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PREFACE This text has been written in accordance with new mathematics syllabus for Class-XII prepared by CBSE in accordance with National Curriculum Framework (NCF) and NCERT guidelines. The text has been written keeping YOU in mind. The main purpose of this text book is to develop mathematical understanding, thinking and reasoning, to acquaint students with different aspects of mathematics used in daily life. Mathematics is an exciting and enjoyable subject when you understand what is going on. Concepts are explained clearly and intutively. I begin each topic with familiar material, explain new concepts in terms of ideas already well understood, offer worked out examples that show in detail how problems are to be solved of all types. Large number of questions of varying difficulty level are solved in this book. Detailed solutions of questions from NCERT book, from Exemplar problems and CBSE question papers set over the years etc. have been solved. Objective type questions, very short answer type questions, short answer type questions, long answer type questions both understanding based and application based are solved. HOTS (Higher Order Thinking Skill) questions are included in this book. Varied and carefully graded exercises containing good questions test student’s deep understanding of mathematical skills are given at the end of each chapter. Review section given at the end of book which contains summary of each chapter that identifies the terms, formulas, rules, and procedures that were introduced in the chapter. Also in review section, practice exercises containing all types of questions are given which can benefit the students to well prepare and build self confidence for final examination. I believe that the mathematics presented here is interesting, useful and worth-studying. Suggestions for further improvement of the book, pointing out printing errors/mistakes which might have crept in, inspite of all efforts will be thankfully received and incorporated in the next edition. Manjeet Singh New Delhi

Note for the Students Mathematical intelligence constitutes abstract thoughts, precision counting, organisational and logical structure. Critical thinking, reasoning and problem solving skills are essential. 1. Mathematics is essentially a practical discipline, in that it is learnt by doing rather than solely by reading about it. To learn mathematics,it is necessary to do mathematics. 2. Develop conceptual understanding of different topics. Try to understand fundamental theory (Mathematical Concepts) of each chapter. If you are uncertain of any concept, go back into the chapter and study it further. 3. Always focus on developing mathematical thinking. Math is one way to generate logical thinking and reasoning among students. Get into the habit of doing some mathematics everyday. 4. Read illustrative examples with great care. Try to understand various procedures which are applied to worked out examples. Use mathematical knowledge and techniques to solve problems given in the exercises. Link conceptual and procedural knowledge. While solving a problem, you must show all of the steps in your working. Marks will be available for showing a correct method and correct answer. 5. Use sketches, diagrams and models to clarify your thinking. Diagrams and graphs can play a major role in understanding mathematics. The purpose of graph is to show numerical facts in visual form so that they can be understood quickly. 6. To make the most of your mathematical abilities, it is worth making effort to develop effective study habits. It develops logical reasoning, abstract thinking and imagination. 7. Mathematics has its own language and symbols. Knowledge of mathematical symbols can be of great use to other sciences. 8. Believe in discussion of mathematics, questioning and justification of thinking. Communication with lecturers and tutors to obtain information or clarification for you to check your understanding and for them to check your understanding. 9. Mathematics help students to imbibe great values: (i) Honesty, (ii) truthfulness, (iii) self-confidence, (iv) self-reliance, (v) open-minded (vi) neatness. 10. Students can be greatly inspired by understanding the contributions made by different mathematicians. (i) Aryabhata (ii) Brahmagupta (iii) Bhaskra (iv) John-Napier (v) Rene Desscartes (vi) Thales (vii) Pythagoras (viii) Plato (ix) Euclid (x) Archimedies (xi) Gelieo (xii) Kepler (xiii) Pascal (xiv) Newton (xv) Leibniz (xvi) Michel Rolle (xvii) Lagrange (xviii) Ramanujam (xix) D’Alembert (xx) Varigon and others had contributed significantly in the field of mathematics. 11. Mathematical Reasoning: The ability to solve problems using thought and reasoning is indispensable to every area of our lives. How well this ability can be used depends on each person’s power of reasoning. The reasoning process leads to conclusion. Inductive and deductive reasoning skills provide a logical framework for reaching conclusions. In mathematical language, there are two types of reasoning: (i) Inductive Reasoning (or Induction): Inductive reasoning is the process of arriving at a general conclusion based on observations of specific examples. That is, a generalisation made on the basis of observed occurrences. The conclusion is called a conjecture. (ii) Deductive Reasoning: Deductive reasoning is the process of proving a specification clusion from one or more general statements. This method of reasoning produces results that are certain within the logical system being developed. A conclusion that is proved true by deductive reasoning is called a theorem.

