B (x2 , y2 , z2 ) isAB=√(x2−x1)

2+( y2− y1)2+(z2−z1)

2

2. Section formula: Coordinates of a point P, which divides the line segment joining two given points A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ) in the ratio m:n(i). internally,

are P (m x2+n x1

m+n,m y2+n y1

m+n,m z2+n z1

m+n ) , the Coordinates of a point Q divides the line segment joining two given points in the ratio m:n

(ii). externally are Q (m x2−n x1

m−n,m y2−n y1

m−n,m z2−n z1

m−n )(iii).coordinate of mid-point are R( x2+x1

2,

y2+ y1

2,

z2+z1

2 )3. Direction cosines of a line :

(i).The direction of a line OP is determined by the angles α , β , γ which makes with OX, OY,OZ respectively. These angles are called the direction angles and their cosines are called the direction cosines.(ii).Direction cosines of a line are denoted by l, m, n; l = cos α ,m =cos β , n=cos γ(iii).Sum of the squares of direction cosines of a line is always 1. l2 + m2 + n2 =1 i.e cos2α + cos2β + cos2γ = 1

4. Direction ratio of a line :(i)Numbers proportional to the direction cosines of a line

are called direction ratios of a line .If a ,b ,c, are , direction ratios of a line, then la=m

b=n

c .

(ii). If a ,b ,c, are , direction ratios of a line , then the direction cosines are

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± a√a2+b2+c2 ,±

b√a2+b2+c2 ±

c√a2+b2+c2

( iii ) . Direction ratio of a line AB passing through the points A(x1, y1, z1) and B (x2, y2, z2) are x2−x1 , y2− y1 , z2−z1

5. STRAIGHT LINE:. (i). Vector equation of a Line passing through a point a and along the direction b, : r = a+μ b,

(ii).Cartesian equation of a Line: x−x1

a=

y− y1

b=

z−z1

c. Where (x1, y1, z1 ) is the given

point and its direction ratios are a,b,c.

6. (i). Vector equation of a Line passing through two points, with position vectors a∧b r=a+μ¿ - a¿

(ii). ).Cartesian equation of a Line:x−x1

x2−x1=

y− y1

y2− y1=

z−z1

z2−z1, two points are (x1,y1) and

(x2,y2).7. ANGLE between two lines (i). Vector equations: r = a1+ λ b1 and r = a2+μ b2,

(ii) ).Cartesian equations: If lines are x−x1

a1=

y− y1

b1=

z−z1

c1 ,

x−x2

a2=

y− y2

b2=

z−z2

c2

cosθ= b1 . b 2|b 1|.|b 2|

(iii). If two lines are perpendicular, then b1.b2=0, i.e. a1a2+b1b2+c1c2=0

(iv) . If two lines are parallel, then b1 =t b2, where t is a scalar. OR b1×b2 = 0, OR a1

a2=

b1

b2=

c1

c2

(v).Ifθ is the angle between two lines with direction cosines ,l1, m1, n1 and l2, m2, n2 then

(a).cosθ=¿¿l1l2 + m1m2 + n1n2 (b). if the lines are parallel, then l1

l2=

m1

m2=

n1

n2

(c). If the lines are perpendicular, then l1l2 + m1m2 + n1n2=0

8.(a).Shortest distance between two skew- lines:(i).Vector equations: r=a1+ λ b1, and: : r=a2+μ b2,

d=|(a 2− a1 ). (b1 ×b2 )|b 1× b2| |.

If shortest distance is zero, then lines intersect and line intersects in space if they are coplanar. Hence if above lines are coplanarIf (a 2−a 1 ) . ( b 1× b 2 )=0

(ii). Cartesian equations: x−x1

a 1= y− y 1

b1= z−z 1

c1, x−x2

a 2= y− y 2

b2= z−z 2

c2

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9.If shortest distance is zero, then lines intersect and line intersects in space if they are coplanar. Hence if above lines are coplanar

|||x2−x1 y2− y1 z2−z1

a1 b1 c1

a2 b2 c2|=0

(b).Shortest distance between two parallel lines: If two lines are parallel, then they are coplanar.Let the lines be : r=a1+ λ b, and: r=a2+μ b,

