PEDAGOGICAL ISSUES

IN

MATHEMATICS

AT

SENIOR SECONDARY LEVELFOR LECTURERS

2012

Chief AdvisorDr. Anita Satia

Director, SCERT

Guidance

Dr. Pratibha SharmaJoint Director, SCERT

Academic Co-ordinatorDr. Anil Kumar Teotia

Sr. Lecturer, DIET Dilshad Garden

Mr. Sanjay KumarLecturer, SCERT

ContributorsDr. Anup Rajput Associate Professor, NCERTDr. Anil Kumar Teotia Sr. Lecturer, DIET Dilshad GardenDr. Satyavir Singh Principal, SNI College, PilanaMr. D.R. Sharma Vice Principal, JNV, MungeshpurDr. R.P. Singh Lecturer, RPVV, Gandhi NagarMr. S.B. Tripathi Vice-Principal, GSSS, Jheel Khurenja

EditorsDr. Anil Kumar Teotia

Sr. Lecturer, DIET Dilshad Garden

Dr. Rashmi AgarwalSr. Lecturer, DIET Dilshad Garden

Mr. Sunil KumarSr. Lecturer, DIET Dilshad Garden

In-charge PublicationMs. Sapna Yadav, Ms. Meenakshi Yadav

Publication TeamSh. Naveen Kumar, Ms. Radha, Sh. Jai Bhagwan

Published by : State Council of Educational Research & Training, New Delhi and printed atEducational Stores, S-5, Bsr. Road Ind. Area, Ghaziabad (U.P.)

Contents

S.No. Chapter Name Page No.

Syllabus for Class-XI 05

Syllabus for Class-XII 08

Designing and Sample Paper for Class-XII Session 2012-13 11

Question Paper-Delhi (2012) 27

Question Paper-Outside Delhi (2012) 39

Question Paper-Outside India(2012) 42

1. Relations and Functions 45

2. Inverse Trigonometric Functions 51

3. Matrices and Determinants 60

4. Continuity and Differentiability 71

5. Application of Derivative 82

6. Integrals 92

7. Application of Integrals 109

8. Differential Equations 114

9. Vectors 121

10. Three Dimensional Geometry 130

11. Probability 139

12. Linear Programming 148

4

MATHEMATICS (Code No. 041)

The Syllabus in the subject of Mathematics has undergone changes from time to time in accordancewith growth of the subject and emerging needs of the society. Senior Secondary stage is a launchingstage from where the students go either for higher academic education in Mathematics or for professionalcourses like Engineering, Physical and Bioscience, Commerce or Computer Applications. The presentrevised syllabus has been designed in accordance with National Curriculum Framework 2005 and asper guidelines given in Focus Group on Teaching of Mathematics 2005 which is to meet the emergingneeds of all categories of students. Motivating the topics from real life situations and other subjectareas, greater emphasis has been laid on application of various concepts.

Objectives

The broad objectives of teaching Mathematics at senior school stage intend to help the pupil:

to acquire knowledge and critical understanding, particularly by way of motivation and visualization,of basic concepts, terms, principles, symbols and mastery of underlying processes and skills.

to feel the flow of reasons while proving a result or solving a problem.

to apply the knowledge and skills acquired to solve problems and wherever possible, by morethan one method.

to develop positive attitude to think, analyze and articulate logically.

to develop interest in the subject by participating in related competitions.

to acquaint students with different aspects of mathematics used in daily life.

to develop an interest in students to study mathematics as a discipline.

to develop awareness of the need for national integration, protection of environment, observanceof small family norms, removal of social barriers, elimination of sex biases.

to develop reverence and respect towards great Mathematicians for their contributions to the fieldof Mathematics.

5

Course Structure Class-XI

One Paper Three Hours Max Marks. 100

Units Marks

I. Sets and Functions 29

II. Algebra 37

III. Coordinate Geometry 13

IV. Calculus 06

V. Mathematical Reasoning 03

VI. Statistics and Probability 12

Total 100

UNIT I : SETS AND FUNCTIONS

1. Sets : (12) Periods

Sets and their representations. Empty set. Finite and Infinite sets. Equal sets. Subsets. Subsets of theset of real numbers especially intervals (with notations). Power set. Universal set.Venn diagrams. Unionand Intersection of sets. Difference of sets. Complement of a set. Properties of complement sets.

2. Relations and Functions : (14) Periods

Ordered pairs, Cartesian product of sets. Number of elements in the cartesian product of two finitesets. Cartesian product of the reals with itself (uptoR × R × R). Definition of relation, pictorialdiagrams, domain, co-domain and range of a relation. Function as a special kind of relation fromone set to another. Pictorial representation of a function, domain, co-domain & range of a function.Real valued functions of the real variable, domain and range of these functions, constant, identity,polynomial, rational, modulus, signum and greatest integer functions with their graphs. Sum, difference,product and quotients of functions.

3. Trigonometric Functions : (15) Periods

Positive and negative angles. Measuring angles in radians and in degrees and conversion fromone measure to another. Definition of trigonometric functions with the help of unit circle. Truthof the identity sin2 x + cos2 x = 1, for all x. Signs of trigonometric functions and sketch of theirgraphs. Expressing sin (x ± y) and cos (x ± y) in terms of sinx, sin y, cosx & cos y. Deducingthe identities like following :

6

tan( )x y =tan tan cot cot 1

,cot( ) ,1 tan tan cot cot

x y x yx y

x y y x

sin sinx y = 2sin cos ,cos cos 2cos cos ,2 2 2 2

x y x y x y x yx y

sin sinx y = 2cos sin ,cos cos 2sin sin .2 2 2 2

x y x y x y x yx y

Identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x. General solution of trigonometricequations of the type sin = sin , cos = cos and tan = tan . Proof and simple applicationof sine and cosine formulae.

UNIT II : ALGEBRA

1. Principle of Mathematical Induction : (06) Periods

Processes of the proof by induction, motivating the application of the method by looking at naturalnumbers as the least inductive subset of real numbers. The principle of mathematical inductionand simple applications.

2. Complex Numbers and Quadratic Equations : (10) Periods

Need for complex numbers, especially, 1, to be motivated by inability to solve every some

of the quadratic equations. Algebraic properties of complex numbers. Argand plane and polarrepresentation of complex numbers. Statement of Fundamental Theorem of Algebra, solution ofquadratic equations in the complex number system. Square-root of a complex number.

3. Linear Inequalities : (10) Periods

Linear inequalities. Algebraic solutions of linear inequalities in one variable and their representationon the number line. Graphical solution of linear inequalities in two variables. Solution of systemof linear inequalities in two variables-graphically.

4. Permutations and Combinations : (12) Periods

Fundamental principle of counting. Factorialn. (n!) Permutations and combinations, derivationof formulae and their connections, simple applications.

5. Binomial Theorem : (08) Periods

History, statement and proof of the binomial theorem for positive integral indices. Pascal’s triangle,

general and middle term in binomial expansion, simple applications.

6. Sequence and Series : (10) Periods

Sequence and Series. Arithmetic progression (A.P.), arithmetic mean (A.M.). Geometric progression(G.P.), general term of a G.P., sum ofn terms of a G.P., Arithmetic and geometric series, infiniteG.P. and its sum, geometric mean (G.M), relation between A.M. and G.M. Sum to n terms of thespecial series :

2 3

1 1 1

, , and .n n n

k k k

k k k

7

UNIT III : COORDINATE GEOMETRY

1. Straight Lines : (09) Periods

Brief recall of 2D from earlier classes. Shifting of origin. Slope of a line and angle between twolines. Various forms of equations of a line : parallel to axes, point-slope from, slope-intercept form,two-point form, intercepts form and normal form. General equation of a line. Equation of familyof lines passing through the point of intersection of two lines. Distance of a point from a line.

2. Conic Sections : (12) PeriodsSections of a cone : circle, ellipse, parabola, hyperbola, a point, a straight line and pair of intersectinglines as a degenerated case of a conic section. Standard equations and simple properties of parabola,ellipse and hyperbola. Standard equations of a circle.

3. Introduction to Three-dimensional Geometry : (08) Periods

Coordinate axes and coordinate planes in three dimensions. Coordinates of a point. Distance betweentwo points and section formula.

UNIT IV : CALCULUS

I. Limits and Derivatives : (18) Periods

Limit of a function. Derivative of a function introduced as rate of change both as that of distance

function and its geometric meaning.0 0

log (1 ) 1lim , lim

xe

x x

x e

x x. Definition of derivative, relate it

to slope of tangent of the curve, derivative of sum, difference, product and quotient of functions.Derivatives of polynomial and trigonometric functions.

UNIT V : MATHEMATICAL REASONING

I. Mathematical Reasoning : (08) Periods

Mathematically acceptable statements. Connecting words/phrases — consolidating the understandingof “if and only if (necessary and sufficient) condition”, “implies”, “and/or”, “implied by”, “and”,“or”, “there exists” and their use through variety of examples related to real life and Mathematics.Validating the statements involving the connecting words-difference between contradiction, converseand contrapositive.

UNIT VI : STATISTICS AND PROBABILITY

1. Statistics : (10) Periods

Measure of dispersion; mean deviation, variance and standard deviation of ungrouped/ groupeddata. Analysis of frequency distributions with equal means but different variances.

2. Probability : (10) Periods

Random experiments : outcomes, sample spaces (set representation). Events : occurrence of events,‘not’, ‘and’ & ‘or’ events, exhaustive events, mutually exclusive events. Axiomatic (set theoretic)probability, connections with the theories of earlier classes. Probability of an event, probabilityof ‘not’, ‘and’ & ‘or’ events.

8

One Paper Three Hours Max Marks. 100

Units Marks

I. RELATIONS AND FUNCTIONS 10

II. ALGEBRA 13

III. CALCULUS 44

IV. VECTORS AND THREE - DIMENSIONAL GEOMETRY 17

V. LINEAR PROGRAMMING 06

VI. PROBABILITY 10

Total 100

The Questions Paper will include questions(s) based on valuesto the extent of 5 marks.

UNIT I : SETS AND FUNCTIONS

1. Relations and Functions : (10) Periods

Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and ontofunctions, composite functions, inverse of a function. Binary operations.

2. Inverse Trigonometric Functions: (12) Periods

Definition, range, domain, principal value branches. Graphs of inverse trigonometric functions.Elementary properties of inverse trigonometric functions.

UNIT-II: ALGEBRA

1. Matrices: (18) Periods

Concept, notation, order, equality, types of matrices, zero matrix, transpose of a matrix, symmetricand skew symmetric matrices. Addition, multiplication and scalar multiplication of matrices, simpleproperties of addition, multiplication and scalar multiplication. Non-commutativity of multiplication

Syllabus Mathematics(041)Class-XII 2012-13

9

of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to squarematrices of order 2). Concept of elementary row and column operations. Invertible matrices and proofof the uniqueness of inverse, if it exists; (Here all matrices will have real entries).

2. Determinants: (20) Periods

Determinant of a square matrix (up to 3 x 3 matrices), properties of determinants, minors, cofactorsand applications of determinants in finding the area of a triangle. Adjoint and inverse of a squarematrix. Consistency, inconsistency and number of solutions of system of linear equations by examples,solving system of linear equations in two or three variables (having unique solution) using inverse ofa matrix.

UNIT-III: CALCULUS

1. Continuity and Differentiability: (18) Periods

Continuity and differentiability, derivative of composite functions, chain rule, derivatives of inversetrigonometric functions, derivative of implicit functions.Concept of exponential and logarithmicfunctions.

Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative offunctions expressed in parametric forms. Second order derivatives. Rolle’s and Lagrange’s Mean

Value Theorems (without proof) and their geometric interpretation.

2. Applications of Derivatives: (10) Periods

Applications of derivatives: rate of change of bodies, increasing/decreasing functions, tangents andnormals, use of derivatives in approximation, maxima and minima (first derivative test motivatedgeometrically and second derivative test given as a provable tool). Simple problems (that illustratebasic principles and understanding of the subject as well as real-life situations).

3. Integrals: (20) Periods

Integration as inverse process of differentiation. Integration of a variety of functions by substitution,by partial fractions and by parts, simple integrals of the following type to be evaluated.

2 2 22 2 2 2 2, , , ,

dx dx dx dx dx

x a ax bx cx a a x ax bx c

2 2 2 22 2

, , ,px q px q

dx dx a x dx x a dxax bx c ax bx c

2 2, ( ) .ax bx c dx px q ax bx c dx

Definite integrals as a limit of a sum, Fundamental Theorem of Calculus (without proof). Basicproperties of definite integrals and evaluation of definite integrals.

4. Applications of the Integrals: (10) Periods

Applications in finding the area under simple curves, especially lines, circles/parabolas/ellipses (instandard form only), Area between the two above said curves (the region should be clearly identifiable).

10

5. Differential Equations: (10) Periods

Definition, order and degree, general and particular solutions of a differential equation. Formation ofdifferential equation whose general solution is given. Solution of differential equations by method ofseparation of variables, homogeneous differential equations of first order and first degree. Solutionsof linear differential equation of the type:

,dy

py qdx

where p and q are functions of x or constant

dx

dy + px = q, where p and q are functions of y or constant

UNIT-IV: VECTORS AND THREE-DIMENSIONAL GEOMETRY

1. Vectors: (12) Periods

Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of avector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point,negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar,position vector of a point dividing a line segment in a given ratio. Scalar (dot) product of vectors,projection of a vector on a line. Vector (cross) product of vectors. Scalar triple product of vectors.

2. Three - dimensional Geometry: (12) Periods

Direction cosines and direction ratios of a line joining two points. Cartesian and vector equation of aline, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of aplane. Angle between (i) two lines, (ii) two planes. (iii) a line and a plane. Distance of a point from aplane.

UNIT-V: LINEAR PROGRAMMING

1. Linear Programming: (12) Periods

Introduction, related terminology such as constraints, objective function, optimization, differenttypes of linear programming (L.P.) problems, mathematical formulation of L.P. problems, graphicalmethod of solution for problems in two variables, feasible and infeasible regions, feasible andinfeasible solutions, optimal feasible solutions (up to three non-trivial constraints).

UNIT-VI: PROBABILITY

1. Probability: (18) Periods

Conditional probability, multiplication theorem on probability. independent events, total probability,Baye’s theorem, Random variable and its probability distribution, mean and variance of random

variable. Repeated independent (Bernoulli) trials and Binomial distribution.

11

Time : 3 Hours Max. Marks : 100

Weightage of marks over different dimensions of the question paper shall be as follows:

A. Weightage to different topics/content units

S.No. Topics Marks

1. Relations and Functions 10

2. Algebra 13

3. Calculus 44

4. Vectors & three-dimensional Geometry 17

5. Linear programming 06

6. Probability 10

Total 100

Note: The Question Paper will include question(s) based on values tothe extent of 5 marks.

B. Weightage to different forms of questions

S.No. Forms of Questions Marks for No. of Total markseach question Questions

1. Very Short Answer questions (VSA) 01 10 10

2. Short Answer questions (SA) 04 12 48

3. Long answer questions (LA) 06 07 42

Total 29 100

C. Scheme of Options

There will be no overall choice. However, internal choice in any four questions of four marks eachand any two questions of six marks each has been provided.

D. Difficulty level of questions

S.No. Estimated difficulty level Percentage of marks

1. Easy 15

2. Average 70

3. Difficult 15

Based on the above design, separate sample papers along with their blue prints and Marking schemes havebeen included in this document. About 20% weightage has been assigned to questions testing higher orderthinking skills of learners.

MATHEMATICS (041)CLASS XII 2012-13

12

S. No. Topics VSA SA LA TOTAL

(a) Relations and Functions - 4(1) -

1. (b) Inverse Trigonometric 2(2) 4(1) - 10(4)

Functions

2. (a) Matrices 2(2) - 6(1)

(b) Determinants 1(1) 4(1) 13(5)

3. (a) Continuity and 1(1) 12(3) -

Differentiability

(b)Application of Derivatives - - 6(1) 44 (11)

(c) Integration - 12 (3)

(d) Applications of Integrals - - 6(1)

(e) Differential Equations 1(1} - 6(1)

4. (a) Vectors 2(2) 4(1) -

(b) 3-Dimensional Geometry 1(1) 4(1) 6(1) 17(6)

5. Linear Programming - - 6(1) 6(1)

6. Probability - 4(1) 6(1) 10(2)

Total 10 (10) 48 (12) 42(7) 100 (29)

The Question Paper will include question(s) based on values to the extent of 5 marks.

SAMPLE QUESTION PAPER MATHEMATICS (041)CLASS XII (2012-13) BLUE PRINT

13

Question Topics Form of Marks RemarksNumber Question

1 Inverse Trigonometric Functions VSA 1

2 Inverse Trigonometric Functions VSA 1

3 Matrices VSA 1

4 Matrices VSA 1

5 Determinants VSA 1

6 Differentiation VSA 1

7 Differential Equations VSA 1

8 Vectors VSA 1

9 Vectors VSA 1

10 3-Dimensional Geometry VSA 1

11 Relations and functions (Type of functions) SA 4

12 Inverse Trigonometric Functions SA 4

13 Determinants SA 4

14 Differentiation SA 4

15 Differentiation SA 4

16 Continuity of Functions SA 4

17 Integration * (Indefinite) SA 4

18 Integration * (Indefinite) SA 4

19 Integration * (Definite) SA 4

20 Vectors SA 4

21 3-Dimensional Geometry* SA 4

22 Probability SA 4

23 Matrices (Solution of System of equation) LA 6

24 Application of Derivative* LA 6

25 Application of Integration LA 6

26 Differential equations (Particular Solution) LA 6

27 3-Dimensional Geometry (Plane) LA 6

28 Linear Programming LA 6

29 Probability* LA 6

*With Alternative question

Questionwise Analysis - Sample Paper

14

General Instructions1. All questions are compulsory.2. The question paper consist of 29 questions divided into three sections A, B and C. Section A comprises of 10

questions of one mark each, section B comprises of 12 questions of four marks each and section C comprises of 07questions of six marks each.

3. All questions : in Section A are to be answered in one word, one sentence or as per the exact requirement of thequestion.

4. There is no over all choice. However, internal choice has been provided in 04 questions of four marks each and 02questions of six marks each. You have to attempt only one of the alternatives in all such questions.

5. Use of calculators in not permitted. You may ask for logarithmic tables, if required.

SECTION-A

Question numbers 1 to 10 carry 1 mark each.

1. Using principal values, write the value of 1 11 12cos 3sin

2 2

2. Evaluate 1 1 1tan [2cos(2sin )]

2

3. Write the value ofx + y + z if

1 0 0 1

0 1 0 1

0 0 1 0

x

y

z

4. If A is a square matrix of order 3 such that |adj A| = 225, find |A|

5. Write the inverse of the matrixcos sin

sin cos

6. The contentment obtained after eating x-units of a new dish at a trial function is given by theFunction C(x) = x3 + 6x2 + 5x + 3. If the marginal contentment is defined as rate of change of (x) withrespect to the number of units consumed at an instant, then find the marginal contentment when threeunits of dish are consumed.

7. Write the degree of the differential equation22 2

2 22 1 0d y d x dy

dx dx dx

8. If a andb are two vectors of magnitude 3 and2

3 respectively such thata b is a unit vector, write

the angle betweena andb .

SAMPLE QUESTION PAPER MATHEMATICS (041)CLASS XII (2012-13)

15

9. If and 7 4a i j k and 2 6 3 ,b i j k find the projection ofa on b .

10. Write the distance between the parallel planes 2x – y + 3z = 4 and 2x – y + 3z = 18

SECTION-B

Question numbers 11 to 22 carry 4 marks each.

11. Prove that the function f: N N, defined byf(x) = x2 + x + l is one - one but not onto.

12. Show that2

1 12

1sin[cot {cos(tan )}]

2

xx

xOR

Solve for x:2

1 1 12 2 2

2 1 23sin 4cos 2 tan

1 1 1 3

x x x

x x x13. Two schools A and B decided to award prizes to their students for three values honesty (x),

punctuality (y) and obedience (z). School A decided to award a total of Rs. 11000 for the three valuesto 5, 4 and 3 students respectively while school B decided to award Rs. 10700 for the three values to4, 3 and 5 students respectively. If all the three prizes together amount to Rs. 2700, then.

i. Represent the above situation by a matrix equation and form Linear equations using matrixmultiplication.

ii. Is it possible to solve the system of equations so obtained using matrices?

iii. Which value you prefer to be rewarded most and why?

14. If x = a( – sin ) andy = a (1–cos ), find2

2

d y

dx

15. If1

2

sin,

1

xy

x show that

22

2(1 ) 3 0d y dy

x x ydx dx

16. The functionf(x) is defined as f(x)

2 , 0 2

3 2 2 4

2 5 , 4 8

x ax b x

x x

ax b x

If f(x) is continuous on [0,8], find the values of a and b

OR

Differentiate2 2

1

2 2

1 1tan

1 1

x x

x x with respect to 1 2cos x

17. Evaluate:2

2

1

1

x xdx

xOR

Evaluate:(1 sin )

(1 cos )x x

e dxx

16

18. Evaluate: 2 2

2

( 1)( 2)

xdx

x x

19. Evaluate:4

0

log (1 + tanx)dx, using properties of definite integrals

20. Let 4 5 , 4 5 and 3 .a i j k b i j k c i j k Find a vectord which is perpendicular to both

a andb and satisfying . 21.d c21. 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(–l, –1, 6)

OR

Find the equation of the perpendicular drawn from the point P(2, 4, –1) to the line

5 3 6.

1 4 9

x y z Also, write down the coordinates of the foot of the perpendicular from P to the

line.

22. There is a group of 50 people who are patriotic out of which 20 believe in non violence. Two personsare selected at random out of them, write the probability distribution for the selected persons who arenon violent. Also find the mean of the distribution. Explain the importance of Non violence inpatriotism.

SECTION-C

Question numbers 23 to 29 carry 6 marks each

23. If A =

1 2 3

2 3 2

3 3 4

, find AA–1. Hence solve the following system of equations:

x + 2y – 3z = –4, 2x + 3y + 2z = 2, 3x – 3y – 4z = 11

24. Find the equations of tangent and normal to the curve7

( 2)( 3)

xy

x x at the point where it cuts

thex-axis

OR

Prove that the radius of the base of right circular cylinder of greatest curved surface area which canbe inscribed in a given cone is half that of the cone.

25. Find the area of the region enclosed between the two circlesx2 + y2 = l and (x – l)2 + y2 = 1

26. Find the particular solution of the differential equation: (x-sin y)dy + (tany)dx = 0 : given thaty = 0whenx = 0

27. Find the vector and Cartesian equations of the plane containing the two lines

2 3 ( 2 5 )r i j k i j k and 3 3 2 (3 2 5 )r i j k i j k

28. A dealer in rural area wishes to purchase a number of sewing machines. He has only Rs. 5760.00 toinvest and has space for at most 20 items. A electronic sewing machine costs him Rs.360.00 and a

17

manually operated sewing machine Rs. 240.00. He can sell an Electronic Sewing Machine at a profitof Rs. 22.00 and a manually operated sewing machine at a profit of Rs.18.00. Assuming that he cansell all the items that he can buy how should he invest his money in order to maximize his profit.Make it as a linear programming problem and solve it graphically. Keeping the rural background inmind justify the Values' to be promoted for the selection of the manually operated machine.

29. In answering a question on a MCQ test with 4 choices per question, a student knows the answer,guesses or copies the answer. Let ½ be the probability that he knows the answer, ¼ be the probabilitythat he guesses and ¼ that he copies it. Assuming that a student, who copies the answer, will becorrect with the probability ¾, what is the probability that the student knows the answer, given thathe answered it correctly?

Arjun does not know the answer to one of the questions in the test. The evaluation process hasnegative marking. Which value would Arjun violate if he resorts to unfair means? How would an actlike the above hamper his character development in the coming years?

OR

An insurance company insured 2000 cyclists, 4000 scooter drivers and 6000 motorbike drivers. Theprobability of an accident involving a cyclist, scooter driver and a motorbike driver are 0.01, 0.03and 0.15 respectively. One of the insured persons meets with an accident. What is the probability thathe is a scooter driver? Which mode of transport would you suggest to a student and why?

18

SECTION-A

1-10 1.7

62.

43. Zero 4. 15 5.

cos sin

sin cos

6. 68 units 7. 2 8.6

9.8

710. 14 1×10 = 10

SECTION-B11. f(x) = x2 + x + 1

Let x1, y

1 N such thatf(x

1) = f(y

1) 1

2 21 1 1 2 1 1 1 1 1 11 1 ( )( 1) 0[As 1 0x x y y x y x y x y for any N] 1

x1 = y

1f is one-one function

Clearly f(x) = 2 1 3x x for x N ½

But f(x) does not assume values 1 and 2 1½f: N N is not onto function

12.1 1

2 2

1 1cos(tan ) cos cos

1 1x

x x1

21 1

2 2

1 1cot sin

1 2

x

x x

2 21 1

2 22

1 1 1sin cot sin sin

2 21

x x

x xx1+1

ORLet x = tan ½

LHS = 1 1 13sin (sin 2 ) 4cos (cos 2 ) 2 tan (tan 2 ) 1½

= 13 2 8 4 2 2 tanx x 1

1 1 12 tan tan

3 6 3x x x 1

Marking Scheme

19

13. (i)

5 4 3

4 3 5

1 1 1

x

y

z

=

11000

10700

2700

5x + 4y + 3x = 11000 1½4x + 3y + 5z = 10700

x + y + z = 2700

(ii ) let A =

5 4 3

4 3 5

1 1 1

|A| = 5(–2) –4(–1) + 3 (1) 1½= – 10 + 4 + 3 = – 3 0

A–1 exists, so equations have a unique solution.(iii) Any answer of three values which proper reasoning will be considered correct. 1

14.2( sin ) (1 cos ) 2 sin

2

dxx a a a

d1

(1 cos ) .sin 2 sin cos2 2

dyy a a a

d1

2

2 sin .cos2 2 cot .