(iv)

SYLLABUS Marks: 80

Units I. Relations and Functions II. Algebra III. IV. V. VI.

Calculus Vectors and Three - Dimensional Geometry Linear Programming Probability Total Internal Assessment

No. of Periods 30 50

Marks 08 10

80 30 20 30 240

35 14 05 08 80 20

UNIT I: RELATIONS AND FUNCTIONS 1. Relations and Functions (15 Periods) Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions, composite functions, inverse of a function. 2. Inverse Trigonometric Functions (15 Periods) Definition, range, domain, principal value branch. Graphs of inverse trigonometric functions. Elementary properties of inverse trigonometric functions. UNIT II: ALGEBRA 1. Matrices (25 Periods) Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. Operation on matrices: Addition and multiplication and multiplication with a scalar. Simple properties of addition, multiplication and scalar multiplication. Non-commutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Concept of elementary row and column operations. Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries). 2. Determinants (25 Periods) Determinant of a square matrix (up to 3 x 3 matrices), properties of determinants, minors, cofactors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. 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 of a matrix. UNIT III: CALCULUS 1. Continuity and Differentiability (20 Periods) Continuity and differentiability, derivative of composite functions, chain rule, derivatives of inverse trigonometric functions, derivative of implicit functions. Concept of exponential and logarithmic functions. Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions 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 and normals, use of derivatives in approximation, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as real-life situations).

(v)

3. Integrals (20 Periods) Integration as inverse process of differentiation.Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them. dx dx dx dx dx ∫ x 2 ± a 2 , ∫ x 2 ± a 2 , ∫ a 2 - x 2 , ∫ ax 2 + bx + c , ∫ ax 2 + bx + c

∫ ax ∫ 

px + q dx, ∫ + bx + c

2

ax 2 + bx + c dx,

px + q ax 2 + bx + c

∫ ( px + q)

dx, ∫

a 2 ± x 2 dx, ∫

x 2 − a 2 dx

ax 2 + bx + c dx

 'HÀQLWH integrals as a limit of a sum, Fundamental Theorem of Calculus (without proof).Basic properties of definite integrals and evaluation of definite integrals. 4. Applications of the Integrals (15 Periods) Applications in finding the area under simple curves, especially lines, circles/parabolas/ellipses (in standard form only), Area between any of the two above said curves (the region should be clearly identifiable). 5. Differential Equations (15 Periods) Definition, order and degree, general and particular solutions of a differential equation. Formation of differential equation whose general solution is given. Solution of differential equations by method of separation of variables, solutions of homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type: dy + py = q, where p and q are functions of x or constants. dx dx + px = q, where p and q are functions of y or constants. dy

UNIT IV: VECTORS AND THREE-DIMENSIONAL GEOMETRY 1. Vectors (15 Periods) Vectors and scalars, magnitude and direction of a vector.Direction cosines and direction ratios of a vector. 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. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors, scalar triple product of vectors. 2. Three - dimensional Geometry (15 Periods) Direction cosines and direction ratios of a line joining two points.Cartesian equation and vector equation of a line, coplanar and skew lines, shortest distance between two lines.Cartesian and vector equation of a plane. Angle between (i) two lines, (ii) two planes, (iii) a line and a plane. Distance of a point from a plane. UNIT V: LINEAR PROGRAMMING 1. Linear Programming (20 Periods) Introduction, related terminology such as constraints, objective function, optimization, different types of linear programming (L.P.) problems, mathematical formulation of L.P. problems, graphical method of solution for problems in two variables, feasible and infeasible regions(bounded and unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints). UNIT VI: PROBABILITY 1. Probability (30 Periods) Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution, mean and variance of random variable.

