12

CBSE PART-I

MATHEMATICS (As per the latest CBSE Syllabus)

K C Sisodia MSc, BEd (Formerly Lecturer in Mathematics) Directorate of Education

Full Marks Pvt Ltd (Progressive Educational Publishers)

New Delhi-110002

Published by:

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Note from the Publishers Full Marks Mathematics-12 has been thoroughly revised and updated as per the latest syllabus and sample question papers issued by CBSE. The book provides complete tutorial support to students in such a way that they can master the method to answer confidently the different types of questions given in their coursebooks developed by the NCERT. The strength of the FULL MARKS series, as a whole, lies in the process that we put in place while creating these support books: (a) Identifying the learning difficulties experienced by a large number of students struggling to perform satisfactorily in the examinations (b) Commissioning experienced writers who have been involved in teaching students who need support (c) Discussing and deciding on the ideal structure to present each concept in the course in a manner that can be easily understood by students (d) Focusing on presenting each concept in learner-friendly chunks so that the students improve step-by-step (e) Trials with students who are self-reliant learners because of non-availability of support in their home environment. (f) Restructuring units on the basis of the trial results We are confident that such efforts will prove the fact that these are Success Books! The way the units are structured is yet another great strength of the series. The book is, thus, a complete, up-to-date, dependable and learner-friendly resource. The support website www.fullmarks.org is an added benefit for the users where one can get a vast platform to share and discuss one’s problems related to one’s subjects with others.

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SYLLABUS One Paper Units I. II. III. IV. V. VI.

Max. Marks: 80

Relations and Functions Algebra 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. Noncommutativity 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 × 3 matrices), properties of determinants, minors, co-factors 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.

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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). 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

∫ x2 ± a2 , ∫ px + q

dx 2

x ±a

∫ ax 2 + bx + c dx, ∫ ∫



2

dx

,∫

2

a −x

2

px + q 2

ax + bx + c

,∫

dx dx ,∫ 2 ax + bx + c ax + bx + c 2

dx, ∫ a 2 ± x 2 dx,∫ x 2 − a 2 dx

ax 2 + bx + c dx, ∫ ( px + q ) ax 2 + bx + c dx

Definite 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.

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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. Binomial probability distribution.

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CONTENTS

1. Relations and Functions .......................................................................................................

9

2. Inverse Trigonometric Functions.......................................................................................

55

3. Matrices ...................................................................................................................................

91

4. Determinants .......................................................................................................................... 144 5. Continuity and Differentiability ......................................................................................... 218 6. Applications of Derivatives ................................................................................................. 297

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QUESTION PAPER DESIGN

Time : 3 Hours

S. No.

Typology of Questions

Max. Marks : 80

Very Short AnswerObjective Type (VSA)

Short Answer (SA)

Long Answer-I (LA-I)

Long Answer-II (LA-II)

1 Mark

2 Marks

4 Marks

6 Marks

Total Marks

1.

REMEMBERING—Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers.

4

1

1

1

16

2.

UNDERSTANDING—Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas.

6

2

3

1

28

3.

APPLYING—Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.

6

2

1

1

20

4.

ANALYSING—Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations. EVALUATING—Present and defend opinions by making judgements about information, validity of ideas, or quality of work based on a set of criteria. CREATING—Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions.

