20 Years Chapterwise Topicwise 2021-2002 JEE Main Solved Papers PDF

SOLVED PAPERS Mathematics SOLVED PAPERS Mathematics Complete Collection of all Questions asked in ONLINE & OFFLINE JEE Main & AIEEE ARIHANT PRAKASHAN (Series), MEERUT ARIHANT PRAKASHAN (Series), MEERUT All Rights Reserved © Publisher Administrative & Production Offices Regd....

SOLVED PAPERS Mathematics SOLVED PAPERS Mathematics Complete Collection of all Questions asked in ONLINE & OFFLINE JEE Main & AIEEE ARIHANT PRAKASHAN (Series), MEERUT ARIHANT PRAKASHAN (Series), MEERUT All Rights Reserved © Publisher Administrative & Production Offices Regd. Office ‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002 Tele: 011- 47630600, 43518550 Head Office Kalindi, TP Nagar, Meerut (UP) - 250002 Tel: 0121-7156203, 7156204 Sales & Support Offices Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Nagpur & Pune. ISBN 978-93-25796-25-6 PO No : TXT-XX-XXXXXXX-X-XX Published by Arihant Publications (India) Ltd. For further information about the books published by Arihant, log on to www.arihantbooks.com or e-mail at [email protected] Follow us on PREFACE JEE Main is a gateway examination for candidates expecting to seek admission in Bachelor in Engineering (BE), Bachelor of Technology (B.Tech) and Bachelor of Architecture (B.Arch) at Indian Institutes of Information Technology (IIITs), National Institutes of Technology (NITs), Delhi Technological University and other Centrally Funded Technical Institutes (CFTIs). JEE Main is also an examination which is like screening examination for JEE Advanced (The gateway examination to India's most reputed Technical Institutes, Indian Institutes of Technology IITs). To make the students well-versed with the pattern as well as the level of the questions asked in the exam, this book contains Chapterwise Topicwise Solutions of Questions asked in Last 20 Years’ Examinations of JEE Main (formerly known as AIEEE). Solutions to all the questions have been kept very detailed and accurate. Along with the indication of level of the exam, this book will also teach you to solve the questions objectively in the examination. We hope this book would be highly beneficial for the students. We would be grateful if any discrepancy or mistake in the questions or answers is brought to our notice so that these could be rectified in subsequent editions. Publisher CONTENTS 1. Sets, Relations and Functions 1-18 2. Complex Numbers and Quadratic Equations 19-49 3. Matrices and Determinants 50-87 4. Permutations and Combinations 88-98 5. Mathematical Induction 99-100 6. Binomial Theorem and Its Simple Applications 101-118 7. Sequences and Series 119-141 8. Limits, Continuity and Differentiability 142-175 9. Applications of Derivatives 176-198 10. Integral Calculus 199-248 11. Differential Equations 249-273 12. Coordinate Geometry 274-342 13. Three-Dimensional Geometry 343-373 14. Vector Algebra 374-393 15. Statistics and Probability 394-424 16. Trigonometry 425-456 17. Mathematical Reasoning 457-467  SYLLABUS UNIT 1 Sets, Relations and Functions Sets and their representation; Union, intersection and complement of sets and their algebraic properties; Power set; Relation, Types of relations, equivalence relations, functions;. one-one, into and onto functions, composition of functions. UNIT 2 Complex Numbers and Quadratic Equations Complex numbers as ordered pairs of reals, Representation of complex numbers in the form a+ib and their representation in a plane, Argand diagram, algebra of complex numbers, modulus and argument (or amplitude) of a complex number, square root of a complex number, triangle inequality, Quadratic equations in real and complex number system and their solutions. Relation between roots and co-efficients, nature of roots, formation of quadratic equations with given roots. UNIT 3 Matrices and Determinants Matrices, algebra of matrices, types of matrices, determinants and matrices of order two and three. Properties of determinants, evaluation of deter-minants, area of triangles using determinants. Adjoint and evaluation of inverse of a square matrix using determinants and elementary transformations, Test of consistency and solution of simultaneous linear equations in two or three variables using determinants and matrices. UNIT 4 Permutations and Combinations Fundamental principle of counting, permutation as an arrangement and combination as selection, Meaning of P (n,r) and C (n,r), simple applications. UNIT 5 Mathematical Induction Principle of Mathematical Induction and its simple applications. UNIT 6 Binomial Theorem and its Simple Applications Binomial theorem for a positive integral index, general term and middle term, properties of Binomial coefficients and simple applications. UNIT 7 Sequences and Series Arithmetic and Geometric progressions, insertion of arithmetic, geometric means between two given numbers. Relation between A.M. and G.M. Sum upto n terms of special series: ∑ n, ∑ n2, ∑ n3. Arithmetico - Geometric progression. UNIT 8 Limit, Continuity and Differentiability Real valued functions, algebra of functions, polynomials, rational, trigonometric, logarithmic and exponential functions, inverse functions. Graphs of simple functions. Limits, continuity and differentiability. Differentiation of the sum, difference, product and quotient of two functions. Differentiation of trigonometric, inverse trigonometric, logarithmic exponential, composite and implicit functions derivatives of order upto two. Rolle's and Lagrange's Mean Value Theorems. Applications of derivatives: Rate of change of quantities, monotonic - increasing and decreasing functions, Maxima and minima of functions of one variable, tangents and normals. UNIT 9 Integral Calculus Integral as an anti - derivative. Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions. Integration by substitution, by parts and by partial fractions. Integration using trigonometric identities. Evaluation of simple integrals of the type dx , dx , dx , dx , x2 ± a2 Öx 2 ± a 2 a2 – x2 Öa 2 – x 2 dx , dx , (px + q) dx , ax 2 + bx + c Öax 2 + bx + c ax 2 + bx + c (px + q) dx , Öax 2 + bx + c Öa 2 ± x 2 dx and Öx 2 – a 2 dx Integral as limit of a sum. Fundamental Theorem of Calculus. Properties of definite integrals. Evaluation of definite integrals, determining areas of the regions bounded by simple curves in standard form. UNIT 10 Differential Equations Ordinary differential equations, their order and degree. Formation of differential equations. Solution of differential equations by the method of separation of variables, solution of homogeneous and linear differential equations of the type dy +p (x) y = q(x) dx UNIT 11 Coordinate Geometry Cartesian system of rectangular coordinates in a plane, distance formula, section formula, locus and its equation, translation of axes, slope of a line, parallel and perpendicular lines, intercepts of a line on the coordinate axes. Ÿ Straight lines Various forms of equations of a line, intersection of lines, angles between two lines, conditions for concurrence of three lines, distance of a point from a line, equations of internal and external bisectors of angles between two lines, coordinates of centroid, orthocentre and circumcentre of a triangle, equation of family of lines passing through the point of intersection of two lines. Ÿ Circles, conic sections Standard form of equation of a circle, general form of the equation of a circle, its radius and centre, equation of a circle when the end points of a diameter are given, points of intersection of a line and a circle with the centre at the origin and condition for a line to be tangent to a circle, equation of the tangent. Sections of cones, equations of conic sections (parabola, ellipse and hyperbola) in standard forms, condition for y = mx + c to be a tangent and point (s) of tangency. UNIT 12 Three Dimensional Geometry Coordinates of a point in space, distance between two points, section formula, direction ratios and direction cosines, angle between two intersecting lines. Skew lines, the shortest distance between them and its equation. Equations of a line and a plane in different forms, intersection of a line and a plane, coplanar lines. UNIT 13 Vector Algebra Vectors and scalars, addition of vectors, components of a vector in two dimensions and three dimensional space, scalar and vector products, scalar and vector triple product. UNIT 14 Statistics and Probability Measures of Dispersion: Calculation of mean, median, mode of grouped and ungrouped data. Calculation of standard deviation, variance and mean deviation for grouped and ungrouped data. Probability: Probability of an event, addition and multiplication theorems of probability, Baye's theorem, probability distribution of a random variate, Bernoulli trials and Binomial distribution. UNIT 15 Trigonometry Trigonometrical identities and equations. Trigonometrical functions. Inverse trigonometrical functions and their properties. Heights and Distances. UNIT 16 Mathematical Reasoning Statements, logical operations and implies, implied by, if and only if. Understanding of tautology, contradiction, converse and contra positive.  EXAM BITES This Pdf Is Downloaded From www.exambites.in Visit www.exambites.in for More Premium Stuffs,Latest Books,Test Papers,Lectures etc. jeeneetadda jeeneetadda_official jeeneetadda VISIT NOW !! Sets, Relations and Functions 1 01 Sets, Relations and Functions TOPIC 1 Ans. (c) n= 97 − 7 + 1 = 16 [Qa n = a + (n − 1)d] Sets Let A = Patient suffering from heart 6 16 ailment and B = Set of patient suffering Hence, sum = [7 + 97] 01 If A = {x ∈ R : | x − 2 | > 1}, from lungs infection 2 Given, n(A) = 89% and n(B) = 98% = 832 QS = n (a + l )  B = {x ∈R : x 2 − 3 > 1} and  n  2   n (A ∪ B) ≥ n (A) + n (B) − n (A ∩ B) C = {x ∈R :| x − 4 | ≥ 2} and Z is the set ⇒ 100 ≥ 89 + 98 − n (A ∩ B ) 04 In a school, there are three types of of all integers, then the number of ⇒ 87 ≤ n(A ∩ B) games to be played. Some of the subsets of the set (A ∩ B ∩ C) C ∩ Z Also, n(A ∩ B) = min { n(A), n(B)} students