(vi)

CONTENTS 7. +PFGſPKVG+PVGITCNU ............................................................................ 9–318 8. &GſPKVG+PVGITCNU ........................................................................... 319–517 9. &KHHGTGPVKCN'SWCVKQPU ................................................................... 518–675 10. 8GEVQT#NIGDTC ................................................................................ 676–788 11. 6JTGG&KOGPUKQPCN)GQOGVT[ ....................................................... 789–906 12. .KPGCT2TQITCOOKPI ..................................................................... 907–948 13. 2TQDCDKNKV[ .................................................................................... 949–1079 O

4GXKGY5GEVKQP ........................................................................... 1080–1172

O

2TCEVKEG2CRGTŌ ...................................................................... 1173–1176

O

2TCEVKEG2CRGTŌ ...................................................................... 1177–1180

(vii)

7

Indefinite Integrals

Introduction The second fundamental problem addressed by calculus is the problem of areas, that is, the problem of determining the area of a region of the plane bounded by various curves. Like the problem of tangents considered earlier, many practical problems in various disciplines require the evaluation of areas for their solution, and the solution of the problem of areas necessarily involves the notion of limits. On the surface, the problem of areas appears unrelated to the problem of tangents. However, we will see that the two problems are very closely related, one is the inverse of the other. Finding an area is equivalent to finding an antiderivative or, as we prefer to say, finding an integral. Y

Y

Slope m = ?

y = f (x) y = f (x) R Area of region under the graph y = f (x) between x = a and x = b

P(x, f (x)) Area = ?

x

X

a

The tangent-line problem motivates differential calculus.

b

X

The area problem motivates integral calculus.

The relationship between areas and antiderivatives is called the fundamental theorem of calculus. It provides a vital connection between the operations of differentiation and integration, one that provides an effective method for computing values of integrals. It turns out that instead of finding the derivative of the function f (x), we need to find a new function F(x) whose derivative is f (x). F′( x) = f ( x) Thus we need to do “reverse of the process of differentiation”. We, therefore, begin with an investigation of antidifferentiation.

9

Antiderivative Since many kinds of problems in many areas, including science and technology, can be solved by reversing the process of finding a derivative or a differential, we shall introduce the basic technique of this procedure in this section. This reverse process is known as antidifferentiation. Let f be defined on [a, b]. Then if there exists a function y = F(x) such that F is continuous on [a, b], differentiable on (a, b), and the derivative of F is f for every x in (a, b), that is, if F′( x) = f ( x), x in ( a, b) Then F is called antiderivative or indefinite integral of f and we write F( x) = ∫ f ( x) dx

This is read as “F(x) is the integral or primitive of f (x) with respect to x” or simply, “F is an antiderivative of f ”. If such a function F exists, then f is said to be integrable and the process of calculating an integral is called integration. The variable x is called the variable of integration and the function f is called the integrand. Constant of Integration. We know that d 3 d 3 ( x ) = 3x 2 ( x + 2) = 3x 2 (ii) (i) dx dx d 3 d 3 ( x − 7) = 3 x 2 ( x + c) = 3x 2 (iii) (iv) dx dx ∴ Integral of 3x2 may be x3, x3 + 2 or x3 – 7 or x3 + c, where c is any arbitrary constant. Thus if F(x) is any integral of f (x), F(x) + c is also an integral of f (x), where c is any arbitrary constant whatsoever. d i.e., if [F( x )] = f ( x ) dx Then, ∫ f ( x) dx = F( x) + c, c being any constant. The arbitrary constant c in the integral F(x) + c is called the constant of integration. Since the result F(x) + c is indefinite on account of indefinite constant, it is called the indefinite integral of f (x). Thus Variable of Integration