22 sin2

ady

dx a½

22 4

22 2

1 1 1 1cosec , . cosec

2 2 2 4 2sin 2 sin2 2

d y d

dx dx aa 1½

15.1

2 1

2

sin1 . sin

1

xy x y x

x½

2

2 2

( 2 ) 11

2 1 1

dy y xx

dx x x1

2(1 ) 1 0dy

x xydx

½

Diff. again we get2

22(1 ) 2 .1 0

d y dy dyx x y x

dx dx dx1

22

2(1 ) 3 0d y dy

x x ydx dx

1

20

16.2 2

lim ( ) lim (3 2) 8x x

f x x 1

As f is continuous at 2 2 4 8 2 4x a b a b ...(i) 1

Similarly as f is continuous atx = 4, 14 = 8a + 5b ...(ii) 1From (i) and (ii),a = 3,b = – 2. 1

OR

2 21

2 2

1 1tan

1 1

x xy

x x, Let 2 cos 2x 1

1 11 cos 2 1 cos 2 cos sintan tan

cos sin1 cos 2 1 cos 21

1 11 tantan tan tan

1 cos 4 4 1

1 2 1 21cos : Let cos

4 2y x z x

1 1

4 2 2

dyy z

dz½+½

17.3

2

1 2 1I

1 ( 1)( 1)

x x xdx x dx

x x x 1

Let2 1 3 1

A B( 1)( 1) 1 1 2 2

x A B

x x x x 1

23 1 3 1I log( 1) log( 1)

2 1 2 1 2 2 2

dx dx xxdx x x c

x x1+1

OR

2

(1 2sin cos1 sin 2 21 cos 2sin

2

x

x

x xex

e dx dxxx 1

= 2

2

1 2sin cos 12 2 cosec cot2 2 22sin

2

x x

x xx x

e dx e dxx

½

= 21cot cosec

2 2 2x xx x

e dx e dx ½

=2 21 1

cot cosec cosec2 2 2 2 2

x x xx x xe e dx e dx 1

21

=2 21 1

cot ( cosec cosec2 2 2 2 2

x x xx x xe e dx e dx C 1

= cot2

x xe C

18. 2 2

2I

( 1)( 2)

xdx

x x

Let 2 ,2 I( 1)( 2) 1 2

dt A Bx t x dx dt dt dt

t t t t1½

Getting A = 1, B = – 1 1

I =1 2

dt dt

t t = log | 1| log | 2 |t t C 1

=2 2log | 1| log | 2 |x x C ½

19. 4 4 4

0 0 0

1 tanI log(1 tan ) ...( ), I log 1 tan , I log 1

4 1 tan

xx dx i x dx dx

x1

or I = 4 4

0 0

2log [log 2 log(1 tan )]

1 tandx x dx

x ...(ii ) 1

Adding (i) and (ii), we get

2I = 4

0log 2 log 2

4dx 1

I = log 28

1

20. Let d xi y j zk

As d a and . 0d b d a and . 0d b 1

. 0 4 5 0d a x y z and . 0 4 5 0d b x y z ...(i) 1

. 21 3 21d c x y z ...(ii) ½

Solving (i) and (ii), we getx = 7,y = – 7, z = – 7 1

7 7 7d i j k ½

21. The equation of plane passing through (3, –1, 2), (5, 2, 4) and (–1, –1, 6) is

3 1 2

3 5 1 2 2 4 0

3 1 1 1 6 2

x y z or

3 1 2

2 3 2 0

4 0 4

x y z

2

22

12x – 16y + 12z = 76 or 3x – 4y + 3z = 19 ...(i) ½Length of from (6, 5, 9) to (i) is

2 2 2

18 20 27 19 6

343 4 31+½

OR

Any point on the line5 3 6

1 4 9

x y z1

is ( 5,4 3; 9 6)

Let this point be Q

dr s of PQ are 7,4 7, 9 7 ½

SincePQ AB ( 7)1 (4 7)4 ( 9 7)( 9) 0

98 = 98 = 1 1The pt. Q is (–4, 1, –3) ½

equation of PQ is2 4 1

6 3 2

x y z and foot of is (–4, 1, –3) 1

22. Let x denote the number of non-violent persons out of selected two. X can take values 0,1,2 non-violent 20: Violent patriotism: 30 ½P(× = 0) = (30 × 29)/50 × 49 = 87/245 ½P(× = 1) = (30 × 20 × 2)/50 × 49 = 120/245 ½P(× = 2) = (20 × 19)/50 × 49 = 38/245 ½Mean = 0 × 87/245 + 1 × 120/245 + 2×38/245 = 198/245 1Importance: In order to have a peaceful environment both the values are required patriotismand non-vilence only patriotism with violence could be very dangerous 1

23. The given matrix is A =

1 2 3

2 3 2 , |A| 6 28 45 67 0

3 3 4

1

A–1 exists

Adj A = 1

6 17 13 6 17 131

14 5 8 , 14 5 867

15 9 1 15 9 1

A 2½

The given system of equations can be written as AX = B

23

Where A =

1 2 3 4

2 3 2 , ,B 2

3 3 4 11

x

y

z

1

X = A–1 B =

6 17 13 4 31

14 5 8 2 267

15 9 1 11 1

1

x = 3,y = –2, Z = 1 ½24. The given curve cuts the x-axis atx = 7, andy = 0 ½

2

2 2 2

7 ( 5 6) ( 7)(2 5)

5 6 ( 5 6)

x dy x x x xy

x x dx x x1½

2

(49 35 6) (0) 1(at 7)

(49 35 6) 20

dyx

dx ½+½

Equation of tangent at the curve at (7, 0) is

10 ( 7)

20y x or 20 7 0x y 1½

Equation of normal to the curve at (7, 0) is

0 20( 7) 20 7 140 0y x x 1½

OR

Let OC =x, VOB ~ B DB 1

( )VO OB h r xB D h

B D DB r

Let S be the curved surface area of cylinder 1

S =2( ) 2

2 2 [ ]h r x h

xh x rx xr r

2

2

2

2 4( 2 ), 0 S

ds h d s hr x

dx r dx r is Maximum 1

0 2ds

r xdx

S is maximum whenx =2

r, i.e., when radius of base of cylinder is half the radius of base of

cone. 1

24

25. Points of inter section are 1 3A ,

2 2 and

1 3D ,

2 21

Area of Shaded region = 2 (Area O ABCO)

=1

12 22

10

2

2 1 ( 1) 1x dx x dx 1

=

11

22 1 2 1

10

2

( 1) 1 1 12 1 ( 1) sin 1 sin

2 2 1 2 2

x x xx x x 2

=1 1 1 13 1 3 1

sin sin ( 1) sin (1) sin4 2 4 2

1

=3 3

4 6 2 2 4 6

=2 3

sq.3 2 units. 1

26. The given different equation can be written as

(cot ) cosdx

y x ydy 1

I.F. =cot log sin sin

y dy ye e y

The solution isx sin y = sin cosy y dy C 1

=1

sin 22

y dy c

or x siny =1

cos 24

y C 1

It is given thaty = 0, whenx = 0 1

1C 0 C 1/ 4

4

x siny = 21 1(1 cos 2 ) sin

4 2y y 1

2x = siny is the reqd. solution. 1

25

27. Here 1 2 3a i j k and 2 3 3 2a i j k 1

2 2 5b i j k and2 3 2 5b i j k

1 2 1 2 5 20 10 8

3 2 5

i j k

n b b i j k 1½

Vector equation of the reqd. plane is11( ). 0 . .r a n or r n a n or

.(20 10 8 ) (2 3 ).(20 10 8 ) 40 10 24 74r i j k i j k i j k 2

.(10 5 4 ) 37r i j kThe Cartesian equation of plane is 10x + 5y – 4z = 37 1½

28. Suppose number of electronic operated machine = x and number of manually operated sewingmachines =y. 2

x + y 20 ...(i)and, 360x + 240y 5760 or 3x + 2y 48 ...(ii)x 0, y 0To maximise Z = 22x + 18y 2Corners of feasible region are A(0, 20), P(8, 12), B(16, 0)Z

A = 18 × 20 = 360, Z

P = 22 × 8 + 18 × 12 = 392, Z

B = 352

Z is maximum at x = 8 andy = 12The dealer should invest in 8 electronic and 12 manually operated machines ½

Keeping the ‘save environment’ factor in mind the manually operated machine should be

promoted so that energy could be saved.29. Let A be the event that he knows the answer, B be the event that he guesses and C be the event

that he copies. 1Then, P(A) = ½, P(B) = ¼ and P(C) = ¼ 1Let X be the event that he has answered correctly.}Also, P(X/A) = 1, P(X/B) = ¼ and P(X/C) = ¾Thus, Required probability P(A/X) = 1 for formula

( / ) ( )

( / ) ( ) ( / ) ( ) ( / ) ( )

P X A P A

P X A P A P X B P B P X C P C

=

11

21 1 1 1 3

12 4 4 4 4

=

12

1 1 32 16 16

1

=2

3

26

If Arjun copies the answer, he will be violating the value of honesty in his character. Heshould not guess the answer as well as that may fetch him negative marking for a wrongguess. He should accept the question the way it is and leave it unanswered as cheating mayget him marks in this exam but this habit may not let him develop an integrity of characterin the long run. 2

ORLet the events defined are E

1: Person chosen is a cyclist

E2: Person chosen is a scoter-driver

E3: Person chosen is a motorbike driver

A: Person meets with an accident ½P(E

1) = 1/6, P(E

2) = 1/3, P(E

3) = 1/2 1

P(A/E1) = 0.01, P(A/E

2) = 0.03, P(A/E

3) = 0.152 1

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

1). P(E

1)/P(A) 1

= 1/2Suggestion: Cycle should be promoted as it is good for ½

i. Health ½ii. No pollution ½

iii. Saves energy (no petrol) ½

27

General Instructions:

(i) All questions are compulsory.(ii) The question paper consists of 29 questions divided into three Sections A, B and C, Section

A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four

marks each and Section C comprises of 7 questions of six marks each.

(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact

requirement of the question.

(iv) There is no overall choice. However, internal choice has been provided in 4 questions of four

marks each and 2 questions of six marks each. You have to attempt only one of the alternatives

in all such questions.

(v) Use of calculators is not permitted.

SECTION-A

Questions numbers 1 to 10 carry 1 mark each.

Q1. If a line has direction ratios 2, –1, –2, then what are its direction cosines? 1

Q2. Find ‘ ’ when the projection of 4a i j k on 2 6 3b i j k is 4 units. 1

Q3. Find the sum of the vectors 2a i j k , 2 4 5b i j k and 6 7 .c i j k 1

Q4. Evaluate :3

2

1dx

x1

Q5. Evaluate (1 ) .x x dx 1

Q6. If =

5 3 8

2 0 1

1 2 3, write the minor of the element a

23. 1

Q7. If2 3 1 3 4 6

5 7 2 4 9 x, write the value ofx. 1

Q8. Simplify :cos sin sin cos

cos sinsin cos cos sin 1

Q9. Write the principal value of 1 11 1cos 2sin

2 2. 1

Q10. Let * be a ‘binary’ operation on N given by a * b = LCM (a, b) for all a, b N. Find 5 * 7. 1

Question Paper-Delhi (2012)

28

SECTION-B

Question numbers 11 to 22 carry 4 mark each.

Q11. If (cos x)y = (cos y)x, find .dy

dx4

OR

If sin y = x sin (a + y), prove that2sin ( )

.sin

dy a y

dx aQ12. How many times must a man toss a fair coin, so that the probability of having at least one head

is more than 80%? 4

Q13. Find the Vector and Cartesian equations of the line passing through the point (1, 2, –4) and

perpendicular to the two lines8 19 10

3 16 7

x y z and

15 29 5.

3 8 5

x y z4

Q14. If , ,a b c are three vectors such thata = 5, b = 12 and c = 13, and 0,a b c find the

value of . . . .a b b c c a 4

Q15. Solve the following differential equation : 4

2 22 2 0.dy

x xy ydx

Q16. Find the particular solution of the following differential equation. 4

2 2 2 21dy

x y x ydx

, given thaty = 1 wherex = 0.

Q17. Evluate : sin sin 2 sin3x x x dx

OR

Evaluate : 2

2

(1 )(1 )dx

x x

Q18. Find the point on the curvey = x3 – 11x + 5 at which the equation of tangent isy = x – 11.

OR

Using differentials, find the approximate value of49.5. 4

Q19. If y = 1 2(tan )x , show that 4

22 2 2

2( 1) 2 ( 1) 2d y dy

x x xdx dx

.

Q20. Using properties of determinants, prove that 6

2

b c q r y z a p x

c a r p z x b q y

a b p q x y c r z

29

Q21. Prove that1 cos

tan , , .1 sin 4 2 2 2

x xx

x6

OR

Prove that 1 1 18 3 36sin sin cos .

17 5 85

Q22. Let A = R – {3} and B = R – {1}. Consider the function f : A B defined by2

( )3

xf x

x.

Show thatf is one-one and onto and hence findf –1. 6

SECTION-C

Questions numbers 23 to 29 carry 6 mark each.Q23. Find the equation of the plane determined by the points A(3, –1, 2), B (5, 2, 4) and C(–1, –1, 6)

and hence find the distance between the plane and the point P(6, 5, 9). 6Q24. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars

(not residing in hostel). Previous year results report that 30% of all students who reside in hostelattain ‘A’ grade and 20% of day scholars attain ‘A’ grade in their annual examination. At the

end of the year, one student is chosen at random from the college and he has an ‘A’ grade, what

is the probability that the student is a hostlier? 6Q25. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours

on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machineB to produce a package of bolts. He earns a profit of` 17.50 per package on nuts and` 7per package of bolts. How many packages of each should be produced each day so as to maximizehis profits if he operates his machines for at the most 12 hours a day? Form the above as alinear programming problem and solve it graphically. 6

Q26. Prove that4

0

( tan cot ) 2.2

x x dx 6

OR

Evaluate3

2

1

(2 5 )x x dx as a limit of a sum.

Q27. Using the method of integration, find the area of the region bounded by the lines 3x – 2y + 1 = 0,2x + 3y – 21 = 0 and x – 5y + 9 = 0. 6

Q28. Show that the height of a closed right circular cylinder of given surface and maximum volume,is equal to the diameter of its base. 6

Q29. Using matrices, solve the following system of linear equations :x – y + 2z = 73x + 4y – 5z = – 5

2x – y + 3z = 12OR

Using elementary operations, find the inverse of the following matrix :

1 1 2

1 2 3

3 1 1

30

Q.No. Value Pooints/Solution 65/1//1 Marks.

SECTION-A

1-10 1.2 1 2

, ,3 3 3

2. = 5 3. 4 j k 4.3

log2 5.

3 2 5 23 2

2 5x x c

6. 2,3M 7 7. 13 8.1 0

0 1 9.2

3 10. 35. 1×10 = 10

SECTION-B

11. (cos ) (cos ) log cos log cosy xx y y x x y 1/2

( sin ) ( sin ). log cos . log cos

cos cos

x dy y dyy x x y

x dx y dx 1+1

(log cos tan ) log cos tandy

x x y y y xdx

1

dy

dx =

log cos tan

log cos tan

y y x

x x y 1/2

OR

sin sin( ) cos cos( ) sin( )dy dy

y x a y y x a y a ydx dx

1

sin( )

cos cos( )

dy a y

dx y x a y 1

sin sin( )sinsin( ) cos .cos( )

sin( )

y dy a yx

ya y dx y a ya y

1

2 2sin ( ) sin ( )

sin( )cos cos( )sin sin

dy a y a y

dx a y y a y y a1

Marking SchemeClass-XII

Mathematics (March 2012)

31

12. Let the coin be tossedn times

P(getting at least one heat) >80

1001

8 8 2 11 P(0) P( ) 1

10 10 10 5o 1

0

0

1 1 1 1 1or or 2 5

2 2 5 2 5

nn n

nC 1

n = 3. 1

13. Let the vector equation of required line bea a b

than a = 2 4i j k

and b = (3 16 7 ) (3 8 5 )i j k i j k 1

= 24 36 72i j k 1

Vector equation of line is

r = ( 2 4 ) (24 36 72 )i j k i j k

}or r = ( 2 4 ) (2 3 6 )i j k i j k 1

and cartesian from is1 2 4

2 3 6

x y z1

14. 20 ( ) 0a b c a b c 1/2

2 2 22( . . . ) 0a b c a b b c c a 1

or 2 2 2| | | | | | 2( . . . ) 0a b c a b b c c a 1

. . .a b b c c a =1

(25 144 169) 1692 . 1

15.

2

2 22 2

2

222 2 0

2 2

y ydy dy xy y x xx xy ydx dx x

1/2

Puttingy

vx

so thaty = vx anddy dv

v xdx dx

1

2 21 1

2 2

dv dvv x v v x v

dx dx1/2

2

22 log

dv dxx c

v x v1

2 2log

log

x xx c y

y x c 1

32

16.2 2 2 2 2 21 (1 )(1 )

dyx y x y x y

dx½

22 (1 )

1

dyx dx

y 1

31tan

3

xy x c 1

x = 0, y = 1 c = /4 1

31tan

3 4

xy x or

3

tan4 3

xy x ½

17. I =1

sin sin 2 sin 3 2sin 3 sin .sin 22

x x xdx x x xdx ½

=1 1

(cos 2 cos 4 )sin 2 (sin 2 cos 2 cos 4 sin 2 )2 2

x x xdx x x x x dx ½

=1 1

sin 4 2cos 4 sin 24 4

xdx x xdx 1

=1 1

cos 4 (sin 6 sin 2 )16 4

x x x dx 1

=1 1 1

cos 4 cos6 cos 216 24 8

x x x c 1

OR

2 2

2 A B C

(1 )(1 ) 1 1

x

x x x x ½

22 A(1 ) (B C)(1 )x x x 1½

0 = A – B, B – C = 0 A + C = 2 A = B = C = 1

2

2

(1 )(1 )dx

x x = 2

1 1

1 1

xdx dx

x x½

=2 11

log |1 | ( 1) tan2

x x x c 1½

18. Slope of tangent,y = x – 11 is 1 ½

3 211 5 3 11dy

y x x xdx

½

If the point is (x1, y

1) then 2

1 13 11 1 2x x 1

x1 = 2 then 1 8 22 5 9y and if 1 2x then 1 19y 1

Since (–2, 19) do not lie on the tangent y = x – 11 ½Required point is (2, – 9) ½

33

OR

Let y = x y y x x ½

dyy x x x

dx

1.

2x x x x

x1

Putting x = 49 and x = 0.5 we get 1

149 (0.5) 49.5

2 49½

149.5 7 7.0357

281

19.1 2 1

2

1(tan ) 2 tan

1

dyy x x

dx x

2 1(1 ) 2 tandy

x xdx

22

2 2

2(1 ) 2 .

1

d y dyx x

dx dx x

22 2 2

2(1 ) . 2 (1 ) 2.d y dy

x x xdx dx

20. Using R1

R1 + R

2 + R

3 we get

LHS =

2( ) 2( ) 2( )a b c p q r x y z

c a r p z x

a b p q x y

1

= 2

a b c p q r x y z

c a r p z x

a b p q x y

1

= 2

a b c p q r x y z

b q y

c r z

1

= 2

a p x

b q y

c r z

1

Using R2

R2 – R

1

R3

R3 – R

1

Using R1

R1 + R

2 + R

3

= RHS R2

– R2

R3

– R3

34

21. 1 1

sincos 2

tan tan1 sin

1 cos2

xx

xx

1

=1 1

2

2sin cos4 2 4 2

tan tan tan4 2

2cos4 2

x xx

x1+1

=4 2

x1

OR

Writing1 18 8

sin tan17 15 and

1 13 3sin tan

5 4 1

LHS =1 1 1 1

8 38 3 7715 4tan tan tan tan

8 315 4 361 ,15 4

1+1

Getting 1 177 36tan cos

36 851

22. Let 1 2, Ax x and 1 2( ) ( )f x f x ½

1 2

1 2

2 2

3 3

x x

x x 1 2 2 1 1 2 1 22 3 2 3x x x x x x x x

x1 = x

21

Hence f is 1 – 1

Let y B,2

( ) 3 23

xy f x y xy y x

x

or x =3 2

1

y

y ½

Since y 1 and3 2

3 A1

yx

yHence f is ONTO 1

and1 3 2( )

1

yf y

y 1

SECTION-C

23. Normal to the plane is AB BCn ½

n = 2 3 2 12 16 12

6 3 2

i j k

i j k 1½

35

Equation of plane is

.(12 16 12 )r i j k = (3 2 ).(12 16 12 )i j k i j k= 76 2

or .(3 4 3 )r i j k = 19 or 3x – 4y + 3z – 19 = 0 }

Distance of plane from the point P(6, 5, 9) is

d =18 20 27 19 6

9 16 9 342

24. Let E1 : selected student is a hostlier

E2

: selected student is a day scholar}

1A : selected student attain ‘A’ grade in exam.

P(E1) =

60

100, P(E

2) =

40

1001

P(A/E1) =

30

100, P(A/E

2) =

20

1001

P(E1/A) = 1 1

1 1 2 2

P(E ) . P(A/E )

P(E ) . P(A/E ) + P(E ) . P(A/E )1

=60 30. 9100 100

40 20 1360 30. .100 100 100 100

1+1

25. Let x package of nuts andy package of bolts be produced each dayLPP is maximise P = 17.5x + 7y 1

subject to x + 3y 12

3x + y 12}

2x 0, y 0 correct graph

vertices of feasible region are A(0, 4), B (3, 3), C (4, 0)Profit is maximum at B(3, 3)i.e. 3 package of nuts and 3 package of bolts 1

36

26. I =

4 4

0 0

sin cos( tan cot )

sin cos

x xx x dx dx

x x1

Putting sin cos ,x x t to get (cos sin )x x dx dt 1

and sin cosx x =21

2

t1

I =0

21

21

dt

t= 1 0

12.[sin ]t 1+1

=1 12(sin 0 sin ( 1) 2.

21

OR

I =3

2

1

(2 5 )x x dx =0

lim [ (1) (1 ) (1 2 ) ... (1 1 )]h

h f f h f h f n h

where ( )f x = 22 5x x and2

hn

or nh = 2. 1

f(1) = 7

f(1 + h) = 2 22(1 ) 5(1 ) 7 9 2h h h

(1 2 )f h = 2 2 22(1 2 ) 5(1 2 ) 7 18 22h h h h 2

(1 3 )f h = 2 2 22(1 3 ) 5(1 3 ) 7 27 2.3h h h h

...........

(1 ( 1) )f n h = 2 27 9( 1) 2.( 1)n h n h

I =2

0

( 1) ( 1)(2 1)lim 7 9 2 .

2 6h

n n n n nh n h h 1

=0

9 1lim 7 ( ) ( )(2 )

2 3hnh nh nh h nh nh h nh h 1

=16 112

14 183 3

1

27. Let AB be 3 2 1 0,x y BC be2 3 21 0x y and AC be 5 9 0x y correct figure : 1

Solving to get A(1, 2), B (3, 5) and C(6, 3) 1½

area of ( ABC) =3 6 6

1 3 1

1 1 1(3 1) (21 2 ) ( 9)

2 3 5x dx x dx x dx 1

37

=

6 63 2 22

1 3 1

1 (21 2 ) ( 9)(3 1)

12 12 10

x xx 1½

=25

7 122

½

=13

2 sq. U. ½

28. Surface area A =22 2rh r (Given) ½

h =2A 2

2

r

r...(1) 1

V =2

2 2 A 2

2

rr h r

r 1

=31

.[A 2 ]2

r r ½

dv

dr =

21[A 6 ]

2r 1

dv

dr = 0 = 2 26 A 2 2r rh r 1

24 2r rh h = 2r = diameter 1

2

2

d v

dr =

1[ 12 ] 0 2

2r h r will give max. volume. 1

29. Given equations can be written as

1 1 2

3 4 5

2 1 3

x

y

z

=

7

5

12 or AX = B 1

a11

= 7, a12

= –19 a13

= – 11

a21

= 1, a22

= – 1 a23

= – 1

a31

= –3, a32

= 11 a33

= 72

–1A =

7 1 31

19 1 114

11 1 7

½

x

y

z

=

7 1 3 7 21

19 1 11 5 14

11 1 7 12 3

x = 2, y = 1,z = 3. 1½

38

OR

Let A =

1 1 2

1 2 3

3 1 1

Writing

1 1 2

1 2 3

3 1 1

= AA

1 0 0

0 1 0

0 0 1

1

c1

c2

1 1 2

2 1 3

1 3 1

= A

0 1 0

1 0 0

0 0 1

½

c2

c2+ c

1

1 0 0

2 3 1

1 4 1

= A

0 1 0

1 1 2

0 0 1

1

c3

c3 –2c

1

c1

c1+ 2c

3

1 0 0

0 1 1

1 2 1

= A

0 1 0

3 3 2

2 2 1

½

c2

c2+ 2c

3

c3

c3+ –c

2

1 0 0

0 1 0

1 2 1

= A

0 1 1

3 3 5

2 2 3½

c1

c1+ c

3

c2

c2+ 2c

3

1 0 0

0 1 0

0 0 1

= A

1 1 1

8 7 5

5 4 3

1

A–1 =

1 1 1

8 7 5

5 4 31

39

General Instructions:

(i) All questions are compulsory.(ii) The question paper consists of 29 questions divided into three Sections A, B and C, Section

A comprises of 10 questions of one mark each, Section B comprises of 12 questions of fourmarks each and Section C comprises of 7 questions of six marks each.

(iii) All questions in Section A are to be answered in one word, one sentence or as per the exactrequirement of the question.

(iv) There is no overall choice. However, internal choice has been provided in 4 questions of fourmarks each and 2 questions of six marks each. You have to attempt only one of the alternativesin all such questions.

(v) Use of calculators is not permitted.

SECTION-A

Questions numbers 1 to 10 carry 1 mark each.Q1. The binary operation * : R × R R is defined asa * b = 2a + b. Find (2 * 3) * 4.

Q2. Find the principal value of 1 1tan 3 sec ( 2).

Q3. Find the value ofx + y from the following equation :

5 3 4 7 62

7 3 1 2 15 14

x

y

Q4. If A T =

3 4

1 2

0 1

and B =1 2 1

1 2 3, then find AAT – BT.

Q5. Let A be a square matrix of order 3 × 3. Write the value of |2A|, where |A| = 4.

Q6. Evaluate :2

2

0

4 x dx

Q7. Given (tan 1)sec ( ) .x xe x xdx e f x c

Write f(x) satisfying the above.

Q8. Write the value of( ). .i j k i j .

Q9. Find the scalar components of the vectorAB with initial point A(2, 1) and terminal point B (–5, 7).

Q10. Find the distance of the plane 3x – 4y + 12z = 3 from the origin.

Question Paper-Outside Delhi (2012)

40

SECTION-B

Questions numbers 11 to 22 carry 4 mark each.Q11. Prove the following :

–1 13 3 6cos sin cot

5 2 5 13

Q12. Using properties of determinants, show that

4

b c a a

b c a b abc

c c a b

Q13. Show thatf : N N, given by

f(x) =1, if is odd

1, if is even

x x

x x

is both one-one and onto.OR

Consider the binary operations * : R × R R and o : R × R R defined asa * b = |a – b| anda ob= a for all a, b R. Show that ‘*’ is commutative but not associative, ‘o’ is associative but not

commutative.

Q14. If x = 1 1sin cos,t ta y a , show that .dy y

dx x

OR

Differentiate2

1 1 1tan

x

x with respect tox.

Q15. If x = a (cost + t sin t) andy = a (sin t – t cost), 0 < t < ,2

find2 2

2 2,d x d y

dt dt and

2

2 .d y

dx

Q16. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground,away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the footof the ladder is 4 m away from the wall?

Q17. Evaluate :2

3

1

| |x x dx

OR

Evaluate :2

0

sin

1 cos

x xdx

x

Q18. Form the differential equation of the family of circles in the second quadrant and touching thecoordinate axes.

41

ORFind the particular solution of the differential equation

2( 1) 1; 0dy

x x ydx

whenx = 2.

Q19. Solve the following differential equation :

( 2(1 ) 2 cot ; 0x dy xy dx x dx x

Q20. Let a = 4 2 ,i j k b = 3 2 7i j k andc = 2 4i j k . Find a vectorp which is perpendicular

to both a andb and . 18.p c

Q21. Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crossesthe XY-plane.

Q22. Two cards are drawn simultaneously (without replacement) from a well-shuffled pack of 52 cards.Find the mean and variance of the number of red cards.

SECTION-C

Questions numbers 23 to 29 carry 6 mark each.Q23. Using matrices, solve the following system of equations :

2 3 3 5x y z , 2 4x y z , 3 2 3x y z .

Q24. Prove that the radius of the right circular cylinder of greatest curved surface area which can beinscribed in a given cone is half of that of the cone.

ORAn open box with a square base is to be made out of a given quantity of cardboard of areac2 square

units. Show that the maximum volume of the box is3

6 3

c cubic units.

Q25. Evaluate :1

2

sin

1

x xdx

xOR

Evaluate :2

2

1

( 1) ( 3)

xdx

x xQ26. Find the area of the region {(x, y) : x2 + y2 4, x + y 2}.

Q27. If the lines1 2 3

3 2 2

x y z

k and

1 2 3

1 5

x y z

k are perpendicular, find the value ofk and

hence find the equation of plane containing these lines.Q28. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin 3 times and notes the number of

heads. If she gets 1, 2, 3 or 4 she tosses a coin once and notes whether a head or tail is obtained. If sheobtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Q29. A dietician wishes to mix two types of foods in such a way that the vitamin contents of the mixturecontains at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitaminA and 1 unit/kg of vitamin C while Food II contains 1 unit/kg of vitamin A and 2 units/kg of vitaminC. It costs 5 per kg to purchase Food I and` 7 per kg to purchase Food II. Determine the minimumcost of such a mixture. Formulate the above as a LPP and solve it graphically.

42

General Instructions:

(i) All questions are compulsory.(ii) The question paper consists of 29 questions divided into three Sections A, B and C, Section

A comprises of 10 questions of one mark each, Section B comprises of 12 questions of fourmarks each and Section C comprises of 7 questions of six marks each.

(iii) All questions in Section A are to be answered in one word, one sentence or as per the exactrequirement of the question.

(iv) There is no overall choice. However, internal choice has been provided in 4 questions of fourmarks each and 2 questions of six marks each. You have to attempt only one of the alternativesin all such questions.

(v) Use of calculators is not permitted.

SECTION-A

Questions numbers 1 to 10 carry 1 mark each.Q1. If the binary operation * on the set Z of integers is defined bya * b = a + b – 5, then write the identity

element for the operation * in Z.

Q2. Write the value of cot (tan–1 a + cot–1 a).

Q3. If A is a square matrix such that A2 = A, then write the value of (I + A)2 – 3A.

Q4. If2 1 10

3 1 5x y , write the value ofx.

Q5. Write the value of the following determinant :

102 18 36

1 3 4

17 3 6

Q6. If 2

1( )x xx

e dx f x e cx

, then write the value off(x).

Q7. If 2

0

3 8a

x dx , write the value of ‘a’.

Q8. Write the value of( ) . ( ) .i j k j k i .

Q9. Write the value of the area of the parallelogram determined by the vectors 2i and3 j .Q10. Write the direction cosines of a line parallel to z-axis.

Question Paper-Outside India (2012)

43

SECTION-B

Questions numbers 11 to 22 carry 4 mark each.

Q11. If4 3 2

( ) ,6 4 3

xf x x

x, show that fof(x) = x for all

2

3x . What is the inverse off ?

Q12. Prove that : 1 1 163 5 3sin sin cos

65 13 5OR

Sovle forx :

1 12 tan (sin ) tan (2sec ),2

x x x

Q13. Using properties of determinants, prove that

32 3 2 4 3 2

3 6 3 10 6 3

a a b a b c

a a b a b c a

a a b a b c

Q14. If ( )m n m nx y x y , prove that .dy y

dx x

Q15. If1cos , 1 1a xy e x show that

22 2

2(1 ) 0.d y dy

x x a ydx dx

OR

If 1 1 0,x y y x 1 1,x x y, then prove that 2

1.

(1 )

dy

dx x

Q16. Show that2

log(1 ) , 12

xy x x

x, is an increasing function ofx throughout its domain.

OR

Find the equation of the normal at the point (am2, am3) for the curveay2 = x3.

Q17. Evaluate : 2 1tanx x dx

OR

Evaluate : 2

3 1

( 2)

xdx

xQ18. Solve the following differential equation :

2

1, 0xe y dx

xdyx x .

Q19. Solve the following differential equation :23 tan (2 )sec 0x xe ydx e y dy , given that whenx = 0, .

4y

44

Q20. If = 3 4 5i j k and 2 4i j k , then express in the form 1 2 , where 1 is parallel

to and 2 is perpendicualr toQ21. Find the vector and cartesian equations of the line passing through the point P(1, 2, 3) and parallel to

the planes . ( 2 ) 5r i j k and . (3 ) 6.r i j k

Q22. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probabilitydistribution of the number of successes and hence find its mean.

SECTION-C

Questions numbers 23 to 29 carry 6 mark each.Q23. Using matrices, solve the following system of equations :

4; 2 3 0; 2x y z x y z x y z

OR

If 1

3 1 1

A 15 6 5

5 2 2

and B =

1 2 2

1 3 0

0 2 1, find (AB)–1.

Q24. Show that the altutude of the right circular cone of maximum volume that can be inscribed in a

sphere of radius R is4R

.3

Q25. Find the area of the region in the first quadrant enclosed by x-axis, the line 3x y and the circle2 2 4.x y

Q26. Evaluate :3

2

1

( )x x dx as a limit of a sum.