play two types of games, is [2021, 27 Aug. Shift-I] ⇒ n (A ∩ B) ≤ 89 but none play all the three games. Ans. (256) ∴ 87 ≤ n (A ∩ B) ≤ 89 Which Venn diagram can justify the A = {x ∈ R : | x − 2 | > 1 } So, n (A ∩ B) ∉{ 79, 81, 83, 85}. above statement? ⇒ A = (− ∞, 1) ∪ (3, ∞) [2021, 17 March Shift-I] B = {x ∈ R : x 2 − 3 > 1 } 03 Let A = {n ∈ N|n2 ≤ n + 10000, }, ⇒ B = (− ∞, − 2) ∪ (2, ∞) B = {3k + 1 | k ∈ N} and C = {2k | k ∈ N}, C = { x ∈ R : | x − 4 | ≥ 2} then the sum of all the elements of ⇒ C = (− ∞, 2] ∪ [6, ∞) the set A ∩ (B − C) is equal to …………. [2021, 27 July Shift-II] (P) (Q) (R) ⇒ A ∩ B ∩C = (− ∞, − 2) ∪ [6, ∞) Ans. (832) (a) P and Q ⇒ (A ∩ B ∩C) C = [−26 ,) (b) P and R ∴ (A ∩ B ∩ C) C ∩ Z = {− 2, − 1, 0, 1, 2, 3, 4, 5} Let A = { n ∈N | n2 ≤ n + 10000 } (c) None of these Number of subsets of (A ∩ B ∩ C) C ∩ Z n2 ≤ n + 10000 = 28 = 256 (d) Q and R n2 − n ≤ 10000 ⇒ n(n − 1) ≤ 100 × 100 Ans. (c) 02 Out of all the patients in a hospital C C C ⇒ A = { 1, 2, 3, ……, 100 } A B A B 89% are found to be suffering from Now, B = {3k + 1 | k ∈N } A B heart ailment and 98% are B = {4, 7, 10, 13, … } suffering from lungs infection. If and C = {2k | k ∈N } K% of them are suffering from both C = {2, 4, 6, 8, … } ailments, then K can not belong to The shaded region of The shaded region The shaded region So, B − C = { 7, 13, 19, ……, 97, … } this Venn diagram of this Venn diagram of this Venn diagram the set [2021, 26 Aug. Shift-I] represents the represent the students represent the students who play all three So, A ∩ (B − C) = { 7, 13, 19, ……, 97 } students who play all who play all three (a) {80, 83, 86, 89} three types of games. type of games. type of games. (b) {84, 86, 88, 90} This form an AP with common difference As now, play all three games, the (c) {79, 81, 83, 85} (d = 6) intersection of all three circles must be (d) {84, 87, 90, 93} ⇒ 97 = 7 + (n − 1)6 zero. 2 JEE Main Chapterwise Topicwise Mathematics 05 Let A = {n ∈ N : n is a 3-digit number} 07 A survey shows that 63% of the people On comparing n = 4 and m − n = 3 B = {9k + 2 : k ∈ N} in a city read newspaper A whereas ∴ m= 7 So, m⋅ n = 28 and C = {9k + l : k ∈ N} for some 76% read news- paper (b) If x%of the l (0 < l < 9) people read both the newspapers, 10 Let X = {n ∈N : 1≤ n ≤ 50}. If If the sum of all the elements of then a possible value of x can be A = {n ∈X : n is multiple of 2} and the set A ∩ (B ∪ C) is 274 × 400, (a) 55 (b) 29 (c) 65 (d) 37 B = {n ∈X : n is a multiple of 7}, then [2020, 4 Sep. Shift-I] then l is equal the number of elements is the Ans. (a) to............ [2021, 24 Feb. Shift-I] smallest subset of X containing Let n(A) = Number of people read Ans. (5) newspaper both A and B is …………. Given, A = { n ∈N : n is a 3-digit number} A = 63% [2020, 7 Jan. Shift-II] B = {9k + 2 : k ∈N } n(B) = Number of people read Ans. (29) C = {9k + l : k ∈N } newspaper Given X = { n ∈N : 1 ≤ n ≤ 50 }, then B = 76% A = { n ∈ X : n is multiple of 2} Q3 digit number of the form3k + 2 are and n(A ∩ B) = Number of people read = {2, 4, 6, 8,......, 50 } {101, 109, … 992} both = x% 100 100 × 1093 and B = { n ∈ X : n = is multiple of 7 ⇒ Sum = [101 + 992] = Q n(A ∪ B) = n(A) + n(B) − n(A ∩ B) 2 2 {7, 14, 21, 28, 35, 42, 49} Q 14, 28, 42∈A and 76 ≤ n (A ∪ B) ≤ 100 Similarly, 3-digit number of the form ∴ Smallest subset of X containing 9k + 5 is ∴ 76 ≤ 63 + 76 − x ≤ 100 elements of both A and B have elements 100 100 × 1099 ⇒ 0 ≤ 63 − x ≤ 24 ⇒ 39 ≤ x ≤ 63 = n(A) + n(B) − n(A ∩ B) [104 + 995] = 2 2 Hence, option (a) is correct. = 25 + 7 − 3 = 29 50 n [Qnumbers are 104, 113, …, 995] 08 Let U X i = U Yi = T , where each X i 11 If A = {x ∈R :| x | < 2} and 100 × 1093 100 × 1099 Their sum = + i =1 i =1 2 2 contains 10 elements and each Yi B = {x ∈R :| x − 2| ≥ 3}, then = 100 × 1096 = 400 × 274 contains 5 elements. If each [2020, 9 Jan. Shift-II] Hence, we can say the value ofl = 5 (a) B − A = R − (−2, 5) element of the setT is an element (b) A − B = [−1, 2) as the second series of numbers of exactly 20 of sets X i ′ s and obtained by setC is of the form 9k + 5. (c) A ∪ B = R − (2, 5) exactly 6 of sets Yi ′ s, then n is ∴Required value ofl = 5 (d) A ∩ B = (−2, − 1) equal to [2020, 4 Sep. Shift-II] Ans. (a) 06 Consider the two sets A = {m∈R : (a) 50 (b) 15 (c) 45 (d) 30 Given sets A = { x ∈ R : | x |< 2}, both the roots of Ans. (d) and B = { x ∈ R : | x − 2 | ≥ 3} x 2 − (m +1) x + m +4= 0 are real} and According to the given information then, A = { x ∈ R : − 2 < x < 2} B= [− 3, 5). Which of the following is number of distinct elements in 50 50 × 10 and not true? [2020, 3 Sep. Shift-I] U Xi = 20 = 25 B = { x ∈ R :(x − 2) ∈ (−∞, − 3] ∪ [3, ∞)} (a) A − B = (− ∞, − 3) ∪ (5, ∞) i=1 {x ∈ R : x ∈ (−∞, − 1] ∪ [5, ∞)} (b) A ∩ B = { − 3} and number of distinct elements in n n× 5 ∴ B − A = R − (−2, 5) (c) B − A = (− 3, 5) U Yi = 6 50 Hence, option (a) is correct. (d) A ∪ B = R i=1 n Ans. (a) Q U Xi = U Yi = T [given] i=1 i=1 12 Two newspapers A and B are Given sets A = { m∈R :both the roots of n× 5 