D =|b× ( a2−a 1 )|b| |

10.General equation of a plane in vector form :- It is given by r . n+d=0 , n is a vector normal to plane.11.General equation of a plane in Cartesian form :- ax+by+cz+d=0 , Where a,b,c are direction ratios of normal to the plane.12.General equation of a plane passing through a point :- if position vector of given point is a then equation is given by ( r−a ) . n=0,n is a vector perpendicular tothe plane.13.General equation of a plane passing through a point :- if coordinates of point are(x , y , z) then equation is a (x−x1 )+b ( y− y1 )+c ( z−z1 )=0, a,b,care direction ratios of a line perpendicular to the plane.14.Intercept form of equation of a plane :-General equation of a plane which cuts off

intercepts a, b and c on x-axis, y-axis, z-axis respectively is xa+ y

b+ z

c=1.

15. Equation of a plane in normal form:- r . n=¿ p,where n is a unit vector along perpendicular from origin and ‛p’ is distance of plane from origin.p is always positive.16.Equation of a plane in normal form :- It is given by lx+my+nz=p, where l ,m,n are direction cosines of perpendicular from origin and ‛p’ is distance of plane from origin. p is always positive.17.Equation of a plane passing through three non-collinear points :- If a , b , c are the position vectors of three non-collinear points, then equation of a plane through three points is given by – ( r−a ) . {( b− a)×(c− a) }=0.18. Equation of a plane passing through three non-collinear points(Cartesian system) :- If plane passing through points (x1 , y1 , z1 ¿ , (x2, y2 , z2 ¿ and (x3 , y3 , z3 ¿ then equation is- ¿

19.If θ is angle between two planes r . n1+d1=0 and r . n2+d2=0 then cosθ=n1 . n2

|n1||n2|

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(i) If planes are perpendicular, then n1 . n2=0 (ii) If planes are parallel, then n1 ×n2=0 or n1= λ n2 , λ is a scalar.20. If θ is angle between two planes a1 x+b1 y+c1 z+d1=0∧a2 x+b2 y+c2 z+d2=0

Then cosθ=a1 a2+b1 b2+c1 c2

√(a12+b1

2+c12)(a2

2+b22+c2

2)(i) If planes are perpendicular ,then a1a2+b1b2+c1c2=0

(ii) If planes are parallel , then a1

a2=

b1

b2=

c1

c2

21. If θ is angle between line r=a+λ m and the plane r . n+d=0 ,then sin θ= m. n|m|.|n|

(i)If line is parallel to plane ,then m .n=0 and (ii)If line is perpendicular to plane , then m× n=0 or m=t n ,t is a scalar.

22 . If θ is angle between line x−x1

a1=

y− y1

b1=

z−z1

c1 and the plane ax+by+cz+d=0 ,then

sin θ=a a1+bb1+cc1

√(a12+b1

2+c12)(a2+b2+c2)

(i)If line is parallel to the plane ,then aa1+bb1+cc1=0

(ii)If line is perpendicular to the plane, then aa1= b

b1= c

c1

23. General equation of a plane parallel to the planer . n+d=0is r . n+λ=0, where λ is a constant and can be calculated from given condition.24. General equation of a plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz +λ=0, where λ is a constant and can be calculated from given condition.25. General equation of a plane (vector form) passing through the line of the intersection of planes

r . n1+d1=0 and r . n2+λ d2=0 is r . (n1+λ n2 )+(d1+ λ d2 )=0 , where λ is a constant and can be calculated from given condition.

26. General equation of a plane(Cartesian form) passing through the line of the intersection of planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 is( a1x+b1y+c1z+d1)+ λ (a2x+b2y+c2z+d2 )=0 , where λ is a constant and can be calculated from given condition.27. Distance of a plane(vector form)r . n+d=0

, from a point with position vector a , is|a . n+d|n| |.

