12

PART 1 & 2

100%

NCERT SOLUTIONS

MATHEMATICS O.P. Chugh N.P. Bali

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SYLLABUS 

Marks: 100 I. II. III. IV. V. VI.

Units No. of Periods Marks Relations and Functions 30 10 Algebra 50 13 Calculus 80 44 Vectors and Three - Dimensional Geometry 30 17 Linear Programming 20 06 Probability 30 10 Total 240 100

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. Binary operations. 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 × 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

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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 reallife 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 , ∫



dx 2

x ±a

2

px + q



,∫

dx 2

a −x

∫ ax 2 + bx + c dx, ∫

2

,∫

dx dx ,∫ 2 ax + bx + c ax + bx + c 2

px + q 2

ax + bx + c

dx, ∫ a 2 ± x 2 dx,∫ x 2 − a 2 dx

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). 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

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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. Repeated independent (Bernoulli) trials and Binomial distribution.

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QUESTION PAPER DESIGN Time : 3 Hours S.  Typology of Questions No.

1.

2.

3.

4.

5.

REMEMBERING (Knowledge based— Simple recall questions, to know specific facts, terms, concepts, principles, or theories; Identify, define, or recite, information) UNDERSTANDING (Comprehension—To be familiar with meaning and to understand conceptually, interpret, compare, contrast, explain, paraphrase, information) APPLICATION (Use abstract information in concrete situation, to apply knowledge to new situations; Use given content to interpret a situation, provide an example, or solve a problem) HIGH ORDER THINKING SKILLS (Analysis & Synthesis—Classify, compare, contrast or differentiate between different pieces of information; Organize and/or integrate unique pieces of information from a variety of sources) EVALUATION (Appraise, judge, and/ or justify the value or worth of a decision or outcome, or to predict outcomes based on values) TOTAL

Max. Marks : 100 Very Short Long Long Short Answer Answer-I Answer-II Answer (SA) (LA-I) (LA-II) Total Weightage (VSA) Marks 2 1 Mark 4 Marks 6 Marks Marks

2

2

2

1

20

20%

1

3

4

2

35

35%

1

–

3

2

25

25%

–

3

1

–

10

10%

–

–

1

1

10

10%

6×6=36

100

100%

1×4=4 2×8=16 4×11=44

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CONTENTS 1. Relations and Functions................................................................ 9 2. Inverse Trigonometric Functions.................................................. 63 3. Matrices.......................................................................................... 85 4. Determinants.................................................................................. 144 5. Continuity and Differentiability.................................................... 221 6. Application of Derivatives............................................................ 307 7. Integrals.......................................................................................... 410 8. Applications of Integrals............................................................... 549 9. Differential Equations.................................................................... 595 10. Vector Algebra............................................................................... 669 11. Three Dimensional Geometry....................................................... 710 12. Linear Programming...................................................................... 760 13. Probability...................................................................................... 817

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QUESTION-WISE BREAK-UP Type of Question VSA SA LA-I LA-II Total

Marks per question 1 2 4 6

Total No. of questions 4 8 11 6 29

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Total Marks 4 16 44 36 100

1

Relations and Functions

Lesson at a Glance 1. A Relation R from a non-empty set A to a non-empty set B is a subset of the cartesian product A × B. i.e., every subset of A × B is a relation from A to B. Thus, (a, b) ∈ R ⇒ (a, b) ∈ A × B. 2. Empty relation or void relation in a set. A relation R in a set A is called empty relation or void relation if R = φ ⊂ A × A, i.e., if no element of A is related to any element of A. Void relation φ is the smallest relation on the set A. 3. (a) Identity relation in a Set A. A relation R in a set A is called Identity relation if (a, a) ∈ R for all a ∈ A and it is denoted by IA. (b) Universal relation in a set. A relation R in a set A is called universal relation if R = A × A, i.e., if each element of A is related to every element of A. 4. Reflexive relation in a set. A relation R in a set A is called reflexive if (a, a) ∈ R for every a ∈ A, i.e., every element of A is Rrelated to itself or aRa for all a ∈ A. 5. Symmetric relation in a set. A relation R in a set A is called symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A i.e., aRb

⇒ bRa for all a, b ∈ A. –1

6. A relation R in a set A is symmetric iff R = R . 7. Transitive relation in a set. A relation R in a set A is called transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A. 8. Equivalence relation in a set. A relation R in a set A is called an equivalence relation if R is reflexive, symmetric and transitive. 9. Definition of function. A relation f from a set A to a set B is a function if every element of A has one and only one image in the set B. Therefore every relation need not be a function, but every function is a relation. 10. A function f from set A to set B is denoted by f : A → B. Set A is called domain of the function and set B is called its co-domain. If B ⊂ R, then f is called a real valued function. The set f(A) = {f(x) : x ∈ A} is called Range of the function and range set f(A) is a subset of co-domain set B.