OR

Evaluate :

4 2

2 20

cos

cos 4sin

xdx

x x

Q27. Find the vector equation of the plane passing through the points (2, 1, – 1) and (–1, 3, 4) and

perpendicular to the planex – 2y + 4z = 10. Also show that the plane thus obtained contains the line

3 4 (3 2 5 ).r i j k i j k

Q28. A company produces soft drinks that has a contract which requires that a minimum of 80 units of thechemical A and 60 units of the chemical B go into each bottle of the drink. The chemicals areavailable in prepared mix packets from two different suppliers. Supplier S had a packet of mix of 4units of A and 2 units of B that costs` 10. The supplier T has a packet of mix of 1 unit of A and 1 unitof B that costs 4. How many packets of mixes from S and T should the company purchase tohonour the contract requirement and yet minimize cost ? Make a LPP and solve graphically.

Q29. In a certain college, 4% of boys and 1% of girls are taller than 1.75 metres. Furthermore, 60% of thestudents in the college are girls. A student is selected at random from the college and is found to betaller than 1.75 metres. Find the probability that the selected student is a girl.

45

Relations and Functions1CHAPTER

Concept Building

We know that a relation R, from nonempty set A to nonempty set B is a subset of A × B. In other wordsR = {(a, b) : aðA and bð B}. If (a, b)ðR we say that a is releated to b under the relation R and we write it asa R b. Further, we know that a relation R in a set A is a subset of A × A. i.e. R A × A.Remark : Let the number of elements in set A =p and number of elements in set B =q, then the numberof elements in A × B =pq. Therefore number of relations from set A to set B = Number of subset ofA × B = 2

pq.

Types of Relations

(i) Empty (or void) relation: A relation R in a set A is called empty relation if no element of A isreleated to any element of A i.e. R = A × A.

(ii ) Universal relation: A relation R in a set A is called universal relation if each element of A is releatedto every element of A i.e. R = A × A.

(iii ) Reflexive Relation: A relation R in a set A is called reflexive relation if each element of A is releatedto itself i.e. if (a, a) R fro all a A.

(iv) Symmetric Relation: A relation R in a set A is called symmetric relation if (a, b) R impliesthat (b, a) R a, b A.

(v) Transitive Relation: A relation R in a set A is called transitive relation if (a, b) R and (b, c)R implies that (a, c) R For all a, b, c A.

(vi) Equivalence Relation:A relation R in a set A is called equivalence relation if it is reflexive, symmetricand transitive.

Example: 1. Let R be a relation in a set A of all triangles in a plane defiend asR = {(a, b) : a is congruent to b}show that R is an equivalence relation.

Thinking process Application

(i) Since each triangle in a plane is congruent (i) Clearly (a, a) R a A.to itself. R is refelexive

(ii) If Ist triangle is congruent to 2nd then 2nd (ii) If (a, b)R, then (b, a) Ris also congruent to Ist. R is symmetric

(iii) If Ist triangle is congruent to 2nd and 2nd is (iii) If (a, b)R & (b, c) Rcongruent to 3rd, then Ist is congruent to 3rd (a, c) R

R is transitiveHence R is equivalence Relation

46

Remark: Let R be an equivalence relation in setx and R dividesx into mutually disjoint subsetsA

i called partitions or subdivisions of A satisfying the conditions:

(i) All element of Aiare releated to each otheri

(ii) No element ofAi is releated to any element ofA

j, i j

(iii) Ai = x andA

iA

j = , i j

The subsetAi are called equivalence classes.

Functions: We know that function is a special type of relation in which each element of the domainassociates with a unique element of the codomain. Range is always a subset of codomain i.e. rangecodomain.

(i) One-one (or injective) Function:A function F : A B is said to be one-one function if no twoelements of A have same image i.e. ifx

1x

2F(x

1) F(x

2) For all x

1x

1A

OR

F(x1) = F(x

2) x

1 = x

2 fer all x

1x

2A.

(ii) Onto (surjective) Function: A function F : A B is said to be onto if for eachb B there alwaysexist a A such that F(a) = b. In other words, F is onto if range of F = codomain of F i.e. ifF(A) = B.

(iii) One-one, onto (bijective) function:A function F : A B is said to be bijective if F is both one-one and onto.

(iv) Many one Function: A function F : A B is said to be many one if it is not a one-one function.

(v) Into function: A function F : A B is said to be a into function if it is not a onto function.

Let F : N N and g : R R be two function defiend as F(x) = 2x andg(x) = 2x. Now we shallshow that I is a bijective function but F not.

For F : N → N For g : R → R

(i) For x1, x

1N (i) For x

1, x

2, R

let f(x1) = f(x

2) let f(x

1) = f(x

2)

2x1 = 2x

2 2x

1 = 2x

2

x1 = x

2x

1 = x

2

F is one-one I is one-one

(ii) Let y N (codomain) (ii) Lety N (codomain)

and letf(x) = y then f(x) = y

2x = y x = 2y R for

x = 2y N everyy R

for somey N everyy R has pre-image

For example fory = 1 N R2

yx

there is nox N (domain) I is onto function

F is not on to Hence I is bijective

Hence F is not bijective Function

47

Example: 2. Show that the function F : N N defined byf(x) = x2 + 1 is not a bijective function

Thinking Process Application

We have to prove that F is one-one and (i) Letx1, x

2N (domain)

onto function. For one-one : If any two such thatf(x1) = f(x

2)

images are equal then their preimages are x12 + 1 = x

22 + 1

also equal. x12 = x

22

For onto : we have to prove that for each x1 = x

2 (� x

1x

2N)

y codomain there existsx domain (ii) Let y N (codomain)

such thatf(x) = y such thatf(x) = y

i.e. x2 + 1 = y

x = 1y

But for y = 1 codomain

x = 0 which does not belong to domain.

Thus y = 1 has no pre-image

F is not onto function

Hence F is not bijective

Composition of Functions: Let A, B and C be three non empty sets. Further let F : A B andg : B C be two functions. Since,f is a functions from A to B, therefore, fer everyx A there exista unique elementf(x) B. Since g is a function from B to C, therefore, fer eachf(x) B there exists aunique elementg{ f(x)} C i.e. For everyx A, there is a unique elementg{ f(x)} C. In other wordswe have a new function from A to C. This new function is called composition of f and g and we denotethis function by gof. Hence gof : A C is defiend by gof (x) = g { f(x)}.

x f x( ) g f x{ ( )}

A B C

gof

F g

Example: 3. Let F : R R and g : R R be defined byf(x) = x + 1 andg(x) = x – 1. Show that

gof = fog = IR, where I

R is the identify function an R

Thinking Process Application

(i) (i) gof(x) = g(f(x))

= g (x+1) = (x +1) –1 = x

= IR.

(ii) (ii) fog(x) = f{ g(x)}

= F(x – 1)

= (x – 1) + 1 = x = IR.

R R RF g

gof

x f x( ) g f x( )

R R RFg

fog

x g x( ) x( ( )f g

48

Remark: (i) It may be possible that gof existes but fog does not exist.

(ii) Composition of functions need not to be commutative i.e. gof and fog may not be equal.

(iii) If gof is one-one then F is also one-one and if gof is onto then g is also on-to.

(iv) If F : A B is a function such that there exist a function g : B A such that gof = Ix and Fog

= Iy. Wherex A and y B, then F is both one-one and on to function.

(v) If F : X Y, g : Y Z and h : Z S then ho (gof) = (hog) of.

Invertible Functions: A function F : X Y is said to be invertible if there exist a functiong : y xsuch that gof = I

x and fog = I

y. The function g is called the inverse of F and is denoted by F–1. In other

words A function F : X Y is inversible if and only if F is one-one and onto function.

Remark: If F : X Y and g : y z be two invertible functions, then gof is also invertible with(gof)–1 = F–1 og–1.

Example: 4. Let F : R R be defined byf(x) = x3 –1 then breve that F–1 exist and find F–1.

We have to prove that F is (i) Letx1, x

2R (domain)

one-one and onto such thatf(x1) = f(x

2)

For one-onewe have to prove that x13 – 1 = x

23–1 x

1 = x

2

if any two images are equal then F is one-one.

their pre images are also equal (ii) Let yR (codomain)For on to we have to prove that for such thatf(x) = yeach y codomain there exists i.e.x3–1 = yx domain such thatf(x) = y x = (1 + y)1/3

(1 + y)1/3 R for every y REvery y codomain has pre-image

in domain.F is onto

F–1 : R R existwe have F–1(y) = (1 + y)1/3

Binary Operations: The term binary means two, therefore binary operation is an operation that operatesany two elements of a set to get a unique element of the same set. We know that when we add two realnumbers we get a unique real number. In this way addition operation is a binary operation.

Let A be a non empty set. Then, a function F : A × A A is called a binary operation on set A.

In other words let * be any operation on a non empty set A, then * is called a binary operation onA if a * b A for everya, b A. In this way set A is closed with respect to operation *.

Properties of Binary Operation: Set * be a binary operation on a non empty set A then

(i) * is called commutative ifa * b = b * a for everya, b A.

(ii) * is called associative ifa * (b * c) = (a * b) * c For everya, b, c A.

(iii) an element e A (if it exists) is called identity element for * if a * e = a = e * a for every a A.

(iv) The element a A is said to be invertible with respect to the binary operation * if there exists anelement b A such that a * b = e = b * a, where e is the identify element on A. The elementb is called inverse of a and is denoted by a–1.

49

Example: 5.Determine whether the function * defined on N by a * b = ab is a binary operation ornot. If yes, determine whether it is commutative and associative. Find the identify (if exists). Find allinvertible elements (if any) and their inverse.

Solution:

Thinking Process Application

* will be a binary operation if given (i) Clearly a * b = abN for all a, b Nset set is closed with respect to *. because product of two Natural Numbers

is again a natural number. * is a binaryoperation.

For commutativity a * b should equal (ii) Here a * b = ab and b * a = ba clearlyto b * a. ab = ab a * b = b * a

* is commutative.For associativity (a * b) * c = a * (b * c) (iii) Here (a * b) * c = (ab) * c = abc &Let e be the identify element a*(b * c) = a * (bc) = abc clearly * isthen a * e = a = e * a associative.

(iv) a * e = a & e * a = ai.e. ae = a & ea = a

e = 1 & e = 1e = 1 N

Let b A be an inverse of element * has 1 as identify on N.a A, a * b = e = b * a a*b = 1 and b*a = 1

b =1

aandb =

1

a

Clearly b N for a = 1 only.1 is the only invertible element On N

for * and its inverse is 1.

Practice Questions

1. Show that the relation R in a set A = {x : x z and 0 x 12} given by R = {(a, b) : |a – b| is a multipleof 4} is an equivalence relation.

2. Show that the function F : N N defined byf(x) = x2 + x is not a bijective function.

3. Let R be a relation on N × N, defiend by (a, b) R (c, d) ad = bc (a, b), (c, d) N × N

show that R is an equivalence relation.

4. Let F : R R be defined byf(x) = 3x + 2. Show that F is invertible. Find F–1.

5. If F : B A is defined by f(x) =3 4

5 7

x

xand g : A B is defined byg(x) =

7 4

5 3

x

x, then show that

fog = IA and gof = I

B where

A = R –3

5and B = R –

7

5.

50

6. Show that F : [–1, 1] R defined byf(x) =2

x

xis one-one but not onto.

7. Let * be a binary operation on Q (rational numbers) defined by a * b = |a – b|. Show that

(i) * is commutative (ii) * is not associative (iii) * does not have identity element.

8. Determine whether the function * defined on R by a * b = a + b – ab is binary operation or not. If

yes, determine whether it is commutative and associative. Find the identity (if it exists). Find allinvertible elements (if any) and their inverse.

9. Check whether the function F : R R defined byf(x) = x2 + 1 is one-one and onto.

10. Let R be a relation on a set A of all human being defined by R = {(a, b) A × A : a is a husbandof b}. Is relation R, reflexive, symmetric and transitive. Explain your answer.

Hints and Answers:

4. F–1(x) =2

.3

x

(i) * is a binary operation

(ii) * is commutative

(iii) * is associative

(iv) e = 0 is the identify on R for *

(v) every a R (a 1) is invertible with respect to * and its inverse is1

a

a9. Neither one-one nor onto.

10. (i) Not refelexive because no one can be a husband of himself.

(ii) Not symmetric because if a is husband of a then b being female can not be husband of anybody.

(iii) Not transitive.

51

Inverse TrigonometricFunction

2CHAPTER

Concept Building

We know that a function F : A B is invertible if and only if it is a one-one and onto function and in thatcase inverse of F is denoted by F–1. All trigonometric functions repeat their values after fixed intervalstherefore trigonometric functions are not one-one functions and hence their inverse do not exist. But, if werestrict their domains they can be made one-one and onto functions and we can obtain their inverses.

Let us consider the case if sine function. If we restrict its domain to interval I where I can be anyone

of the intervals3

, , , , ,2 2 2 2 2 2 and so on, then sine function becomes one-one and onto. So

inverse of sine function can exist and is denoted by sin–1. sin–1 : [–1, 1] IFor different choices of I we have different inverse functions called as branches of inverse sine

function. If we choose I = ,2 2 then the inverse function sin–1 : [ 1,1] ,

2 2 is called a principal

value branch of the inverse sine function. Similarly we have principal value branch for other five inversetrigonometric functions.

List of Principal Value Branches and the Domain of Inverse Trigonometric Functions:Functions Domain Range (Principal value Branch)

y = sin–1x 1 1x 22y

y = cos–1x 1 1x 0 y

y = tan–1x x2 2

y

y = cot–1x x 0 < y <

y = sec–1x1

1

x

x

2

02

y

y

y = cosec–1x1

1

x

x

02

02

y

y

52

Properties of inverse trigonometric functions :

P1

(i) 1sin (sin )x x, ,2 2

x

(ii ) 1sin(sin )x x , [ 1,1]x

(iii ) 1cos (cos )x x, [0, ]x

(iv) 1cos(cos ) ,x x [ 1,1]x

(v) 1tan (tan ) ,x x ,2 2

x

(vi) 1tan(tan ) ,x x x R.

(vii) 1cot (cot ) , (0, )x x x

(viii ) 1cot(cot ) , .x x x R

P2

(i) 1 11sin cosec ,x

x |x| 1

(ii ) 1 11cos sec ,x

x |x| 1

(iii ) 1 11tan cot ,x

xx > 0

P3

(i) 1 1sin ( ) sin ,x x x [–1, 1]

(ii ) 1 1tan ( ) tan ,x x x R

(iii ) 1 1cosec ( ) cosec ,x x |x| 1

(iv) 1 1cos ( ) cos ,x x x [–1, 1]

(v) 1 1sec ( ) sec ,x x |x| 1

(vi) 1 1cot ( ) cot ,x x x R

P4

(i)1 1sin cos , [ 1,1]

2x x x

(ii )1 1tan cot , R.

2x x x

(iii ) –1 1cosec sec , | | 12

x x x

P5

(i) 1 1 1tan tan tan , 11

x yx y xy

xy

53

(ii ) 1 1 1tan tan tan , 11

x yx y xy

xy

P6

(i) 1 12

22 tan tan ,| | 1

1

xx x

x

(ii ) 1 12

22 tan sin ,| | 1

1

xx x

x

(iii )2

1 12

12 tan cos , 0

1

xx x

x

Procedure for Finding Principal Value of Inverse Trigonometric Functions :Step 1: Write the given inverse trigonometric function with the point at which principal value is to be

determined. Keep the given point within brackets. For example to find principal value of1 1

sin2

we writed as 1 1sin

2.

Step 2: If the number within the bracket is negative then use a suitable part of proporty P3 to adjust

negative sign. For example1 1 1 11 1 1 1

sin sin and cos cos2 2 2 2

Step 3: In place of the number within the bracket write an angle in the form of that trigonometric function

whose inverse is written outside the bracket. For example,1 1 11 1sin sin sin sin .

2 2 6

Step 4: Use a suitable part of P1 to get rid of the trigonometric function and its inverse from the problem.

The result so obtained is the required principal value.

For example : 1 11 1sin sin sin sin .

2 2 6 6

Example 1:Find the principal value of tan–1(–1).

Thinking Process Application

Since the value within the bracket is negative tan–1 (–1) = –tan–1 (1)

therfore we use P3 (ii) to adjust negative sign. =1tan tan

4

In place of 1 we can wrire tan/4. Using = . ,4 2 2

P1(V) we get required value.

54

Example 2:Find the principal value of 1 1cot

3.

Thinking Process Application

Since the value within the bracket is negative 1 11 1cot cot

3 3

therefore we use P3(vi) to adjust negative sign. = 1cot cot

3

In place of1

3 we can writecot

3. Using =

3

P1 (vii) we get rid of cot–1 & cot and finally =

2

3

we get required value.

Principal values of the Functions of the type F–1 {f(θ)} :

Case 1 : If Principal value branch of the given inverse trigonometric function, then use a suitableresult from P

1 to get the required value for example.

1cos cos6 6 where 0,

6

Case 2 : If there is a negative sign with then use any one of the following results :

sin sin ,cos( ) cos , tan( ) tan

sec( ) sec ,cot( ) cot ,cosec( ) cosec

Now adjust the negative sign by using a suitable part of propertiy P3.

For example : 1 17sin sin

8

=1 117 17

sin sin sin sin8 8

=17

8 when ,

2 2

Case 3 : When > 2 as as in the above example, replace by some suitable value less than 2. In the

above example17

28 8

and sin 2 sin 88

.

55

Principal value in the above example can be found out as 1 17sin sin

8

= 1sin sin8 8

which ,2 2

Example 3:Evaluate 1 7cos cos

3

Thinking Process Application

Here is negative1 7

cos cos3

therefore we can use cos(–0) = cos . =1 7

cos cos3

7[0, ]

3 therefore

7

3 can be written as2

3= 1cos cos 2

3

Now we can use P1(iii) to get rod of cos–1 = 1cos cos

3

and cos. In this way we get required = [0, ]3

principal value.

Value of Function of the Type F[g–1(θ)]Step A. Convert the function g–1 into F–1 using following property :

1 1 1 1sin cos tan cotP B P B

H H B P

=1 1sec cos

H H

B P where P, B and H

denotes perpendicular, base and hupotenuse respectively.

56

Step 2 : Use a suitable part of P1 to get rid of trigonometric function and its inverse trigonometric function.

The result so obtained is the required value.

Example 4: Evaluate 1 8tan cos

17

Thinking Process Application

1 8cos

17 is to be converted into

1 18 15cos tan

17 8

1tanx

y by using 1 1cos tan

B P

H B

Using P1(vi) we can get rid of tan & tan–1.

1 18 15tan cos tan tan

17 8 =

15

8.

Values of combination of Inverse Trigonmetric Functions of the Form – –1 11 2F g ( ) ± g ( )θ θ

Step 1:Convert all the inverse trigonometric functions into inverse tangent functions.

Step 2:Use a suitable property to get the required result.

Example 5 :Evalute 1 13 4cot sin sec

4 3

Thinking Process Application

sec–1 can be coverted into1 14 3

sec cos3 4

cos–1 by using sec–1(x) = 1 1cos

x1 13 4

cot sin sec4 3

Now 1 1sin ( ) cos ( ) 2x x , therefore = 1 13 3cot sin cos

4 4there is no need to convert sin–1 and

= cot 2sec–1 into tan–1.

Example 6 :Evalute 1 13 12tan sin cos

5 13

57

Thinking Process Application

First of all we have to convert sin–1 and cos–11 13 3

sin tan5 4

1 112 5cos tan

13 12

Now1 13 12

sin cos5 13

1 1tan ( ) tan ( )x y formula can be used = 1 15 3tan tan

12 4

= 1

5 3412tan351 12 4

Now 1tan(tan )x formula can be used to = 1 56tan

33

get the required value 1 13 12tan sin cos

8 13

= 1 56tan tan

33= 56/33.

Example 7 :Solve the equation : 1 1 2tan 2cot

3x x

Thinking Process Application

Since 1 1tan cot 2x x 1 1 2tan 2cot

3x x

12cot x should be written as 1 1cot cotx x1 1 1 2

tan cot cot3

x x x

1cot3 2

x

1cot6

x

1cot ( ) cotx y x y cot6

x

3x .

into tan–1

58

Writing the Inverse Trigonometric Function in Simplest From :

(i) Write the given inverse trigonometric function in the form1F [ ( )]g x .

(ii) Now using suitable substitution/Trigonometric identify try to convert g(x) into f(x).

(iii) Finally using suitable property from P1 we get the simplest from of the given inverse trigonometric

function.

Example 8 :Write the following in simplest from

1

2 2tan ,| | .

xx a

a x

Thinking Process Application

If we replacex by a sin the term within the Substitutingx = a sin

bracket will convert into the form tan and i.e., 1sin .x

athen using the property 1tan (tan ) we get

wil lead to the simplest form.1 1

2 2 2 2

sintan tan

sin

x a

a x a a

=1 sin

tancos

a

a

= 1tan (tan )

=1sin .

x

a

Practice Questions

1. Write principal value of 1 1cos .

2

2. Write the principal value of 1cosece 2.

3. Show that 1 1

2tan sin .

1

xx

x

4. Write 1 3sin (3 4 )x x in the simplest form.

5. Evaluate 1cosec cosec .4

6. Solve forx : 1 1 1 8tan ( 1) tan ( 1) tan .

31x x

59

7. Prove that : 1 1 163 1 3cos 2 tan sin .

65 5 5

8. If 1 1 1tan tan tan ,a b c prove thata b c abc

9. Prove that : 1 1 sin 1 sincot , 0,

2 41 sin 1 sin

x x xx

x x

10. Evaluate the following :

1 13 5cos sin sin .

5 13

Hints and Answers

1.3

2.4

3. Hint : Substitutex = tan 4. 13sin x (Hint, substitutex = sin )

5.4

6.1

8,4

9. Hint : Rationalise the denominater and then use Trigonometric formulas.

10.33

.65

60

Matrices and Determinants3CHAPTER

Concept Building

An ordered rectangular arrangement of numbers enclosed by capital bracket [ ] is named as matrix. Pluralof matrix is named as matrices. These numbers are called elements of the matrix. Matrix is donoted bycapital letters of the English alphabet and its elements are denoted by small letters Horizontal lines of thenumbers are called rows of the matrix and vertical lines are called columns of the matrix. A matrix havingm rows andn columns is called a matrix of orderm b y n, written asmxn. In general a matrix of ordermxnis written as :

A =

11 12 13 1

21 22 23 2

31 32 33 3

1 2 3

1 2 3

n

n

n

i i i in

m m m mn

a a a a

a a a a

a a a a

a a a a

a a a a

é ùê úê úê úê úê úê úê úê úë û

(OR)A = [a

iy]

m×n where 1� i � m, 1� j � n, andm, n� N.

Types of Matrices—A matrix A = [aiy]

m×n is said to be a :

(i) Row matrix if m = 1 i.e. matrix having only one row(ii ) Column matrix if n = 1 i.e. matrix having only one column(iii ) Zero/Null matrix if each of its elements is zero.(iv) Square matrix if m = n i.e., if the number of rows is equal to the number of column(v) Diagonal matrix, Itm = n anda

ij = 0 wheni � j

i.e. if all its non diagonal elements are zero.

For example, A =

15 0 0

0 –5 0

0 0 7

é ùê úê úê úë û

is a diagonal matrix order 3.

(vi) Scalar matrix, ifm = n, aij = 0 wheni � j anda

ij = k wheni = j i.e. if its non diagonal elements are

equal

For example, B =

2 0 0

0 2 0

0 0 2

é ùê úê úê úê úë û

is a scalar matrix of order 3.

61

(vii) Unit/Identity matrix, if m = n, aij = 0 wheni = j anda

ij = 1 wheni � j i.e. if the non diagonal elements

are 1 only

For example A =

1 0 0

0 1 0

0 0 1

é ùê úê úê úë û

is a unit matrix of order 3.

Equality of Matrices—Two matrices A and B are said to be equal if they are of the same order

and each element of A is equal to the corresponding elements of B

For example A =0 2

3 5

é ùê úë û

and B =0 3

3 5

é ùê úë û

are

not equal matrices while

C =1 2

3 5

é ùê úë û

and D =1 2

3 5

é ùê úë û

are equal.

Example 1: Find the values of x and y if

2

3

x y

y

+é ùê úë û

=6 2

4 3

é ùê úë û

Thinking Process

Since two matrices are given equal therefore,

their corresponding elements must be equal

i.e. x + y = 6 and y = 4

Scalar Multiplication of a Matrix—Let A = [aij]

m×n be a matrix and k be any scalar, then KA is

another matrix which is obtained by multiplying each element of A by the scalar k. i.e. KA [Kaij]

m×n

Addition of Matrices—It A = [aij] and B = [b

ij] be two matrices of the same order say m×n, then

the sum of the two matrices is defined as a matrix C = [cij]

m×n where c

ij = a

ij + b

ij� “ values of

i and j.

For example if

A =1 –1 3 1 –1 2 –3

, B , C2 0 5 2 0 1 5

é ù é ù é ù= =ê ú ê ú ê ú

ë û ë û ë û

then A + B =4 0

7 2

é ùê úë û

and A + C is not defined

Properties of Matrix Addition

(i) Matrix addition is commutative i.e. if A and B are two matrices of the same order then A + B =B + A.

(ii ) Matrix addition is associative i.e. if A, B and C are three matries of the same order then(A + B) + C = A + (B + C)

Application

Solving x + y = 6

and y = 4

We get x = 2

� x = 2, y = 4 Ans.

62

(iii ) Existence of additive identity—Let A = [aij]

m×n by any matrix then there always exists a null matrix O

of orderm × n such that A + O = A = O + A.(iv) Existence of additive inverse—For any matrix

A = [aij]

m×n there exists a unique matrix

–A = [–aij]

m×n such that A + (–A) = 0 = (–A) + A.

Difference of Two Matrices—Let A = [aij]

m×n and B = [b

ij]

m×n be two matrices then we define

the difference of matrices A and B as

A – B = A + (–B) = [aij – b

ij]

m×n for all i and j

Multiplication of Matrices—The broduct of two matrices A and B is defiend if the number of

columns of A is equal to the number of rows of B.

Let A = [aij]

m×n and B = [b

jk]

n×b be two matrices, then the product of A and B is a matrix C of

order m × b i.e. AB = C = [cik]

m×b where c

ik is obtained by taking ith row of A and kth column of

B, multiplying them element wise and taking the sum of these products.

Example 2 : Find the product of A and B, if exists

Where A =4 5 6

0 1 2

é ùê úë û

and B =

1 2

1 0

1 2

é ùê úê úê úë û

Thinking Process

Here A is a matrix of order 2 × 3 and B is a

matrix of 3 × 2. Therefore product of A and

B is defiend and it will be a matrix of 2 × 2

i.e., AB =11 12

21 22

c c

c c

é ùê úë û

Now to get c11

we take Ist

Row of A and Ist column of B multiply them

element wise and take the sum of these

products

Application

c11

= 4 × 1 + 5 × 1 + 6 × 1

= 4 + 5 + 6 = 15

c12

= 4 × 2 + 5 × 0 + 6 × 2

= 8 + 0 + 12 = 20

c21

= 0 × 1 + 1 × 1 + 2 × 1

= 0 + 1 + 2 = 3

c22

= 0 × 2 + 1 × 0 + 2 × 2

= 0 + 0 + 4 = 4

� AB =15 20

3 4

é ùê úë û

Properties of Matrix Multiplication

(i) In general AB� BA. i.e. matrix multiplication is not commutative.(ii ) (AB)C = A(BC) i.e. matrix multiplication is associative.(iii ) A(B + C) = AB + AC (or) (A + B)C = AC + BC.

i.e. in matrices multiplication dustributes over addition.(iv) For every square matrix A there exists an identity matrix (I) of the same order such thus

AI = A = I.A.

63

Transpose of a Matrix—Let A = [aij]

m×n be any matrix, then the matrix obtained by interchanging

the rows and columns of A is called transpose of A and it is denoted by AI or AT i.e.

TA or A¢ = [aji]

n×m

For example if A =0 1 2

3 7 –5

é ùê úë û

, then AT =

0 3

1 7

2 –5

é ùê úê úê úë û

Properties of Transpose

(i) (A1)1 = A (ii ) (KA) 1 = KA1 (where K is any scalar)(iii ) (A + B)1 = A1 + B1 (iv) (AB)1 = B1A1.Symmetric and Skew Symmetrix Matrices—A square matrix A = [a

ij] is called a symmetrix matrix

if A1 = A i.e. [aij] = [a

ji] for all values of i and j. and it is called a skew symmetric matrix if

A1 = –A i.e. [aji] = –[a

ij] For all values of i and j.

Theorem 1: For any square matrix A, A + A1 is always symmetrix matrix and A – A1 is always

skew symmetric matrix.

Theorem 2: Every square matrix can be expressed as a sum of symmetric and skew symmetric

matrices

i.e. A =1 1

2 2

A A A Aæ ö æ ö+ -+ç ÷ ç ÷è ø è ø

Example 3: Express A =3 5

1 –1

é ùê úë û

as the sum of a symmetric and skew symmetric matrices.

Thinking Process

To express A as a sum of symmetric and

skew symmetric matrices we have to find

out

1

2

A A+ and

1

2

A A-

Application

A1 =3 1

5 –1

é ùê úë û

1

2

A A+ =

6 61

6 –22

é ùê úë û

1

2

A A- =

0 41

–4 02

é ùê úë û

A =3 3 0 2

3 –1 –2 0

é ù é ù+ê ú ê ú

ë û ë û

64

Elementary Operations (Transformations) of a Matrix

There are six transformations in a matrix which are known as elementary transformations. Three of these

transformations are on rows and other three are on columns. They are as following:

(i) Interchange of any two rows i.e. Ri R

j

(ii ) Multiplication of all elements of any row by a non zero scalar i.e. Ri� kR

i

(iii ) Addition to the elements of any row, the corresponding elements of any other row multiplied by a nonzero scalar i.e. R

i� R

i + kR

j

(iv) Interchange of any two columns i.e.ci c

j

(v) Multiplication of all elements of any column by a non zero scalar i.e.ci� kc

i

(vi) Addition to the elements of any column, the corresponding elements of any other column multipliedby a non zero scalar i.e.,c

i� c

i + kc

j

Invertible Matrices—Let A be a square matrix of order m. If there exists another square matrix

B of the same order m such that AB = BA = Im, then B is called inverse of A and it is denoted

by A–1. In that case A is said to be invertible

Note—

(i) If B is the inverse of A, then A is also inverse of B.