∫ f ( x) dx = F( x) + c Integrand

10

MATHEMATICS–XII

Constant of integration

The expression

∫ f ( x) dx is read as the antiderivative of

f with respect to x. Thus the

differential dx serves to indentify x as the variable of integration. The term indefinite integral is a synonym for antiderivative.

Remember 1. In this text, whenever we write ∫ f ( x) dx = F( x) + c , it means that F is an antiderivative of f. 2. Basic Integration Rules: The inverse nature of integration and differentiation can be verified by substituting F′(x) for f (x) + C in this definition to obtain integration is the “Inverse” of differentiation.

Moreover, if

∫ f ( x) dx

∫ F´( x)dx = F( x) + C d ⎡ f ( x) dx ⎤ = = F( x) + c, then ⎦ dx ⎣ ∫

f ( x)

Differentiation is the ‘inverse’ of integration. These two equations allow us to obtain integration formulae directly from differentiation formulae, as shown in the following summary.

Fundamental Formulae or Standard Forms The differential coefficients of elementary functions are known to us. And as integration is the inverse process of differentiation, the various formulae for the derivatives of functions make it possible for us to know the corresponding formulae for integrals. They are listed below for ready reference. Since 1.

Therefore

d ⎛ x n +1 ⎞ n ⎜ ⎟ = x , x ≠ −1 dx ⎜⎝ n + 1 ⎟⎠

In particular,

1.

d ( x) = 1 ⋅ x 0 = 1 dx

n ∫ x dx =

xn + 1 + c, (if n ≠ −1) n +1

In particular ∫ 1⋅ dx = ∫ x° dx = x

2.

d ⎡ (ax + b) n + 1 ⎤ n ⎢ ⎥ = (ax + b) dx ⎣ a (n + 1) ⎦

2.

n ∫ (ax + b) dx =

3.

d 1 ( log| x |) = dx x

3.

∫ x dx = log | x | +c

4.

d ⎡ log | ax + b | ⎤ 1 = ⎢ ⎥ dx ⎣ a ⎦ ax + b

4.

∫ ax + b = a log| ax + b | + c

5.

∫e

6.

ax + b ∫ e dx =

d x (e ) = e x dx d ax + b 6. (e ) = ae ax + b dx

5.

(ax + b)n +1 + c (If n ≠ −1) a(n + 1)

1

dx

x

1

dx = e x + c eax +b a INDEFINITE INTEGRALS

11

Since

Therefore

ax +c log a

7.

d x ( a ) = a x log a dx

7.

x ∫ a dx =

8.

d (sin x ) = cos x dx

8.

∫ cos x dx = sin x + c

9.

d {sin( ax + b)] = a ⋅ cos ( ax + b) dx

9.

∫ cos(ax + b)dx = a sin(ax + b) + c

10.

d (cos x) = − sin x dx

10.

∫ sin x dx = − cos x + c

11.

d [cos ( ax + b)] = − a sin(ax + b) dx

11.

∫ sin (ax + b) dx = − a cos (ax + b) + c

12.

d (tan x) = sec ² x dx

12.

∫ sec

2

x dx = tan x + c

13.

d [tan (ax + b)] = a sec ² (ax + b) dx

13.

∫ sec

2

(ax + b) dx =

14.

d (cot x ) = − cosec 2 x dx

14.

∫ cosec

2

15.

d [cot (ax + b )] = − a cosec² (ax + b ) 15. dx

∫ cosec

2

1

1

1 tan (ax + b) + c a

x dx = − cot x + c (ax + b) dx 1 = − cot (ax + b) + c a

16.

d (sec x ) = sec x tan x dx

16.