10

11.

12.

13.

14.

15.

16.

17.

MATHEMATICS–XII Types of functions (i) One-one (or Injective) Function: A function f : X → Y is said to be one-one function if distinct elements have distinct images. i.e. for every pair x1, x2 ∈ domain X s.t. f (x1) = f (x2) ⇒ x1 = x2 (ii) Onto (or Surjective) Function: A function f : X → Y is said to be an onto function if Range set f (X) = Co-domain Y. i.e. if every y ∈ co-domain Y, there exists. x ∈ domain X s.t. y = f (x). (iii) Bijective Function: A function f : X → Y is said to be bijective if f is both one-one and onto. (iv) Identity Function: A function f : A → A is said to be an identity function if f (x) = x V x ∈ A. Composition of Two Functions. Let f : A → B and g : B → C be two functions. Then the composition of f and g (or composite of f and g) denoted by gof, is defined as the function gof : A → C given by (gof)(x) = g(f (x)), for all x ∈ A. Invertible function. A function f : X → Y is said to be invertible if there exists a function g : Y → X such that gof : X → X is identity fuction = Ix and fog : Y → Y is identity fuction = IY –1 The function g is called inverse of f and is denoted by f The function f is invertible (i.e. has an inverse) iff f is one-one and onto. Binary Operation on a set. Let A be a non-empty set. A binary operation, usually denoted by *, on A is a function from A × A into A i.e., * : A × A → A. The unique element of the set A associated by * with the ordered pair (a, b) ∈ A × A is denoted by a * b. Commutative Binary Operation on a set. A binary operation * on a set A is said to be Commutative if a * b = b * a for all a, b ∈ A. Associative Binary Operation on a set. A binary operation * on a set A is said to be associative if (a * b) * c = a * (b * c), for all a, b, c ∈ A. Identity element of a binary operation. An element e ∈ A is said to be the identity element for the binary operation * on set A if a * e = a = e * a for all a ∈ A. Inverse of an element w.r.t. a binary operation. An element b ∈ A is said to be the inverse of an element a ∈ A w.r.t. a binary operation * if a * b = e = b * a.

RELATIONS

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TEXTBOOK QUESTIONS SOLVED Exercise 1.1 (Page No. 5-7) 1. 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} (Important) (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) Given: Set A = {1, 2, 3, ..., 13, 14} Relation R in the set A is defined as R = {(x, y) : 3x – y = 0} i.e., R = {(x, y) : – y = – 3x i.e., y = 3x} ...(i) Is R reflexive? Let x ∈ A. Putting y = x in (i), x = 3x Dividing by x ≠ 0 (x ∈ A here is ≠ 0), 1 = 3 which is impossible. Therefore (x, x) ∉ R and hence R is not reflexive. Is R symmetric? Let (x, y) ∈ R, therefore by (i) y = 3x Interchanging x and y in (i), x = 3y which is not true y and ≠ 3y) 3 For example, 2 ∈ A, 6 ∈ A and 6 = 3(2) (i.e., y = 3x) but 2 ≠ 3(6) = 18. ∴ (2, 6) ∈ R but (6, 2) ∉ R ∴ R is not symmetric. Is R transitive? Let (x, y) ∈ R and (y, z) ∈ R. Therefore by (i), y = 3x and z = 3y. To eliminate y, Putting y = 3x in z = 3y, z = 3(3x) = 9x ∴ By (i), (x, z) ∉ R. ∴ R is not transitive. (ii) Given: Relation R in the set of natural numbers defined as (... y = 3x ⇒ x =