(ii ) Only square matrix passesses inverse. A rectangular matrix does not possess inverse becausge forthe product AB and BA to be defiend, it is necessary that the matrices A and B should be squarematrices of the same order;

(iii ) Inverse of a square matrix, if it exists, is unique.

(iv) If A and B are invertible matrices of the same order

then (AB)–1 = B–1.A–1

(v) (A–1)–1 = A

(vi) (A–1)1 = (A1)–1

Inverse of Matrix by Elementary Operations

(i) To find the inverse of a matrix by elementary operations, We can use either elementary row transfor-mations or elementary column transformations, but both type of operations can not be used simulta-neously.

(ii ) Let A = PQ where A, P, Q are square matrices of the some order then the effect of any elementary rowoperation an A is same as applying this elementary row operation on P and keeping Q unchanged.Similarly the effect of any elementary column operation on A is same as applying this elementarycolumn operation on Q and keeping P unchanged. In view of above discussion, we conclude that tofind the inverse of a matrix A using elementary row operations we write A = IA and apply a sequenceof row operations on A = IA till we get I = BA. This matrix B will be inverse of A. Similarly, to find A–1 using column operations we write A = AI and apply a sequence of column operations on A = AI tillwe get I = AB. This matrix B will be inverse of A.

Remark—In case after applying one or more elementary operations on A = IA (or A = AI), If

we obtain all zeros in one or more rows (or columns) of the matrix A on LHS, then A–1 does not

exist.

65

Algorithm to Find Inverse of a Matrix Using Elementary Row Operations (or columnoperations)

Let A =

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

é ùê úê úê úë û

be a square matrix of order 3. To find

A–1 we can proceed as following :

For row operations write A = IA

Step 1: On LHS make a11

= 1 using any of the three row transformations

Step 2: Make a21

and a31

each zero with the keep of a11

= 1 and using suitable row operations

Step 3: Make a22

= 1, using any row operations.

Step 4: Make a12

= a32

= 0 with the help of a22

= 1 and using suitable row operations.

Step 5: Make a33

= 1, using a suitable row operation.

Step 6: Make a13

= a23 = 0, by using a33 = 1 and suitable row operations

In this way matrix A on LHS converts into a unit matrix and matrix I on RHS converts into a

matrix (say B). This matrix B is the inverse of A.

For column operations—Write A = AI

Step 1: On LHS, make a11

= 1 using a suitable column operation.

Step 2: Make a12

= a13

= 0 by using a11

= 1 and suitable column operations

Step 3: Make a22

= 1, using a suitable column operation.

Step 4: Make a21

= a23

= 0 by using a22

= 1 and suitable column operations

Step 5: Make a33

= 1, by using a suitable column operation.

Step 6: Make a31

= a32

= 0 by using a33

= 1 and suitable column operations

In this may matrix A on LHS converts into a unit matrix and matrix I on RHS converts into a

matrix (say B). This matrix B is the inverse of matrix A.

Example 4: Find the inverse of the following matrix using elementary row operations

A =

0 1 2

1 2 3

3 1 1

é ùê úê úê úë û

Thinking Process

For row transformation A = IA

On LHS

S t e p 1 : To make a11

= 1 we can interchange

R1 and R

2

Step 2: Here a21

is already zero and a31

is to

be converted into zero by using a11

= 1

Application

0 1 2

1 2 3

3 1 1

é ùê úê úê úë û

=

1 0 0

0 1 0

0 0 1

é ùê úê úê úë û

A.

Applying R1� R

2.

1 2 3

0 1 2

3 1 1

é ùê úê úê úë û

=

0 1 0

1 0 0

0 0 1

é ùê úê úê úë û

A.

Applying R3 R

3 – 3R

1.

66

Step 3: No need of step 3 because a22

is

already 1.

Step 4: a12

and a32

are to be made zero by

using a22

= 1

Step 5: To make a33

= 1 we can divide R3 by

2.

Step 6: a13

and a23

are to be made zero each

with the help of a33

= 1.

We have got

I = BA

Therefore B is the inverse of A.

1 2 3

0 1 2

0 –5 –8

é ùê úê úê úë û

=

0 1 0

1 0 0

0 –3 1

é ùê úê úê úë û

A

By applying R1� R

1 – 2R

2 and R

3 = R

3 + 5R

2.

1 0 1

0 1 2

0 0 2

-é ùê úê úê úë û

=

2 1 0

1 0 0

5 3 1

-é ùê úê úê ú-ë û

A

By applying R3� R

3/2

1 0 1

0 1 2

0 0 1

-é ùê úê úê úë û

=

2 1 0

1 0 0

5 / 2 3/ 2 1/ 2

-é ùê úê úê ú-ë û

A.

By applying R1� R

1 + R

3 and R

2� R

2 – 2R

3.

1 0 0

0 1 0

0 0 1

é ùê úê úê úë û

=

1 2 1 2 1 2

4 3 1

5 2 3 2 1 2

-é ùê ú- -ê úê ú-ë û

A

Hence A–1 =

1 2 1 2 1 2

4 3 1

5 2 3 2 1 2

-é ùê ú- -ê úê ú-ë û

Determinant—To every square matrix A we can associate a number called determinant of the

matrix A and it is denoted as |A|. Only square matrices have determinants.

Determinant of a Square Matrix of Order 1: Let A = [a11

] be a square matrix of order 1, then

we define |A| = a11

Determinant of a square matrix of order 2:

Let A =11 12

21 22

a a

a a

é ùê úë û

be a square matrix of order 2, then we define

|A| =11 12

21 22

a a

a a = a

11a

22 – a

21a

12

Determinant of a square matrix of order 3

Let A =

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

é ùê úê úê úë û

be a square matrix of order 3

67

then we define |A| =

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

é ùê úê úê úë û

22 23 21 2311 12

32 33 31 33

a a a aa a

a a a a-

=21 23

1331 33

a aa

a a

Remark : Determinant can be expanded using any row or column.

Properties of Determinants—

1. If each element in a row (or column) of a determinant is zero, then the value of the determinant is zero.2. If any two rows (or columns) of a determinant are identical, then the value of the determinant us zero.3. If any two rows (or columns) of a determinant are interchanged then sign of the determinant changes

i.e. Ri � R

j (or c

i andc

j) = |A| = – |A|.

4. The value of the determinant remains unchanged if its rows and columns are interchanged.i.e., |A1| = |A|

5. If each element of a row (or column) of a determinant is multiplied by a constantk, then its value getsmultiplied by K. In other words we can take out any common factor from any row (or column) of adeterminant i.e.

1 1 1

2 2 2

3 3 3

ka kb kc

a b c

a b c =

1 1 1

2 2 2

3 3 3

| A |

a b c

k a b c k

a b c

=

In general |kAn×n

| = kn |An×n

|

6. If some or all elements of a row (or column) of a determinant are expressed as sum of two (or more)terms, then th determinant can be expressed as sum of two (or more) determinants.

7. The value of the determinant remains same if we apply the operations Ri� R

i + kR

j (or) c

i� c

i + kc

j

Remark: (i) |AB| = |A||B|

(ii) |A–1| =1

A

Area of a trangle using determinants—Area of a �ABC with vertices A(x1y

1), B(x

2, y

2), C(x

3y

3)

is given by

1 1

2 2

3 3

11

12

1

x y

x y

x y

Minors and cofactors—Minor of an element aij of a determinant is the determinant obtained by

deleting the ith row and jth column in which element aij lies. It is denoted by m

ij. cofactor of an

element aij of a determinant is denoted by A

ij and is defiend as A

ij = (–1)i+j m

ij.

Adjoint of a square matrix—Let A = [aij]n×n

be a square a matrix, then adjoint of A, is defiend

68

as the transpose of the matrix [Aij]n×n

where aij is the cofactor if the element aij and it is denoted

as adj A.

A(adj A) = (Adj A) A = |A| I.

|adj A| = |A|n–1 where A is a square matrix of order n.

Singular and non singular matrices—A square matrix A is said to be singular if |A| = 0 and non

singular of |A| � 0.

It A and B are non singular matrices of the same order then AB and BA are also non singular

matrices of the same order

A square matrix is invertible if |A| � 0

i.e., It A is a non singular matrix.

and A–1 =A

A

adj

Solution of a system of linear equations using inverse of matrix

A system of linear equations in three variables can be expressed in a matrix equation. For example

a1x + b

1y + c

1z = d

1

a2x + b

2y + c

2z = d

2

a3x + b

3y + c

3z = d

3

�

1 1 1

2 2 2

3 3 3

a b c x

a b c y

a b c z

é ùé ùê úê úê úê úê úê úë ûë û

=

1

2

3

d

d

d

é ùê úê úê úë û

� AX = B

where A =

1 1 1 1

2 2 2 2

3 3 3 3

, X , B

a b c x d

a b c y d

a b c z d

é ù é ùé ùê ú ê úê ú

= =ê ú ê úê úê ú ê úê úë ûë û ë û

By solving the matrix equation Ax = B or X = A–1B, we get the solution of the given system of

equations.

Remark:

(i) If |A| � 0, then the system of equations has unique solution and hence it consistent.(ii ) If |A| = 0 and (adj A) B � 0 (zero matrix), then the system of equations has no solution and hence

it is inconsistent(iii ) If |A| = 0 and (adjA)B = 0 (zero matrix), then the system of equations may be either consistent or

inconsistent according of the system has either insinitely many solutions or no solution.

Practice Questions

1. Construct a 2 × 3 matrix A = [aij] wherea

ij = 3i – j

2. If2 1 10

3 1 5x y

-é ù é ù é ù+ =ê ú ê ú ê ú

ë û ë û ë û, find x andy.

69

3. Express the following matrix as a sum of a symmetric and skew symmetric matrix.

1 3 5

6 8 3

4 6 5

é ùê ú-ê úê ú-ë û

4. Given that A =3 2

4 2

-é ùê ú

-ë û and I =

1 0

0 1

é ùê úë û

Find the real numberk such that

A2 – KA + 2I = 0

5. If A =2 3

3 4

-é ùê úë û

, show that AA2 – 6A + 17I = 0.

Hence find A–1.

6. Using properties of determinants show that

b c c a a b

q r r p p q

y z z x x y

+ + +

+ + +

+ + +

= 2

a b c

p q r

x y z

7. Show thatx = 2 is one of the roots of the equation

6 1

2 3 3

3 2 2

x

x x

x x

- -

- -

- +

= 0. Find other roots also.

8. Find the inverse of the following matrix using elementary transformations:

2 3 3

2 2 3

3 2 2

-é ùê úê úê ú-ë û

9. Solve the following system of linear equations using inverse of matrix.

2x + 3y + 4z = 8, 3x + y – 2z = 2, 4x – y – 5z = –9

10. If A =

1 1 2 2 0 1

0 2 3 and B 9 2 3

3 2 4 6 1 2

- -é ù é ùê ú ê ú

- = -ê ú ê úê ú ê ú- -ë û ë û

Find AB and hence solve the following system of equationsx – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2

70

Hints and Answers

1.2 1 0

5 4 3

é ùê úë û

2. x = 3,y = –4

3.

1 3 2 1 2 0 9 2 9 2

3 2 8 9 2 9 2 0 3 2

1 2 9 2 5 9 2 3 2 0

-é ù é ùê ú ê ú- + - -ê ú ê úê ú ê ú-ë û ë û

4. 1 5. 1 4 31A

3 217- é ù= ê ú

-ë û

7. 1, –3 8.

2 5 0 3 5

1 5 1 5 0

2 5 1 5 2 5

-é ùê ú-ê úê ú-ë û

9. x = 1,y = – 2, z = 3 10. Hint AB = I� A–1 = B.x = 0,y = 5,z = 3.

71

Continuity andDifferentiability

4CHAPTER

Concept Building

Continuity of a function

By continuity of a functiony = f(x) in a given interval (a, b), we mean there is no break in the curveof y = f(x), between pointsa andb. Mathematically we define continuity of a function it a point likethis.

Let f be a real function defined on the subset of real bumbers and ‘C’ be a point in the domain

of f then f is said to be continuous, atx = C if

lim ( )x c

f x�

= f(x)

we may say like this

lim ( )x c

f x�

= f(x) = lim ( )x c

f x�

i.e. L.H.L. = f(C) = R.H.L.A real functionf is said to be continous in a given open or closed interval if it is continuous at every

point of the interval.A real functionf is said to be continuous if it is continuous at every point in the domain of f.If a function is not continuous then it is called discontinuous function.

Example 1: show thatf(x) =sin

, 0

1, 0

xif x

xx if x

���

�� � ��

is continuous atx = 0

Solution: At x = 0

LHL =0

lim ( )x

f x� =

0

sinlimx

x

x!" =

0

sin( )lim

0h

h

h#

$

$ =

0

sinlimh

h

h%

&

& = 1

RHL =0

lim ( )x

f x' =

0lim( 1)x

x()� =

0lim(0 1)h

h*

+ + =0

lim 1h

h*

+ = 1

f(0) = 0 + 1 = 1.Here LHL = f(0) = RHL, f is continuous atx = 0.Example 2: Find the value ofk if a function f defined as

f(x) =

3

sin ( 1); 02

tan sin; 0

k x if x

x xif x

x

π-. /00

120 3

04

is continuous atx = 0.

72

At x = 0

f(0) = sin (0 1)2

kπ

= sin2

kπ

= k.

LHL =0

lim ( )x

f x5

=0

lim sin ( 1)2x

k xπ

6 =

0lim sin (0 1)

2hk h

π7

=0

lim sin (1 )2h

k hπ

6 = k. 1 = k.

RHL =0

lim ( )x

f x8 = 3

0

tan sinlimx

x x

x9

= 30

tan(0 ) sin(0 )lim

(0 )h

h h

h: =

30

tanh sinhlimh h;

=30

tanh(1 cosh)limh h<

=

2

20 0

2sintanh 2

lim limh h

h

h h= =

> ?@ AB C

=

2

02

sin 1 121.2 lim × =4 2

2

h

h

hD

E FG HG HG HI J

As f is continuous functionK LHL = f(0) = RHL

L k =1

2

Practice Question

Discuss the continuity of the function at the given points.

1. f(x) = 3

5 4, if 0 11.

4 3 , if 1 2

x xat x

x x x

M NOPQ

M MR

2. f(x) =1 1

when 00.

2 when 0

x xx

at xxx

STU

VWU VX

3. f(x) = 2 , 00.1 1 2

2, 0

xx

at xx xx

YZ[

\]^[ \_

4. f(x) =1

sin ; 00.

0; 0

x xat xx

x

`ab

cdb ce

73

5. f(x) =2

3

1 cos 4; 0

120.

1; 0

sin 2

x

xx

xat x

ex

s

fghh

ijkh l

hm

6. It the functionf(x) =

3 , f 1

11; if 1

5 2 ; if 1

ax b i x

x

ax b x

n opq

rsq t uv

is continuous atx =1, find the values ofa andb.

7. If f(x) =

1 1if 1 0

2 1if 0 1

2

kx kxx

xx

xx

w x y yy z {||

}x| z z

| y~

is continuous atx = 0, then find the value ofk.

8. Find the value of constantsa andb if

f(x) =

4if 4

4

if 4

4if 4

4

xa x

x

a b x

xb x

x

��� �� �

��� ��

� �� � �

���

is continuous ata = 4.

9. Given thatf(x) =

2

1 cos 4if 0

if 0.

if 016 4

xx

xa x

xx

x

��� �

��

���� �� � ��

If f is continuous atx = 0, find the value ofa.

10. Determine the values ofa, b, c if the functionf(x) =2

3/ 2

sin( 1) sinif 0

C if 0

if 0

a x xx

xx

x bx xx

bx

� � ���

����

�� �� �

��

is continuous at

x = 0.

Answers

1. Continuous 2. Discontinuous 3. Discontinuous4. Continuous 5. Continuous 6.a = 3, b = 2

7.1

2

�8. a = 1, b = –1 9. a = 8

10. a =3

2

�, C =

1

2 andb may be arbitrary..

74

Derivative of a function

The derivative of a functionf at x = C (wheref is a real function and ‘C’ is the point in the domain

of f ) is denoted asf ′(C) and defined as

f ′(C) =C

( ) (C)lim

Cx

f x f

x�

provided this limit exists.

Left hand derivative L.H.D. Lf ′(C) =0

(C ) ( )limh

f h f C

h�

Right hand derivative R.H.D Rf ′(C) =0

(C ) ( )limh

f h f C

h�

The derivative off is said to exist atx = C ifLHD or Lf ′(C) = RHD or Rf ′(C)The process of finding the derivative of a function is called differentiation.

The derivative off is denoted asf ′(x), or � �( ) .d

f xdx

If y = f(x) thendy

dx or y  = f ′(x).

Example 1: Show thatf(x) = 2x2 + x + 5 is differentiable atx = 1Solution: Here f(x) = 2x2 + x + 5At x = 1

LHD or Lf ′(1) =0

(1 ) (1)limh

f h f

h¡

¢ ¢

¢

=£ ¤

2

0

2(1 ) (1 ) 5 2 1 5limh

h h

h¥

¦ §¨ ©

=2

0

2 4 2 1 5 2 1 5limh

h h h

hª

=0

lim(5 2 )h

h«

= 5

RHD or Rf ′(1) =0

(1 ) (1)limh

f h f

h¡

=£ ¤

2

0

2(1 ) (1 ) 5 2 1 5limh

h h

h¥

¦ §¨ ©

=2

0

2 4 2 1 5 2 1 5limh

h h h

hª

=0

lim(5 2 )h

h«

= 5As LHD = RHD,¬ f is derivable atx = 1 andf  (1) = 5.Example 2: Show thatf(x) = |x – 2| is not derivable at x = 2.At x = 2

Solution : LHD or Lf ′(2) =0

(2 ) (2)limh

f h f

h­

® ®

®

75

=0

2 2 2 2limh

h

h¯

=0

lim 1h

h

h°±

RHD or Rf ′(2) =0

(2 ) (2)limh

f h f

h²

=0

2 2 2 2limh

h

h³

=0

lim 1h

h

h°±

As Lf ´(2) µ Rf ´(2)¶ f(x) = |x – 2| is not derivable at x = 2

Note : Every differentiable function is continuous but the converse is not true.

Practice Questions

Check the derivability of the following.(1) f(x) = x2 + x – 1 at x = 3

(2) f(x) =2 3

3 2

x

x

·

·at x = 2

(3) f(x) = |x + 2| at x = –2

(4) f(x) = [x] at x = 1(5) f(x) = |x + 1| + |x – 1| at x = –1

Derivative of Composite function

If f be a composition of two functionsu & vi.e. f = v 0 ui.e. f(x) = v[u(x)]i.e. f(x) = v(t) where t = u(x)

¶ f ′(x) = v´(t)dt

dx

=dv dt

dt dxThis is also calledchain rule

Example: Finddy

dx wherey = sin xe .

Solution: y = sin xe

dy

dx = sin xd

edx

= cos x xde e

dx

76

= cos x x de e x

dx

=cos

2

x xe e

x

Derivative of Implicit function

A function in two variablesx andy i.e. f(x, y) = 0 where any one of the variable is not expressible interms of the other variable is called implicit function.

Example: Finddy

dx if x2 + xy + y2 = 0.

Solution: x + xy + y2 = 0

Differentiating w.r.t.x we get,

2x + .1 2dy dy

x y ydx dx

= 0

( 2 )dy

x ydx

= –(2x + y)

dy

dx =

2– .

2

x y

x y

Derivative of Inverse trigonometric function:

Example: If y = sin–1

2

4

2

1

x

x then find

dy

dxSolution: Let x2 = tan = tan–1x2

y =2

–1

4

2sin

1

x

x

= –1

2

2 tansin

1 tan

= sin–1(sin 2 ) = 2 = 2tan–1x2

dy

dx =

24

2( )

1

dx

x dx =2

4.

1

x

xDerivation of logarithmic function :

logd

xdx

=1

x

log ( )a

df x

dx =

1log . ( ).

( ) a e f xf x

77

Example: If y = 2log 1x x , finddy

dx.

Solution: y = 2log 1x x

dy

dx = 2

2

11

1

dyx x

dxx x

=2 2

1 11 2

1 2 1x

x x x

=2 2

2 2

1 1

1 1

x x

x x x =

2

1

1x

Derivative of exponential function.

xde

dx = ex ;

xda

dx= axloga

Example: Finddy

dx wherey = x8. 8x

Solution: y = x8. 8x

dy

dx = 8 88 8x xd d

x xdx dx

= x8.8xlog8 + 8x.8x7

= x7.8x[xlog8 + 8]Logarithmic Differentiation

Example: If y = xsinx + (sinx)x then finddy

dx

Solution: Let u = xsinx and v = (sinx)x

logu = logxsinx logv = log(sinx)x

logu = sinxlogx logv = xlog sinxDiff. w.r.t. x, Diff. w.r.t. x,

1 du

u dx =

sinlog cos

xx x

x

1 dv

v dx =

coslogsin .1

sin

x xx

xdu

dx =

sinlog .cos

xu x x

xdv

dx = cot logsinv x x x

= sin sinlog .cosx x

x x xx

dv

dx = (sin ) cot logsinxx x x x

We are given that

y = xsinx + (sinx)x

78

y = u + v

dy

dx =

du dv

dx dx

dy

dx = sin sin

log cosx xx x

x + (sin ) cot logsinxx x x x

Derivative of functions in Parametric form.A function expressed between two variablesx andy in the form.x = f(t), y = g(t)

is said to be parametric form witht as a parameter. Heredy

dx is given as

dy

dx =

/

/

dy dt

dx dt

Example: Finddy

dx if x = et sint; y = et cost

Solution: We are given thatx = etsint y = etcost

dx

dt = etcost + etsint

dy

dt = et(–sint) + etcost

= et(cost + sint) = et(cost – sint)

dy

dx =

/

/

dy dt

dx dt =

(cos sin )

(cos sin )

t

t

e t t

e t t =

cos sin

cos sin

t t

t t

Second order DerivativeLet y = f(x)

thendy

dx = f (x) is called fist order derivative

andd dy

dx dx =

2

2

d y

dx = f (x) = y = y

2 is called second order derivative.

Example: If y = Ae2x + Be3x, prove that2

2 5 6 0d y dy

ydx dx

.

Solution: Given thaty = Ae2x + Be3x

thendy

dx = 2Ae2x + 3Be3x

2

2

d y

dx = 4Ae2x + 9Be3x

Hence2

2 5 6d y dy

ydx dx

= 4Ae2x + aBe3x – 5(2Ae2x + 3Be3x) + 6(Ae2x + Be3x)

= 0.Example: If y = (tan–1x)2, show that

(x2 + 1)2y2 + 2x(x2 + 1)y

1 = 2

Solution: Given that y = (tan–1x)2

79

y1 = 1

2

12 tan

1x

xi.e. (1 + x2)y

1 = 2tan–1x

Differentiating both side, w.r.t.x we get,

(1 + x2)y2 + 2xy

1 = 2

2

1 x

(1 + x2)2y2 + 2x(1 + x2)y

1 = 2

Rolle’s Theorem

Let f : [a, b] R be continuous on [a, b] and differentiable on (a, b) such thatf(a) = f(b) wherea andb are some real numbers. Then These exists somec in (a, b) such thatf (c) = 0.

Mean Value Theorem

Let f : [a, b] R be continuous function on [a, b] and differentiable on (a, b). Then these exists somec in (a, b) such that

f (c) =( ) ( )

.f b f a

b a

Example: Verify Rolle’s Theorem for the function f defined asf(x) = cos 24

x on 0,2

Solution: We are given

f(x) = cos 24

x .

(i) f, being a cosine function, is continuous in0,2

(ii ) f (x) = –sin 2 .24

x

¸ f (x) is finite in 0,2

f is derivable on 0,2

(iii ) f(0) = 0

2f = 0

f(0) = .2

f

All the three conditions of Rolle’s Theorems are satisfied, therefore, Rolle’s Theorem is applicable

By Rolle’s Theorem, f (c) = 0

– 2sin 2 C4

= 0

80

2 C4

= 0

C =4

0,2

Hence Rolle’s Theorem is verified.

Example: Verify Mean value theorem for the function defined byf(x) = x(x – 2) on [0, 4]

Solution: Given function isf(x) = x(x – 2)

(i) f, being a polynomial function, is continuous in [0, 4](ii ) f (x) = 2x – 2.

f (x) is finite in (0, 4), therefore,f is derivable in (0, 4)Now f(0) = 0(0 – 2) = 0

f(4) = 4(4 – 2) = 8

By Mean value theorem,f (C) =(4) (0)

4 0

f f

2C – 2 =8 0

4 0

2C – 2 = 22C = 4C = 2 (0, 4)

Hence Mean value theorem is verified.

Practice Question

Find the derivative of the following1. tan–15x 2. log(1 + e2x)3. sin3(tanx) 4. log

5log

5x

Finddy

dx in the following.

5. ax2 + 2hxy + by2 = 0 6. ex+y + ex–y = 4.

Finddy

dx in the following.

7.2 2

–1

2 2

1 1tan

1 1

x x

x x8.

2–1 1 1

tanx

x

9. 2log 1x x 10.1 sin

log1 sin

x

x

11. y = x x x ¹ 12. y =xxxº

13. y = xlogx + (logx)x 14. y =1 1

x

xxx

81

15. If y = Aemx + Benx then prove that2

2 ( )d y dy

m n mnydx dx

= 0.

16. If y =–1sina xe , then prove that

22 2

2(1 )d y dy

x x a ydx dx

= 0.

Verify Rolle’s Theorem for the function defined as

17. f(x) = exsinx on [o, ] 18. f(x) = x3 – 7x2 + 16x – 12 on [2, 3]

Verify Mean Value Theorem for the function defind as

19. f(x) = x(x – 1)(x – 2) on [0, 4] 20. f(x) = by2

( )

x ab

x a b on [a, b]

82

Application of Derivatives5CHAPTER

Concept Building

1. Rate of change of quantities with respect to other quantities.

dy

dx represents the rate of change ofy with respect tox

Let y = f(x)

thendy

dt = f (x)

dx

dt

wheredy

dt represents the rate of change ofy

dx

dt represents the rate of change ofx.

If C(x) is cost function then

Marginal cost =Cd

dx or C(x).

If R(x) is revenue function then

Marginal Revenue =Rd

dx or R(x).

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

vertex lower most. Its semi vertical angel is –1 1tan

2. Water is poured into it at a constant rate of 5m3/

minute. Find the rate at which the level of the water is rising at the instant when the depth of the waterin the tank is 10m.

Solution: Let V, r andh be the volume, radius and depth of the water cone at any time ‘t’Here it is given that

V

dt

d = 5m3/minute.

= –1 1tan

21

tan2

Nowr

h =

1

2r =

2

h

So V = 21

3r h =

21

3 2

hh =

3

12

h

83

Vd

dt = 2

4

dhh

dt

5 = 2

4

dhh

dt

dh

dt =

2

20

h

10h m

dh

dt = 2

20

(10) =

1/ min.

5m

Hence level of water is rising at the rate of1

/ min5

m at the instant when depth of water is 10m.

Practice Questions

1. A particle moves along the curvey = 321

3x . Find the points on the curve at which they–coordinates

are changing twice at fast asx–coordinates.

2. A balloon which always remains spherical is being infleted by pumping in gas at the rate of1800cm3/s. Find the rate at which the radius of the balloon is increasing when the radius of theballoon is 15cm.

3. A man 2m high walks at a uniform speed of 5km/hr away from a lam ppost 6m high. Find the rateat which his shadow is increasing.

4. The total cost C(x) associated with the production ofx units of an item is given byC(x) = 0.005x3 – 0.005x2 + 25x + 10000.Find the marginal cost when 10 units are produced.

5. A car starts from a point P at timet = 0 second, and stops at the point Q. The distancex in metrescovered by it int seconds is given by

x = 2 23

tt

Find the time taken by it to reach Q and also find distance between P and Q.2. Increasing and Decreasing Function: Strictly increasing function:

A function f is said to be increasing function on(a, b) if x

1 < x

2f(x

1) < f(x

2) x

1, x

2(a, b)

graphicallyf is increasing if the graph ofy = f(x)moves up asx moves to the right.

84

Strictly decreasing function:A functionf is saidto be decreasing function on (a, b) if x

1 < x

2

f(x1) > f(x

2) for all x

1, x

2(a, b).

Defunction: Let f be continuous on [a, b] anddifferentiable on (a, b)

(i) If f (x) > 0, x (a, b) then f is strictlyincreasing function on (a, b).

(ii ) If f (x) < 0, x (a, b) thenf is strictly decreasing functionon (a, b).

Example : Find the intervals for which the function defined byf(x) = x3 – 12x2 + 36x + 17 is strictlyincreasing or strictly decreasing.

Solution: We have f(x) = x3 – 12x2 + 36x + 17f (x) = 3x2 – 24x + 36

= 3(x2 – 8x + 12)= 3(x – 2)(x – 6)

Let f (x) = 03(x – 2)(x – 6) = 0

x = 2, 6Required intervals are (– , 2), (2, 6), (6, )

Intervals f ′(x) = 3(x – 2)(x – 6) Sign of f ′(x) Nature of f

(– , 2) f (x) = 3(–) (–) = +Ve f (x) > 0 Strictly increasing

(2, 6) f (x) = 3 (+) (–) = –Ve f (x) < 0 Strictly decreasing

(6, ) f (x) = 3(+) (–) = +Ve f (x) > 0 Strictly increasing

Hencef is strictly increasing in (– , 2) (6, 8) and strictly decreasing in (2, 6).