∫ sec x tan x dx = sec x + c

17.

d (cosec x ) = − cosec x cot x dx

17.

∫ cosec x

18.

∫

d 1 −1 18. dx (sin x) = 1 − x2

dx 1− x

2

−1

cot x dx = − cosec x + c = sin −1 x + c

19.

−1 d (cos −1 x) = dx 1 − x2

19.

∫

20.

d 1 (tan −1 x) = dx 1 + x2

20.

∫ 1+ x

21.

d −1 (cot −1 x) = dx 1 + x2

21.

∫ 1 + x2 dx = cot

12

MATHEMATICS–XII

1 − x2 1

2

−1

dx = cos −1 x + c

dx = tan −1 x + c −1

x+c

Since

Therefore

22.

d 1 (sec −1 x) = dx | x | x2 − 1

22.

∫| x|

23.

−1 d (cosec−1 x) = dx | x | x2 − 1

23.

∫| x|

1 x −1 2

−1 x −1 2

dx = sec −1 x + c

dx = cosec−1 x + c

An Important Note. If we know the integral of f (x), then the integral of f ( ax + b) is obtained by replacing x by ax + b and dividing the result by a. If a function to be integrated is in any of the above forms, then we can directly write down its integral. But, in other cases, we have to bring the function in the above form. For this purpose, some theorems and rules of integration are necessary.

Two General Theorems Theorem 1. The integral of the product of a constant and a function is equal to the product of the constant and integral of the function i.e.,

∫ k f ( x) dx = k ∫ f ( x) dx, ‘k’ being a constant. Proof: Let

d

∫ f ( x)dx = φ( x), then dx [φ( x)] = f ( x)

d d [ k φ( x )] = k ⋅ [ φ ( x )] dx dx [' The derivative of the product of a constant and a function is equal to the product of the constant and the derivative of the function] ⎡ d ⎤ = k f ( x) ⎢' dx [φ( x )] = f ( x ) ⎥ ⎣ ⎦

Now

∫

∴ By definition,

∫

k ⋅ f ( x) dx = k ⋅φ( x) = k ⋅ f ( x) dx.

Theorem 2. The integral of the sum or the difference of two functions is equal to the sum or difference of their integrals i.e.. Proof: Let and

∫ [ f ( x) ± f ( x)] dx = ∫ f ( x) dx ± ∫ f ∫ f ( x)dx = φ ( x ) ∫ f ( x)dx = φ ( x ) 1

−

and Now,

2

1

1

1

2

1

2

( x) dx

1 4 cos 2 x − e3 x = f1(x) 2 3

d [φ2 ( x)] = f2(x) dx d d [φ1 ( x) ± φ2 ( x)] = [φ1 ( x) ± φ2 ( x)] = f (x) ± f (x) 1 2 dx dx INDEFINITE INTEGRALS

13

[' The derivative of the sum or difference of two functions is equal to the sum or difference of their derivatives.] ∴ By definition of the integrals of a function

∫ [ f ( x) ± f ( x)] dx = φ ( x) ± φ ( x) = ∫ f ( x) dx ± ∫ f 1

2

1

2

1

2

( x) dx.

We can extend this theorem to a finite number of functions and can have the following result:

∫ [ f1 ( x) ± f 2 ( x) ± ... ± f n ( x)] dx = ∫ f1 ( x) dx ± ∫ f 2 ( x) dx ± ... ± ∫ f n ( x) dx.