published in a city. It is known that ⇒ 25 = ⇒n = 30. x 2 − (m + 1) x + m + 4 = 0 are real} and 6 25% of the city population reads A B = [− 3, 5) and 20% reads B while 8% reads 09 Set A has m elements and Set B has Q Roots of x 2 − (m + 1) x + m + 4 = 0 are real, both A and B. Further, 30% of m∈R n elements. If the total number of those who read A but not B look ∴ D ≥ 0 ⇒(m + 1) 2 − 4(m + 4) ≥ 0 subsets of A is 112 more than the into advertisements and 40% of ⇒ m2 − 2m − 15 ≥ 0 total number of subsets of B, then those who read B but not A also ⇒ m − 5m + 3m − 15 ≥ 0 2 the value of m ⋅ n is..... look into advertisements, while [2020, 6 Sep. Shift-I] ⇒ m (m − 5) + 3 (m − 5) ≥ 0 50% of those who read both A Ans. (28.00) and B look into advertisements. ⇒ (m + 3) (m − 5) ≥ 0 ⇒ m∈ (− ∞, − 3] ∪ [5, ∞) It is given that n(A) = m and n(B) = n Then, the percentage of the and 2m = 2n + 112. population who look into ∴ A = (− ∞, − 3] ∪ [5, ∞) ∴ A − B = (− ∞, − 3) ∪ [5, ∞) [Qnumber of subsets of set A and B are advertisements is 2m and 2n respectively] (a) 13.5 (b) 13 A ∩ B = { − 3}, B − A = (− 3, 5) and A∪B = R ⇒ 2m − 2n = 24 (7) (c) 12.8 (d) 13.9 Hence, option (a) is correct. ⇒ 2n (2m − n − 1) = 24 (23 − 1) [2019, 9 April Shift-II] Sets, Relations and Functions 3 n (C ∩ A) =  Ans. (d) 140  A B = 14  10  Let the population of city is 100. Then, n(A) = 25, n(B) = 20 and n(A ∩ B) = 8 (numbers divisible by both 2 and 5) n (A ∩ B ∩ C) =  140  A B U C =4  30  From the Venn diagram, it is clear that (numbers divisible by 2, 3 and 5) if A − B ⊆ C, then A ⊆ C. and n (A ∪ B ∪C) 17 8 12 n(U)=100 Now, if (A − C) ⊆ B, for this statement = Σn(A) − Σn(A ∩ B) + n (A ∩ B ∩C) the Venn diagram. = (70 + 46 + 28 ) − (23 + 9 + 14) + 4 = 102 A ∴ Number of students who did not opt Venn diagram B any of the three courses = Total students − n (A ∪ B ∪C) So, n(A ∩ B ) = 17 and n(A ∩ B) = 12 = 140 − 102 = 38 C According to the question, Percentage of the population who look into From the Venn diagram, it is clear that 15 Let S = {1, 2, 3,... , 100}. The number advertisement is A ∩ B ≠ φ, A ∩ B ⊆ C and A –C = φ ⊆ B but A ⊆ B. of non-empty subsets A of S such  30  × n(A ∩ B )  +  × n(A ∩ B)  40 that the product of elements in A is =  100    100  14 In a class of 140 students even, is 2019, 12 Jan. Shift-I]  50  numbered 1 to 140, all even (a) 2 50 (2 50 − 1) (b) 2 50 − 1 + × n(A ∩ B)  numbered students opted  100  (c) 2 50 + 1 (d) 2 100 − 1 Mathematics course, those whose Ans. (a)  30   50  × 17  +  × 12 +  40 = × 8 number is divisible by 3 opted  100   100   100  Given, set S = { 1, 2, 3,...,100 }. Total Physics course and those whose number of non-empty subsets of = 5.1 + 4.8 + 4 number is divisible by 5 opted S = 2 100 − 1 = 13.9 Chemistry course. Then, the Now, numbers of non-empty subsets of number of students who did not S in which only odd numbers {1, 3, 5, … , 13 Let A,B and C be sets such that opt for any of the three courses is 99} occurs = 2 50 − 1 φ ≠ A ∩ B ⊆ C. Then, which of the [2019, 10 Jan. Shift-I] So, the required number of non-empty following statements is not true? (a) 42 (b) 102 (c) 38 (d) 1 subsets of S such that product of (a) B ∩ C ≠ φ Ans. (c) elements is even. (b) If (A − B) ⊆ C, then A ⊆ C A B = (2 100 − 1) − (2 50 − 1) (c) (C ∪ A) ∩ (C ∪ B) = C = 2 100 − 2 50 = 2 50 (2 50 − 1). (d) If (A − C) ⊆ B, then A ⊆ B 16 Let Z be the set of integers. If A = {x ∈ Z : 2 ( x + 2 ) ( x − 5 x + 6 ) = 1} and 2 C [2019, 12 April, Shift-II] Ans. (d) B = {x ∈ Z : − 3 < 2x − 1 < 9}, then the Let A be the set of even numbered number of subsets of the set A × B, Key Idea Use Venn diagram for operations students then of sets. is [2019, 12 Jan. Shift-II]  140  n(A) = = 70 (a) 212 (b) 218 According to the question, we have the  2  (c) 215 (d) 210 following Venn diagram. ([.] denotes greatest integer function) Here, A ∩ B ⊆ C and A ∩ B ≠ φ Ans. (c) Let B be the set of those students whose number is divisible by 3, then Given, set A = { x ∈ Z :2 (x + 2 ) (x 2 − 5 x + 6 ) = 1} C n(B) =  140  Consider,2 (x + 2 ) (x − 5 x + 6 ) = 1 = 2º 2 = 46 A B  3  ⇒ (x + 2) (x − 3) (x − 2) = 0 ([.] denotes greatest integer function) ⇒ x = −2, 2, 3 Let C be the set of those students whose ⇒ A = { −2, 2, 3} number is divisible by 5, Also, we have set n(C) =  140  B = { x ∈ Z : − 3 < 2x − 1 < 9} A∩B then = 28  5  Consider, −3 < 2x − 1 < 9, x ∈ Z Now, from the Venn diagram, it is clear ([.] denotes greatest integer ⇒ −2 < 2x < 10, x ∈ Z function) that ⇒ −1 < x < 5, x ∈ Z n(A ∩ B) =  140  B ∩ C ≠ φ, is true Now, = 23 ⇒ B = {0, 1, 2, 3, 4}  6  Also, (C ∪ A) ∩ (C ∪ B) = C ∪ (A ∩ B) = C So, A × B has 15 elements. (numbers divisible by both 2 and 3) ∴ Number of subsets of A × B = 215. n(B ∩C) =  is true. 140  =9 [Qif n(A) = m, the number of possible If (A − B) ⊆ C, for this statement the Venn  15  diagram is subsets = 2m ] (numbers divisible by both 3 and 5) 4 JEE Main Chapterwise Topicwise Mathematics 17 Let S = {x ∈R : x ≥ 0 and Ans. (b) After pre-multiply by P −1 and post-multiply by P, 2| x − 3| + x ( x − 6) + 6 = 0.Then,S According to the question, let’s consider we get option (b) (2, 3) and (3, 4) satisfy (a) is an empty set. [JEE Main 2018] 0 < | x − y | ≤ 1 but (2, 4) does not satisfy it. P −1 AP = B (b) contains exactly one element. So, (B, A) ∈ R for matrix P −1. (c) contains exactly two elements. 