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28. Distance of a plane(Cartesian form)ax+by+cz+d=0, , from a point (x1,y1,z1) is

|a x1+b y1+c z1+d

√a2+b2+c2 |.FLOW CHART

(a) To find shortest distance between two skew lines1 Find a1, a2, b1 &b2

2 Find a2−a1

3 Findb1×b2

4 Find |b1× b2|5 ( b1× b2 ) . (a2−a1 )

6 Distance = |(b1× b2 ) . (a2− a1 )|b1× b2| |

(b) To find Coordinates of foot of perpendicular from origin to the plane:1. Direction ratio of any line perpendicular to the given plane.2. Equation of the line through origin and perpendicular to the given plane. 3. Coordinates of any point(P) on the line.4. Find the value of λ, by putting the coordinates of P in the plane.5. Find required coordinates of foot of perpendicular

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63

(c) To find coordinates of image of a point in the plane

1. Direction ratios of any line (PP’) perpendicular to the given plane.

2. Equation of the line PP’ through a given point (P) and perpendicular the given plane.

3. Coordinates of any point on the plane (say P’)

4. Coordinates of midpoint (M) of PP’

5. Find the value of r ,as M will satisfy the plane

6. Then find coordinates of P’ Images of point P

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QUESTIONS ON 3-D.LEVEL-1

6) The equation of a line is given by4−x

2= y+3

3= z+2

6 . Write the direction cosines of a line

parallel to given line. Ans:¿27

, 37

, 67>¿

7) Find the equation of the plane passing through the point (-1,3, 2) and perpendicular to the planes x +2y +3z = 5 and3x +3y + z = 0.

8) Find the co-ordinates of the point where the line through (3, 4, 1) and (5, 1, 6) crosses the xy-plane .

9) Find the shortest distance between the following pair of lines :x−1

2= y−2

3= z−3

4; x−2

3= y−4

4= z−5

5Ans: 1

√6

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LEVEL-2

7. Find the Coordinate of foot of the perpendicular from origin to the plane 3x+4y-5z=7

8. Q 1: Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the

origin, then

9. Find the distance of the point (−1 ,−5 ,−10 ) from the point of intersection of the line r=(2 i− j+2 k )+λ ¿ ) and the plane r . ( i− j+ k )=5. ANS :13 units .

10. Find the distance between the point P(6,5,9 ) and the plane determined by the points A

(3 ,−1,2 ),B(5,2,4 ) and C(−1 ,−1,6 ). ANS:6

√34units.

11. Find the equation of the plane passing through the points (3,4,1), (0,1,0) and is parallel to line

x+3

2= y−3

7= z−2

5 Ans:8 x−13 y+15 z+13=0

12. Find the values of p so that the lines :1−x

3=7 y−14

2 p= z−3

2 and 7−7 x

3 p= y−5

1=6−z

5

are at right angles Ans: 7011

13. Find the shortest distance between the lines x+1=2 y=−12 z and x= y+2=6 z−6 Ans: 2

14. Find the shortest distance between the following pair of lines :x−1

2= y+1

3=z ; x+1

5= y−2

1, z=2

Ans: 3

√59

LEVEL-3

10. Find the equation of the plane which is perpendicular to the plane 5 x+3 y+6 z+8=0 and which contains the line of intersection of the planes x+2 y+3 z−4=0 and 2 x+ y−z+5=0 .

Ans :51 x+15 y−50 z+173=0

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11. Find the equation of a plane which is at a distance of 7 units from the origin and which is

normal to the vector 3 i¿

+5 j¿

−6k¿

. Ans.:- r→

.(3i¿

+5 j¿

−6 k¿

)−7√70=0

12. Find a unit vector perpendicular to the plane of the triangle ABC, where the coordinates of its vertices are A (3 ,−1,2 ) , B (1 ,−1 ,−3 ) and C (4 ,−3,1 ).

Ans: 1

√165(10 i+7 j−4 k )

13. Find the image of the point (1,6,3 ) in the line x1= y−1

2= z−2

3.

14. Find the image of the point (1,3,4 ) in the plane x− y+z=5.

15. Find the vector and the Cartesian equations of the plane passing through the intersection of the planes r . ( i+ j+k )=6 and r . (2 i+3 j+4 k )=−5 and the point(1 ,1 , 1 ).