Example: Find the intervals in which the functionf(x) = sin4x + cos4x, 0 x2

increasing or

decreasing.Solution: Here f(x) = sin4x + cos4x

f (x) = 4sin3x cosx – 4cos3sinx= –4sin xcosx (cos2x – sin2x)= –2sin2x.cos2x = –sin4x

For increasing function,f (x) > 0 –sin4x > 0

sin4x < 0 < 4x < 24 2

x

Thus f(x) is increasing in ,4 2

For decreasing function,f (x) < 0

85

–sin4x < 0 sin 4x > 0 0 < 4x <

0 < x <4

Thus f(x) is decreasing in 0,4

Practice Questions

Find the intervals in which the following functions are strictly increasing or decreasing.1. f(x) = 2x3 – 15x2 + 36x + 102. f(x) = (x – 1)(x + 2)2

3. f(x) = log(1 + x) –1

x

x4. f(x) = sinx – cosx on [0, 2 ]

5. f(x) = 33

1x

x; x 0

6. f(x) =4sin 2 cos

2 cos

x x x x

x

7. f(x) = x4 – 2x2.8. Show that the functionf given byf(x) = tan–1 (sinx + cosx), x > 0, is always an strictly increasing

function in 0,4

and strictly decreasing function in , .4 2

9. Find the intervals in which the function given byf(x) = sin3x, x 0,2

is

(a) strictly increasing (b) strictly decreasing10. Find the reast value of ‘a’ such that the function f given byf(x) = x2 + ax + 1 is strictly increasing

on (1, 2).3. Tangents and Normals

(i) Slope of the tangent to the curvey = f(x) is f (x) and theslope of the tangent to the curvey = f(x) at the point(x

0, y

0) = f (x

0).

(ii) Slope of normal to the curvey = f(x) is1

( )f x

Slope of Normal to the curvey = f(x) at (x0, y

0) =

0

1

( )f x

(iii) Equation of tangent at (x0, y

0) to the curvey = f(x) is given by

y – y0 = f (x

0)(x – x

0)

(iv) Equation of Normal of (x0, y

0) to the curvey = f(x) is given by

y – y0 = 0

0

1( )

( )x x

f x

86

(v) If tangent is parallel tox–axis then f (x) = 0 and then equation of tangent at (x0, y

0) is given

by y = y0.

In this case, the equation of the tangent at (x0, y

0) is given byx = x

0.

(vi) If tangent is perpendicular to the x-axis thenf (x) , that is normal is parallel tox-axis.Example: Find the equation of tangents to the curvey = 4x3 – 3x + 5 which are perpendicular to

the line 9y + x + 3 = 0.Solution: Let (x

1, y

1) be the point of contact of tangent and the curve

y1 = 4x

13 – 3x

1 + 5 ...(5)

slope of given line =1

9Slope of tangent = 9.

Now y = 4x3 – 3x + 5

dy

dx = 12x2 – 3

Slope of tangent1 1( , )x y

dy

dx = 12x

12 – 3.

12x12 – 3 = 9

x12 = 1 or x

1 = ±1.

If x1 = 1, y

1 = 4 – 3 + 5 = 0

If x1 = –1, y

1 = –4 + 3 + 5 = 4.

Points of contact are (1, 6) and –1, 4)

Equation of tangent at (1, 6) isy – 6 = 9(x – 1) y = 9x – 3

Equation of tangent at (–1, 4) is

y – 4 = 9(x + 1) y = 9x + 13Example: Find the points on the curvey = x3 at which the slope of the tangent is equal to the

y-coordinate.Solution: Given curve is y = x3 ...(1)

dy

dx = 3x2

We are given that, slope of tangent =y-coordinate3x2 = y ...(2)

From (1) & (2) we get,x3 – 3x2 = 0 x2(x – 3) = 0 x = 0, 3.When x = 0, then y = 0When x = 3, then y = 27

Then points are (0, 0) and (3, 27).

Practice Questions

1. Find the equation of the tangent to the curvex y = a at the point2 2

,4 4

a b .

87

2. Find the equation of the tangent and normal to the curve.

x = 1 – cos ; y = – sin at = .4

3. Find the point on the curvey = x3 – 3x, where the tangent is parallel to the chord joining the points(1, –2) and (3, 18).

4. For the curvey = 4x3 – 2x5; find all the points at which the tangent passes through origin.5. Find the points on the curvex2 + y2 – 2x –4y + 1 = 0 where the tangents to the curve are parallel

to y–axis.

4. ApproximationConsider a functionf : D R and DCR and lety = f(x)

Let P(x, y) & Q(x + x, y + y) be two neighbouringpoints on two curve.

y = f(x + x) – f(x)The differential ofx is denoted bydx and is defined by

dx = x.The differential ofy is denoteddy and is defined by

dy = f (x)dx

or dy = .dy

xdx

In case wheredx = x is relatively small when compared withx, dy is a good approximation ofy and denoted asdy y

i.e. if y = f(x)then approximate value off(x + x) is given by

f(x + x) = f(x) + f (x) x.

Example: By using differentials, find approximate value of0.037

Solution: Let y = x andx = 0.04 andx + x = 0.037x = (x + x) – x = 0.037 – 0.040 = –0.003

y + y = x x

y = –x x y = –x x x

= 0.037 – 0.04

y = 0.037 – 0.2

0.037 = 0.2 + y ...(1)

Now y =dy

xdx

=1

. .2

dx x x

dx x

=1

( 0.003)2 0.04

=0.003

2 0.2 =

0.003

0.4 = –0.0075 ...(2)

From (1) & (2) we get, approximate value of0.037 = 0.2 – 0.0075 = 0.1925

88

Example: Find the approximate value off(2.01),where f(x) = 4x2 + 5x + 3Solution: Let f(x) = 4x2 + 5x + 3Here x = 2 and x = 0.01.

f(2) = 4(2)2 + 5 (2) + 3 = 29f (x) = 8x + 5.

Approximate value off(x + x) is given byf(x + x) = f(x) + f (x). x

f(2.01) = f(2) + f (2) × 0.01= 29 + [8(2) + 5] × 0.01= 29 + 0.21= 29.21

Practice Questions

Use differentials to find the approximate values of the followings.

1. 26 2. 100.02

3. 3 26.5 4. (80)1/4

5. If the radius of a circle is 8cm and it increases by 3% find the approximate increase in its area.5. Maxima and Minima.

First Derivative Test.

Let f be a differentiable function defined on the interval I and letx = a I then(a) x = a is a point of local maxima if

(i) f (a) = 0 and(ii) f (x) changes sign from positive to negative asx increases through ‘a’ i.e. f (x) > 0 at every

point sufficiently close to and the left ‘a’ and f (x) < 0 at every point sufficiently close to andright of ‘a’.

(b) x = a is a point of local minima if(i) f (a) = 0 and

(ii) f (x) change sign from negative to positive asx increases through ‘a’ i.e. f (x) < 0 at everypoint sufficiently close to and the left of ‘a’ and f (x) > 0 at, every point sufficiently close toand to the right of ‘a’.

(c) If f (a) = 0 andf (x) does not change sign asx increases through ‘a’ i.e. f (x) has the same signin the complete neighbourhood of ‘a’ then ‘a’ is neither a point of local maxima and nor a pointg

to local minima. Infact such a point is called a point of inflexion.Second Order derivative test:Theorem: Letf be a real valued function having second order derivative at C such that

(i) f (C) = 0 andf (C) > 0 thenf has a local minimum value at C.(ii) f (C) = 0 andf (C) < 0 thenf has a local maximum value at C(iii) f (C) = 0 andf (C) = 0 then second order derivative test fail and go to first order derivative

test inorder to decide whether it is a point of inflexion or not.

89

Example: Find both the maximum and the minimum value of 3x4 – 8x3 + 12x2 – 48x + 1 on theinterval [1, 4]

Solution: Let f(x) = 3x4 – 8x3 + 12x2 – 48x + 1f (x) = 12x3 – 24x2 + 24x – 48

Let f (x) = 012x3 – 24x2 + 24x – 48 = 0

x3 – 2x2 + 2x – 4 = 0(x – 2)(x2 + 2) = 0

x = 2.f(1) = 3(1)4 – 8(1(3 + 12(1)2 – 48(1) + 1 = –40

f(2) = 3(2)4 – 8(2)3 + 12(2)2 – 48(2) + 1 = –59

f(4) = 3(4)4 – 8(4)3 + 12(4)2 – 48(4) + 1 = 257

Maximum value off on [1, 4] = –59

Maximumu value off on [1, 4] = 257.Example: Find the points of local maxima or local minima for the functionf(x) = sinx – cosx,

0 < x < 2Solution: We havef(x) = sinx – cosx 0 < x < 2

f (x) = cosx + sinxLet f (x) = 0 cosx + sinx = 0 tanx = –1.

x =3 7

,4 4

Now f (x) = –sinx + cosx

3

4f = –

3 3 1 1sin cos 0

4 4 2 2

f is local maximum atx =3

.4

Now, f7

4 =

7 7– sin cos

4 4 =

1 1

2 2 = 2 > 0

f is local minimum atx =7

4

Local Max. value =3

4f =

3 3sin cos 2

4 4

Local Min. value =7

4f =

7 7sin cos 2

4 4Example: Find the point on the curvey2 = 4x which is nearest to the point (2, –8)

Solution: Let P(t2, 2t) be the point nearest to Q(2, –8) on the curve.

y2 = –4x ...(1)Now PQ2 = (t2 – 2)2 + (2t + 8)2

= t4 – 4t2 + 4 + 4t2 + 32t + 64= t4 + 32t + 68

90

Let PQ2 = f(t)f(t) = t4 + 32t + 68

f (t) = 4t3 + 32 = 4(t3 + 8)Let f (t) = 0 t3 + 8 = 0 t = –2

f (t) = 12t2

f (–2) = 12(–2)2 = 48 > 0.f is minimum att = –2

PQ is minimum.The point P is (4, –4) which is nearest to Q(2, –8).

Example: A right circular cylinder is inscribed in a right circular cone. Show that the curved surfacearea of the cylinder is maximum when the diameter of the cylinder is equal to the radius of the baseof the cone.

Solution: Let h be the height andr be the radius of the base of the cylinder inscribed in a rightcircular cone of height H and R be the radius of the base.

As OCD ~ OEB

CD

EB =

OC

OE

R

r =

H

H

h

r =R

(H )H

h ...(1)

Let S(h) be the surface area function of the cylinderS(h) = 2 rh.

S(h) =R

2 (H )H

h h

=22 R

(H )H

h h

S (h) =2 R

(H 2 )H

h

and S (h) =2 R

( 2).H

Now Let S(h) = 0.

2 R(H 2 )

Hh = 0

H – 2h = 0 2h = H h =H

2H

S2

=4 R

H < 0

91

Curved surface area is Maximum whenh =H

2 putting H = 2h in ...(1)

r =R 1

(2 ) R2 2

h hh

R = 2rRadius of the cone = diameter of the cylinder

Hence curved surface area of the cylinder is maximum when its diameter is euqal to the radius ofthe base of the cone.

Practice Questions

1. Find two numbers whose sum is 15 and the square of one multiplied by the cube of the other ismaximum.

2. Show that a closed right circular cylinder of given total surface areas and the maximum volumeV is such that its height is equal to the diameter of the base.

3. show that the semi-vertical angle of the right circular cone of given slant height and maximumvolume is –1tan 2.

4. Show that the semi vertical angle of a right circular cone of given surface area and maximum volume

is –1 1sin .

35. Show that the height of a right circular cylinder of maximum volume inscribed in a right circular

cone of heighth and radiusr is .3

h

6. A wire of length 28cm is to be cut into two pieces. One of the pieces is to be made into a squareand the other into a circle. What should be the lengths of the two pieces so that the combined areaof the square and circle is minimum?

7. A rectangle is inscribed in a semi-circle of radiusr with one of its side on the diameter of the semi-circle. Find the dimensions of the rectangle so that its area is maximum. Find also the area.

8. Show that the attitude of a right circular cone of maximum. Curved surface which can be inscribed

in a sphere of radius R is4R

.3

9. Show that a right circular cylinder which is open at the top and has a given surface area will havethe greatest volume if its height is equal to the radius of the base.

10. Prove that the right circular cone of maximum volume which can be inscribed in a sphere of radius

‘a’ has attitude equal to4

.3

a

11. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is8

27 of

the volume of the sphere.12. Show that the volume of the greatest cylinder which can be inscribed in a right circular cone of

height h and the semi-vertical angle is 3 24tan .

27h

92

Integrals6CHAPTER

Concept Building

Antiderivatives (or primitives).

If [F( )] ( )d

x f xdx

» , then F(x) is called antiderivative or integral off(x). or a prinmitive off(x) with

respect tox.

Again [F( ) ] ( )d

x c f xdx

¼ ½

So, F(x) + c is also antiderivative off(x) where C is a constant which may assume infinite number ofvalues, therefore, F(x) + c is called indefinite integral off(x) w.r.t. x and is written as

( ) ( )f x dx F x c¾ ¿

The process of finding an integral of a given function is known as integration and it is the inverseprocess of differentiation.

Rules of Integration

If ( )f x dx = F(x) + c

then K ( )f x dx = ( ) F( )k f x dx k x cÀ Á wherek is any constant.

and  Ã1 2( ) ( )f x f x dxÄ = 1 2F ( ) F ( )x x CÅ Æ

Also, Ç È1 2( ) ( )f x f x dxÉ ÊË = 1 2F ( ) F ( )x x CÌ ÍÎ Ï whereÐ, Ñ are constants.

Standard results of integration

1dx x cÒ Ó

1

, 11

nn x

x dx c nn

ÔÕ Ö × Ø

Ö

1( )( ) ; 1

( 1)

nx a b

ax b dx c nn a

ÙÚÚ Û Ú Ü Ý

Ú

1logdx x c

xÞ ß

1 log | |ax bdx c

ax b a

ßÞ ß

ß

x xe dx e cà ámx n

mx n ee dx c

m

ââ ã ä

log

xx

a

aa dx cå æ

log

mx nmx n a

a dx cm a

çç è é

93

sin cosxdx x cê ë ìcos( )

sin( )ax b

ax b dx ca

í îî ï î

cos sinxdx x cñ òsin( )

cos( )ax b

ax b dx ca

óó ô ó

2sec tanxdx x cõ ö2 tan( )

sec ( )ax b

ax b dx ca

÷÷ ø ÷

2cosec cotxdx x cù ú û2 cot( )

cosec ( )ax b

ax b dx ca

ü ýý þ ý

sec tan secx xdx x cÿ ðsec( )

sec( ) tan( )ax b

ax b ax b dx ca

�� � � �

cosec cot cosecx xdx x c� � �cos ( )

cosec( )cot( )ec ax b

ax b ax b dx ca

� �� � � �

tan log | cos | log | sec |xdx x c x c� � log | sec( ) |

tan( )ax b

ax b dx ca

�� � �

cot log | sin |xdx x cÿ ðlog | sin( ) |

cot( )ax b

ax b dx ca

ýý þ ý

sec log | sec tan |xdx x x c � �log | sec( ) tan( )

sec( )ax b ax b

ax b dx ca

� � �� � �

cosec log | cosec cot |dx x c� � �log | cosec( ) cot( )

cosec( )ax b ax b

ax b dx ca

� � �� � �

1 1

2

1sin or cos

1dx x c x c

x

� �� � � ��

1 12

1tan or cot

1dx x c x c

x� �� � � �

�

1 –1

2

1sec or cosec

1dx x c x c

x x

� ! " !"

Substitution Method.

# $# $

1F( )

( ) F ( )1

nn x

f x x dx cn

%

& ''

( )log | ( ) |

( )

f xdx f x c

f x( )

Following are some substitution which are useful in evaluating integrals.Expression Substitutions

2 2a x* or 2 2a x+ tan / cotx a a,

2 2a x- or 2 2a x. sin / cosx a a,

94

2 2x a/ or 2 2x a0 sec / cosecx a a1

a x

a x

2

3 or

a x

a x

3

2cosx a4

l Special Integrals

2 2

1dx

a x5 =

1sinx

ca

6 7

2 2

1dx

x a8 = 2 2log | |x x a c9 9 9

2 2

1dx

x a: = 2 2log | |x x a c; < ;

2 2

1dx

x a= =

11tan

xc

a a> ?

2 2

1dx

x a@ =

1log

2

x ac

a x a

AB

B

2 2

1dx

a x@ =

1log

2

a xc

a a x

BB

A

l Integral by using by parts Method.

To evaluate the integral of the type ( ). ( )f x g x dx let f(x) be the first function andg(x) be the second

function. Then

( ) ( )f x g x dx = ( ) ( ) ( ). ( )f x g x dx f x g x dx dxC DEF G

stating in word, we spell it like this.Integral of the product of two functions– first function × integral of the second function– Integral of [differential of first × integral of the second function]

l To evaluate the integral of the type H I( ) ( )xe f x f x dxJ write the given integral as

K L( ) ( ) ( ) ( ) .x x xe f x f x dx e f x dx e f x dxM N M

Integrating by parts only the first parts, keeping the second as it is, we get

O P( ) ( ) . ( )x xe f x f x dx e f x cQ R Q

l To evaluate integral of the type 2

1dx

ax bx c? ? where 2ax bx cS S can not be factorised into linear

factors make the cofficient ofx2 unity by taking the coefficient ofx2 common. Complete the square byadding and subtracting the square of the half the coefficient ofx. Then the integral reduces to any oneof the three standard forms.

l To evaluate integral of the type 2

px qdx

ax bx c

T

T T

95

or 2

px qdx

ax bx c

U

U U we write

px+ q = A2( ) .

dax bx c B

dxV V V

Equate from both sides, the coefficient ofx and the constant terms to find the values of A and B.Separate the integral by putting the value of A and B which reduces to the any one of the forms abovestated.

l Integration by partial fractions.In some cases we find integral by using partial fraction of the following types

Form of the rational fraction Forms of its partial fraction

,( )( )

px qa b

x a x b

WX

Y Y

A B

x a x bZ

[ [

2( )

px q

x a

\

] 2( )

A B

x a x a\

] ]

2

( )( )( )

px qx r

x a x b x c

^ ^

_ _ _

A B C

x a x b x cZ Z

[ [ [

2

2( )( )

px qx r

x a x b

` `

a a2( )

A B C

x a x b x bb b

c c c

2

2( )( )

px qx r

x a x bx c

d d

e d d 2

A Bx C

x a x bx c

ff

g f f

where 2x bx ch h can not be factorised further

In above case A, B and c are real numbers which will be determined suitably.l Some Special Integral

22 2 2 2 11

sin2 2

a xa x dx x a x c

aij k j l l

22 2 2 2 2 21

log2 2

ax a dx x x a x x a cm n m m m m m

22 2 2 2 2 21

log2 2

ax a dx x x a x x x c] o ] ] \ ] \

l Definite IntegralLet f be a continuous function defined on the closed interial [a, b] then the definite integral

of f(x) w.r.t. x from a to b (the lower limit and the upper limit respectively is denoted by( )b

af x dx

Further, let F(x) be the antiderivative off(x), then definite integral in evaluated as

( ) F( ) F( ) F( )b b

aaf x dx x b ap p q

96

l Definite integral as a limit of sums

To evaluate definite integral ( )b

af x dxas a limit of sums, anyone of the following forms can be used.

r s0

( ) lim ( ) ( ) ( 2 ) ... { ( 1) }b

a hf x dx h f a f a h f a h a n h

tu v v v v v v v w where

b ah

n

xy

z {0

( ) lim F( ) ( 2 ) ( 3 ) ... ( )b

a hf x dx h a h f a h f a h f a nh

|} ~ ~ ~ ~ ~ ~ ~ ~ where

b ah

n

xy

l Properties of definite integrals

P1

: ( )b

af x dx = ( )

b

af t dt

P2

: ( )b

af x dx = ( )

a

bf x dx�

P3

: ( )b

af x dx = ( ) ( )

c b

a cf x dx f x dx� wherea < c < b

P4

: ( )b

af x dx = ( )

b

af a b x dx� �

P5

:0

( )a

f x dx =0

( )a

f a x dx�

P6

:2

0( )

af x dx =

0 0( ) (2 )

a af x dx f a x dx� �

P7

:2

00

2 ( ) ; if (2 ) ( )( )

0 ; if (2 ) ( )

aa f x dx f a x f x

f x dxf a x f x

� � ��� �� � � ��

P8

:2

00

2 ( ) ; if ( ) ( )( )

0 ; if ( ) ( )

aa f x dx f x f x

f x dxf x f x

� � ��� �� � � ��

Example : Evaluate

(i)2

2

sin

(1 cos )

xdx

x�(ii)

sec

sec( ) tan

xdx

x x�(iii) 2(tan cot )x x dx�

Solution: (i)2

2

sin

(1 cos )

xdx

x� =

2

2

1 cos

(1 cos )

xdx

x

�

�

=1 cos

1 cos

xdx

x

�

�

=

2

2

2

2sin2 tan

22cos2

xx

dx dxx

� �� � �

� �

=2sec 1

2

xdx  ¡

¢£ ¤¥ ¦

97

= 2 tan2

xx c§ ¨

(ii)sec

sec tan

xdx

x x© =

sec (sec tan )

(sec tan )(sec tan )

x x xdx

x x x x

ª

« ª

=2sec sec tan

1

x x xdx

¬

= 2sec sec tanxdx x x dx­

= tan secx x c® ¯

(iii) 2(tan cot )x x dx°

= 2 2(tan cot 2 tan cot )x x x x dx± ±

= 2 2(sec 1 cosec 1 2)x x dx² ³ ² ³

= 2 2(sec cosec )x x dx´

= tan cotx x c® ¯

Example : Findsin

sin ( )

xdx

x x aµ

Solution :sin

sin ( )

xdx

x x aµ

=sin[( ) ]

sin( )

x a adx

x a

¶ ·

¶

=sin( )cos cos( )sin

sin( )

x a a x a adx

x a

¸ ¹ ¸

¸

= [cos sin cot( ]a a x a dxº »

= cos sin log | sin( |x a a x a c¼ ½ ¼

Example : 2 2 2 2

sin cos

sin cos

x xdx

a x b x¾

Solution : Let I = 2 2 2 2

sin cos

sin cos

x xdx

a x b x¾

let 2 2 2 2sin cosa x b x t¼ ¿2 2[ .2sin cos .2cos sin ]a x x b x x dx dtÀ Á

2 22( )sin cosa b x x dx dt Ã

2 2sin cos

2( )

dtx x dx

a bÄ

¶

98

Å I = 2 2

1 1

2( )dt

a b tÆ

= 2 2

1log | |

2( )t c

a bÇ

È

=2 2 2 2

2 2

1log | sin cos |

2( )a x b x c

a bÉ É

Ê

Example : Evaluate 2 1tanx x dxË

Solution : 2 1tanx x dxË

= 1 2tan .x x dxÌ

=3 3

12

1tan . .

3 1 3

x xx dx

xÍ Î

Ï

=3 3

12

1tan

3 3 1

x xx dx

xÐ Ñ

Ò

=3

12

1tan

3 3 1

x xx x dx

xÓ Ô Õ

Ö Ö× ØÙÚ Û

=3 2

12

1 1 2tan .

3 3 2 6 1

x x xx dx

xÜ Ý Þ

Þ

=3 2

1 21tan log |1 |

3 6 6

x xx x cß à á á á

Example : Evaluate1 1

1 1

sin cos

sin cos

x xdx

x x

â â

â âÈ

Ç

Solution : Let I =1 1

1 1

sin cos

sin cos

x xdx

x x

â â

â âÈ

Ç

=

1 1sin sin2

2

x xdx

ã ãä åæ æç èé ê as

1 1sin cos2

x xë ëì í

=12

2sin2

x dxîï ñòó ô

õ ö

= 14sin 1x dx dx÷ ø

= 1

4I x cù ú ...(i)

99

Let 1sin xû = t ü x = sin2t

ü dx = 2sint cost atü dx= sin2t dt

ý I1 = 1sin x dxþ

= .sin 2t t dt

=cos 2 cos 2

1.2 2

t tt dtÿ ÿð � ð �

ÿ� � � �� � � �

=cos 2 sin 2

2 4

t t t� �

=2 21 1

(1 2sin ) .2sin 1 sin2 4

t t t t� � � �

=11 1

sin [1 2 ] 12 2

x x x x �

Hence I =1 24 2 1 1

sin2 2

xx x x x c� � �� �� � � �� �� �

� �� �

Example : Evaluate :2 1

( 1)( 2)( 3)

xdx

x x x

�

� � �

Solution : Let2 1

( 1)( 2)( 3) 1 2 3

x A B C

x x x x x x

�� � �

� � � � � �

ý 2 1 ( 2)( 3) ( 1)( 3) ( 1)( 2)x A x x B x x C x x� � � � � �

Putting, x = 1, 1 = A(3) (–2) ü A =1

6!

Putting, x = –2, –5 = B(–3)(–5) ü B =1

3"

Putting, x = 3, 5 = C(2) (5)ü C =1

2

ý2 1

( 1)( 2)( 3)

x

x x x

�

� � � =

1 1 1 1 1 1

6 1 3 2 2 3dx dx dx

x x x �

�

=1 1 1

log | 1| log | 2 | log | 3 |6 3 2

x x x c# # # $ $ # $

Example : Evaluate :2

2 2( 1)( 4)

xdx

x x% %

Solution :2

2 2( 1)( 4)

xdx

x x% %

100

=2 2

2 2

1 4( 1) ( 4)

3 ( 1)( 4)

x xdx

x x

& ' &

& &

= 2 2

1 4 1

3 4 1dx

x x( )

*+ ,- -. /

= 2 2

4 1 1 1

3 4 3 1dx dx

x x0

1 1

=1 14 1 1

. tan tan3 2 2 3

xx c2 23 4

=1 12 1

tan tan3 2 3

xx c5 56 7

Example : Evaluate :3 2q

dx

x x8 9

Solution :3 2q

dx

x x: ; =

22 9 1 3 13

4 4 2 2

dx dx

x x x<

=

> ? > ?@ A @ @ @B C B CD E D E

=2 2

3 3 1log

2 2 2x x cF G F G F GH I H H IJ K J K J K

L M L M L M

=23

log 3 22

x x x cN O

P Q P Q QR ST U

Example : Evaluate22 sin

.1 cos

xxe dx

x

V

V

Solution : Let I =22 sin

1 cosxx

e dxx

W

W

=2 sin 2

.(2)1 cos 2

tte dt

t

4

4

1Put

2 2

xt dt dx

X YZ [ Z\ ]

^ _

= 2

2(1 sin cos )2

2costt t

e dtt

`

= 22 (tan sec )te t t dta

=2

I II2 tan 2 sect te t e t dtb b

= 2 22 tan . sec 2 sect t tt e e t dt e t dtc de fg h

101

= 2 22 tan 2 sec 2 sect t te t e t dt e t dti j

= 2 tante t ck

= 22 tan2

x xe cl

Example : Evaluate : 2

5 2

3 2 1

xdx

x x

m

n n

Solution : The given integral is of the form2

( ),

px qdx

ax bx c

o

o o therefore, we express.

p2(5 2) A (1 2 3 ) B A(2 6 ) B

dx x x x

dxq r s s s r s s

Equating the coefficients ofx and constant term both the sides, we get 6A = 5 and2A + B = 2

or A =5

6 and B =

11

3

t

p 2

(5 2)

(1 2 3 )

xdx

x x

u

v v = 2 2

5 2 6 11

6 (1 2 3 ) 3 3 2 1

x dx

x x x x

wx

w w w w

= 1 2

5 11I I (say)

6 3y ...(i)

I n I

1 putting (1 + 2x + 3x2) = t, so that (2 + 6x) dx = dt

p I1 = 2

1 1log | | log | 3 2 1|dt

t c x x ctz { z { { { ...(ii)

and I2 = 22 2

2

12 13 2 1 3 1 233 3 3 3

dx dx dxxx x

x x

| |} } ~ � ~ �~ �} }� � } } � �� �� � � � � �

Putting1

3x t� �� �� �

� � so thatdx= dt

we get I2 = 2

2

1

3 23

dt

t� �

� � �� �

= 12

1tan

2 23

3 3

tc� �

=1

2

1 3tan

2 2

tc� �

= 12

13

1 3tan

2 2

xc�

� ��� �

� � � ...(iii)

102

Using (2) & (3) in (1), we get

2

5 2

(1 2 3 )

xdx

x x

�

� � =

2 15 11 (3 1)log 3 2 1 tan

6 3 2 2

xx x c  ¡¡ ¡ ¢ ¡

Example: Evaluate6 7

( 5)( 4)

xdx

x x

£

¤ ¤

Solution : This integral is of the form2

( )px qdx

ax b x c

¥

¥ ¥

Let us express 6x + 7 =2A ( 9 20) B

dx x

dx¦ § §

= A (2x – 9) + B

Equating the coefficient ofx and constant terms from both sides, we get6 = 2A and 7 = – 9A + B

Which gives A = 3 and B = 34

6 7

( 5)( 4)

xdx

x x

¨

© © =

2 2

(2 9)3 34

9 20 9 20

x dxdx

x x x x

ª«

ª « ª «...(1)

= 1 23I 34I¬

In I1, putting 2 9 20x x t­ ® ¯ so that (2x – 9) dx + dt

° I1 =

2

(2 9)

9 20

xdx

x x

±

± ²= 12

dtt c

t³ ¡

= 212 9 20x x c´ µ µ ...(2)

Now consider I2 =

2 29 20 9 12 4

dx dx

x xx

¶· ¸ ¹ º

· ·» ¼½ ¾

Putting9

,2

x t¿ À

Á ÂÃ ÄÅ Æ

so thatdx = dt

° I2 =

22 1

2

dt

t¹ º

· » ¼½ ¾

= 22

1log

4t t cÇ È Ç

=2

2

9 9 1log

2 2 4x x c

É Ê É ÊË Ì Ë Ë ÌÍ Î Í Î

Ï Ð Ï Ð

=2

2

9log 9 20

2x x x cÑ Ò¢ ¡ ¢ ¡ ¡Ó Ô

Õ Ö...(3)

103

Substituting (2) and (3) in (i), we obtain

6 7

( 5)( 4)

xdx

x x

×

Ø Ø =

2 296 9 20 34 log 9 20

2x x x x x cÙ ÚÛ Ü Ü Û Û Û Ü ÜÝ Þ

ß à

[wherec = c1 + 34c

2]

Example : Evaluate2

0

sin .cos .x x dx

Solution : Let t = sinx, then = cosx dx

whenx = 0, t = 0 and when ,2

x á t = 1

â

2

0

sin .cos .x x dx =

113 2

00

2.