Remarks:

∫

∫

1. If k is a constant, then k f ( x ) dx = k f ( x ) dx

∫

2. [ f1 ( x) ± f 2 ( x)] dx =

∫ f ( x) dx ± ∫ f ( x) dx 1

2

SOLVED EXAMPLES Example 1. Find an antiderivative (or integral) of the following by the method of inspection: [NCERT] (i) sin 2x, (ii) cos 3x, (iii) e2x, (iv) (ax + b)2, (v) sin 2x – 4e3x Sol.(i) We look for a function whose derivative is sin 2x d ( − cos 2 x) = − ( − 2sin 2 x) = 2sin 2 x We know that dx 1 d d ⎛ 1 ⎞ ( − cos 2 x ) = ⎜ − cos 2 x ⎟ ⇒ sin 2x = 2 dx dx ⎝ 2 ⎠ Hence, antiderivative of sin 2x is −

1 cos 2 x or 2

∫

sin 2 x dx = −

1 cos 2 x + C . 2

(ii) We look for a function whose derivative is cos 3x. We know that

⇒

d (sin 3x) = 3 cos 3x dx 1 d d ⎛1 ⎞ (sin 3x ) = ⎜ sin 3 x ⎟ cos 3x = 3 dx dx ⎝ 3 ⎠

Hence, an antiderivative of cos 3x =

1 sin 3 x or 3

1

∫ cos 3x dx = 3 sin 3x + C

(iii) We look for a function whose derivative is e 2x Note that

d 2x 1 d 2x d ⎛1 ⎞ (e ) = 2e 2 x ⇒ e 2 x = (e ) = ⎜ e 2 x ⎟ dx 2 dx dx ⎝ 2 ⎠

Hence, an antiderivative of e2x is

14

MATHEMATICS–XII

1 2x e or 2

∫

1 e2 x dx = e2 x + C 2

(iv) We look for a function whose derivative is (ax + b)2 d d ( ax + b )3 = 3( ax + b) 2 × ( ax + b) dx dx = 3(ax + b)2 × a = 3a (ax + b)2 1 d d ⎧1 3 3⎫ ⇒ (ax + b) 2 = 3a dx ( ax + b ) = dx ⎨ 3a ( ax + b ) ⎬ ⎩ ⎭ 1 Hence, an antiderivative of (ax + b) 2 is (ax + b)3 3a 1 ( ax + b)3 (ax + b) 2 dx = or 3a (v) We look for a function whose derivative is sin 2 x − 4e3 x d (cos 2 x) = − 2sin 2x dx 1 d d ⎛ 1 ⎞ (cos 2 x ) = ⎜ − cos 2 x ⎟ ⇒ sin 2x = − 2 dx dx ⎝ 2 ⎠

∫

d ⎛ 4 3x ⎞ d ⎛ 4 3x ⎞ 4 3x 3x 3x ⎜ e ⎟ = × 3e = 4e ⇒ 4e = ⎜ e ⎟ dx ⎝ 3 ⎠ dx ⎝ 3 ⎠ 3

Also

1 4 cos 2 x − e3 x . 2 3 1 4 3 x (sin 2 x − 4e3 x ) dx = − cos 2 x − e + C 2 3

Hence, antiderivative of sin 2 x − 4e3 x is −

∫

or

Example 2. Write down the integral of (i ) x 2 (ii ) x − 9 (iii ) 1 (iv)

∫

Sol. (i )

x2 + 1 1 + C = x3 + C 2 +1 3

x 2 dx =

∫

∫

(iii ) 1⋅ dx = x 0 dx = (iv)

∫

(v )

∫

(vi )

∫

1 (vi ) x x2

x− 9 + 1 1 +C=− 8 +C −8 8x

(ii ) x − 9 dx =

∫

x (v )

x dx =

∫

1 x2

x0 + 1 +C= x+C 0 +1

dx =

1 +1 x2

+C=

3 2 x

3

+C=

1 3 +1 2 2 1 x− 2 + 1 x− 1 1 dx = +C= +C=− +C − 2 +1 −1 x² x −2 x 3 dx

2 − +1 3

2 2 x +C 3

1 1

x3 = = + C = 3x 3 + C 2 1 − +1 3 3 x

INDEFINITE INTEGRALS

15

−2 3

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