Ans:20 x+23 y+26 z−69=0 r . (20 i+23 j+26 k )=6916. Find the shortest distance between the lines whose vector equations are:

r=(1−t ) i+( t−2 ) j+ (3−2t ) k∧ r=( s+1 ) i+ (2 s−1 ) j−(2 s+1 ) k Ans:8√2929

17. . Find the distance of the point (1 ,−2,3 )¿ the plane x-y+z=5 measured ‖ to the

linex+12

= y+33

= z+1−6 . ANS: other point (

97

,−117

, 157

¿ , distance=1unit

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Study Module of Linear programming problems

LINEAR PROGRAMMINGSCHEMATIC DIAGRAM

Topic concept Degree of Importance

Linear Programming (i)Introduction **(ii)Some solved problems

***

(iii) Diet Problem ***

iv) Manufacturing Problem

***

(v) Allocation Problem

**

(vi) Transportation Problem

*

vii) Miscellaneous Problems

**

Introduction:

Linear programming problems: A Linear Programming Problem is one that is concerned with finding the optimal value (maximum or minimum value) of a linear function (called objective Function) of several variables (say x and y), subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints).The term linear implies that all the mathematical relations used in the problem are linear relations while the term programming refers to the method of determining a particular plan of action.

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Objective function : Linear function Z = ax + by, where a, b are constants, which has to be maximised or minimized is called a linear objective function.

Constraints: The linear inequalities or inequations or restrictions on the variables of a

linear programming problem are called constraints. The conditions x ≥ 0, y ≥0 are

called non-negative restrictions.

Optimization problem: A problem which seeks to maximise or minimise a linear

function (say of two variables x and y) subject to certain constraints as determined by

a set of linear inequalities is called an optimisation problem. Linear programming

problems are special type of optimisation problems.

Feasible region: The common region determined by all the constraints including

non-negative constraints x, y of a linear programming problem is called the feasible

region (or solution region) for the problem.

Optimal (feasible) solution: Any point in the feasible region that gives the optimal

value (maximum or minimum) of the objective function is called an optimal solution.

IMPORTANT SOLVED PROBLEMS

Q1. A dietician wishes to mix together two kinds of foods X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units vitamin B and 8 units of vitamin C. The vitamin contents on one kg. food is given below :

Food Vitamin A Vitamin B Vitamin C

XY

12

22

31

One kg. of food X costs Rs. 16 and one kg. of food Y costs Rs. 20. Find the least cost of the mixture which will produce a required diet?

Sol. Let x kg and y kg food of two kinds of foods X and Y to be mixed in a diet respectively.

The contents of one kg. food of each kind as given below:

Food Vitamin A Vitamin B Vitamin C Cost

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X 1 2 3 16

Y 2 2 1 20Minimum

Requirement 10 12 8

The above L.P.P. is given as

Minimum, Z = 16x + 20 y

subject to the constraints

x + 2y ≥10, 2x + 2y ≥ 12,

3x + y ≥ 8, x, y ≥ 0

L1 : x + 2y = 10 L2 : x + y = 6 L3 : 3x + y = 8

A B C D E F

Corner points Z = 16x + 20 yA (10,0) 160F (0, 8) 160

69

X 10 0Y 0 5

x 2 0y 2 8

X 6 0Y 0 6

70

G (1, 5) 116H (2,4) 112

Here the cost is minimum at H (2,4)

Since the region is unbounded therefore Rs. 112 may be or may not be the minimum value of C.

For this draw of inequality

16x + 20y < 112

i.e. 4x + 5y -< 28

L : 4x + 5 y = 28

x 7 2y 0 4

Clearly open half plane has no common point with the feasible region so minimum value of Z is Rs. 112.

Q2. An aero plane can carry a maximum of 200 passengers. A profit of Rs. 1000 is made on each executive class ticket and a profit of Rs. 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?

SOL. Let the number of executive class ticket = x

And the number of economy class tickets = y

Given, maximum capacity of passengers = 200

∴ x + y ≤ 200

Atleast 20 seats of executive class are reserved.

∴ x ≥ 20

Also atleast 4x seats of economy class are reserved

∴ y ≥ 4x

Therefore, above L.P.P. is given as

Maximum P = 1000x + 600y \

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subject to the constraints

x + y ≤ 200 ,

x ≥ 20,

y ≥ 4x or 4x – y ≤ 0

x ≥ 0, y ≥ 0

L1 : x + y = 200 L2 : 4x + y = 0

A B C D

L3 : x = 20

(Maximum)

∴ here profit is maximum at F (40,160)

∴ 40 tickets of executive class and 160 tickets of economy class to sold to get maximum profit and maximum profit is Rs. 136000.