4

1

1

1

16

20×1=20

6×2=12

6×4=24

4×6=24

80

TOTAL

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1

Relations and Functions Syllabus

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

Facts that Matter Relations Types of Relations  Empty Relation: A relation R on a set A is said to be an empty relation if no element of A is related to any element of A. For example: Let A = {2, 3, 4} and R = {(a, b) : a, b ∈ A and ab > 20 } is an empty relation.  Universal Relation: A relation R in a set A is said to be a universal relation if each element of A is related to every element of A. For example: Let A = {1, 2, 3} and R = {(a, b) : a, b ∈ A and a + b < 10} is a universal relation.  Reflexive Relation: Let A be any non-empty set then a relation R on A is said to be reflexive if (a, a) ∈ R for every a ∈ A. For example: Let A = {1, 2, 3} then R = {(1, 1), (2, 2), (3, 3), (1, 2), (3, 1)} is a reflexive relation.  Symmetric Relation: Let A be any non-empty set then a relation R on A is said to be symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A × A For example: Let A = {2, 3, 4} then R = {(2, 2), (3, 4), (4, 3), (2, 4), (4, 2)} is a symmetric relation.  Transitive Relation: Let A be any non-empty set then a relation R on A is said to be transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A. For example: Let A = {1, 2, 3} then R = {(1, 1), (1, 2), (2, 1), (2, 3), (1, 3)} is a transitive relation.  Equivalence Relation: A relation R on a set A is said to be an equivalence relation if it is simultaneously a reflexive, symmetric and transitive relation. For example: Let A = {1, 2, 3, 4, 5} then R = {(a, b): a, b ∈ A and |a – b| is even} is an equivalence relation.  The inverse of an equivalence relation is also an equivalence relation. Example 1. Show that the relation R defined by (a, b) R (c, a) ⇔ ad = bc on the set N × N is an equivalence relation. Sol. Here (a, b) R (c, d) ⇒ ad = bc, a, b, c, d ∈ N Now (a, b) R (a, b) ⇒ ab = ba, (a, b) ∈ N × N So R is reflexive relation. Now (a, b) R (c, d) ⇒ ad = bc, (a, b), (c, d) ∈ N × N ⇒ (c, d) R (a, b) ⇒ cb = da ⇒ bc = ad So R is symmetric relation. Now (a, b) R (c, d) ⇒ ad = bc, (a, b), (c, d) ∈ N × N and (c, d) R (e, f) ⇒ cf = de, (c, d), (e, f) ∈ N × N 9

Multiplying adcf = bcde = af = be ⇒ (a, b) R (e, f) So R is transitive relation. Hence R is an equivalence relation. Example 2. Let A be the set of all students of class X in a school and R be the relation, having the same sex on A. Prove that R is an equivalence relation. Do you think, co-education may be helpful in a child development and why? [VBQ] Sol. (i) Since x R x ∀ x ∈ A [Q  Male is related to male and female is related to female i.e. same sex] ∴ R is reflexive (ii) Let x R y [x and y are of the same sex ∴ y and x are of the same sex] ∴ yRx Thus, R is symmetric (iii) Let x R y and y R z [x and y are of the same sex, y and z are of the same sex] ∴ x and z are of the same sex Thus, x R y and y R z ⇒ x R z ∴ R is transitive. Hence R is an equivalence relation. Co-education is very helpful since it leads to the balanced development of the children. Example 3. Let A = {1, 3, 5, 7} and relation R is defined in set A as follows: (a) R = {(x, y) : x, y ∈ A, xy ≥ 50} (b) R = {(x, y) : x, y ∈ A, x + y ≤ 14} (c) R = {(x, y) : x, y ∈ A, xy = x2} Sol. Here A = {1, 3, 5, 7} A × A = {(1, 1), (1, 3), (1, 5), (1, 7), (3, 1), (3, 3), (3, 5), (3, 7), (5, 1), (5, 3), (5, 5), (5, 7), (7, 1), (7, 3), (7, 5), (7, 7)} (a) R = {(x, y) : x, y ∈ A, xy ≥ 50} R= {}=φ Hence R is an empty relation. (b) R = {(x, y) : x, y ∈ A, x + y ≤ 14} R = {(1, 1), (1, 3), (1, 5), (1, 7), (3, 1), (3, 3), (3, 5), (3, 7), (5, 1), (5, 3), (5, 5), (5, 7), (7, 1), (7, 3), (7, 5), (7, 7)} = A×A Hence R is a universal relation. (c) R = {(x, y) : x, y ∈ A, xy = x2} R = {(1, 1), (3, 3), (5, 5), (7, 7)} R = {(x, x) : x ∈ A} Here R is an identity relation. Example 4. (a) If A = {1, 2, 3, 4}, define relation on A which have properties of being: (i) reflexive, transitive but not symmetric. (ii) reflexive, symmetric and transitive. (iii) symmetric but neither reflexive nor transitive (b) Let R be a relation on the set of natural number N as follows. R = {(x, y) : x ∈ N, y ∈ N, 2x + y = 41}. Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive. [NCERT Exemplar] Sol. (a) Given: A = {1, 2, 3, 4} (i) Let R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2)} ∴ R is reflexive and transitive but R is not symmetric [Q (2, 1) ∉ R] (ii) Let R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (2, 3), (1, 3)} R is reflexive, symmetric and transitive (iii) Let R = {(1, 1), (2, 2), (3, 1), (1, 3)}. So, R is symmetric 10