12

MATHEMATICS–XII

R = {(x, y) : y = x + 5 and x < 4} i.e., R = {(x, x + 5) and x = 1, 2, 3} ∴ In Roster form; R = {(1, 1 + 5), (2, 2 + 5), (3, 3 + 5)} i.e., R = {(1, 6), (2, 7), (3, 8)} Domain of R is the set of x co-ordinates of R. i.e., the set {1, 2, 3}. R is not reflexive because (1, 1) ∉ R, (2, 2) ∉ R, (3, 3) ∉ R R is not symmetric because (1, 6) ∈ R but (6, 1) ∉ R. R is transitive because there are no two pairs of the type (x, y) and (y, z) ∈ R = {(1, 6), (2, 7), (3, 8)}; so we should have no reason to expect (x, z) ∈ R. Remark 1. Please note carefully and learn that relation R in the above question is transitive. Remark 2. Whenever set A is a small finite set, it is always better to write R in roster form. (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} defined as R = {(x, y) : y is divisible by x}. ∴ In roster form; R = {(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)} R is reflexive because (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) all ∈ R, i.e., (x, x) ∈ R for all x ∈ A. R is not symmetric because (2, 4) ∈ R but (4, 2) ∉ R as 2 is not divisible by 4. Is R transitive? Let x, y, z ∈ A such that (x, y) ∈ R and (y, z) ∈ R. ∴ By definition of R in this question, y is divisible by x and z is divisible by y. ∴ There exist natural numbers m and n such that y = mx and z = ny. To eliminate y: Putting y = mx in z = ny, we have z = n(mx) = (nm)x ∴ z is a multiple of, x, i.e., z is divisible by x. ∴ (x, z) ∈ R. ∴ R is transitive. (iv) Given: Relation R in the set Z of all integers defined as R = {(x, y) : (x – y) is an integer} ...(i) Is R reflexive? Let x ∈ Z. Putting y = x in (i), x – x = 0 is an integer which is true. ∴ (x, x) ∈ R for all x ∈ Z ∴ R is reflexive. Is R symmetric? Let x ∈ Z, y ∈ Z, and (x, y) ∈ R. ∴ By (i) (x – y) is an integer. i.e., – (y – x) is an integer and hence y – x is an integer. ∴ (y, x) ∈ R. ∴ R is symmetric. Is R transitive? Let x ∈ Z, y ∈ Z and z ∈ Z such that (x, y) ∈ R and (y, z) ∈ R. ⇒ By (i), x – y is an integer and y – z is also an integer. Adding to eliminate y, x – y + y – z = Integer + Integer i.e., x – z is an integer. ⇒ (x, z) ∈ R by (i). ∴ R is transitive.

RELATIONS

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FUNCTIONS

13

(v) R is a relation in the set A of human beings of a town. (a) Given: R = {(x, y) : x and y work at the same place} ...(i) Is R reflexive? Let x ∈ A. Putting y = x in (i), x and x work at the same place which is true (... x and x is just one person x) Is R symmetric? Let x ∈ A, y ∈ A and (x, y) ∈ R. ∴ By (i), x and y work at the same place. i.e., same thing as y and x work at the same place. Therefore by (i), (y, x) ∈ R. Is R transitive? Let x ∈ A, y ∈ A, z ∈ A such that (x, y) ∈ R and (y, z) ∈ R. ∴ By (i), x and y work at the same place. Also By (i), y and z work at the same place. Therefore, we can say that x and z also work at the same place. ∴ By (i), (x, z) ∈ R. Therefore, R is transitive. (b) Same solution as of part (a) (Replace the word “work” by “live” and “place” by “locality” in the solution of part (a)). (c) R = {(x, y) : x is exactly 7 cm taller than y) ...(i) Is R reflexive? Let x ∈ A. Putting y = x in (i), we have x is exactly 7 cm taller than x, which is false. (... No body can be taller than oneself.) Is R symmetric? Let x ∈ A, y ∈ A and (x, y) ∈ R. Therefore, by (i), x is exactly 7 cm taller than y. ∴ y is exactly 7 cm shorter than x. ∴ By (i), (y, x) ∉ R. ∴ R is not symmetric. Is R transitive? Let x ∈ A, y ∈ A, z ∈ A such that (x, y) ∈ R and (y, z) ∈ R. ∴ By (i), x is exactly 7 cm taller than y and y is exactly 7 cm taller than z. ∴ x is exactly (7 + 7) = 14 cm (and not 7 cm) taller than z. ∴ By (i), (x, z) ∉ R. ∴ R is not transitive. (d) Given: R = {(x, y) : x is wife of y} ...(i) Is R reflexive? Let x ∈ A. Putting y = x in (i), we have x is wife of x which is false. (... No lady can be wife of herself.) ∴ R is not reflexive. Is R symmetric? Let x ∈ A, y ∈ A and (x, y) ∈ R. ∴ By (i), x is wife of y. Hence y is husband (and not wife) of x. ∴ (y, x) ∉ R. ∴ R is not symmetric.

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MATHEMATICS–XII

Is R transitive? There can’t be any three x, y, z ∈ A (Set of Human beings) such that both (x, y) ∈ R and (y, z) ∈ R. (... (x, y) ∈ R ⇒ x is wife of y (⇒ Husband i.e., Man) and hence (y, z) will never belong to R as no man y can be wife of any human being z). Hence we no reason to expect that (x, z) ∈ R. ∴ R is transitive. (e) Given: R = {(x, y) : x is father of y} ...(i) Is R reflexive? Let x ∈ A. Putting y = x in (i), we have x is father of x which is false. (... No body can be father of oneself) Is R symmetric? Let x ∈ A, y ∈ A such that (x, y) ∈ R. ∴ By (i), x is father of y. Hence y is son or daughter (and not father) of x. ∴ (y, x) ∉ R. ∴ R is not symmetric. Is R transitive? Let x, y, z ∈ A such that (x, y) ∈ R and (y, z) ∈ R. ∴ By (i), x is father of y and y is father of z. Hence x is grandfather (and not father) of z. ∴ (x, z) ∉ R. ∴ R is not transitive. 2. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b 2 } is neither reflexive nor symmetric nor transitive. Sol. Relation R = {(a, b) : a, b are real and a ≤ b2} ...(i) Is R reflexive? Let a be any real number. Putting b = a in (i), a ≤ a2 which is not true for any positive real number less than 1. 2