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X 0 200Y 200 0

X 0 50Y 0 200

Corner Points P = 1000x + 600yE (20, 80) 68000

F ( 40, 160) 136000G (20, 180) 128000

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Q3. A factory manufactures two types of screws, A and B; each type requiring the use of two machines an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit Rs. 7 and screws B at a profit of Rs. 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.

Sol. Let number of packages of screws A produced = x

And number of packages of screws B produced = y

The number of minutes for producing 1 unit of each item is given below:

Screw Automatic Machine

Hand operated Machine

Profit

A 4 6 7B 6 3 10

Time available

240 240

Therefore, the above L.P.P. is given as

Maximise, P = 7x + 10 y subject to the constraints.

4x + 6y ≤ 240 ; 6x + 3y ≤ 240

i.e. 2x + 3y ≤120 : 2x + y ≤80, x, y≥0

L1 ; 2x + 3y = 120 L2 : 2x + y = 80 A B C D

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(maximum)

Here profit is maximum at E (30,20)

∴ Number of packages of screws A = 30

Number of packages of screws B = 20

Maximum profit = Rs. 410.

Flow Chart

Step 1. Write the given informations in the tabulated form.

Step2. Form the L.P.P model of the problem.

Step3. Draw all the constraints by converting them in to equations.

Now we solve the L.P.P. by CORNER POINT METHOD which has the following steps

Step 1. Find the feasible region of the linear programming problem bounded by all the constraints and determine its corner points (vertices) either by inspection or by solving the two equations of the lines intersecting at that point.

73

X 60 0Y 0 40

x 40 0y 0 80

Corner points P = 7x + 10yO (0,0) 0C (40,0) 280B (0,40) 400E (30, 20) 410

74

Step 2. Evaluate the objective function Z = ax + by at each corner point. Let M and m, respectively denote the largest and smallest values of these points.

Step 3. (i) When the feasible region is bounded, M and m are the maximum and minimum values of Z.

(ii) In case, the feasible region is unbounded, we have:

Step 4. (a) M is the maximum value of Z, if the open half plane determined by

ax + by > M has no point in common with the feasible region. Otherwise, Z

has no maximum value.

(b) Similarly, m is the minimum value of Z, if the open half plane determined by ax + by < m has no point in common with the feasible region. Otherwise, Z has no minimum value

ASSIGNMENTS(i) LPP and its Mathematical Formulation

LEVEL I1. A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires at least 240 units of calcium, at least 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A?(ii) Graphical method of solving LPP (bounded and unbounded solutions)

LEVEL ISolve the following Linear Programming Problems graphically:1. Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.2. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.3. Minimise Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

(iii) Diet Problem

LEVEL II1. A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1,400 calories. Two foods X and Y are available at a cost of Rs. 4 and Rs. 3 per unit respectively. One unit of the food X contains 200 units of vitamins, 1 unit of mineral and 40 calories, whereas one unit of food Y contains 100 units of vitamins, 2 units of minerals and 40 calories. Find what combination of X and Y should be used to have least cost? Also find the least cost.

2. A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2

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units/kg of vitamin A and 1 unit/kg of vitamin C.Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate this problem as a linear programming problem to minimise the cost of such a mixture. In what way a balanced and healthy diet is helpful in performing your day-to-day activities (iv) Manufacturing Problem

LEVEL II1.A company manufactures two articles A and B. There are two departments through which these articles are processed: (i) assembly and (ii) finishing departments. The maximum capacity of the assembly department is 60 hours a week and that of the finishing department is 48 hours a week. The production of each article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing. If the profit is Rs. 6 for each unit of A and Rs. 8 for each unit of B, find the number of units of A and B to be produced per week in order to have maximum profit.

2. A company sells two different produces A and B. The two products are produced in a common production process which has a total capacity of 500 man hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The demand in the market shows that the maximum number of units of A that can be sold is 70 and that for B is 125. Profit on each unit of A is Rs. 20 and that on B is Rs. 15. How many units of A and B should be produced to maximize the profit? Solve it graphically. What safety measures should be taken while working in a factory?