3t dt tã äå æ ç

=3 2 3 22

1 03è éêë ì

=2

3

Example : Evaluate4

0

sin cos.

9 16sin 2

x xdx

x

í

í

Solution : Let I =4

0

sin cos

9 16sin 2

x xdx

x

î

î

=4

0

sin cos

16[1 1 2sin cos ]

x xdx

a x x

ï

ï ñ ï

=

4

2 20

(sin cos )

9 16[1 (sin cos 2sin cos )]

x x dx

x x x x

ò

ò ó ò ó ô

=

4

20

(sin cos )

9 16[1 (sin cos ) ]

x x dx

x x

õ

õ ö ö

Put, sin cos ,x x t÷ ø then (cos sin )x x dx dtù ø

whenx = 0, t = sin 0 – cos 0 = – 1

whenx =4

, t = sin4

– cos4

= 0

â I =0 0 0

2 2 2 21 1 19 16(1 ) 25 16 5 (4 )

dt dt dt

t t tú ú úû û

ü ý ý ý

=0

1

1 1 5 4 1 1log log1 log

2 5 4 5 4 40 9

t

t þ

ÿ ð �� �� � � � �� � � ��

=21 1 1

log 9 log3 log3.40 40 20

� �

104

Example : Prove that4

0log(1 tan ) log 2

8d� �

Solution : Let I =4

0log(1 tan )d� ...(i)

=4

0log 1 tan

4d

� �� �� �� �� �

� �� �

=4

0

1 tanlog 1

1 tand

! "#$ %#& '

=4

0

2log

1 tand

( )* +,- .

=4 4

0 0log2 log(1 tan )d d/ �

0 I = 40log 2 [ ] I1 2 by using ...(i)

or 2I = log 243 45 67 8

I = log 28

Hence4

0log(1 tan ) log 2

8d9 :

Example : Evaluate3 2

1| sin |x x dx

;

Solution :3 3

1 ,2 2

x x< = = > < = =

Now, 1 0x? @ @ A 0 sin( ) 0x xB C C D E ( 1 0)xF G GH

A | sin( ) | sin( )x x x xI

Again 0 < x < 1A 0 xJ J

A sin( ) 0x K

A sin( ) 0x x L

and | <x <3/2A3

2x MN

A sin( ) 0x O

A sin( ) 0x x P

A | sin |x x = sin( )x xQ

thus, | sin |x x = sin( ), 1 1x x xR S S

and, | sin |x x =3

sin( ),12

x x xT U U

03 2

1| sin |x x dx

; =

1 3 2

1 1| sin | | sin |x x dx x x

VW

105

=1 3 2

1 1sin sinx x dx x x dx

XY Z

=1 3 2

0 12 sin( ) sin( )x x dx x x dx[ ...(i)

...(\ x sin x is even function)

Now sinx x dx =1

cos cosx

x x dx]

^

= 2

1cos sin

xx x

_` ...(ii )

from (i) & (ii) we get

3 2

1| sin |x x dx

a =

1 3/ 2

2 20 1

1 12 cos sin cos sin

x xx x x x

b c b cd e d d ef g f gh i h i

= 2 2 2

1 1 3 3 1 3 1 12 cos sin 0 cos sin cos sin

2 2 2

jk lm n m nk o k k o k k op qr s r st u t uv w

=2

1 1 12 0 0 ( 1) 0

x yz { | }~ � ~ � � ~� �� �� �

� � � �� �

=2 2

2 1 1 3 1�� � �

Example : Evaluate4 2

1( )x x dx� as the limit of sums.

Solution : Let 2( ) ; 1, 4f x x x a b� � � �

� nh = b – a = 4 – 1 = 3

By using definite integral as a limit of sums

as ( )b

af x dx = � �

0lim ( ) ( ) ... ... { ( 1) }h

h f a f a h f a n h�

� � � � � � �

�4

1( )f x dx =

0lim [ (1) (1 ) (1 2 ) ... {1 ( 1) }]h

h f f h f h f n h�

� � � � � � � �

�4 2

1( )x x dx� = 2 2 2

0lim (1 1 {(1 ) (1 )} {(1 2 ) (1 2 )} ...h

h h h h h�

� � � � � � � � � � � ��2[{1 ( 1) } {1 ( 1) }]n h n h� �     �  

=2 2 2 2

0lim 1 (1 ) (1 2 ) ... {1 ( 1) }h

h h h n h¡

¢ £¤ ¤ ¤ ¤ ¤ ¤ ¤ ¥¦ §

¨ ©0

lim 1 (1 ) (1 2 ) ... {1 ( 1) }h

h h h n hª

« ¬ ¬ ¬ ¬ ¬ ¬ ¬ «

=2 2 2

0lim 1 (1 2 ) (1 4 4 ) ...h

h h h h h­

® ¯ ¯ ¯ ¯ ¯ ¯ ¯°2 2{1 2( 1) ( 1)n h n h ±² ² ³ ² ³ ´

0lim [1 1 ... term) { 2 ... ( 1) }]h

h h h n hµ

¶ · · · · · · · ¶

=2 2 2 2 2

0lim 2 {1 2 3 ... ( 1)} {1 2 3 ...( 1) }h

h n h n h n¸

¹ º» » » » » ¼ » » » ¼½ ¾

0lim [ {1 2 3 ... ( 1)}]h

h n h n¿

À Á Á Á Á Á À

106

= 2

0

( 1) ( 1)(2 1)lim

2 6h

n n n n nh h h

Â

à à ÃÄ ÅÆÇ È

É Ê

=0

( ) ( )(2 )lim

2 6h

nh h nh nh nh h nh hh

Ë

Ì Ì ÌÍ ÎÌ ÏÐ Ñ

Ò Ó

=3 3 3 3 6 9 27

92 6 2 2

Ô Ô ÔÕ Ö Õ Ö

Example : Evaluate4 2

0( )xx e dx× as limit of sums

Solution : Here 2( ) , 0, 4xf x x e a bØ Ù Ø Ø

4 0 4.nh b aÚ Û Ú Û ÚBy using definite integral as a limit of sum as

( )b

af x dx = Ü Ý

0lim ( ) ( ) ( 2 ) ... { ( 1) }h

h f a f a h f a h a n hÞ

ß ß ß ß ß ß ß à

4 2

0( )xx e dxá =

2 4 2( 1)

0lim 1 ( ) (2 ) ... ... {( 1) }h h n h

hh h e h e n h e â

ãä åæ æ æ æ æ æ æ ç æè é

=2 4 2( 1) )

0lim (1 ... {1 2 3... ( 1)}h h n h

hh e e e h nâ

ãä åæ æ æ æ æ æ æ æ çè é

=2

20

1( 1) ( 1)lim

1 2

nh

hh

e nh nh

eê

ë ìí íîï ñ

íò ó

=2

20 0

1 ( )lim lim

1 2

nh

hh h

e nh nh hh

eô ô

õ ö÷ ÷õ öøù ú ù ú÷ û üû ü

=8

2

0

1 4(4 0)1 2

limh

h

e

e

hý

þ þÿ

þ

=8 1 16

2 2

e ð�

=815

2

e�2

0

1lim 2

h

h

e

h�

� ���� �

�

Practice Problems

Evaluate the following integrals

1. 2 2

1

sin cosdx

x x2.

3 3

2 2

sin cos

sin cos

x xdx

x x

�

3. 4sin x dx 4.5log 3log

3log log

x x

x x

e edx

e e

ç

ç

107

5. 2 2 2 2

sin 2

sin cos

xdx

a x b x 6. 2 2 2 2

1

sin cosdx

a x b x

7. 2

( 1)

sin

x

x

x edx

x e

�8.

1

sin( )cos( )dx

x x�� ��

9. 3

1

sin cosdx

x x10.

1 cos sintan

cos sin

x xdx

x x� �� �� �

�� �

11. 2

1

3 13 10dx

x x� �12.

1

( 1)( 2)dx

x x� �

13. 2

2 1

4 3

xdx

x x

�

� �14. 2

3

2 5

xdx

x x

�

� �

15. 2( 1) ( 2)

xdx

x x !16. 2 31

xdx

x x x" " "

17.1sin

xdx

a x#

$18. 2( 3) 3 4x x x dx% & &

19.2

2

( 1)

( 1)

xx edx

x

'

'20. 2

1 1

log (log )dx

x x

( )*+ ,

- .

21. Evaluate3 2

1(2 )x x dx/ as a limit of sums.

22. Evaluate5 2

2( 1)x x dx/ / as a limit of sums.

23. Evaluate4 2 1

2

xe dx0 as a limit of sums.

24. 2

30

cos

cos sin2 2

xdx

x x1 2%3 45 6

25.2

4 40

sin 2

sin cos

xdx

x x7

26.2

2sin | |x dx

8 27.2 2

0[ ]x dx

28.22

0

sin

sin cos

xdx

x x%29.

1

20

log(1 )

1

xdx

x

9

9

30.0

tan

sec tan

x xdx

x x:

Answers

1. tan cotx x c; < 2. sec cosecx x c= >

3.3 sin 2 sin 4

8 4 32

x x xc? @ @ 4.

3

3

xc%

108

5.2 2 2 2

2 2

1log sin cosa x b x c

a bA A

B6.

11 tantan

a xc

ab bC D E

FG HI J

7. cot( )xxe cK L 8.1 sin( )

logcos( ) sin( )

xc

x

MNO

P MN MP

9.2tan

log | tan |2

xx cQ Q 10.

2

4 2

x xcR S

11.1 3 2

log17 5

xc

x

TU

U12.

3log ( 1)( 2)

2x x x cV W V V W

13. 2 22 4 3 3log 2 4 3x x x x x cX X Y X X X X X

14. 21 2 1 6log | 2 5 | log

2 6 1 6

xx x c

x

Z ZZ Z [ [

Z [

15.2 1 2

log | 11 log | 2 |9 3( 1) 9

x x cx

R R R S SR

16.2 11 1 1

log |1 | log |1 | tan2 4 2

x x x c\] ^ ^ ^ ^ ^

17.1 1tan tan

x xx ax a c

a a_ _` a a

18.2 3 2 2 11 2 7 2

(3 4 ) 3 4 sin3 2 2 7

x xx x x x cbc c

d d d c d d c c

19.1

1x x

e cx

ef ghi j

hk l20. log

xc

xS

21.64

322.

105

223.

34( 1)

2

ee m

24. 2 2n 25.2

26. 2

27. 5 2 3R R 28.1

log( 2 1)2

S 29. log 28

30. ( 2)2

o

109

Application of Integrals7CHAPTER

Concept Building

Area Under the Curve

Let y = f(x) be a continuous function defined on [a, b]

l The area bounded by the curvey = f(x) above thex-axis and between the linesx = a, x = b is given by

( )b b

a ay dx f x dxp

l If the curve between the linesx = a, x =b lies belowthex-axis, then the required area is given by.

(i) ( ) ( )b b b

a a ay dx ydx f x dxq r q r q

OR

(ii) ( )b

af x dx

l The area bounded by the curvex = f(y) above they-axis and the linesy = c, y = d is given by

( )d d

c cx dy f y dys

110

l If the curve between the linesy = c, y = d lies belowthey-axis (to the left ofy-axis) then the area is givenby.

(i) ( ) ( )d d d

c c ex dy x dy f y dyt u t u t

or

(ii) ( )d

cf y dy

l The area enclosed by the curvesy = f(x) andy =g(x) betweenx = a andx = b [as given in figure]

v w( ) ( )b

af x g x dxx

Example : Sketch the graph ofy = |x + 1|.

Evaluate1

3| 1|x dx

yz . What does this value

represent on the graph?Solution : We have y = |x + 1|

{ we havey = x + 1 for =1, 1 0

( 1), 1 0

x x

x x

| | }~�� | | ��

x � – 1

=1, 1

1, 1

x x

x x

� � ���� � � ��

andy = – x – 1 for x < – 1

for drawing the graph ofy = x + 1, we plot the points C(–1, 0) and B(1, 2) and get CB which lies on the

right of x = –1. For drawing the graph of y = –x –1, we plot the points C(–1, 0) and get the line CE

which lies on the left ofx = – 1.

Now,1

3| 1|x dx

yz =

1 1

3 1( 1) ( 1)x dx x dx

�

� �� � � �

=

1 12 2

3 1

( 1) ( 1)

2 2

x x�

� �

� � � �� �� �� � � �� � � �

=2 2 2 21 1

[( 1 1) ( 3 1) [(1 1) ( 1 1) ]2 2

�� � � � � � � � � �

=1 1[(1 4)] (4) 2 2 4

2 2

�� � � � � .

This value represents the area of the shaded region as shown in fig.

111

Example :Find the area of the region in the first quadrant enclosed byx-axis andx = 3y by the 2 2 4.x y� �

Solution : The given equation are

x = 3y ...(i)

and 2 2x y� = 4 ...(ii)

from (i) and (ii), we get3y2 + y2 = 4

or 4y2 = 4or y = ± 1

But we are to find the shaded area, so we takey = 1

� The co-ordinates of the point of intersection of (i) & (ii) are P( 3,1).

The area of the shaded region is the sum of the area of the region OPA and PAB

� Area =3 2 2

0 3

14

3x dx x dx  ¡

=

3 222 1

30

1 44 sin

2 2 2 23

x x xx ¢£ ¤ £ ¤

¥ ¦ ¥§ ¨ § ¨© ª© ª

=1 11 3 2 2 3 3

0 0 2sin 2sin2 2 2 2 23

« «¬ ­¬ ­® ¯ ° ® ° ¯ ¯± ²± ²

³ ´ ³ ´

= 1 13 3 32sin 1 2sin

2 2 2µ µ¶ · ·

= 2 22 3 3¸ ¹

Hence required area =3

sq. units.

Example : Find the area of the region enclosed between the two circles2 2 1x yº » and (x – 1)2 + y2 = 1

Solution : The given circles are x2 + y2 = 1 ...(i)and ( x – 1)2 +y

2 = 1 ...(ii)

(i) and (ii) intersect at A1 3

,2 2

¼ ½¾ ¿À Á

and B1 3

,2 2

 ÃÄÅ ÆÇ È

from (i) 21y xÉ Ê and

from (ii) 21 ( 1)y xË Ì Ì

� Required area =1 12 2 2

10 2

2 1 ( 1) 1x dx x dxÍ ÎÏ Ï Ð ÏÑ ÒÓ Ô

112

=

1 122 21 1

10 2

( 1) 1 ( 1) 1 1 12 sin ( 1) sin

2 2 2 2

x x x xx xÕ Õ

Ö ×Ø Ø Ø ØÙ ÚÛ Ø Û Û

Ù ÚÙ ÚÜ Ý

= 1 1 1 1

1 3 1 31 1 1 1 1 12 2 2 22 sin sin ( 1) sin 1 sin

2 2 2 2 2 2 2 2Þ Þ Þ Þ

ß àá â á âã ãã ãä å åæ çä è éê ëè éæ çì ä ä ì ä äí î ã ãã ãæ çï ñ ò óò óô õ

=3 3

4 6 2 2 4 6ö ÷

ø ø ø ù ø øú ûü ý

=3 3

4 6 2 2 4 6þ þ ÿ ÿ þ þ

=2 3

3 2

ð ��� �

� � sq. units.

Example : Using integration, find the area of the triangular region whose vertices are P(1, 0), Q (2, 2) andR(3, 1)

Solution : Equation of line PQ

0

2 0

y�

� =

1

2 1

x�

�� y = 2x – 2

Equation of line QR

2

1 2

y

=

2

3 2

x

� y = – x + 4Equation of line PRArea of�ABC = Area under line PQ + Area under line QR – Area under line PR

=2 3 3

1 2 1

1(2 2) ( 4)

2

xx dx x dx dx

�� � �

=

3 32 2221

2 1

( 4) ( 1)( 1)

2 4

x xx

� � �� � �

�

=1 1

[1 0] [1 4] [4 0]2 4

� � � � �

=3 3

1 12 2

� � �

Practice Questions

1. Find the area common to the circlex2 + y2 = 16 and the parabolax2 = 6y.

2. find the area of the region bounded by | 1|y x� � andy = 1.

113

3. AOB is the region of ellipse2 2

2 21

x y

a b� � in the first quadrant, where OA = a, OB = b, then find the

area between the arc AB and chord AB of the ellipse.

4. Find the area of the larger region bounded byy = cosx, y = x + 1 andy = 0.

5. F i n d t h e a r e a e n c l o s e d b e t w e e n t h e p a r a b o l a 4 y = 3x2 and the line 2y = 3x + 12.

6. Find the area bounded by the curvex2 = 4y, the linex = 2 and thex-axis.

7. Using integration, compute the area bounded by the linesy = x + 1,x + 2y = 2 and 2x + y = 7.

8. Using integration, find the area of the triangle ABC whose vertices are A(2, 3), B(4, 7) and C(6, 2).

9. Using integration, find the area aboutx-axis and included betweenx2 + y2= 8x and the parabolay2 = 4xandx-axis.

10. Using integration, find the area of the region bounded by the parabolay2 = 4x and the circle 4x2 + 4y2

= 9.

Answers

1.32 4 3

3 3

� ��� �

� �7. 6

8. 9 9. [8 3 ]3

�

10. 12 9 9 1sin

6 8 4 3!" #

114

Concept Building

Differential Equation

l A equation involving derivative (derivatives) of the dependent variable with respect to independentvariable (variables) is called a differential equation.

l A differential equation involving derivative (derivatives) of the dependent variable with respect toone independent variable is called an ordinary differential equation.

Order of Differential Equation

l Order of a differential equation is defined as the order of the highest order derivative occuring in thedifferential equation.

Degree of Differential Equation

l Degree of a differential equation is defined if it is a polynomial equation in its derivatives.l Degree (when defined) of a differential equation is the highest power (positive integer only) of the

highest order derivative in it.

Solution of Differential Equation

l A function which satisfies the given differential equation is called it’s solution. The solution which

contains as many arbitrary constants as the order of the differential equation is called a general solu-tion and the solution free from arbitary constants is called particular solution formation of differentialequation.

l To form a differential equation from a given function we differentiate the function successively asmany times as the number of arbitary constants in the given function and then eliminate the arbitaryconstant.

Methods of Solving Differential Equation

1. Variable separable methodl Variable separable method is used to solve such an equation in which variables can be separated

completely i.e. terms containingy should remain withdy and term containingx should remainwith dx.

2. Solution of Homogeneous Differential EquationA function f(x, y) is said to be a homogeneous function of degree n iff($x

1$y) = $nf(x,y) for some non

zero. Constants$.

Differential Equation8CHAPTER

115

A differential equation which can be expressed in the form ( , ) or ( , )dy dx

f x y g x ydx dy= = where

f(x, y) andg(x, y) are homogeneous function of degree zero is called a homogeneous differentialequation.

3. Solution of Linear Differential Equation of First Order

l A differential equation of the formdy

py Qdx+ = where P, Q are constants or functions ofx is

called a first order linear differential equation. Standard solution here intergrating factor =pdx

eò

and solution ispdx pdx

ye Qe dxò ò=ò + C.

l A differential equation which can be written in the form 1 1

dxP y Q

dy+ = is also called a first order

linear differential equation and in this case P1, Q

1 should be constants or functions ofy only.

Solution : Here I.F. =1p dy

eò and solution is1 1

1

p dy p dyye Q e dy cò ò= +ò

Initial Value Problems

A first order differential equation together with an initial condition is called an initial value problem andthe solution of an initial value problem is called a particular solution.Example : Find the order and degree of each of the following differential equation.

(i)2

2 9 0d y

ydx

+ = (ii )32

2 2d y dy

ydx dx

æ ö+ =ç ÷è ø

(iii )

3 52

2 0d y dy

ydx dx

æ ö æ ö+ + =ç ÷ç ÷ è øè ø

(iv) sin 0dy dy

ydx dx

æ ö+ + =ç ÷è ø

Solution :(i) Order = 2 ; degree = One

(ii ) Order = 2 ; degree = One(iii ) Order = 2 ; degree = Three(iv) Order = 1 ; degree is not defined

Example : Find the differential equation of the family of curves.y = ex (A cosx + B sinx).

Solution : We are giveny = ex (A cosx + B sinx) ...(i)

differentiating (i) w.r.t.,x we get

dy

dx= ex (–A sin x + B cosx) + ex (A cosx + B sinx)

= y + ex (–A sin x + B cosx) ...[using (i)]

%dy

ydx- = ex (–A sin x + B cosx) ...(ii )

116

Differentiating (ii ) w.r.t. x, we get

2

2

d y dy

dx dx- = ( Acos Bsin )xe x x- - + ex (–A sin + B cos x) ...(iii )

from (i), (ii ), (iii ) we get2

2

d y dy

dx dx- =

dyy y

dx- + -

&

2

2 2 2 0d y dy

ydx dx

- + = is the required differential equation.

Example : Solve the differential equation (1 +e2x)dy + ex (1 +y2)dx = 0 given thaty = 1, whenx = 0.Solution : The given differential equation is (1 +e2x)dy + ex(1 + y2) dx = 0

or 2 21 1

x

x

dy e dx

y e+

+ += 0 Or 2 21 1

x

x

dy e dx

y e+

+ +ò ò = C1

& tan–1 y + I1= C

1 ...(i)

When I1= 21

x

x

e dx

e+òPutex = t, thenex dx = dt

I1= 221

dxc

t+

+ò

= tan–1 t + c2

= tan–1 ex + c2

...(ii)& from (i) and (ii )

tan–1 y + tan–1 ex = c1 – c

2 = c(say)

when x = 0, y = 1& tan–1 1 + tan–1 e0 = cor tan–1 1 + tan–1 1 =c

&4 4

p p+ = c or c =

2

p

Example : Find the perticular solution of the differential equation.

2 2(1 ) (1 ) 0,dy

x ydx

+ + + = given that

y = 1 whenx = 0.Solution : Its &

The given differential equation can be written as

2 21 1

dy dx

y x+

+ += 0

or 2 21 1

dy dx

y x+

+ +ò ò = 0

117

or tan–1 y + tan–1 x = c ...(i)Now y = 1, whenx = 0' tan–1 1 + tan–1 0 =c

or tan–1 tan4

pé ùê úë û

+ tan–1 (tan 0) =c

or4

p= c ...(ii )

from (i) and (ii) tan–1 y + tan–1 x =4

p

or 1tan1 4

y x

xy- é ù+ p

=ê ú-ë û

or tan 11 4

y x

xy

+ p= =

-

Thusy + x = 1 – xy is the particular solution of the given equation.Example : Solve the differential equation.

dyy x

dx- =

dyx y

dx+

Solution : We are given

dyy x

dx- =

dyx y

dx+

or ( )dy

y xdx

+ = y – x

'dy

dx=

y x

y x

-

+

(dy

dx=

1

1

y

xy

x

-

+

...(i)

Thus (i) is a homogeneous equation asdy

dx is a function of

y

xæ öç ÷è ø

let y = vx

'dy

dx=

dvv x

dx+

Substitution in (i), we get

dzz x

dx+ =

1

1

zx x v

zx y v

- -=

+ +

'dz

xdx

=21 1

1 1

v vv

v v

- - -- =

+ +

' 2

1

1

vdv

v

+

+=

dx

x

-

118

or 2 2

1

1 1

v dxdv c

v v xé ù

+ + =ê ú+ +ë ûò ò

or2 11

log |1 | tan log | |2

v v x c-+ + + =

or2

12

1log 1 tan log | |

2

y yx c

x x- æ ö

+ + + =ç ÷è ø

or2 2

12

1log tan log | |

2

x y yx c

x x-+ æ ö

+ + =ç ÷è ø

Example : Solve the differential equation cot sin 2 .dy

y x xdx+ =

Solution : The given differential equation is cot sin 2 .dy

y x xdx+ =

Here P = cotx and Q = sin 2x

) I.F. =cos

cot log|sin |sin sinx

dxxdx xxe e e xòò = = =) The general solution is

y · sinx = f sin 2x · sinx dx + c ...[ .I.F. Q.IF ]y dx\ =ò

= 2sin cos sinx x xdx c× +ò

= 22 sin cosx xdx c+ò

=32sin

3

xc+

) y sinx =32sin

3

xc+ is the required solution.

Example : Solve the differential equationyey dx= (y3 + 2x ey)dy.

Solution : The given differential equation isyey dx= (y3 + 2x ey)dy

ordx

dy=

3 2 y

y

y xe

ye

+

ordx

dy=

2 2yy e xy

- +

or2dx

xdy y- = y2e–y

119

This is a linear differential equation and

* I.F. =2

22log log 2

2

1dyy yye e e y

y

--- -ò

= = = =

* The solution is 22 2

1 1yx y e dyy y

- é ù× = ê ú

ë ûò

or 2

x

y = y ye dy e c- -= - +ò

Thusx = –y2e–y + cy2 is the required solution.

Practice Question

1. Determine order and degree (if defined) of the following differential equation.

(i)2

2 0d y dy

ydx dx

+ + =

(ii)32

2 0d y dy

ydx dx

æ ö+ + =ç ÷è ø

(iii) cos 0dy dy

dx dxæ ö

+ =ç ÷è ø

(iv) 2( ) 0yy y e ¢+ + =¢¢¢ ¢¢

2. Form the differential equation of the family of circles in the first quadrant and touching coordinateareas.

Solve the following differential equations

3.2 2 2 21

dxx y x y

dx= + + +

4.2 2(1 ) (1 )

dyy x x y

dx- = +

5. (x3 + y3)dy – x2y dx = 0

6.2 2dy

x y x ydx- = +

7. 2 2 0x x

y yye dx y xe dyæ ö+ - =ç ÷è ø

8. 2 tan sindy

y x xdx+ =

9. xy log x + 2 logx = (x – 2)

10. ydx – (x + 2y2)dy = 0

120

Answers

1. (i) order : 2 degree : One(ii) order : 2 degree : One(iii) order : 1 degree not difined(iv) order : 3 degree not defined

2. (x – y)2 [(y )2 + 1] = (x + yy )2

3. tan–1 y =3

3

xx c+ +

4. (1 – x2) (1 +y2) = c

5.3

3 log3

xy c

y- =

6. 2 2 2y x y cx+ + =

7. 2ex/y + log | y | = 28. y = cosx + c cos2 x

9.2

log logdy

x x y xdx x+ =

10. x = 2y2 + cy

+,+

Vectors9CHAPTER

Concept BuildingScalar quantities :—Those quantities which have only magnitude and no direction, are called scallers,e.g. length, mass, time.Vector Quantities :—Those quantities which have magnitude as well as direction are called vectors. e.g.Force, velocity.

Vector: A directed line segment is called a vector e.gAB---.

represents AB a directed line segment (fig.)having A as initial point and B as terminal point.

If a point O is fixed as the origin in plane andp is any points theOP///0

is called the position vectorof P with respect to O.

It 1, 2, 3 are the angles made a vectorOP///0

with OX, OY, OZ then cos1, cos2, cos3 are called directioncosmos and denoted byl, m, n.

e2 + m2 + n2 = 1Illustration 1: Show that the direction consines of a vector equally inclined to the axes OX, OY

and OZ are1 1 1

, , .3 3 3

Solution: Let angle made by the given vector with OX, OY and OY be1. This direction cosmesof given vector are

cos1, cos2, cos34 cos21 + cos21 + cos21 = 15 3cos21 = 1

5 cos2a =1

35 cos1 =

1

3

This direction cosmes of the given vector are1 1 1

, ,3 3 3

Type of Vectors : (i) A vector whose magnitude is zero is called a null vector or zero vector.

(ii ) A vector whose magnitude is one unit is called a unit vector, denoted bya .

677

(iii ) Co-initial vector: Two or more vectors having the same initial point are called co-initial vectors.(iv) Collinear vectors: The vectors which have same support are called collinear vectors.(v) Like vectors: The vectors which have same direction are called like vectors.

(vi) Unlike vectors: The vectors which have opposite direction are called unlike vectrors.(vii) Equal vectors: Two vectors are equal if they have same magnitude and the direction.

(viii ) Negative of a vector: A vector with same magnitude but opposite direction is called negativevector of the given vector.

Vector in two dimensions: Let r8 be the position vector of a point P(x, y)

(i) Then ˆ ˆOPr xi yj9 9 :;;;<<

(ii ) ˆxi and yj are known as component vectors ofr=

alongx andy-axis andx andy known as

components ofr8

alongx andy-axis.

(iii ) Magnitude of r8

, is given by, 2 2r x y> ?@

(iv) Unit vector alongr=

, is given by,ˆ ˆr xi yj

rr r

AB B

CC

C C

i.e., ˆ ˆ.x y

r i jr r

D EF

F F

(v) If G is angle whichr=

makes withx-axis then cosG = x

rH , sinG = .

y

rI

(vi) cosG, sinG are known as direction cosines of.rJ

(vii) For ˆ ˆ;r xi yjK LM

x andy are knwon as direction ratios ofr=

, i.e., components of a vector are

direction ratios ofr=

, but converse may or may not be true.Vector in three dimensions:

If r8

is position vector of point P(x, y, z)

(i) Then ˆˆ ˆr xi yj zkN O OP

ˆ ˆ,xi yj and ˆzk are component vectors andx, y andz are

components of vectorsr8

alongx, y andz-axis.