21 Let N be the set of natural numbers So, R is a symmetric relation. (d) contains exactly four elements. and a relation R on N be defined by For transitive relation, Ans. (c) R = {(x, y) ∈N × N : x 3 − 3x 2 y − xy 2 Let ARB and BRC We have, 2 | x − 3 | + x ( x − 6) + 6 = 0 + 3y 3 = 0}.Then the relation R is So, A = PBP −1 and B = PCP −1 [2021, 27 July Shift-II] Let x −3= y Now, A = P (PCP −1 ) P −1 ⇒ x =y+3 (a) symmetric but neither reflexive nor transitive. ⇒ A = (P) C (P −1 ) 2 ⇒ A = (P) 2.C.(P 2 ) −1 2 ∴ 2 | y | + (y + 3)(y − 3) + 6 = 0 ∴(A, C) ∈ R for matrix P 2. (b) reflexive but neither symmetric nor ⇒ 2 |y | + y2 − 3 = 0 transitive. ∴R is transitive relation. ⇒ | y |2 + 2 | y | − 3 = 0 (c) reflexive and symmetric, but not Hence, R is an equivalence relation. ⇒ (| y | + 3)(| y | − 1) = 0 transitive. ⇒ | y | ≠ −3 ⇒ | y | = 1 (d) an equivalence relation. 23 Let A = {2, 3, 4, 5,.... , 30} and ‘−~ ’ be ⇒ y=±1 Ans. (b) an equivalence relation on A × A, ⇒ x − 3 = ± 1 ⇒ x = 4, 2 Given, relation R on N is defined by defined by (a, b) −~ (c, d), if and only if ⇒ x = 16, 4 R = {(x, y) ∈N × N : x 3 −3x 2 − xy 2 + 3y 3 = 0 } ad = bc. Then, the number of x 3 − 3x 2 y − xy 2 + 3y 3 = 0 18 If X = (4 n − 3n − 1: n ∈ N) and ordered pairs, which satisfy this ⇒ x 3 − xy 2 − 3x 2 y + 3y 3 = 0 Y = {9 (n − 1) : n ∈ N}, where N is the ⇒ x ( x 2 − y 2 ) − 3y ( x 2 − y 2 ) = 0 equivalence relation with ordered set of natural numbers, then X ∪ Y ⇒ (x − 3y) (x 2 − y 2 ) = 0 pair (4, 3) is equal to [2021, 16 March Shift-II] is equal to [JEE Main 2014] ⇒ (x − 3y) (x − y) (x + y) = 0 (a) 5 (b) 6 (c) 8 (d) 7 (a) N (b) Y − X (c) X (d) Y Now, x − x =0 Ans. (d) Ans. (d) ⇒ x = x, ∀ (x, x) ∈N × N Q X = {4n − 3n − 1 : n ∈N } A = {2, 3, 4, 5, …, 30 } So, R is a reflexive relation. a = bc X = {0, 9, 54, 243,...} [put n = 1, 2, 3,...] But not symmetric and transitive ∴ (a, b) R (4, 3) Y = {9 (n − 1) : n ∈N } relation because, (3, 1) satisfies but (1, 3) does not. Also, ⇒ 3a = 4b Y = {0, 1, 18, 27, K } a =   b 4 It is clear that X ⊂ Y. (3, 1) and ⇒ (1, −1) satisfies but (3, − 1) does not.  3 ∴ X ∪Y = Y Hence, relation R is reflexive but neither b must be a multiple of 3,b can be symmetric nor transitive. (3, 6, 9, … 30). 19 If A, B and C are three sets such that Also, a must be less than or equal to 30. A ∩ B = A ∩ C and A ∪ B = A ∪ C, 22 Define a relation R over a class of (a, b) = (4, 3), (8, 6), (12, 9), (16, 12), (20, 15) then [AIEEE 2009] n × n real matrices A and B as “ARB, (24, 18), (28, 21) (a) A = C (b) B = C if there exists a non-singular matrix ⇒7 ordered pairs (c) A ∩ B = φ (d) A = B P such that PAP −1 = B”. Then which Ans. (b) of the following is true ? 24 Let R = {(P, Q) |, P and Q are at the Given, A ∩ B = A ∩ C and A ∪ B = A ∪ C [2021, 18 March Shift-II] same distance from the origin} be a ∴ B =C (a) R is symmetric, transitive but not relation, then the equivalence class reflexive. of (1, –1) is the set TOPIC 2 (b) R is reflexive, symmetric but not [2021, 26 Feb. Shift-I] transitive. (a) S = {(x, y) | x 2 + y2 = 4} Relations (c) R is an equivalence relation. (b) S = {(x, y) | x 2 + y2 = 1} (d) R is reflexive, transitive but not (c) S = {(x, y) | x 2 + y2 = 2} 20 Which of the following is not symmetric. (d) S = {(x, y) | x 2 + y2 = 2} correct for relation R on the set of Ans. (c) Ans. (d) real numbers ? [2021, 31 Aug. Shift-I] (a) (x, y) ∈ R ⇔ 0 < | x | − | y | ≤ 1 is neither For reflexive relation, Let P (a, b) and Q (c, d) are any two points. transitive nor symmetric. ∀ (A, A) ∈ R for matrix P. Given, OP = OQ (b) (x, y) ∈ R ⇔ 0 < | x − y | ≤ 1 is symmetric ⇒ A = PAP −1 is true for P = 1 i.e. a 2 + b 2 = c 2 + d 2 and transitive. So, R is reflexive relation. (c) (x, y) ∈ R ⇔ | x | − | y | ≤ 1 is reflexive but For symmetric relation, Squaring on both sides, not symmetric. Let (A, B) ∈ R for matrix P. a2 + b2 = c2 + d2 … (i) (d) (x, y) ∈ R ⇔ | x − y | ≤ 1 is reflexive and R = {((a, b), (c, d)) :a 2 + b 2 = c 2 + d 2 } ⇒ A = PBP −1 symmetric. Sets, Relations and Functions 5 R (x, y), S (1, − 1), OR = OS (equivalence (a) B ⊂ A Now, if Y contains any 2 elements, then class) (b) A ⊂ B these 2 elements cannot be used in any This gives OR = x 2 + y 2 and OS = 2 (c) A ∩ B = φ (an empty set) way to construct Z, because we want (d) neither A ⊂ B nor B ⊂ A Y ∩ Z = φ. ⇒ x2 + y2 = 2 Ans. (b) And from the remaining 3 elements ⇒x 2 + y 2 = 2 (Squaring on both sides) which are not present inY,23 subsets can ∴ S = {(x, y) : x 2 + y 2 = 2} We have, |a − 5 | < 1 and |b − 5 | < 1 be made each of which can be equal to Z ∴ −1 < a − 5 < 1 and −1 < b − 5 < 1 and stillY ∩ Z = φwill be true. 25 If R = {(x, y): x, y ∈ Z , x 2 + 3y 2 ≤ 8} is a ⇒ 4 < a < 6 and 4 < b < 6 Hence, total number