Q3.A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?

LEVEL III

1.A manufacture makes two types of cups, A and B. Three machines are required to manufacture the cups and the time in minutes required by each is as given below:

Type of Cup MachinesI II III

A 12 18 6B 6 0 9

Each machine is available for a maximum period of 6 hours per day. If the profit on each cup A is 75 paise, and on B it is 50 paise, show that the 15 cups of type A and 30 cups of type B should be manufactured per day to get the maximum profit.(v) Allocation Problem

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LEVEL II1. Ramesh wants to invest at most Rs. 70,000 in Bonds A and B. According to the rules, he has to invest at least Rs. 10,000 in Bond A and at least Rs. 30,000 in Bond B. If the rate of interest on bond A is 8 % per annum and the rate of interest on bond B is 10 % per annum , how much money should he invest to earn maximum yearly income ? Find also his maximum yearly income.

Q2 A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

LEVEL III

1. An aeroplane can carry a maximum of 250 passengers. A profit of Rs 500 is made on each executive class ticket and a profit of Rs 350 is made on each economy class ticket. The airline reserves at least 25 seats for executive class. However, at least 3 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?

Answers(i) LPP and its Mathematical Formulation

LEVEL I

1. Z = 6x + 3y, 4x + y ≥ 80, x + 5y ≥115, 3 x+2 y ≤150 x, y ≥0

(ii) Graphical method of solving LPP (bounded and unbounded solutions)

1. Minimum Z = – 12 at (4, 0), 2. Maximum Z =

1945,

1920at

19235

3. Minimum Z = 7 at

21,

23

(iii) Diet Problem

LEVEL II

1. Least cost = Rs.110 at x = 5 and y = 30

2. Minimum cost = Rs.380 at x = 2 and y = 4

(iv) Manufacturing Problem

LEVEL II

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1. Maximum profit is Rs. 120 when 12 units of A and 6 units of B are produced

2. For maximum profit, 25 units of product A and 125 units of product B are produced

and sold.

3. 800 dolls of type A and 400 dolls of type B; Maximum profit = Rs 16000

(v) Allocation Problem

LEVEL II

1. Maximum annual income = Rs. 6,200 on investment of Rs. 40,000 on Bond A and

Rs. 30,000 on Bond B.

Q2. 200 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150000.

LEVEL III

1. For maximum profit, 62 executive class tickets and 188 economy class ticket should be sold.

PROBABILITY:INTRODUCTION:Topic Concepts Degree of Importance Reference from NCERT

Book Vol.IIProbability (i) Conditional Probability *** Article 13.2 and 13.2.1

Solved Ex. 1 to 6Ex, 13.1 Q.N.-1,5 to 15

(ii) Multiplication theorem on probability

** Article 13.3solved Ex. 8 & 9Ex. 13.2 Q.N.-2,3,13,14, 16

(iii) Independent events *** Article 13.4Solved Ex. 10 to 14Ex 13.2 Q.N.-1,6,7,8,11

(iv) Baye’s Theorem, Partition of a sample space and theorem of total probability

*** Article 13.5, 13.5.1, 13.5.2Solved Ex. 15 to 21Ex. 13.3 Q.N.-1 to 12Misc. Ex. Q.N. 13 to 16

(v) Random Variables & *** Articles 13.6, 13.6.1, 13.6.2

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Probability distribution Mean & Variance of Random Variables

13.6.3Solved Ex. 24 to 29Ex 13.4 Q.No.-1,4 to 15

(vi) Bernaulli’s trails and Binomial distribution

*** Articles 13.7, 13.7.1, 13.7.2Solved Ex. 31 & 32Ex. 13.5 Q.N.- 1 to 13

Concept Mapping:

Conditional Probability:P ( E/F )=P ( E ∩ F )P ( F )

,P(F)≠ 0

Multiplication Theorem: P ( E ∩ F )=P ( E ) . P (F /E )

If E and F are independent events then P ( E ∩ F )=P ( E ) . P (F ) and vice versa