(ii ) Magnitude of r=

, is given by, 2 2 2r x y zQ R RS

(iii ) If T, U, V are the angles which vectorr8

makes withx, y andz-axis, then its direction cosinescosT, cosU, cosV are

cos ; cos ; cos .x y z

r r rW X Y Z[ [ [

(iv) Unit vector alongr8

, is given by,,

ˆ ˆˆ ˆ ˆ ˆˆ . (cos ) (cos ) (cos )ˆ ˆ ˆ

x y zr i j k i j k

r r r\ ] ] \ ^ ] _ ] `

abc

Illustration 2: Find the components of vectorvd

whose magnitude is5 3 and which makes an angleof 120° with positive directions ofx-axis.

Solution: Let components ofvd

along X axis and Y axis bex andy then ˆ ˆv xi yje fg

using figure

OM

5 3 = cos120° and

MP

5 3 = sin120°

h x = 5 3 cos120i

and y = 5 3 sin120j

Þ x =1 5 3

5 32 2

kl mk no pq r

and y =3 15

5 32 2

s tuv w

x y

Hence the required component ofvz

are 5 32

{ and 152

Thus vd

= 5 3 15ˆ ˆ2 2

i j|

}

Illustration 3: If ˆ ˆ2a i j~ ��

and ˆ ˆ2 2b i j~ ��

; find unit vector parallel to the vector3 2a b���

Solution: ˆ ˆ3 3 6a i j� ��

and ˆ ˆ2 4 4b i je fg

h 3 2a b��� = ˆ ˆ ˆ ˆ(3 6 ) (4 4 )i j i j� � � = ˆ ˆ10i j� �

Unit vector parallel to(3 2 )a b�gg

=3 2

3 2

a b

a b

�

�

��

��

=ˆ10 1 10ˆ ˆ

1 100 101 101

i ji j

� � �� �

�Addition of vectors:

Triangle law of vector addition

It two vectors AB����

and BC����

are represented inmagnitude and direction by any two sides of triangle

then their sumAC����

is represented by third side taken

in reverse order.

Parallelogram law of vector addition

If any two vectorsad

and b�

are are representedin magnitude and direction by the two adjacent sidesof a parallelograms then their sumc

d is representedby the diagonal of the parallelogram which iscoixitial with the given vector.

���

Properties of vector addition: Let a�

and b�

be any two vectors (i) a b b a� � �� �� �

(ii ) ( )a b c� ��� �

= ( )a b c� ��� �

(iii ) 0a�  

= 0a a� �    

Multiplication of vector by a scalar: Let ¡ be a scalar anda¢ be any vector then a£

¤ is defined

as a vector having the support as that ofa¢

and having magnitude as¥ times the magnitude ofa¢

and

having same or opposite direction ofa¢

, a acording as¡ is positive of negative.

Remarks: Let a�

andb�

be any two vectors andk andm be any scalars. This (i) ( )ka ma k m a¦ § ¦¨ ¨ ¨

(ii ) ( )k ma kma©ª ª

(iii ) ( )k a b ka kb« ¬ «­ ­­ ­

Section formula: Let A and B be two points represented by position vectorsOA®®® 

andOB¯¯ °

respectivelywith respect to origin of refrence.

Internal DivisionLet C be a point dividing the line AB in the ratiom : n

P.V of C is given by

OB OAOC

m n

m n

±²

±

³³³ ³³³³³³

If C is and point then

OB OCOC

2

µ¶

··· ······

External DivisionLet C be the point dividing the line joiningexternally in ratiom : n

Position vector of C is given by

OB OAOC

m n

m n

¹º

¹

»»»¼ »»»¼»»»¼

Illustration 4: If ˆˆ ˆ 2a i j k½ ¾ ¾¿

and ˆˆ ˆ3 2b i j kÀ Á ÂÃ

then find ( 2 ).(3 )a b a bÄ ÅÆ ÆÆ Æ

Solution: 2a bÇÈÈ

= ˆˆ ˆ ˆ( 2 ) (6 4 2 )i j k i j kÉ É É É Ê

= ˆˆ ˆ7 5 0i j kË Ë

and 3 2a bÌ� = 3 ˆˆ ˆ ˆ( 2 ) (3 2 )i j k i j kË Ë Í Ë Í

= ˆˆ ˆ0i j kÎ Î

Ï ( 2 ).(3 )a b a bÐ ÑÒ ÒÒ Ò

= ˆ ˆˆ ˆ ˆ ˆ(7 5 0 ).(0 )i j k i j kË Ë Ë Ë

= 7(0) + 5(1) + 0(1)= 5

Illustration 5: Find the projection of ˆˆ ˆ7 4a i j kÓ Ô ÕÖ

on ˆˆ ˆ2 6 3b i j k× Ø ØÙ

ÚÛÜ

Thinking Process

Find .a bÝÝ

Find | |bÞ

Projection of aß and b

à =

.

| |

a b

b

áá

á

Appliction

.a bÝÝ

= ˆ ˆˆ ˆ ˆ ˆ(7 4 ).(2 6 3 )i j k i j kâ ã â â

= 7(2)+1(6) – 4(3) = 8

| |bä

= 2 2 3(2) (6) (3)å å

= 7

Projection of aæ and b

à

=8

7

Product of Vectors: 1. Scalar Product2. Vector Product

1. Scalar (or dot) Product of Vectors: Let aß

and bà

be two non-zero vectors inclined at an angle

ç. Then the scalar product or dot productaæ

and bè

is represented bya béêê

a bëìì = cosa b í

ÞÞ

Properties of dot product:

(1) a bëìì is scalar quantity (2)( . ).a b c

ää ä is not defined

(3) a bëìì

= .b aî îï

(4) .( . ) . .a b c a b a cñ òä ää ä ä ä ä

(5) .( ) ( ). ( . )a b a b a bó ô ó ô óõ õ õõ õ õ

(6) . 0 0, 0 ora b a b a bö ÷ ö öø ø øøø ø ø

(7) 2 2. | |a b a añ ñää ä ä

(8) Projection ofaß

along bà

is.

| |

a b

b

áá

á and projection vector ofaß

along bà

is.

| |

a bb

b

ù úû üý þ

ÿÿ ÿÿ

(9) Cosç =.

,| || |

a b

a b

ðððð ç is the angle betweena

ß and b

à.

(10) If ˆˆ ˆ, ,i j k are unit vectors alongx, y, z axis then 2 2 2ˆˆ ˆi j k� � ; ˆ ˆ. 0,i j � ˆˆ. 0,i k � ˆ ˆ. 0k i �

(11) If 1 1 1ˆˆ ˆa a i a j a k� � �

� and 1 1 1

ˆˆ ˆb b i b j b k� � �

, then .a b

= a1a

2 + b

1b

2 + c

1c

2

If a b��

then . 0a b �

� a1a

2 + b

1b

2 + c

1c

2 = 0

2. Vector (or cross) Product of two vectors:Let aß

and bà

be two non-zero vectors inclined at anangleç. Then, the vector or cross product ofa

ß and b

à represented bya

ß×bà

.

aß

×bà

= ˆsin .a b n���

|aß

×bà

| = |aß

×bà

|sinç

sinç =a b

a b

�����

���

Case Result

a�

or b�

= 0 a�

×b�

= 0

a�

and b�

are like vectors or a�

×b�

= ab is sin 0° n�

= 0�

� = 0° or � = �

a�

and b�

are perpendicular or a b���

= ab

� = 90°

Properties of Vector Product:

Let a�

, b�

and c� are three vectors

(1) a�

×b�

= – ( )b a�

(2) ( )a b c a b a c! " # ! " !$ $$ $ $ $ $

(3) ( ) ( ) ( )ma b m a b a mb% & % & %' ' '' ' '

If 1 1 1ˆˆ ˆa a i b j c k( ) )

* = 2 2 2

ˆˆ ˆ .b a i b j c k+ , ,-

Then

a b.//

= 1 1 1

2 2 2

ˆˆ ˆi j k

a b c

a b c

Remarks: (1) As ˆˆ ˆ, ,i j k are mutually perpendicular vectors. Theis ˆ ˆˆ ˆ ˆ ˆi i j j k k0 1 0 1 0 = 0 and ˆˆ ˆ ,i j k2 3ˆˆ ˆ,j k i2 3 ˆ ˆ ˆk i j4 5

(2) A unit vector 6 7n perpendicular to each of the vectora� and b

� is n =

a b

a b

8

8

9999

(3) The area of triangle having adjacent sidesa� and b

� is

1

2a b:

;;

(4) The area of parallelograms having adjacent sidesa� and b

�is a b.

//

(5) The area of quadrilateral having diagonala�

and b�

is1

2a b:

;;

Illustration 5: If ˆ ˆˆ ˆ ˆ ˆ2 3 , 2 3 5 .a i j k b i j k< = > < > =??

Then find a b.//

Solution:

a b.//

= 1 2 3

2 3 5

i j k

@

@

ABC

= i{–2(–5)–3(3)}– j {1(–5)–2(3)}+k{1(3)–2(–2)}

= ˆˆ11 7i j kD D

Illustration 6: Find a vector of magnitude 3, which is perpendicular to both the vectors ˆˆ ˆ3 4i j kE F andˆˆ ˆ6 5 2i j kG H

Solution:Thinking Process Application

Fing vector product of two vectors a bIJJ =

ˆˆ ˆ

3 1 4

6 5 2

i j k

K

K

= ˆˆ ˆ18 18 9i j kL M

Find magnitude ofa bIJJ N Oa bP

QQ = 2 2 2(18) ( 18) (9)R S R = 27

Required vector =a b

a b

T

T

UUUU Vector =

ˆˆ ˆ18 18 9

27

i j kV W

=2 2 1 ˆˆ ˆ3 3 3

i j kX Y

Scalar Triple Product: The scalar Triple product of three vectorsaZ

, b[

and cZ

denoted by a b c\ ]^ _`` `

is

equal to the dot product of the first vector by the cross product of remaining two in order.

i.e. a b c\ ]^ _`` `

= N O a b N O. . .a b c b c a c a bP c P c PQ Q QQ Q Q Q Q Q

Properties of Scalar Triple Product:(i) The scalar triple product of vectors does not change of the order of its factors are circularly rotated.

i.e. a b c\ ]^ _`` `

= b c a\ ]^ _` ` `

= c a b\ ]^ _`` `

(ii ) If any vector out of three is equal to any other vector with multiplication of a scalar quantity thenthe value of scalar triple product is zero.

(iii ) a b c\ ]^ _`` `

Gives as the volume of a parallelopiped where three sides parallelopiped are represented

,a bdd and c

Z.

Illustration 6: Find the volume of the parallelopiped whose adjacent sides are represented by ,a bdd

andcZ

where aZ

= ˆˆ ˆ3 2 5 ,i j kH G b[

= ˆˆ ˆ2 2i j ke f and cZ

= ˆˆ ˆ4 3 2i j kG G

Solution: Volume of parallelopied = a b c\ ]^ _`` `

=

3 2 5

2 2 1

4 3 2

g

g

g

= 3(4 + 3) + 2(4 – 4) + 5(6 + 8)

= 91 cubic unit

hij

Questions for Practice

(1) If ˆ ˆˆ ˆ ˆ ˆ2 3 , 3 2 ,a i j k b i j kk l m k m lnn

show thata bopp

and a bqrr

are perpendicular to each other..

(2) Express the vector ˆˆ ˆ5 2 5a i j ks t uv

as sum of two vector such that one is parallel to the vectorˆˆ3b i kw x

y and the other is perpendicular to.b

z

(3) Let ˆˆ ˆ ˆ, 3a i j b j k{ | { |}}

and ˆˆ7 ,c i k~ ��

find a vectord�

which is perpendicular to botha�

and b�

and 1c d� �zz

(4) Write the value of ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( )i j k j i k k i j� � l � � l � �

(5) I fa�

and b�

are two unit vectors and� is the angle between then, show that1

sin2 2

a b�

� ���

(6) Prove that

� �2

a b���

= � �222

–a b a b� �� �� �

Hint and Solution

(1) Given that ˆ ˆˆ ˆ ˆ ˆ2 3 , 3 2a i j k b i j k� � � � � ���

Now a bopp

= ˆ ˆˆ ˆ ˆ ˆ2 3 3 2i j k i j k� � � � �

= ˆˆ ˆ4i j k� �

and a bqrr

= ˆ ˆˆ ˆ ˆ ˆ2 3 3 2i j k i j k� � � � �

= ˆˆ ˆ2 3 5i j k� � �

  ¡ ¢ ¡ ¢a b a b£ ¤ ¥¦ ¦¦ ¦ = ˆ ˆˆ ˆ ˆ ˆ4 2 3 5i j k i j k§ ¨ © ¨ § ¨

= 4(–2) + 1(3) + (–1)(–5)

= –8 + 3 + 5 = 0

Hence a bª««

and a b¬««

are perpendicular

(2) Let 1 2a a a­ ®¯ ¯ ¯

where1az is perallel tob

�and 2a

° is perpendicular tob

± 1az

= Ab²

± 1az

= ˆˆ(3 )i k³ ´ ± 1az

= ˆˆ3 Ai kµ ¶

Also 2a°

= 2 1a a·¸ ¸

± 2a°

= ¹ ºˆ ˆˆ ˆ ˆ5 2 5 3i j k i k» ¼ » ½ ¼ ½

± 2a°

= ¾ ¿ ˆˆ2 (5 )5 3 i j k� � � À� À

2a bÁÁ

= 2a bÂÃÃ

= 0

± Ä Å Æ Çˆˆ ˆ2 (5 )5 3 3i j k i kÈ É ÈÊ ËÈ Ê É = 0

± (5 – 3Ì) (3) + (–2) (0) + (5 – Ì)1 = 0± Ì = 2

ÍÎÏ

Hence 1aÐ

= 3(2)i + 2k = 6i + 2k

and 2aÑ

= [5 – 3(2)]i + (–2)j + (5 – 2)k

= ˆˆ ˆ2 3i j kÒ Ó

(3) Let dÔ

= ˆˆ ˆai bj ckÕ Õ

a dÖ

× a dØÙÙ

= 0 = a(1) + b(–1) + (10) = 0 × a – b = 0 ...(i)

b dÚ Ú

× a dÛÜ

= 0 × a(0) + b(3) + c(–1) = 0

× 3b – c = 0 ...(ii )

and 0 dÝÚÚ

= 0 × 7(a) + 0(b) + (–1)c = 0× 7a – c = 0 ...(iii )

Solving (i), (ii ) and (iii ) we get

a =1

,4

b =1

,4

c =3

4

Þ dÔ

=1 1 3 ˆˆ ˆ4 4 4

i j kß ß

(4) à á â ㈈ ˆˆ ˆ ˆˆ ˆˆi j k i jj k i kä å ä å ä ææ æ

= ˆ ˆˆ ˆ ˆ ˆ( )i i j j k kç è ç é è ç = 1 – 1 + 1 = 1

(5) Consider2

ˆa bê = ë ì2

ˆa bê

= 2 2ˆ ˆˆ ˆ2a b a bí î ï

=22 ˆ ˆ2 cosˆ ˆa ab bñ ò ó

×2

ˆa bê = 1 + 1 – 2cosô = 2(1 – cosô)

× 2ˆa bê = 24sin

2

õ

× sin2

ö =

1ˆˆ

2 a b÷

(6) LHS = ë ì2

ˆa bø

=2

ˆsin na b ùúú

= 2 2ˆsin na b ûüü

=22 2sina b ý

þþ

=22 2(1 cos )a b ÿ ð

��

=2 22 2 2cosa ab bò ó

� �� �

= � �222

a b a bê �� �� �

= R.H.S

130

Three Dimensional Geometry10CHAPTER

Concept Building

Before studying straight line and plane in three dimensional geometry, let us understand some basic concept.(i) Distance between two given points P(x

1, y

1, z

1) and Q(x

2, y

2, z

2) is

|P�| = 2 2 22 1 2 1 2 1( ) ( ) ( )x x y y z z� � � � �

(ii) Direction ratios of a line having two points (x1y

1z

1) and (x

2, y

2, z

2) are

x2 – y

1, y

2 – y

1, z

2 –z

1

(iii) Angle between two lines whose direction ratios area1, b

1, c

1 anda

2, b

2, c

2 is given by

cos� = 1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

a a b b c c

a b c a b c

(a) If lines are perpendicular thena1a

2 + b

1b

2 + c

1c

2 = 0

(b) If line are parallel, then 1 1 1

2 2 2

a b c

a b c

Exmaple : If co-ordinates of A, B, C and D are (1, –1, 2), (3, 1, –2), (4, 1, 3) and (1, –2, 3) respectively.

Then find AB. Also find the angle betwen AB and CDSolution :

Thinking Process Application

a = 2 2 22 1 2 1 2 1( ) ( ) ( )x x y y z z� � � � � AB = 2 2 2(3 1) {1 ( 1)} ( 2 2)� � � � � � �

= 2 2 2(2) (2) ( 4) 4 4 16 24� � � � �

First find direction ratios then apply the dr’s of AB = 2, 2, –4 or 1, 1, – 2

formula dr’s of CD = –1, –3, 0 or 1, 3, 0

let � be the angle between AB and CD then cos� = 1 2 1 2 1 2

2 2 2 2 2 21 1 2 2 2

1 3 0 4

6 10 60

a a b b c c

a b a a b c

� � � �� �

� � � �

� 1 2cos

15� � �

� � � �� �

131

Straight Line

(i) Vector equation of a straight line passing through a fixed point

with position vectora� and parallel to a given vectorb

� is

r a b� � ��� � where is a parameter..

(ii) Cartesion equation (symmetrical form) of the straight linepassing through a fixed point (x

1y

1z

1) having direction ratiosa,

b, c is givne by 1 1 1x x y y z z

a b c

! ! !" "

(iii) The parametric equations of the straight 1 1 1x x y y z z

a b c

! ! !" " are 1 1 1, ,x x a y y b z z c# $ # % $ # % $

where is parameter.

(iv) The co-ordinates of any point on the line 1 1 1x x y y z z

a b c

! ! !" " are (x

1 + a , y

1 + b , z

1 + c ) where

&R.Example 2 :Find vector and cortesion equations of the line which is passing through the point (1, 2, 3) and

parallel to the vector ˆˆ ˆ2 5 7i j k' ( . Also find parametric form of the line.

Solution :

Thinking Process

1. Postion vector of the fined point (1, 2, 3)is needed.

2. Equation of the line isr a y) * +, , ,

3. For cortesion equation, convert vectorequation into costesion from

4. Parametric from of the line isx = x1 + a ,

y = y1 + b , z = z

1 + c

Application

Let a- be the position vector of (1, 2, 3) then

ˆˆ ˆ2 3a i j k. / /0

ˆ ˆˆ ˆ ˆ ˆ( 2 3 ) (2 5 7 )r i j k i j k1 2 2 2 3 4 25

Here (x1y

1z

1) is (1, 2, 3) anda : b : c is (2, –5, 7). So

equation is1 2 3

2 5 7

x y z6 6 67 7

6Parametric equation isx = 2 + 1,y = –5 + 2,z = 7 + 3

(v) Vector equation of a straight line passing through

the given points with position vectorsa-

andb�

is

( )r a b a8 9 : ;<< < <

where is a parameter

(vi) Cartesian equation of the straight line passingthrough two given points A(x

1y

1z

1) and B(x

2y

2z

2) is

given by

1

2 1

x x

x x

=

= = 1 1

2 1 2 1

y y z z

y y z z

> >?

> >

132

(vii) Angles between two straight lines

(i) Let 1 1r a b@ A BCC C

and 2 2r a bD E FGG G

be the equations of straightlines then angle between these two lines will be same as thatof the angle between1b

H and 2b

I be cause1b

H and 2b

I are parallel

vector to the straight lines.If J be the angle between both the lines then

cosJ =1 2

1 2

.

| | | |

b b

b b

K K

K K

(ii) If1 2. 0b b @C C

, then cosJ = 0 L b1 M b

2 L both lines are

perpendicular to each other.

(iii) If1 2b bN OP P

, thus

cosJ =2

2 2 22

2 2 2

. | |1

| | . | | | |

b b b

b b b

Q QR R

Q Q

S S S

S S S

L cosJ = 1L J = 0°L 1 2||b bC C

L both lines are parallel to each otherExample 3 :Write Cartesian equation of the following line given in vector forms

rT = ˆ ˆˆ ˆ ˆ ˆ(2 4 ) ( )i j k i j kU V U W V V

Solution :

Thinking Process Applications

Direction ratios and point through Direction ratios are 1, –1, –1 and point is (2, 1, –4)

which the line passes are required2 1 4

1 1 1

x y zX X YZ Z

X Xequation can be given by

1 1 1x x y y z z

a b c

[ [ [\ \

Example 4: Find the angle between the pair of lines given by ˆ ˆˆ ˆ ˆ ˆ(3 2 4 ) ( 2 2 )r i j k i j k] U V U W U U^

andˆˆ ˆ ˆ ˆ(5 2 ) (3 2 6 )r i j i j k@ _ A ` A A

C

Solution :

Thinking Process Application

Identify1ba

and2bC

1ba

= ˆˆ2 2i j kA A

2bC

= ˆˆ3 2 6i j kE E

use the formula cosJ =2 2 2 2 2

ˆ ˆˆ ˆ ˆ ˆ( 2 2 ).(3 2 6 )

1 (2) (2) (3) (2) (6)

i j k i j kb b b b

b b b

1 2

1 2

.cos

| || |

b b

b bc d

e e

e e cosJ =3 4 12 19

3.7 21

f fg

J =1 19

cos21

h i jk lm n

133

(ii) Let lines are 1 1 1

1 1 1

x x y y z z

a b c

o o op p and 2 2 2

2 2 2

x x y y z z

a b c

q q qr r and angle between them iss

then

coss = 1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

a a b b c c

a b c a b c

t t

t t t t

if a1a

r thencos 0 90u v w u v x

y lines are perpendicular

Shortest Distance between two lines :Let l1 and l

2be two skew lines given by 1 1r a bz { |

} }} and

2 2r a b~ ���

respectively, where1a�

and 2a�

are position vectors of points onl1 andl

2 there shortest distance

between two given points is given by

1 1 2

1 2

( ).( )

| |

a a b b

b b

� �

�

� �� �

� �

Remark : If two lines are intersecting, then shortest distance between then is zero. i.e.

2 1 1 2( ).( )a a b b� �� �� � = 0

Shortest Distance Between Two Parallel Lines

Let l1 andl

2 be two parallel lines given by 1r a b� � �

� � and 2r a b� � �

� �� respectively. Then shortest distance

between then is given by

2 1.( )

| |

b a a

b

�� � �

�

Shortest distance between two skew lines in Cartesian form : Let 1 1 1

1 1 1

x x y y z z

a b c

� � �� � and

2 2 2

2 2 2

x x y y z z

a b c

q q qr r are two skew lines then shortest distance between them is given by

2 1 2 1 2 1

1 1 1

2 2 2

x x y y z z

a b c

a b c

� � �

Remark : If

2 1 2 1 2 1

1 1 1

2 2 2

x x y y z z

a b c

a b c

� � �

= 0 then lines are intersecting.

Exmaple 5 :Find the shortest distance between the lines ˆˆ ˆ ˆ(4 ) ( 2 3 )r i j i j k� � � � � ��

and ˆˆ ˆ( 2 )r i j k� � ��

ˆˆ(2 4 5 )i j k  ¡   ¢ .

134

Solution :

Thinking Process Application

Find 2 1a a£¤ ¤

and1 2b b¥¦ ¦

2 1a a£¤ ¤

= ˆˆ ˆ ˆ ˆ( 2 ) (4 )i j k i j§ ¨ § §

= ˆˆ3 2i k© ª

1 2b b¥¦ ¦

=

ˆˆ ˆ

ˆ ˆ1 2 3 2

1 1 2

i j k

i j« ¬ «

«

use formula,d =2 1 1 2

1 2

( ).( )

| |

a a b b

b b

­ ®

®

¯ ¯¯ ¯

¯ ¯ d = 2 2

ˆˆ ˆ ˆ( 3 2 ).(2 )

(2) (1)

i k i j° ± °

±

d =6

5

²

³ d =6

5

Example 6 :Show that the lines5 7 3

4 4 5

n y z´ ´ µ¶ ¶

´ and

8 4 5

7 1 3

x y z· · ·¸ ¸ interest each other..

Solution :

Thinking Process Application

use2 1 2 1 2 1

1 1 1

2 2 2

x x y y z z

a b c

a b c

¹ ¹ ¹

d =

3 3 8

4 4 5

7 1 3

º

º

for finding the shortest distance = 3 (12+5) + 3 (12 + 35) + 8 (–24)

= 51 + 141 – 192 = 0

If shortest distance is zero then » lines intersect each otherlines intersect

Plane

Various equations of the plane :(i) Equation of a plane passing through a given point whose position vector

is a¼

and perpendicular to a given vectorn¼ is given by . .r n a n½

¾ ¾ ¾ ¾ or

( ). 0r a n¿ ÀÁ Á Á

.

135

(ii) Equation of a plane when a unit vectorn perpendicular to the plane is given and its perpendiculardistance d from the origin is also given then the equation of the plane is given by

ˆ.r n dÂÃ

(iii) If unit vector ˆˆ ˆn li mj nkÄ Å Å wherel, m, n are direction cosines and p, the perpendicular distance

from origin to normal are given then equation of the plane islx + my + nz = p

Exmaple 7 :Find vector equation of the plane which is at a distance of5

14 unit from the origin and its

normal vector from the origin is ˆˆ ˆ2 2 3i j kÆ Ç . Also find its cartesian form.

Solution :

Thinking Process Application

Convert normal into a unit vector ˆˆ ˆ2 2 3n i j kÈ É ÊË

ˆˆ ˆ2 2 3

14

i j kn

Ì ÍÎ

Ï

required equation is ˆ.r n dÐÑ

ˆˆ ˆ(2 2 3 ) 5.

14 14

i j kr

Ì ÍÎ

Ï

Cartesian form is ˆˆ ˆ ˆ( ).( )xi yj zk n dÒ Ò Ó

ˆ ˆˆ ˆ ˆ ˆ( ).( 2 3 ) 5

14 14

xi yj zk i j kÔ Ô Õ ÔÖ

2 3 5x y z× Ø Ù

Angle between line and Plane

Angle between line r a bÚ Û ÜÝÝ Ý and plane .r n dÂ

à à is

complementary to the angle between line and normal to the plane.Let Þ be the angle between line and plane andß the angle

between line and normal to the plane

cos(90 – Þ) =.

| || |

b n

b n

à à

à à

sinÞ =.

| || |

b n

b n

à à

à à

(i) If b ná âã ã

, then.

sin 1 90.

n n

n n

äå æ æ ç å æ è

ä

é é

é é

ê line is perpendicular to the plane

(ii) If . 0 sin 0 0b në ì í ë ì í ë ì line is parallel to the plane.

136

Angle between two planes

The angle between two planes is defined as the angle between their normals.

L e t î be the angle between normal1nï

and 1nï

of the plane 1 1.r n dñï ï

and 2 2.r n dòó

respectively..

Then according to definition angle between planes is

cosî =1 2

1 2

.

| | | |

n n

n n

ô ô

ô ô

Remarks : (i) If 1 2.n nõ õ

= 0 then both planes are perpendicular

(ii) If 1 2n nö ÷ø ø

, then both planes are parallel

(iii) The angle between two planes is always taken as acute angle

If 1 1 1 1 0a x b y c z dù ù ù ú and 2 2 2 2 0a x b y c z dû û û ü be two planes andî is the angle between then

wherea1, b

1, c

1 anda

2, b

2, c

2 are direction ratios of normal to the planes then

cosî =1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

a a b b c c

a b c a b c

ý ý

ý ý ý ý

Remarks : (i) If 1 2 1 2 1 2 0a a b b c cþ þ ÿ then planes are perpendicular to each other..

(ii) If 1 1 1

2 2 2

a b c

a b cð ð , then given planes are parallel to each other..

Example 8 :Show that the line ˆ ˆˆ ˆ ˆ ˆ(2 2 3 ) ( 4 )r i j k i j k� � � � � � �� is parallel to the plane ˆˆ ˆ( 5 ) 5r i j k� � � �� .

Also find the distance between themSolution :

Thinking Process Application

r a b� �� � is parallel to ˆˆ ˆ2 2 3a i j k� �� and ˆˆ ˆ 4b i j k� � �

�

.r n d�� � when . 0b n�

� � ˆˆ ˆ5n i j k� � �� andd = 5

.b n� � = ˆ ˆˆ ˆ ˆ ˆ( 4 ).( 5 )i j k i j k� � � �

= 1 – 5 + 4 = 0

� line is parallel to the plane

distance between line and distance =ˆ ˆˆ ˆ ˆ ˆ(2 2 3 ).( 5 ) 5

ˆˆ ˆ| 5 |

i j k i j k

i j k

� �

=.

| |

a n d

n

!" "" =

2 10 3 5

27

# $ #

=10

27

137

Determination of Plane under given Condition

(i) Equation of plane in intercept from is 1x y z

a b c% % & wherea, b, c are intercepts made by the plane on

thex-axis,y-axis andz-axis.