of ways to relation on the set of integers Z, Now, 4(a − 6) 2 + 9(b − 5) 2 ≤ 36 construct sets Y and Z such that then the domain of R −1 is (a − 6) 2 (b − 5) 2 Y ∩Z=φ ⇒ + ≤1 [2020, 2 Sep. Shift-I] 9 4 (a) { −1,0,1} = 5 C 0 × 25 + 5 C 1 × 25 − 1 + K + 5 C 5 × 25 − 5 Taking axes as a-axis and b-axis (b) { − 2, − 1,1,2 } = (2 + 1) 5 = 35 (6, 7) (c) { −2, − 1,0,1,2 } b Method 2 (d) {0, 1} Q (6, 6) Since,Y ⊆ X, Z ⊆ X, hence we can only P Ans. (a) use the elements of X to construct sets Given relation, (0, 5) (3, 5)(4, 5) (6, 5) (9, 5) Y and Z. R = {(x, y) : x, y ∈Z, x 2 + 3y 2 ≤ 8} S R (6, 4) Every elements in X (say a) has four For, y 2 = 0, x 2 = 0, 1, 4 For, y 2 = 1, (6, 3) options (as far as going to Y and Z is x 2 = 0, 1, 4 concerned). For, y 2 = 4, x 2 ∈φ a (a) a ∈Y , i.e., a is present in Y and a ∈ Z, ∴ Range of R is possible values of y i.e., ‘ a ’ is present in Z. = { − 1, 0, 1} The set A represents square PQRS inside (b) a ∈Y , i.e., a is present in Y and a ∉Z, set B representing ellipse and hence i.e., ‘ a ’ is not present in Z. ∴ Domain of R −1 = Range of R = { − 1, 0, 1} A ⊂ B. (c) a ∉Y , i.e., a is not present in Y and 26 Let R1 and R2 be two relations defined a ∈ Z, 28 Let X = {1, 2, 3, 4, 5}. The number of i.e., a is present in Z. as follows different ordered pairs (Y , Z) that (d) a ∉Y , i.e., a is not present in Y and R1 = {(a, b) ∈R 2 : a 2 + b 2 ∈Q} can formed such that Y ⊆ X , Z ⊆ X a ∉Z, and R2 = {(a, b) ∈R 2 : a 2 + b 2 ∉Q}, and Y ∩ Z is empty, is [AIEEE 2012] i.e., a is not present in Z. where Q is the set of all rational (a) 5 2 (b) 35 Analysis of the above 4 cases numbers. Then [2020, 3 Sep. Shift-II] (c) 2 5 (d) 5 3 (a) Ifa is present in Y and also in Z, then (a) R1 and R2 are both transitive. Ans. (b) it will be certainly present inY ∩ Z. (b) Neither R1 nor R2 is transitive. Given A set X = { 1, 2, 3, 4, 5} (b) Ifa is present in Y but not present in (c) R 1 is transitive but R 2 is not transitive. Z, then it will not be present in To find The number of different (d) R 2 is transitive but R 1 is not transitive. Y ∩ Z. ordered pairs (Y , Z) such that Y ⊆ X, Z ⊆ X Ans. (b) and Y ∩ Z = φ.Since,Y ⊆ X, Z ⊆ X, hence (c) Ifa is not present in Y but present in Given relation, we can only use the elements of X to Z, then also it will not be present in R 1 = {(a, b) ∈ R 2 :a 2 + b 2 ∈ Q } construct sets Y and Z. Y ∩ Z. Let (a, b) ∈ R 1 ⇒a 2 + b 2 ∈ Q and (b, c) ∈ R 1 Method 1 (d) If a is not present in both of Y and ⇒ b2 + c 2 ∈Q Number of ways Z, then it will not be present in ∴a 2 + 2b 2 + c 2 ∈ Q , but we can not say n ( Y ) Number of ways to make Z such Y ∩ Z. to make Y that Y ∩ Z = φ that a 2 + c 2 ∈ Q , so (a, c) ∉R 1 We wantY ∩ Z = φ to which only case (a) ∴R 1 is not transitive. 0 5 C0 25 is not favourable and remaining cases, And another given relation 5 i.e., cases (b), (c) and (d) are favourable. R 2 = {(a, b) ∈ R 2 :a 2 + b 2 ∉Q } 1 C1 24 Hence, for every elements ‘ a ’ in X, Let (a, b) ∈ R 2 ⇒a 2 + b 2 ∉Q 2 5 C2 23 there are 3 favourable ways such that and (b, c) ∈ R 2 ⇒b 2 + c 2 ∉Q 3 5 C3 22 Y ∩ Z = φ. ∴a 2 + 2b 2 + c 2 ∉Q , but we can not say that a 2 + c 2 ∉Q , so (a, c) ∉R 2 5 ⇒ Total number of ways 4 C4 21 ∴R 2 is not transitive. = 3 × 3 × 3 × 3 × 3 = 35 5 5 C5 20 [as there are 5 elements in X and each of Hence, option (b) is correct. them have 3 options to go or not to go to Let us explain anyone of the above 6 Y and Z] 27 Two sets A and B are as under rows say third row. In third row, A = {(a, b) ∈R × R : | a − 5| < 1and 29 Let R be the set of real numbers. Number of elements inY = 2 | b − 5| < 1} ∴Number of ways to selectY = 5 C 2 ways Statement I A = {(x, y) ∈R × R : y − x B = {(a, b) ∈R × R:4(a − 6) + 9(b − 5) 2 2 Because any 2 elements of X can be part is an integer} is an equivalence ≤ 36}. Then, [JEE Main 2018] of Y. relation on R. 6 JEE Main Chapterwise Topicwise Mathematics Statement II 30 Consider the following relations 31 Let R be the real line. Consider the B = {(x, y) ∈R × R : x = αy for some R = {(x , y) | x , y are real numbers and following subsets of the plane R × R rational number α} is an x = wy for some rational number S = {(x, y) : y = x + 1 and 0 < x < 2 } equivalence relation on R.   m p T = {(x, y) : x − y is an integer} w} ; S =   ,  m, n, p and q are (a) Statement I is true, Statement II is  n q  Which one of the following is true? [AIEEE 2008] true; Statement II is not a correct integers such that n, q ≠ 0 and explanation of Statement I (a) T is an equivalence relation on R but qm = pn}. Then, [AIEEE 2010] (b) Statement I is true, Statement II S is not. (a) R is an equivalence relation but S is is false (b) Neither S nor T is an equivalence not an equivalence relation. (c) Statement I is false, Statement II relation on R. is true (b) Neither R nor S is an equivalence (c) Both S and T are equivalence (d) Statement I is true, Statement II is relation. relations onR. true; Statement II is a correct (c) S is an equivalence relation but R is (d) Sis an equivalence relation onR but explanation