n

Mathematics–XII

R is not reflexive because (3, 3), (4, 4) ∉ R R is not transitive because (3, 1) and (1, 3) ∈ R but (3, 3) ∉ R (b) R = {(x, y) : x, y ∈ N, 2x + y = 41} = {(x, y) : y = 41 – 2x} y∈N ⇒ 41 – 2x > 0 41 ⇒ x< and x ∈ N 2 ⇒ x = 1, 2, 3,..., 20 ...(i) 1 ≤ x ≤ 20 ⇒ – 2 ≥ – 2x ≥ – 40 ⇒ 41 – 2 ≥ 41 – 2x ≥ 41 – 40 ⇒ 39 ≥ y ≥ 1 ⇒ 1 ≤ y ≤ 39 and y ∈ N and y is odd. ⇒ y = 1, 3, 5,..., 39 ...(ii) From (i) and (ii), we get Domain of R = {1, 2, 3, 4,..., 20} and Range of R = {1, 3, 5, 7, 9,..., 39} Now (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6),..., (13, 13) ∉ R ∴ R is not reflexive (1, 39) ∈ R but (39, 1) ∉ R ⇒ R is not symmetric (11, 19) ∈ R and (19, 3) ∈ R but (11, 3) ∉ R ∴ R is not transitive. Hence R is neither reflexive, nor symmetric and nor transitive.

NCERT TEXTBOOK QUESTIONS SOLVED EXERCISE 1.1 Q1. Determine whether each of the following relations are reflexive, symmetric and transitive: (i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0} (ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A= {1, 2, 3,4, 5, 6} as R = {(x, y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer} (v) Relation R in the set A of human beings in a town at a particular time given by (a) R = {(x, y} : x and y work at the same place} (b) R = {(x, y} : x and y live in the same locality} (c) R = {(x, y} : x is exactly 7 cm taller than y} (d) R = {(x, y} : x is wife of y} (e) R = {(x, y} : x is father of y} Sol. (i) Here A = {1, 2, 3, ..., 13, 14} R = {(x, y) : 3x – y = 0}

R = {(1, 3), (2, 6), (3, 9), (4, 12)} Now 1 ∈ A but (1, 1) ∉ R. So R is not reflexive relation. (1, 3) ∈ R but (3, 1) ∉ R. So R is not symmetric relation. (1, 3) ∈ R and (3, 9) ∈ R but (1, 9) ∉ R. So R is not transitive relation. Thus R is neither reflexive nor symmetric nor transitive. (ii) Here R = {(x, y) : y = x + 5 and x < 4} R = {(1, 6), (2, 7), (3, 8)} A = {1, 2, 3} Now 1 ∈ A but (1, 1) ∉ R. So R is not reflexive relation. (1, 6) ∈ R but (6, 1) ∉ R. So R is not symmetric relation. In R, (a, b) ∈ R but there is no ordered pair (b, c) ∈ R. So R is not transitive relation. Thus R is neither reflexive nor symmetric nor transitive. (iii) Here A = {1, 2, 3, 4, 5, 6} R = {(x, y) : (x, y) ∈ A, y is divisible by x} Relations and Functions

n 11

A = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)} Now 1, 2, 3, 4, 5, 6 ∈ A ∴ (1, 1), (2, 2) (3, 3), (4, 4), (5, 5), (6, 6) ∈ R So R is reflexive relation. (1, 2) ∈ R but (2, 1) ∉ R So R is a not symmetric relation. (1, 1) ∈ R and (1, 2) ∈ R ⇒ (1, 2) ∈ R. (2, 2) ∈ R and (2, 4) ∈ R ⇒ (2, 4) ∈ R. (3, 3) ∈ R and (3, 6) ∈ R ⇒ (3, 6) ∈ R and so on. So R is transitive relation. Thus R is reflexive and transitive but not symmetric. (iv) Here R = {(x, y) : x – y is an integer, x ∈ Z, y ∈ Z} Now x – x = 0 is an integer ⇒ (x, x) ∈ R. So R is reflexive relation. x – y is an integer than y – x is also an integer. (x, y) ∈ R ⇒ (y, x) ∈ R. So R is a symmetric relation. x – y is an integer, y – z is an integer then x – z is also an integer. ∴ (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R. So R is a transitive relation. Thus R is reflexive, symmetric and transitive. (v) Here A is a set of human beings in a town at a particular time. (a) R = {(x, y) : x and y work at the same place} Now place of work of both x and x is same ∴ (x, x) ∈ R So R is reflexive relation. Place of work of x and y is same as place of work of y and x ∴ (x, y) ∈ R = (y, x) ∈ R So R is symmetric relation. Place of work of x and y and place of work of y and z are same then place of work x and z is same. ∴ (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R So R is a transitive relation. Thus R is reflexive, symmetric and transitive. 12