Bayes Theorem: If E1 ,E2 and E3 are three events of sample space S and

E1∪ E2∪ E3 = S & pairwise disjoint sets. A is any event of nonzero probability. Then

P (E1/A )=P (E1 ) P (A /E1 )

P (E1 ) P (A /E1 )+P (E2 ) P ( A /E2 )+P (E3 ) P ( A /E3 )Probability Distribution: The probability distribution of a random variable X is the system of numbers

X : x1 x2 x3 ... xn

P(X) : p1 p2 p3 … pn

Where, pi¿0 ,∑i=1

n

pi=1 , i=1,2 ,… ,n

Mean of a random variable X i.e. E(X) = μ = ∑i=1

n

x i p i , i=1,2 , …, n

Variance of a random variable X i.e. σ 2x=E (X 2)−[ E(X) ]2

Standard Deviation: σ x=√Variance Binomial Distribution: B (n, p)

P(X = r) = nCr pr qn−r , r=0 , 1 ,…,nand q = 1-p, where p is the probability of success.Solved Examples:

1. A biased die is twice as likely to show an even number as an odd number. The die is rolled three times. If occurrence of an even number is considered a success, then write the probability distribution of number of successes. Also find the mean number of successes.Which human value is violated in this case.

Solution:

P(odd number) = 1

1+2=1

3 P (even number) = 2

1+2=2

3Here occurrence of an even number is considered a success. Let the number of success is a random variable x and can take values 0, 1, 2 or 3.The probability distribution of number of successes is as below:

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P(x = 0) = P (no success) = P (FFF) =

13×1

3×1

3= 1

27

P(x = 1) = P (one success) = P(SFF, FSF, FFS) = 3( 2

3×1

3×1

3 )= 627

P (x = 2) = P (two success) = P(SSF, SFS, FSS) = 3( 2

3×2

3×1

3 )=1227

P (x = 3) = P (three success) = P (SSS) = ( 2

3× 2

3×2

3 )= 827

X = xi 0 1 2 3P(x) = pi 1/27 6/27 12/27 8/27

Mean number of successes = ∑ x i pi

= (0× 1

27 )+(1× 627 )+(2×12

27 )+(3× 827 )=54

27=2

Having unbiased is violated in this case.

Q.2 Probabilities of solving a specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.

Solution:P(A) = 1/2 = prob. that A will solve the problemP(B) = 1/3 = prob. that B will solve the problem

(i) Probability that the problem is solved= 1 - prob. that none of them solve the problem

= 1−P ( A ) .P ( B ) =1−(1−1

2 )(1−13 )=1−( 1

2×2

3 )=23

(ii) Probability that exactly one of them will solve the problem

=

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

¿12×(1−1

3 )+(1−12 )×1

3=(12 ×2

3 )+(12 ×13 )=3

6=1

2

Q.3 Two cards are drawn simultaneously without replacement from a well shuffled pack of 52 cards. Find the mean and variance of the number of aces.

Solution:Let x denote the number of aces in a draw of two cards. x is a random variable which can assume the values 0, 1 and 2

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P( x=0 )=P(no ace )=48C2

52C2=

48×472×152×512×1

=188221

P( x=1 )=P(one ace and one non−ace )

=4 C1×48 C1

52C2=4×48×2

52×51=32

221

P( x=2 )=P( two aces)=4 C2

52 C2=4×3

52×51=1

221The probability distribution of x is

x or xi 0 1 2P(x) = pi 188/221 32/221 1/221

Mean = E(x) = ∑ x i pi

= (0×188

221 )+(1×32221 )+(2× 1

221 )= 213

E (x2)=∑ x i2 p i=(0×188

221 )+(1×32221 )+(4×1

221 )=36221

Var ( x )=E ( x2 )−(E ( x ) )2=36221

−(213

2)=4002873

Q.4 A family has 2 children. Find the probability that both are boys, if it is known that(i) at least one of the children is a boy (ii) the elder child is a boy

Solution:S= {B1 B2 ,B1 G2 ,G1 B2 ,G1 G2}

(i) at least one of the children is a boy

A = Both the children are boys {B1 B2 }B = At least one of the children is a boy= {B1 B2 ,B1 G2 , G1 B2}

Required probability =

P( AB )=P ( A∩B )

P(B)=

14

34

=13

(ii) The elder child is a boy

A = Both the children are boys = {B1 B2 }B = elder child is a boy = = {B1 B2 , B1 G2 }

Required probability =

P( AB )=P ( A∩B )

P(B)=

14

24

=12

Q.5 Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of

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certain drugs reduces his chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random, suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga. In a student life state any one point how yoga and meditation influence.