1 1 1 1 For example, for a = , ≤   = (i.e., a ≤ a2) is not true 2 2 4  2 1 1 as > . ∴ R is not reflexive. 2 4 Is R symmetric? Let us take a = 1 and b = 2. Now a = 1 ≤ 22 = 4 (= b2) is true. ∴ By (i), (a, b) ∈ R.1414 But b = 2 > 1 2 = 1 (a 2 ) i.e., b is not less than equal to a 2 , therefore (b, a) ∉ R. ∴ R is not symmetric. Is R transitive? Let us take a = 10, b = 4, and c = 2 (All three are real numbers) Now by (i), (a, b) = (10, 4) ∈ R (... a = 10 ≤ b2 (= 42) is true) Again by (i), (b, c) = (4, 2) ∈ R (... b = 4 ≤ c2 (= 22) is true) (4 ≤ 22 ⇒ 4 ≤ 4 ⇒ Either 4 < 4 or 4 = 4 But 4 = 4 is true). But (a, c) = (10, 2) ∉ R [... a = 10 > 22 (= b2)] ∴ R is not transitive. ∴ R is neither reflexive, nor symmetric nor transitive. Remark. It may be noted that 4 ≤ 4 is true. Also 4 ≥ 4 is true. 8 ≥ 5 is true and 3 ≤ 6 is true.

RELATIONS

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15

3. 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. Given: Set {1, 2, 3, 4, 5, 6} = set A (say) Relation R = {(a, b) : b = a + 1} = {(a, a + 1) : a ∈ A} ...(i) Putting a = 1, 2, 3, 4, 5, 6 (given) in (i). Roster form of relation R is {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)} ...(ii) R is not reflexive. [... By (ii), (1, 1) ∉ R, (2, 2) ∉ R] (... We know that for relation R to be reflexive, (a, a) ∈ R for all a ∈ A) R is not symmetric because by (ii), (1, 2) ∈ R but (2, 1) ∉ R. R is not transitive because (x, y) = (1, 2) ∈ R, (y, z) = (2, 3) ∈ R but (x, z) = (1, 3) ∉ R. 4. Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric. Sol. Given: Relation R in the set of real numbers is defined as R = {(a, b) : a ≤ b} ...(i) Is R reflexive? Putting b = a in (i), we have a ≤ a which is true. [... a ≤ a ⇒ Either a < a or a = a and out of the two a = a is true) R is not symmetric because by (i), (1, 2) ∈ R as a = 1 ≤ b (= 2) but (2, 1) ∉ R because b (= 2) > 1(= a) (i.e., b ≤ a is not true). Is R transitive? Let a, b, c be three real numbers such that (a, b) ∈ R and (b, c) ∈ R. ∴ By (i), a ≤ b and b ≤ c. Therefore a ≤ c and hence by (i), (a, c) ∈ R. ∴ R is transitive. 5. Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is reflexive, symmetric or transitive. Sol. Relation R = {(a, b) : a, b are real and a ≤ b3} ...(i) Is R reflexive? Let a be any real number. Putting b = a in (i), a ≤ a3 which is not true for any positive real number less than 1. For example, for a =

1 1 1 , ≤   2 2  2

3

=

1 (i.e., a ≤ a3) is not true 8

1 1 > . ∴ R is not reflexive. 2 8 Is R symmetric? Let us take a = 1 and b = 2. Now a = 1 ≤ 23 = 8 (= b3) is true. Therefore by (i), (a, b) ≤ R. Now, b = 2 > 13 (= 1) (a3) i.e., b is not less than or equal to a3. Therefore, (b, a) ∉ R. ∴ R is not symmetric. Is R transitive? Let us take a = 10, b = 4, c = 2 (All three are real numbers) Now by (i), (a, b) = (10, 4) ∈ R (... a = 10 ≤ b3 (= 43 = 64 is true) Again by (i), (b, c) = (4, 2) ∈ R (... b = 4 ≤ c3 (= 23 = 8) is true. But (a, c) = (10, 2) ∉ R (... a = 10 > 23 (= c3 = 8)). as

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