n

Mathematics–XII

(b) R = {(x, y): x and y live in the same locality} Now x and x both live in the same locality ∴ (x, x) ∈ R So R is reflexive relation. x and y live in the same locality then y and x also live in the same locality ∴ (x, y) ∈ R = (y, x) ∈ R So R is symmetric relation. x and y live in the same locality and y and z live in the same locality then x and z live in the same locality. ∴ (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R So R is a transitive relation. Thus R is reflexive, symmetric and transitive. (c) R = {(x, y): x is exactly 7 cm taller than y} Now x cannot be 7 cm taller than x. So R is not reflexive relation. If x is exactly 7 cm taller than y, then y can not be taller than x. ∴ (x, y) ∈ R ⇒ (y, x) ∉ R. So R is not symmetric relation. x is exactly 7 cm taller than y and y is exactly 7 cm taller than z then x cannot be 7 cm taller than z. So R is not transitive relation. Thus R is neither reflexive nor transitive nor symmetric. (d) R = {(x, y): x is wife of y} Now x cannot be wife of x. So R is not reflexive relation. x is wife of y then y is not wife of x. So R is not symmetric relation. If x is wife of y but y cannot be wife of anybody. So R is not transitive relation. Thus R is neither reflexive nor symmetric nor transitive. (e) R = {(x, y); x is a father of y} Now x cannot be father of himself. So R is not reflexive relation. x is father of y than y cannot be father of x. So R is not symmetric relation. x is a father of y and y is a father of z then x cannot be father of z.

So R is not transitive relation. Thus R is neither reflexive nor symmetric nor transitive. Q2. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b2} is neither reflexive nor symmetric nor transitive. Sol. Here R = {(a, b) : a ≤ b2} 1 1 Now ∈ R but is not less than equal 2 2 to  1   2

2

1 1 ∴  , ∉ R  2 2 So R is not a reflexive relation. Let (2, 9) ∈ R then (9, 2) ∉ R because 9 is not less than or equal to (2)2. So R is not symmetric relation.  1 , − 4 1 2 2  ∈ R because 2 ≤ (– 4)  −4, − 1  ∈ R because − 4 ≤  − 1    2 2

2

1 1 but  ,−  ∉ R 2 2  So R is not transitive relation. Thus R is neither reflexive nor symmetric nor transitive. Q3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive. Sol. Here A = {1, 2, 3, 4, 5, 6} R = {(a, b): b = a + 1} = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} Now 1, 2, 3, 4, 5, 6 ∈ A but (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R So R is not reflexive relation. (1, 2) ∈ R but (2, 1) ∉ R So R is not symmetric relation. (3, 4) ∈ R and (4, 5) ∈ R but (3, 5) ∉ R So R is not transitive relation. Thus R is neither reflexive nor symmetric nor transitive. Q4. Show that the relation R in R defined as R = {(a, b) : a ≤ b} is reflexive and transitive but not symmetric. [CBSE 2019] Sol. Here R = {(a, b) : a ≤ b} Let b = a then (a, b) ∈ R because a ≤ b. So R is reflexive relation.