of Statement I not an equivalence relation. T is not. Condition for equivalence (d) R and S both are equivalence Ans. (a) relation A relation which is relations. Since, (1, 2) ∈S but (2, 1) ∉S symmetric, reflexive and Ans. (c) transitive is equivalence relation. So, S is not symmetric. Given, relation R is defined as [AIEEE 2011] Hence, S is not an equivalence relation. R = {(x , y) | x , y are real numbers and Ans. (b) x = wy for some rational number w} Given, T = {(x, y) : (x − y) ∈l } Statement I (i) Reflexive xRx ⇒ x = wx Now, x − x = 0 ∈I, it is reflexive relation. A = {(x, y) ∈ R × R : y − x is an integer} ∴ w = 1 ∈rational number Again now, (x − y) ∈I (a) Reflexive The relation R is reflexive. y − x ∈I, it is symmetric relation. xRx : (x − x) is an integer. i.e., true (ii) Symmetric xRy ⇒ / yRx as 0 R 1 Let x − y = I 1 and y − z = I 2 ∴ Reflexive But 1R0 ⇒1 = w ⋅ (0) Now, x − z = (x − y) + ( y − z) = I 1 + I 2 ∈I So, T is also transitive. (b) Symmetric which is not true for any rational number. Hence,T is an equivalence relation. xRy : (x − y) is an integer. ⇒− ( y − x) is an integer. The relation R is not symmetric. 32 Let W denotes the words in the ⇒ ( y − x) is an integer. Thus, R is not equivalence relation. English dictionary define the ⇒ yRx Now, for relation S which is defined as relation R by  m p  ∴ Symmetric S =  ,  m, n, p and q ∈integers R = {(x, y) ∈W × W :the words (c) Transitive  n q  x and y have atleast one letter in xRy and yRz such that n, q ≠ 0 and qm = pn} common}. Then, R is [AIEEE 2006] ⇒(x − y) is an integer and ( y − z) is an m m (i) Reflexive R ⇒ mn = mn [true] (a) reflexive, symmetric and not integer. n n transitive ⇒(x − y) + ( y − z) is an integer. The relationS is reflexive. (b) reflexive, symmetric and transitive m p ⇒(x − z) is an integer. (ii) Symmetric R ⇒ mq = np (c) reflexive, not symmetric and n q transitive ⇒ xRz p m ⇒ np = mq ⇒ R (d) not reflexive, symmetric and transitive ∴ Transitive q n Hence, A is an equivalence relation. Ans. (a) The relationS is symmetric. m p p r Let W = {CAT ,TOY , YOU,...} Statement II (iii) Transitive R and R n q q s Clearly, R is reflexive and symmetric but B = {(x, y) ∈ R × R : x = αy for some not transitive. rational numberα} ⇒ mq = np [Q CAT RTOY , TOY R YOU ⇒ / CAT R YOU ] 1 If α = , then for reflexive, we have and ps = rq 2 33 Let R = {(3, 3), (6, 6), (9, 9), (12, 12), ⇒ mq⋅ ps = np⋅rq 1 xR x ⇒x = x , which is not true, ⇒ ms = nr (6, 12), (3, 9), (3, 12), (3, 6)} be a 2 m r ∀ x ∈ R − {0 }. ⇒ = relation on the set ∴Bis not reflexive on R. n s A = {3, 6, 9, 12}. The relation is m r ⇒ R [AIEEE 2005] Hence,Bis not an equivalence relation n s on R. (a) reflexive and symmetric only. The relationS is transitive. (b) an equivalence relation. Hence, statement I is true, statement II Hence, the relation S is equivalence (c) reflexive only. is false. relation. (d) reflexive and transitive only. Sets, Relations and Functions 7 Ans. (d) ⇒log 1 ≤ log (3 − 2 sin x + 2 cos x) Also f (3) = f (2 + 1) = f (2) + f (1) 5 5 Since, for every elements of A, there ≤ log 5 5 = f (1 + 1) + f (1) exists elements (3, 3), (6, 6), (9, 9), (12, 12) ⇒0 ≤ f (x) ≤ 2 f (3) = f (1) + f (1) + f (1) ∈R ⇒ R is reflexive relation. ⇒f (x) ∈ [0, 2] ⇒ 9 = 3f(1) ⇒ f(1) = 3 Now, (6, 12) ∈R but (12, 6) ∉R, so it is not a ∴ f (2) = f (1 + 1) = f (1) + f (1) = 3 + 3 = 6 symmetric relation. 36 The domain of the function Hence,f (2) ⋅f (3) = 6⋅9 = 54 Also, (3, 6), (6, 12) ∈R ⇒(3, 12) ∈R  3x 2 + x − 1  x − 1 ∴ R is transitive relation. f (x) = sin − 1   + cos − 1   38 The domain of the function  ( x − 1)  2  x + 1 1 + x 34 Let R = {(1, 3), (4, 2), (2, 4), (2, 3), cosec − 1   is is [2021, 31 Aug. Shift-II]  x  (a) 0,  (b) [− 2, 0] ∪  ,  (3, 1)} be a relation on the set 1 1 1 [2021, 26 Aug. Shift-II] A = {1, 2, 3, 4}. The relation R is  4   4 2  (a)  − 1, −  ∪ (0, ∞) 1 (a) a function. [AIEEE 2004] (c)  ,  ∪ {0 } 1 1 (d) 0,  1  2   4 2   2  (b)  − , 0  ∪ [1, ∞) (b) transitive. 1 (c) not symmetric. Ans. (c)  2  (d) reflexive.  3x 2 + x − 1  −1  x − 1   1  f (x) = sin−1   + cos   (c)  − , ∞ − {0} Ans. (c)  (x − 1)  2  x + 1  2  Given, R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} is a (d)  − , ∞  − {0 } x−1 x −1 1 relation on the set A = { 1, 2, 3, 4}. −1 ≤ ≤ 1 ⇒−1 − 1 ≤ − 1≤ 1 − 1  2  x+1 x+1 (a) Since, (2, 4) ∈ R and (2,3) ∈ R. So, R is −2 1 Ans. (d) not a function. ⇒ −2 ≤ ≤0 ⇒ 0≤ ≤1 x+1 x+ 1 1+ x f (x) = cosec −1  1+ x  (b) Since, (1, 3) ∈ R and (3, 1) ∈ R but  ≥1 ⇒ 1≤ x + 1< ∞  x  x (1, 1) ∉R. So, R is not transitive. (c) Since, (2, 3) ∈R but (3, 2) ∉R. So, R is ⇒ 0≤ x< ∞ Clearly , x ≠ 0 not symmetri(c) ⇒ x ∈ [0, ∞) …(i) | 1 + x |2 ≥ | x |2 (d) Since, (1, 1), (2, 2), (3, 3), (4, 4) ∉R. So, 3x 2 + x − 1 1 + x 2 + 2x ≥ x 2 and −1 ≤ ≤1 R is not reflexive. (x − 1) 2 2x + 1 ≥ 0 1 ⇒ − (x − 1) 2 ≤ 3x 2 + x − 1 ≤ (x − 1) 2 , x ≠ 1 x≥ − 2 TOPIC 3 ⇒ − (x 2 − 2x + 1) ≤ 3x 2 + x − 1  1  and 3x 2 + x − 1 ≤ x 2 − 2x + 1 So, x ∈ − , ∞ − {0 } Domain-range, Odd-even  2  ⇒ 4x 2 − x ≥ 0 and Periodic Functions and 2x 2 + 3x − 2 ≤ 0 39 Let {S = 1, 2, 3, 4, 5, 6, 7}. Then the ⇒ x (4x − 1) ≥ 0 number of possible functions 35 The range of the function, and (x + 2)(2x − 1) ≤ 0 f : S → S such that f (m ⋅ n) = f (m) ⋅ f (n)   3π  f (x) = log 3 + cos  + x ⇒ x ∈ (−∞, 0] ∪  , ∞  1 for every m, n ∈ S and m ⋅ n ∈ S is 5  4   4  equal to …………. π π + cos  + x + cos − x  1 [2021, 27 July Shift-I] and x ∈ −2, 4  4   2  Ans. (490)  3π    1 1 S = {1, 2, 3, 4, 5, 6, 7} ⇒ x ∈ (−2, 0] ∪ , …(ii) − cos  − x  is  4 2  4  f :S → S [2021, 01 Sep. Shift-II] Domain of f in Eq. (i) ∩ Eq. (ii) f (m ⋅ n) = f (m) f (n)  1 1 m, n ∈ S ⇒m, n ∈ S (a) (0, 5) (b) [−2,2] ∴ x ∈ {0 } ∪ ,  4 2   1  If mn ∈ S ⇒mn≤ 7 (c)  , 5  (d) [0, 2] So, (1 ⋅ 1, 1 ⋅ 2,..., 1 ⋅ 7) ≤ 7  5  37 Let f : N → N be a function such that (2 ⋅ 2, 2 ⋅ 3) ≤ 7 Ans. (d) f (m + n) = f (m) + f (n) for every m, When m = 1, f (n) = f (1) ⋅ f (n) ⇒f(1) = 1  3π   π  n∈N. If f (6) = 18, then f (2) ⋅f (3) is When m = n = 2, f (x) = log 5  3 + cos + x  + cos + x    4  4  equal to [2021, 31 Aug. Shift-II] f (2) = 1 ⇒ f (4) = 1 or π 3π  f (4) = f (2) f(2) =  + cos − x  − cos − x   (a) 6 (b) 54  f (2) = 2 ⇒ f (4) = 4 4   4  (c) 18 (d) 36 When, m = 2, n = 3 = log (3 − 2 sin x + 2 cos x) Ans. (b) 5  When, f(2) = 1, f (m + n) = f (m) + f (n), m, n ∈N  Q− 2 ≤ − 2 sin x + 2 cos x ≤ 2  f(3) = 1 to 7 ∴ f (3 + 3) = f (3) + f (3) f (6) = f (2) f(3)  ⇒1 ≤ 3 − 2 sin x + 2 cos x ≤ 5 When, f(2) = 2, ⇒ f (6) = 2f (3) = 18 [Q f (6) = 18] f(3) = 1 or 2 or 3 8 JEE Main Chapterwise Topicwise Mathematics And f(5), f(7) can take any value (1-7) 1  i.e. x ∈ (−∞, − 1] ∪ [1, ∞) − {integers} (A) ∩ (B) = x ∈  ,1 [Qf (5) = f (1) ⋅f (5) ≤ 7 and f (7) = f (1) ⋅f (7) ≤ 7 }  2  i.e. all non-integers except the interval The possible combination is 3 [−1, 1] ∴ α+β= f(1) = 1 f(1) = 1 2 (here, −1 and 1 are included in except f(2) = 1 f(2) = 2 case, because of −1 and 1 are integers). f(3) = (1 − 7) f(3) = (1 − 3) 42 Let [x] denote the greatest integer ≤ x, where x ∈R. If the domain of 44 If the functions are defined as f(4) = 1 f(4) = 4 the real valued function f (x) = x and g (x) = 1 − x , then what f(5) = (1 − 7) f(5) = (1 − 7) is the common domain of the f (6) = f (3) f (6) = f (3) | [x]| − 2 f (x) = is (−∞, a) ∪ [b, c) following functions? f(7) = (1 − 7) f(7) = (1 − 7) | [x]| − 3 f + g, f − g, f / g, g / f ,g − f , where So, total = (1 × 1 × 7 × 1 × 7 × 1 × 7) ∪ [u , ∞) , a < b < c, then the value of (f ± g) (x) = f (x) ± g (x), (f / g)(x) = f (x) + (1 × 1 × 3 × 1 × 7 × 1 × 7) a + b + c is [2021, 20 July Shift-I] g (x) = 490 (a) 8 (b) 1 [2021, 18 March, Shift-I] 40 If [x] be the greatest integer less (c) −2 (d) −3 (a) 0 ≤ x ≤ 1 (b) 0 ≤ x < 1 100  (−1) n n Ans. (c) (c) 0 < x < 1 (d) 0 < x ≤ 1 than or equal to x , then Σ   | [x] | − 2 | [x] | − 2 Ans. (c) n=8  2  f (x) = ≥0 | [x] | − 3 | [x] | − 3 Given,f (x) = x and g (x) = 1 − x is equal to [25 July 2021, Shift-II] Let | [x] | = t ∴Domain of f (x) = D 1 is x ≥ 0 (a) 0 (b) 4 + – + i.e. D 1 : x ∈ (0, ∞) (c) −2 (d) 2 2 3 and domain of g (x) = D 2 is 1 − x ≥ 0 Ans. (b) t ∈ (–∞, 2] ∪ (3, ∞) ⇒ x≤ 1 We have, 3 i.e. D 2 : x ∈ (− ∞ 1] 100  (− 1) n n  2 ∑  2  (Q[x] is the greatest integer As, we know that, the domain of f + g, f − g, g − f will be D 1 ∩ D 2 as well as n = 8  f function) –3 –2 1 2 3 the domain for is D 1 ∩ D 2 except all Substitute the values of n g | [x] | = 3 ⇒x ∈ [− 3, − 2) ∪ [3, 4) those value(s) of x, such that g (x) = 0. = + [− 4.5] + + [− 5.5] g Domain of x = [−∞, − 3) ∪ [− 2, 3) ∪ [4, ∞) Similarly, for is D 1 ∩ D 2 but f (x) ≠ 0. + K + [− 49.5] + a = −3 f = 4 − 5 + 5 − 6 + K − 50 + 50 b = −2 Hence, common domain for =4 f g c =3 (f + g), (f − g),   ,   and (g − f) will be g  f  41 If the domain of the function ∴ a + b + c = − 3 + (− 2) + 3 = − 2 0 < x < 1. cos −1 x 2 − x + 1 f (x) = is the interval 43 The real valued function cosec −1 x 45 A function f (x) is given by −1  2x − 1 sin   f (x) = , where [x] denotes 5x  2  x − [x] f (x) = x , then the sum of the 5 +5 (α, β) , then α + β is equal to the greatest integer less than or series [2021, 22 July Shift-II] equal to x, is defined for all x 2 3  39 f   + f   + f   + K + f   3 1 1 (a) (b) 2 (c) (d) 1 belonging to [2021, 18 March Shift-I] 2 2  20  20  20  20 (a) all reals except integers Ans. (a) (b) all non-integers except the interval [ is equal to [2021, 25 Feb. Shift-II] cos−1 x 2 − x + 1 − 1,1] 29 49 f (x) = (c) all integers except 0, − 1, 1 (a) (b)  2x − 1  2 2 sin−1   (d) all reals except the interval [−1, 1] 39 19  2  (c) (d) x 2 − x + 1 ≥ 0 and x 2 − x + 1 ≤ 1 Ans. (b) 2 2 ⇒x ∈ R, x (x − 1) ≤ 0 cosec−1 x Ans. (c) Given,f (x) = 0≤ x≤ 1 …(i) x − [x] 5x Given,f (x) = , then, −1  2x − 1  π 5 +5x ⇒ 0 < sin  < cosec−1 x  2  2 ⇒ f (x) = 52 − x 2x − 1 {x} f (2 − x) = 2 −x ⇒ 0< 0  x ≠ 1 integers Sets, Relations and Functions 9 5x + 5 Ans. (c) 49 For a suitable chosen real constant This gives,f (x) + f (2 − x) = =1 5x + 5 x 2 −` x − 2 a, let a functin f : R − {a} → R be Given, g (x) = , f (x) = sin−1 x  1   1   1   39  2x 2 −` x − 6 a −x ⇒ f   + f 2 −  = f  ?