(ii) Equation of the plane passing through three given points having position vector, ,a b c'' ' is given by

( )( )( )r a b a c a( )* * *+ ,-- - - - -

= 0

or . /( ). ( ) ( )r a b a c a0 0 1 022 2 2 2 2

= 0

Cartesian equation of the plane passing three given points (x1, y

1, z

1) (x

2, y

2, z

2) and (x

3, y

3, z

3) is given by

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

x x y y z z

x x y y z z

x x y y z z

3 3 33 3 33 3 3

= 0

Example 9 : If a plane makes intercept of lengtha, b andc onx-axis, y-axis andz-axis respectively then

prove the equation of the plene is1x y z

a b c

44 4 = 1

Solution : Let equation of the required plane is Ax + By + Cz + D = 0. As plane passes through (a, 0, 0),(0, b, 0) and (0, 0,c).

So, we have Aa + D = 05 A =D

a

6

Bb + D = 0 5 A =D

b

6

C C + D = 05 A =D

c

6

5D D D

Dx y z

a b c7 7 7 8 = 0

5 D 1x y z

a b c9 :; < < ;= >? @ = 0

5x y z

a b cA A = 1

Example 10 :Find equation of the plane passing through the intersection of two planes3 2 4x y zB C B = 0

and 2 0x y zD D E F and the point (2, 2, 1)

Solution : Equation of the plane passing through the intersection of two plane3 2 4 0x y zG H G I and

2 0x y zD D E F is (3 2 4) ( 2) 0x y z x y zJ K J K L K K J M ...(i)

Since plane (i) passes through (2, 2, 1)

N 6 2 2 4 (2 2 1 2)O P O P Q P P O = 0

138

R 22 3 0

3

ST U V W U V

Putting the value ofX in (i) we get

2(3 2 4) ( 2)

3x y z x y z

YZ [Y \ Y \ \ \ Y] ^_ ` = 0

R 7 5 4 8x y za b a = 0

Questions for Practice

1. If the lines1 2 3

3 2 2

x y z

k

c c cd d

c and

1 1 6

2 1 5

x y z

k

e e ef f

e are perpendicular fidn the value ofk.

2. If a line makes anglegh ih j with the direction of the axes, then write the value of 2 2 2sin sin sink l m l n .

3. Find the angle between, the line ˆ ˆˆ ˆ ˆ ˆ(5 4 ) (2 3 )r i j k i j ko p p q r p qs and the plane ˆˆ ˆ.(3 4 ) 5 0r i j kt t t uv .

4. Find the equation of the plane through the points (2, –3, 1) and (5, 2, –1) and perpendicular to the

plane 4 5 2 0.x y za b b w

5. Find the distance of the point (2, 3, 4) from the plane3 2 2 5 0x y zx x x y measured parallel to the

line3 2

3 6 2

x y zz {| | .

6. Find the angle between the planes whose vector equations are ˆˆ ˆ.(2 2 3 ) 5r i j k} ~ �� andˆ ˆ.(3 3 5 ) 3r i j k� � �� .

7. Find the equation of the plane passing through the point (1, 1, 1) and containing the linesˆ ˆˆ ˆ ˆ ˆ( 3 3 5 ) (3 5 )r i j k i j k� � � � � � � �� . Also show that the plane contains the line ˆˆ ˆ( 2 5 )r i j k� � � ��

ˆˆ ˆ( 2 5 )i j k� � � � .

8. Find vector and Cartesian equations of the line through the point (5, 2, –4) when are parallel to the

vector ˆˆ ˆ3 2 8 .i j k� �9. Find the image of the point (1, 3, 4) in the plane 5.x y z� � �

10. Find shortest distance between the lines ˆˆ ˆ(1 2 ) (2 3 ) (3 4 )r i j k� � � � � � � � �� and ˆ(2 3 )r i� � ��

ˆˆ4(1 ) 5(1 )j k� � � � � � � .

Answers

1. k =10

7

�2. 2 3.

1 5sin

2 91� � �  ¡¢ £ 4. 4x y z¤ ¤ ¥ 5. 7

6. 1 15cos

731¦ § ¨© ª« ¬

7. ˆˆ ˆ.( 2 2 )r i j k� �� = 0 8.

5 2 4ˆ ˆˆ ˆ ˆ ˆ.(5 2 4 ) (3 2 8 ),3 9 8

x y zr i j k i j k

­ ­ ®® ­ ® ¯ ® ­ ° °

­±

9. (3, 1, 6) 10.1

16 units.

139

Probability11CHAPTER

Concept Building

Let E and F be two events associated with the sample space of a rendom experiment. If it is given thatevent F has already accured then the probability of E is called the conditional probability of E and it isdenoted by P(E/F).

It is given by, P(E/F) =P(E F)

P(F)

² whereP(F) ³ 0.

For example : It we toss three coins simultaneously and E is the event of getking “atleast two heads”

and F be the event first coin shows tail. Suppose we are givne that F occurs then what is probabiltiy ofE i.e. we have to find out P(E/F).Here, s = {HHH, HHT, HTT, TTH, THH, HTH, THT}

E = {HHH, HHT, THH, HTH}F = {THH, THT, TTH, TTT}

with the information of occurrence of F, were sure that the cases in which first coin does not result intoa tail should not be considered while finding the probability of E. In this may set F can be consideredas sample space for event E.´ Probability of E given that F has occurred

=No. of element of set E favorable to F

Total number of elements of F

i.e. P(E/F) =1

4

=No. of elements common of E and F

No. of elements in F

=(E F)

(F)

n

n

µ

Dividing N & D by number of elements in swegel

P(E/F) =

(E F)P(E F)( )

(F) P(F)( )

n

n sn

n s

¶¶

·

Similarly P(F/E) =P(E F)

P(F)

²

140

Properties of Conditional ProbabilityP

1 : P(S/F) = P(F/F) = 1

P2 : P[A¸ B/F) = P(A/F) + P(B/F) – P (A ¹ B/F)

P3 : P(E1/F) = 1 – P(E/F).

Example 1 :A family has two children. Find the probability of having both children boys given that atleastone of them is a boy.

Thinking Process

First of all we have to find out sample space ofthe random experiment.Now events E and F are to be found out.after defining these three sets we can find outconditional probability of E provided F hasoccrued.

Application

Let b denotes boy and g denotes girlThen s = {bb, bg, gb, gg}E = {bb}F = {bb, bg, gb}

Here, P(F) =3

4& P(E¹ F) = 1/4

º P(E/F) =1/ 4 1

3/ 4 3»

Multiplication theorem on Probability : Let two cards are drawn one after another from a well shuffdedpack of playing card and suppose we have to find the probability of the event “a king and a queen”, means

we have to find out the the probability of “E ¹ F” where E denotes geting king and F denotes geting queen.

Using the formulas for P(E/F) or P(F/E) we can find the prbabiltiy of P(E¹F)º P(E¹ F) = P(E/F) . P(F) where P(F)¼ 0

OR= P(F/E) . P(E) where P(E)¼ 0.

Similarly : P(E¹ F¹ G) = P(E) . P(F/E) . P(G/E¹F)Example 2 : An urn contains 6 block balls and 5 green balls. Two ball are drawn from the urn one afteranother without replacement. Find the probability of getting both balls green.

Thinking Process

Both balls are green means first ball is greenand 2nd ball is also green. So we have to usemultiplication theorem.Now when first ball is green, then the 2nd ballto be green is nothing but it is conditionsprobability of F givne that E has occurred.

ApplicationLet E : First ball is green

F : 2nd ball is greenClearly

P(E) =5

11after Ist draw there are 6 black and 4 green ballsleft in the urn.

Here, P(F/E) =4 2

10 5½

º P(E¹F) = P(E) . P(F/E)

=5 2

11 5¾

=2

11

141

Independent Events :In the above example occurrence of the event E has affected the probability of eventF b e c a u s e b a l l a r e d r a w n w i t h o u t r e p l a c e m e n t a n d w h e n t h e I

st ball is green then for 2nd ball to be green,number of balls changes from 5 to 4 and 11 to 10. In this case events are dependents.

In the above example if the balls are drawn with replacement then occurence of E does not affect theoccurence of F because if Ist drawn ball is green then for 2nd ball to be green, number of balls in the urndonot change because after Ist draw the drawn ball is kept back in the urn. In this case events are independent.In this case

P(E¿ F) = P(E) . P(F).Hence two envents E and F associated with the same random experiment are said to be independent if

P(E¿ F) = P(E) . P(F).Remark :(i) For three events A, B and C to be

independent : P(A¿ B) = P(A) . P(B)P(B¿ C) = P(B) . P(C)P(A¿ C) = P(A) . P(C)

and P(A¿ B ¿ C) = P(A) . P(B). P(C)(ii) Two events A and B are said to be dependent if

P (A¿ B) À P(A) . P(B)(iii) Mutually exclusive events and independent event are different terms. The term independent is defind

in terms of probability of events where as mutually exclusive is defined in terms of subset of smaplespace. Moreover mutually exclusive events never have an out-come common but independent eventsmay have common outcome.Independent Experiments :Two events are said to be independent if for every pair of events E andF where E is associated with the first experiment and F with the 2nd experiment, the probability of thesimultaneous occurence of the events E and F when the two experiments are performed is the theproduct of P(E) and P(E) calcualted separately on the basis of two experiments i.e.

P(E¿F) = P(E) . P(F)Example 3 : A die is thrown. If E as event ‘number obtained is a multiple of 3’ and F is the event ‘the

number appearing is even’. Check whether E and F are independents.

Thinking Process

First of all we have to write sample space ofthe random experiment.To find P(E) we have to write E,To find P(F) we have to write FTo find P(E¿F), we have to write (E¿F)

Application

s = {1, 2, 3, 4, 5, 6}

E = {3, 6} Á P(E) =1

3

F = {2, 4, 6}Á P(F) =1

2

E¿F = {6} Á1

P(E F)6

 Ã

clearly P(E¿ F) = P(E) . P (F).Ä E and F are independent event.

142

Bayets Theorem :Consider two bags, bag I contain 5 white and 4 red balls and bag II contains 6 white and9 red balls. One ball is drawn at random form one of the bags. We can find the probability that the balldrawn is of a particular colour, if we are given the bag from which ball is drawn. But can we find theprobability that the ball drawn is from a particular bag if the colour of the ball drawn is given? Here wehave to find the reverse probability of a particular bag to be selected. When an event occured after it isknown. Famous Mathematician, Johan Bayes solved the problem of finding reverse probability by usingconditional probability. The formula developed by him is know as Bayets theorem.

Some terms related to Buyers theorem :

(i) Partition of a sample space :Let E1, E

2......E

n be the events of a sample space S, then they are said to

represent a partitionof the smaple space s if(a) E

iÅ E

j = Æ, i Ç j, i, j = 1, 2, 3, .....n

(b) E1È E

2È .......È E

n = S.

(c) PEi > 0 for alli = 1, 2, 3, ...........n.

For example any non empty Event E and its compleemnt E1 form a partition of the sample space S,since they satisfy EÅ E1 = Æ and EÈ E1 = S.

(ii) Theorem of total Probability : Let E1, E

2, E

3 ..........., E

n be a partition of the samle space S and

suppose that each of the events E1, E

2, E

3................E

n has non zero probability of occurence. Let A be

any event associated with S, thenP(A) = P(E

1).P(A/E

1) + P(E

2) . P(A/E

2) +....+ P(E

n) . P(A/E

n)

=1

P(E ).P(A/E )n

i iiÉ

B ayes Theorem :Let E1, E

2 ........, E

n be n non empty events which conistitute a partition of S and A

is any envent of non zero probability, then

EP AiÊ Ë

Ì Í =

1

( ). ( / )

P(E )P(A/E )

i in

i ii

P E P A E

Î

for anyi = 1, 2, 3 ............n.

Example 4 :There are two bags. Bag I contains 3 red and 4 black ball while bag II contains 5 red and 6black balls. One ball is drawn at random from one of the bags and it is found to tbe red. Find the probabilitythat it was drawn from bag II.

Thinking Process

First we have to choose two partitions of thesample space.

Now we have to choose an event of the samplespace.

Application

Let E1 : Choosing bag I

E2 : Choosing bag II

Ï P(E1) = P(E

2) =

1

2.

Let A : Drawing red ballP(A/E

1) : Probability of drawing red ball form

Bag I= 3/7

143

We have to find the reverse probability usingthe conditional probability of event A. For thispurpose we will use Bayes theorem.

P(A/E2) = Probability of drawing red ball from

bag II = 5/11.

P(E2/A) = 2 2

1 1 2 2

P(E ).P(A/E )

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

=(1/2)(5/11)31 1 572 2 11

Ð ÑÐ Ñ Ð Ñ Ð ÑÒÓ ÔÕ Ö Õ Ö Õ ÖÕ Ö

=35

68

Random Variable and its Probability Distribution

Let us consider the random experiment of tossing of three coins simultaneously. The sample space of theexperiment is S = {HHH, HHT, HTT, TTT, TTH, THH, HTH, THT}. Let us take of the number of heads inthis experiment. We can assign to each outcome of the experiment a single real number. This single realnumber may vary with different outcomes to the experiment. Here it is a variable. Also its value dependsupon the outcome of a random – experiment and hence it is called a random variable. A random variable is

usually denoted by x. Here X(HHH) = 3, X(HHT) = 2, X(HTT) = 1, X(TTT) = 0, X (TTH) = 1, (THH) = 2,X(THT) = 1, X(HTH) = 2.

Hence a random variable is a function (say x) whose domain is the set of outcomes (or sample space)of a random experiment and its codomain is the set of real numbers.

Remark : More than one random variables can be defined on the same sample space.

Probability distribution of a random Variable :As we have discussed earlier a random variable X can take different real values, say, 0, 1, 2, ............ wefind the probabilities corresponding to different possible real values of X i.e. we find

P (X = 1), P(X = 2), P(X = 3) ...................., P(X = r).Values of the random variable and the corresponding probabilities are then put in the tabular from as :

Xi

x1

x2

x3

............. xn

P(x) P1

P2

P3

............. Pn

Such a description giving the values of the random variable along with corresponding probabilities iscalled probability distribution of the random variable.

Remark : (i) If xi is one of the possible values of a random variable X, the statment X =x

i is true at

some point (s) of the sample space. Hence the probability that X takes valuexi is always non zero. i.e.

P(x = xi) × 0.

(ii) For all possible value of X, all elements of the sample space are covered. Hence, the sum of all theprobabilities in a probability distibution must be one.

Mean of a random variable :Let X be a random variable whose possible valuesx1, x

2, x

3, .......x

n

occur with probabilities P1, P

2, P

3, ............P

n respectively. The mean of X, denoted byØ, is the number

144

1

n

i it

x pÙ

. In other words the mean ofx is the weighted average of the possible values of x, each value

is being weighted by its probability with which is occurs. Mean of a random variablex is also called theexpectation ofx, denoted by E(x).

Variance of a random variable :Variance of a random variable measures the spread or variability inthe values of the random variable. It is given by

Ú2x = var(x) =

2

2

1 1

( ) ( )n n

i i i ii i

x P x x P xÛ Û

Ü ÝÞ ß àá â

=22( ) ( )x xã äå æ ç where E(x2) =

21

1

( ( )n

i

x P xè

If the variance is small, then the values of the random variable are close to the mean.Example 5: Two cards are successively drawn without replacement (of simultaneously) from a well

shuffled pack of 52 cards. Find the mean and variance fo the number of kings.

Thinking Process

To find mean and variance of a random variable,we have to find the probability distribution. Whenwe draw two cords there are three possibilities,both of the cards are king, only one card is king,none of the two cards is king. So values assignedto the random variable are 0, 1, 2.

To find mean we needéx

1P(x

i)

Application

Let X be the random variable “Number of Kings”.

X can have values 0, 1, 2.Now P(x = 0) = P(no king)

= 2

2

48 188

52 221

c

cê

P(x = 1) = P(one king)

= 1

2

14 48 32

52 221c

c

cëì

P(r = 2) = P(both king)

=2

24 1

52 221c

cí

Probability distribution is

X 0 1 2

P(x)188

221

32

221

1

221

î Mean (ï) = é(x) = éx1P(x

1)

=88 32 1

0 1 2211 221 221

ñ ò ñ ò ñ

=34

.221

145

Bernoulli trials : Suppose we toss a coin four times. Each toss is called a trial of the random expriment. Inthis case number of trials is four. Each trial has two out comes namely success or failure i.e. If we considergetting a head as success then getting a tail is automatically failure. In each of the trials the probability ofsuccess or failure remains constent. Also outcome of any trial as independent of outcome of any other trial.Such triall are called Bernoulli trials.

Hence, trials of a random experiment are called Bernoulli trials if they satisfy the following conditions:(i) Threre should be a finite number of trials

(ii) Trials should be independent(iii) Each trial has two outcoms : sucess or failure(iv) The probability of success or failure remains the same in each trial.

Binomial Distribution : If we consider a random experiment with large number of trials. For exampletossing a coin 15 times. If getting a head is considered as a success and we want to find the probability of5 successes. In this case it is very difficult to write the sample space and then finding the required probability.In such cases a formula known as Binomial distribution is used to find the required probability.

Let X be a random variable which takes values 0, 1, 2, ..............r, ..........n thenP(x = r) = Probability of r successes

= n n r rrC bóô

Wheren is the number of Bernoulli trialsP is the probability of success in each trialõ is the probability of failure in each trial.The Binomial distribution of variable X is given by the following table with parametensn andp.

X 0 1 2 3 ..... r ..... n

P(x) 0n nC ö 1

1 Pn nC ÷ø 2 22 Pn nC q ÷ 3 3

3 Pn nC q ù ..... Pn n r rrC ÷ø ..... Pn n

nC

P(x) is called the probability function of the Binomial distribution. A Binomial distribution withn-Bennouli trials and probability of success in each trial as P, is denoted by B(n, p)Example 6 :A Fair coin is tossed 10 times. Find the probability of the following when getting a head isconsidered as success (i) exactely six successes (ii) at least six successes.

Now2 2 2

1

188 32( ) 1

221 221ix P x oú û ü ý ü

2 1 362

221 221þ ÿ ð

Variance = � �22 2

1 1P( ) P( )x i ix x x x� � ø � ø

=2

36 34

221 221� �

� � �

= 2

6800

(221)

To find variance we need� & õxi2 (Px

i)

146

Thinking Process

Here number of Bernoulli trials is 10 and proceedfurther were need p i.e. probability of successprobability of atleast six successes means we haveto find the sum of probabilites of six or more thansix successes.

Application

Here n = 10, p =1

2, E =

1

2

P(x = 6) =

10 6 610

6

1 1

2 2C

�� � � �� � � �� � � �

=

1010

6

1 105

2 512C

� � �� �� �

P(x � 6) = P(x = 6) + P(x = 7) + P(x = 8) + P(x = 9)+ P(x = 10)

=10 10 10

10 10 106 7 8

1 1 1

2 2 2C C C

� � � � � �� � ! ! !" # " # " #

10 1010 10

9 10

1 1

2 2C C

$ % $ %& &' ( ' () * ) *

= + ,10

10 10 10 10 106 7 8 9 10

1

2C C C C C

- ./ / / / 0 12 3

=193

512

Practice Questions

1. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the bagsis selected at random and a ball is drawn from the bag which is found to be red. Find the probabilitythat the ball is drawn from the first bag.

2. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Theprobability of an accident with scooter, car and truck are 0.01, 0.03 and 0.15 respectively. One of theinsured persons meet with an accident. What is the probability that he is a scooter driver?

3. In a factory which manufactures belts, machines A, B and C manufacture respectively 25%, 35% and40% of the belts. Of their outputs 5, 4 and 2 percent are respectively desective belts. A belt is drawn atrandom from the product and is found to be defective. What is the probability that it is manufacturedby machine B?

4. Find the mean and variance of the number obtained on a throw of an unbiased die.5. Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. Find the

probability distribution of the number of queens.6. From a lot of 30 bulbs which include 6 desectives, a sample of 4 bulbs is drawn at random with

replacement. Find the probability distribution of the number of defective bulbs.7. A pair of dice is thrown 4 times. If getting a doublet is considered a success. Find the probability of

two successes.

147

8. Five cards are drawn successively with replacement from a well shuffled deck of 52 cards. What is theprobability that :(i) all five cards are spades?(ii) only 3 cards are spades?(iii) none is a spade?

9. A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize are1

.100

What is the probability that he would win a prize. (i) at least one (ii) exactely once (iii) atlest twice?10. Suppose a girl throws a die. If she gets 5 or 6 on the die, she tosses a coin three times and notes the number

& heads. If she gets 1, 2, 3 or 4 she tosses a coiin once and notes whether a head or tail is obtained. Ifshe obtains exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Hint and Answers

1.2

32.

1

523.

28

694. Mean =

21

6, variance =

35

12

5. X 0 1 2 6. X 0 1 2 3 4

P ( x)144

169

24

169

1

169P(x)

256

625

256

625

96

625

16

625

1

625

7.25

2168. (i)

1

1024 , (ii)

45

512 , (iii)

243

1024

9. (i)50

991

1004 5

6 7 89 : (ii)49

1 99

2 100; <= >? @ (iii)

49149 99

1100 100

A BC D EF G 10.

8

11

148

Linear Programming12CHAPTER

Concept Building

The term linear implies that all the mathematical relations used in a problem are linear relations while theterms programming refers to the method of determining a particular programme or plan of action. Linearprogramming is technique to find optimal value (maximum or minimum) of a linear function of a numberof variable (sayx andy) subject to a number of linear inequalities in the variables involved and the variablestake non negative value only. A problem which seeks to maximise or minimise a linear function is calledoptimisation problem or linear programming problem.Objective Function : A linear function of the formz = ax + by where a, b are constants, which is to bemaximised or minimised is called objective function.Constraints : The linear inequalities or restrictions on the variables of a linear programming problem arecalled constraints. The conditions of two typeax + byH 0 orax + byI 0 wherea andb both can not be zero,are called linear inequalities or constraints.For Example : A furniture dealer deals in only two types of items, tables and chairs. He has 50,000 toinvest and has storage space of at most 60 pieces. A table cost` 2500 and a chair` 500. He estimates thatfrom the sale of one table he can make a profit of` 250 and that from the sale of one chair a profit of` 75.He want to know that how many tables and chairs he should buy from the available money so as to maximisehis profit, assuming that he can sell all the items which he buys.

Let us try to formulate this problem mathematically.Let x be the number of tables andy be the number of chairs that the dealer buys. Obviouslyx andy

must be non negative i.e.x H 0, y H 0. ....(i)Again dealer is constrained by the maximum amount he can invest and the maximum number of items

he can store.J 250x + 500y I 5000 or 5x 7I 100 (money restriction) andx + y I 60. (Storage restriction) ...(iii)Here (i), (ii) & (iii) are constraints of the linear programming problem. Further dealer wants to invest

in such a way so as to maximise his profit sayz, which stated as a function ‘f’ x andy is given byz = 250x + 75y ...(iv)

Herez is objective function of the problem.Hence the mathematical formulation of the given problemis : maximisez = 250x + 75ysubject to the constrains

5x + y I 100x + y I 60

x > 0,y H 0.

149

Graphical Method to solve a Linear Programming Problem

Step 1 : Draw the graph using constraints. In theabove problem graph using constraints is givenbelow.

The graph of this system (shaded region)consists of the points common to are halfplanes determined by the constrains (i), (ii)& (iii). Each point in this region representsa feasible choice open to the dealer forinvesting in tables and chairs. The region,therefore, is called the feasible region forthe problem. Every point of this region iscalled a feasible solution to the problem.Thus we have.

Feasible Region :The common regiondetermined by all the constraints including non negative constraintsx, y, K 0 of a linear programmingproblem is called the feasible region or solution region for the problem in the above graph OABC isthe feasible region for the problem.The region other them feasible region is called infeasible region.Feasible Solution :The points within and on the boundary of the feasible region represents feasiblesolutions of the problem. Every point within and on the boundary of the feasible region representsfeasible solution of the problem.Optimal Solution : Any point in the feasible region that gives the optimal value (maximum orminimum) of the objective function is called an optimal solution.Theorem :When an objective function has optimal value, then optimal value of the objective functionmust occur at a corner point of the feasible region.

Step 2 :Compute the value ofz at corner pointsi.e, 0(0, 0), A (20, 0), B(10, 50, C(0, 60)Let M and m respectively denote the largest and smallest values of z at these corner points.As in the above example the feasible region is bounded therefore M is the maximum value of z and mis the minimum value ofz.If Feasible region is Unbounded :Let z = ax + by, then(i) M is the maximum value ofz if open half plane determined byax + by > M has no point in

common with the feasible region, otherwisez has no maximum value.(ii) m is the minimum value ofz if open half plane determined byax+ by< m has no point in common

with the feasible region. Otherwisez has no minimum value.Remark : (i) A feasible region is said to be bounded if it can be enclosed within a circle. Otherwise,it is called unbounded.(ii) If the optimal (maximum or minimum) value of the objective function occurs at two vertices

(corners) of the feasible region then every point on the line segment joining these points will givethe same optimal value.

Example : Solve the following problem graphically.minimisez = 50x + 70y

150

Subject to the constraints 2x + y L 8x + 2y M 10

x, yM 0

Practice Questions

1. A dealer has 15000 only for a purchase of rice and wheat. A bag of rice costs` 1500 and a bag ofwheat costs 1200. He has a storage capacity of ten bags only and he gets a profit of ` 110 and 80per bag of rice and wheat respectively. Formulate it as a linear programming problem to get themaximum profit and solve it graphically.

2. A house wife wishes to mix two types of food x and y is such a way that the vitamin contents of themixture contains at least 8 units vitamin A and 11 units vitamin B. Foodx costs 60 per kg and foody costs 80 per kg. Foodx contains 3 units per kg of vitamin A and 5 units per kg of vitamin B whilefood y contains 4 units per kg of vitamin A and 2 units per kg of vitamin B. Formulate it is a linearprogramming problem to minimize the costs of the mixture and solve it graphically.

3. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machineB to produce a packgage of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce

Thinking Process

To solve the given LPP first of all we have to drawgraph using constraints.

As it is clear from the graph that the feasible regionis unbounded, so to find out optimal value i.e.minimum is this case of the objective function wehave to form a new constraint using Z and theminimum value of z obtained at corner points ofthe feasible region.

Now we shall draw the graph of the constraints5x + 7y < 38

Now we have to check whether the resulting halfplane determined by the above constraint has anypoint common with the reasible region.

Application

HereZ at A (0, 8) = 560Z at B (2, 4) = 380 minimumZ at C(10, 0) = 500.Now inequality (constraint) is

50x + 70y < 380i.e. 5x + 7y < 38It is clear from the graph that the open half plandetermined by the in equality 5x + 7y < 38 has nopoint in common with the feasible region. Thus380 is the minimum value of Z at B (2, 4)

151

a package of bolts. He earns a profit of` 17.50 per package on nuts and` 7.00 per package on bolts.How many packages of each should be produced each day so as to maximise his profit, is he operateshis machines for at most 12 hours a day? Solve it as a LPP.

4. One kind of cake requires 200 gm of flour and 25 gm of fat and another kind of cake requires 100 gmof flour and 50 gm of fat. Find the maximum number of cakes which can be made from 5 kg of flourand 1 kg of fat assuming that there is no shortage of the other ingredients used in the cakes.

5. There are two types of fertilizers F1 and F

2. Fertilizer F

1 consists of 10% nitrogen and 6% phosphoric

acid and F2 consists 5% nitrogen and 10% phosphoric acid. After testing the soil conditions a farmer

finds that he needs atleast 14 kg of nitrogen and 14 kg of phosphorite acid for his crop. If F1 costs` 6

per kg and F2 costs 5 per kg, determine how much of each type of fertilizer should be used so that

nutrient requirements are met at a minimum cost. What is minimum costs? Solve it as a LPP.

6. A diet is to contain at least 80 unit of vitamin A and 100 units of minerals. Two foods F1 and F

2 are

available. Food F1 costs 4 per unit and F

2 costs 6 per unit. One unit of food F

1 contains 3 units of

vitamin A and 4 unit of minerals while one unit of food F2 contains 6 units of vitamin A and 3 units of

minerals. Find the minimum cost for diet that consists of mixture of these two foods and also meetsthe minimal nutritional requirements. Solve it as a LPP.

7. Solve the following problem graphically :

maximise z = 8000x + 12000 y

Subject to the constraints

3x + 4y N 60,x + 3y N 30, xO 0, y O 0.

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

Food Vitamin A Vitamin B Vitamin C

X 1 2 3

Y 2 2 1

If one kg of foodx costs 16 and one kg of food y costs` 20. Find the least costs of the mixture whichwill produce the required diet.

9. A manufacturer makes two types of toys A and B. Three machines are needed for this purpose andtime (in minutes) required for each toy on the machine as is given below :

Types of Toyes Machines

I II III

A 12 18 6

B 6 0 9

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is` 7.50 and that on each toy of type B is` 5. How many toys of each type be manufactured in a day toget maximum profit? Solve it is a LPP.

10. Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supplyto 3 ration shops D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of

152

transportations per quintal from godowns to the shops are given in the following table

Transpertation cost per quintal (in Rupees)

From/To A B

D 6 4

E 3 2

F 2.50 3

How should the supplies be transported in order that the transportation cast is minimum? Find theminimum cost. Solve it as a LPP.

Hints and Answers

1. Ten bags of rice only and maximum profit =` 1100.

2. Minimum cost = 160, at all points on the line segment joining8

,03

P QR ST U and

12,

2P QR ST U .

3. Three packages of nuts and 3 packages of bolts, maximum profit =` 73.50.4. Maximum number of cakes = 30 of one kind and 10 cakes of another kind.5. 100 kg of F

1 and 80 kg of F

2, minimum cost = 1000.

6. Minimum cost = 1000.Hint : minimisez = 4x + 6y subject tot he constraints 3x + 6y V 80, 4x + 3y V 100,x V 0, y V 0.

7. Maximum profit = 1,68,000.8. Minimum cost = 112.9. 15 toys of type A and 30 toys of type B.10. From A : 10, 50, 40 units from B : 50,0,0 units to D, E and F respectively. Minimum cost =` 510.

q q q