Let (a, b) ∈ R ⇒ a ≤ b ⇒ (b, a) ∉ R Since both a ≤ b and b ≤ a Can’t hold true together. So R is not symmetric relation. Let (a, b) ∈ R and (b, c) ∈ R ⇒ a ≤ b and b ≤ c ⇒ a ≤ c ⇒ (a, c) ∈ R. So R is transitive relation. Thus R is reflexive and transitive but not symmetric. Q5. Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is reflexive, symmetric or transitive. Sol. Here R = {(a, b) : a ≤ b3} 1 1 1 1 Let a = then  ,  ∉ R because  2 2 2 2 1 is not less than equal to    2

2

So R is not reflexive relation. Let (1, 2) ∈ R then (2, 1) ∉ R because 2 is not less than equal to (1)3. So R is not symmetric relation. Let (9, 3) ∈ R then (3, 2) ∈ R then (9, 2) ∈ R because 9 is not less than equal to (2)3. So R is not transitive relation. Thus R is neither reflexive nor symmetric nor transitive. Q6. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive. Sol. Let A = {1, 2, 3} R = {(1, 2), (2, 1)} Now 1 ∈ A but (1, 1) ∉ R. So R is not reflexive relation. (1, 2) ∈ R and (2, 1) ∈ R. So R is symmetric relation. (1, 2) ∈ R and (2, 1) ∉ R but (1, 1) ∉ R. So R is not transitive relation. Thus R is symmetric but neither reflexive nor transitive. Q7. Show that the relation R in the set A of all the books in library of a college given by R {(x, y) : x and y have same number of pages} is an equivalence relation. Sol. Here R = {(x, y) : x and y have same number of pages} Now obviously, (x, x) ∈ R Relations and Functions

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So R is reflexive relation. (x, y) ∈ R ⇒ (y, x) ∈ R because number of pages in both the boxes is same So R is symmetric relation. (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R because number of pages in book x and z is same So R is transitive relation. So R is reflexive, symmetric and transitive. Thus R is an equivalence relation. Q8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even} is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}. Sol. Here A = {1, 2, 3, 4, 5} R = {(a, b) : |a – b| is even} = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5)} Now 1, 2, 3, 4, 5 ∈ A ⇒ (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) ∈ R Also |a − a| = 0, which is even ⇒ (a, a) ∈ R So R is reflexive relation. If |a − b| is even, |b − a| is also even. So (a, b) ∈ R ⇒ (b, a) ∈ R So R is symmetric relation. If (a, b) ∈ R and (b, c) ∈ R ⇒ |a − b| is even and |b − c| is even. Say a − b = 2k1 and b − c = 2k2 Where k1 and k2 are integers. Then a − c = (a − b) + (b − c) = 2k1 + 2k2 = 2(k1 + k2) ⇒ |a − c| is even ⇒ (a, c) ∈ R So R is transitive relation. Now the elements of {1, 3, 5} are related to each other because |1 – 3| = 2; |3 – 5| = 2, |1 – 5| = 4 All numbers are even numbers. The elements of {2, 4} are related to each other because |2 – 4| = 2, which is even number. No element of set {1, 3, 5} is related to any element of {2, 4} because |1 – 2| = 1; |3 – 2| = 1; |5 – 2| = 3; 14

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|1 – 4| = 3; |3 – 4| = 1; |5 – 4| = 1, which are not even numbers. Q9. Show that each of the relations R in the set A = {x ∈ Z; 0 ≤ x ≤ 12} given by (i) R = {(a, b) : |a – b| is a multiple of 4} (ii) R = {(a, b) : a = b} is an equivalence relation. Find the set of all elements related to 1 in each case. Sol. Here A = {x ∈ Z; 0 ≤ x ≤ 12} = {0, 1, 2, 3, ..., 12} (i) R = {(a, b) : |a – b| is a multiple of 4} = {(0, 0), (0, 4), (0, 8), (0, 12), (1, 1), (1, 5), (1, 9), (2, 2), (2, 6), (2, 10), (3, 3),(3, 7), (3, 11), (4, 4), (4, 8), (4, 12), (5, 5), (5, 9), (6, 6), (6, 10), (7, 7), (7, 11), (8, 8), (8, 12), (9, 9), (10, 10), (11, 11), (12, 12)} Now 0, 1, 2, 3, ..., 12 ∈ A then (0, 0), (1, 1), (2, 2), ...., (12, 12) ∈ R So R is a reflexive relation. (1, 5) ∈ R then (5, 1) ∈ R because |5 – 1| = |1 − 5|. So R is symmetric relation. (2, 6) ∈ R and (6, 10) ∈ R then (2, 10) ∈ R because |2 – 10| = 8 which is multiple of 4. So R is transitive relation. Thus R is an equivalence relation. The set of elements related to 1 is {1, 5, 9}. (ii) R = {(a, b) : a = b} = {(0, 0), (1, 1), (2, 2), ..., (12, 12)} Now 0, 1, 2, 3, ...., 12 ∈ A then (0, 0), (1, 1), (2, 2), ..., (12, 12) ∈ R So R is a reflexive relation. (2, 2) ∈ R then (2, 2) ∈ R So R is symmetric relation. (5, 5) ∈ R and (5, 5) ∈ R then (5, 5) ∈ R So R is transitive relation. Thus R is an equivalence relation. The set of elements related to 1 is {1}. Q10. Give an example of a relation. which is (i) Symmetric but neither reflexive nor transitive (ii) Transitive but neither reflexive nor symmetric (iii) Reflexive and symmetric but not transitive (iv) Reflexive and transitive but not symmetric (v) Symmetric and transitive but not reflexive.