Solution:Let E1 and E2 be events of selection of meditation and yoga and prescription of medicine respectively.Let A = event of having heart attack.

We have P (E1 )=P (E2)=

12

P(AE1 )=(40−30

100 ×40)%=28100

P(AE2 )=(40−25

100×40)%=30

100

Required probability = P ( E1

A )

=

P (E1 )×P( AE1 )

P (E1 )×P( AE1 )+P (E2 )×P( A

E2 )=

12×28

10012×28

100+ 1

2×30

100

=2858

=1429

yoga and meditation improves our physical and mental health.

Q.6 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?Which mode of transport would you suggest to students and why?

Solution:Let E1, E2 and E3 are the events of selection of a scooter driver, car driver and truck driver respectively. Let A = event that the insured person meets with an accident.

P (E1 )=200012000

=16

P ( E2 )=400012000

=13

P (E3)=600012000

=12

P( AE1 )=0 .01 P( A

E2)=0 .03 P( AE3 )=0 . 15

Required probability = P ( E1

A )

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=

P (E1)×P(AE1 )

P (E1)×P(AE1 )+P (E2)×P(A

E2 )+P (E3)×P(AE3 )

=

16×0 .01

16×0 .01+1

3×0 .03+1

2×0 . 15

¿0 . 010 . 01+0 .06+0 . 45

=0. 010. 52

=152

Cycle should be suggested as it is good for (i) health (ii) no pollution (iii) saves energy (no fuel).

Practice ProblemLevel-1

1. If P(A) = 0.3, P(B) = 0.2 find P( B

A ), if A and B are mutually exclusive events.

2. A coin is tossed thrice and all the 8 outcomes are equally likely:E: the first throw results in headF: the last throw results in tailAre the events independent?

3. Given P(A) = 1/4 , P(B) = 2/3 and P(AUB) = 3/4. Are the events independent?4. If A and B are independent events, find P(B) if P(AUB) = 0.60 and P(A) = 0.35.5. Two cards are drawn with replacement from a well shuffled pack of 52 cards. Find the

probability distribution of the number of spades.6. 4 defective apples are accidentally mixed with 16 good ones. Three apples are drawn at

random from the mixed lot. Find the probability distribution of the number of defective apples.

7. A random variable X is specified by the following distribution:X 2 3 4P(X) 0.3 0.4 0.3

Find the mean and variance of distribution.

Level-21. A dice is thrown twice and sum of numbers appearing is observed to be 6. What is the

conditional probability that the number 4 has appeared at least ones?2. The probability of A hitting a target is 3/7 and that of B hitting is 1/3. They both fire at

their target find the probability that:(a) at least one of them will hit the target(b) only one of them will hit the target

3. A company has two plants to manufacture bicycles. The first plants manufactures 60% of the bicycle and second plant 40%. Out of that 80% of the bicycles are rated of standard quality at the first plant and 90% of the standard quality at the second plant. A bicycle is picked up at random and found to be standard quality. Find the probability that it comes from the second plant.

Level-3

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1. A class consists of 80 students, 25 of them are girls and 55 are boys. 10 of them rich and remaining poor; 20 of them are fair complexioned. What is the probability that selecting a fair complexioned rich girl.

2. Two integers are selected from integers 1 to 11. If the sum is even, find the probability that both numbers are odd.

3. Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

4. A coin is biased so that the head is 3 times as likely to occur as a tail. If the coin is tossed twice, find the probability distribution of the number of tails.

5. The mean and Variance of a binomial distribution are 4/3 and 8/9 respectively. Find P(X 1).6. A card is drawn from a pack 52 cards is lost. From the remaining cards of the pack, two

cards are drawn and are found to be both hearts. Find the probability of the lost card being a heart.

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