Sol. (i) R = {(1, 2), (2, 1)} R is not reflexive because (1, 1) ∉ R. R is symmetric because (1, 2) ∈ R ⇒ (2, 1) ∈ R. R is not transitive because (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∉ R (ii) R = {(1, 2), (2, 3), (1, 3)} R is not reflexive because (2, 2) ∉ R, (1, 1) ∉ R, and (3, 3) ∉ R. R is not symmetric because (1, 2) ∈ R but (2, 1) ∉ R R is transitive because (1, 2) ∈ R and (2, 3) ∈ R ⇒ (1, 3) ∈ R (iii) R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}. R is reflexive because (1, 1), (2, 2), (3, 3) ∈ R. R is symmetric because (1, 2) ∈ R ⇒ (2, 1) ∈ R. R is not transitive because (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R. (iv) R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 2), (2, 3), (1, 3)}. R is reflexive because (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) ∈ R. R is not symmetric because (1, 2) ∈ R but (2, 1) ∉ R. R is transitive because (1, 2) ∈ R and (2, 3) ∈ R ⇒ (1, 3) ∈ R. (v) R = {(2, 3), (3, 2), (1, 1), (1, 3), (3, 1)}. R is not reflexive because (3, 3) ∉ R. R is symmetric because (2, 3) ∈ R ⇒ (3, 2) ∈ R. R is transitive because (1, 3) ∈ R and (3, 1) ∈ R ⇒ (1, 1) ∈ R. Q11. Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre. Sol. Let O be the origin R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin} = {(P, Q) : OP = OQ} Now let OP = x then (x, x) ∈ R since OP = OP.

So R is reflexive relation. Let OP = OQ = x then (x, x) ∈ R ⇒ (x, x) ∈ R. So R is symmetric relation. Let OP = OQ = x and OQ = OR = x then (x, x) ∈ R and (x, x) ∈ R ⇒ (x, x) ∈ R. So R is transitive relation. Thus R is an equivalence relation. The distance of all points related to P, from the origin is same as OP. Since a circle is the locus of all points equidistant from a fixed point (here 0), the set of the points related to P is a circle passing through P with O as the fixed point (centre). Q12. Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1 is similar to T2} is equivalence relation. Consider three right angled triangles T1 with sides 3, 4, 5; T2 with sides 5, 12, 13; and T3 with sides 6, 8, 10. Which triangles among T1, T2 and T3 are related? Sol. Here R = {(T1, T2) : T1 is similar to T2} = {(T1, T2) : T1 ∼ T2} where T1, T2 ∈ A We know that every triangle is similar to itself ⇒ (T1, T1) ∈ R for all T1 ∈ A So R is reflexive relation. (T1, T2) ∈ R where T1, T2 ∈ A ⇒ T1 ∼ T2 then T2 ∼ T1 ⇒ (T2, T1) ∈ R So R is symmetric relation. (T1, T2) ∈ R and (T2, T3) ∈ R where T1, T2, T3 ∈ A ⇒ T1 ∼ T2 and T2 ∼ T3 ⇒ T1 ∼ T3 ⇒ (T1 , T3) ∈ R So R is transitive relation. Thus R is an equivalence relation We know that two triangles are similar if their sides are proportional. We see that sides of triangle T1 are proportional to the sides of triangle T3. Thus T1 is realted to T3. Q13. Show that the relation R defined in the set A of all polygons as R = {(P1, P2) : P 1 and P 2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angled triangle T with sides 3, 4 and 5? Sol. Here R = {(P1, P2) : P1 and P2 have same number of sides} where P1, P2 ∈ A Relations and Functions

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