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Class 12, , Shiv Das, , 2021-22, , CBSE, , Term 1, (Nov-Dec), , s, Q, MC, , Chapterwise, QUESTION BANK, , MATHEMATICS, , Competency Based Questions, • Case Studies with MCQs, • Objective Type MCQs, , Chapterwise, • Definitions & Formulae, • Questions & Answers, , OMR Based, Sample Question Paper, FOLLOW US, R.P.A., , shivdasbooks

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© All, , • No part of this book may be reproduced or copied in any form or by any means without the, written permission of the publishers., , Rights, Reserved, , • The publishers have taken all possible, precautions in publishing this book, yet if any, mistake has crept in, the publishers shall not be, responsible for the same., • All disputes shall be subject to the jurisdiction, of court at Delhi only., , PUBLISHED BY SHIV DAS & SONS, C-8, Sector 6, Noida, Uttar Pradesh-201 301, India, +91-9910055604, 7840040400, 0120-4197490, 4305480, [email protected] • www.shivdas.in, Trade contact: [email protected], , Printed at:, First Impression, Corporate Services Pvt. Ltd., E-114, Sector-63, Noida, Uttar Pradesh-201301

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During the course of the year if there are any updates, in the Curriculum due to the Pandemic, we will, provide all the updates for free with the help of the, given Barcode., , FOR LATEST CBSE, 2022 UPDATES, Scan The QR CODE, , FREE DOWNLOADS, AND OTHER CONTENT, Scan QR Code, , Scan The QR CODE, , FOLLOW US, , Shivdasbooks

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C ontents, , Term–1, , Latest Syllabus (Issued by CBSE), , ..., , (vi), , Competency Based Questions (CBQs), Chapter 1, , Relations and Functions, , ..., , C-1, , Chapter 2, , Inverse Trigonometric Functions, , ..., , C-7, , Chapter 3, , Matrices, , ..., , C-13, , Chapter 4, , Determinants, , ..., , C-23, , Chapter 5, , Continuity and Differentiability, , ..., , C-29, , Chapter 6, , Applications of Derivatives, , ..., , C-35, , Chapter 12, , Linear Programming, , ..., , C-43, , Chapterwise CBSE Examination Questions, Chapter 1, , Relations and Functions, , ..., , 1, , Chapter 2, , Inverse Trigonometric Functions, , ..., , 17, , Chapter 3, , Matrices, , ..., , 33, , Chapter 4, , Determinants, , ..., , 49, , Chapter 5, , Continuity and Differentiability, , ..., , 74, , Chapter 6, , Applications of Derivatives, , ..., , 106, , Chapter 12, , Linear Programming, , ..., , 144, , ..., , OMR-1, , OMR Based Sample Question Paper, ✽ • •• • ✽, , (iv)

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CBSE BOARD EXAM 2021-22, NEW PATTERN, (CBSE Circular no. Acad-51/2021) dated 5th July, 2021, , TERM-1, , 50% Syllabus for Nov-Dec. 2021, Questions Types (90 Minutes), , CASE BASED, MCQs, , MULTIPLE CHOICE, QUESTIONS, , ASSERTION-REASON, TYPE MCQs, , ✰, , CBSE to send Question Papers with marking scheme., , ✰, , Exams will be conducted in the school., , ✰, , OMR sheet to be used for Evaluation., , ✰, , Term-1 marks will add to the final result., , Internal Assessment, ✰ 3 Periodic Tests, ✰ Student Enrichment, ✰ Portfolio, ✰ Practical, ✰ Speaking listening activities, ✰ Projects, (v)

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Latest Syllabus (Issued by CBSE) (24, , TH, , July 2021), , MATHEMATICS (2021-22), COURSE STRUCTURE—CLASS XII, , Theory, TERM-I, Units, , Marks, , I., , Relations and Functions, , 08, , II., , Algebra, , 10, , III., , Calculus, Linear Programming, , 17, 05, , Total, , 40, , Internal Assessment, , 10, , Total, , 50, , V., , UNIT I: RELATIONS AND FUNCTIONS, 1. Relations and Functions:, , Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions., 2. Inverse Trigonometric Functions:, Definition, range, domain, principal value branch., , UNIT II: ALGEBRA, 1. Matrices:, 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, Invertible matrices (Here all matrices will have real entries)., 2. Determinants:, Determinant of a square matrix (up to 3 × 3 matrices), minors, co-factors and applications of determinants, in finding the area of a triangle. Adjoint and inverse of a square matrix. 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:, Continuity and differentiability, derivative of composite functions, chain rule, derivative 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., 2. Applications of Derivatives:, Applications of derivatives: increasing/decreasing functions, tangents and normals, 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)., (vi)

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UNIT V: LINEAR PROGRAMMING, 1. Linear Programming:, 13 Periods, Introduction, related terminology such as constraints, objective function, optimization, different types of, linear programming (L.P.) problems, graphical method of solution for problems in two variables, feasible, and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions, (up to three non-trivial constraints)., Internal Assessment, , 10 Marks, , Periodic Test, , 5 marks, , Mathematics Activities: Activity file record + Term end Assessment of one, activity & Viva, , 5 marks, , Note: For activities NCERT Lab Manual may be referred, , (vii)

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(vi)

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Chapter: One, , Relations and Functions, ,  Multiple Choice Questions (MCQs), 1. Let A = {1, 2, 3, 4}. Let R be the equivalence, relation on A × A defined by (a, b) R (c, d), if a + d = b + c. Then the equivalence class, [(1, 3)] is, , 6. Which of the following functions form Z into Z, bijections?, (a) f (x) = x3, , (b) f (x) = x + 2, , (c) f (x) = 2x + 1, , (d) f (x) = x2 + 1, , s, a, iv D, , Sh, , 5. If f, f is, (a), (b), (c), (d), , : R → R be defined by f(x) =, one-one, onto, bijective, f is not defined, , 2, , x ∀ R, then, x, , 7. If f : R → R be the function defined by f (x) =, x3 + 5, then f–1(x) is, (a) (x + 5)1/3, , (b) (x – 5)1/3, , (c) (5 – x)1/3, (d) (5 – x), 8. If f : A → B and g : B → C be the bijective, functions, then (gof)–1 is, (a) f–1og–1 , , (b) fog, , g–1of–1 , , (d) gof, , (c), , 3x + 2, , 3, , 9. If f : R –   → R be defined by f (x) =, ,, 5x − 3, 5, then, (a) f–1(x) = f (x), , (b) f–1(x) = –f (x), , (c) fof (x) = –x, , (d) f–1(x) =, , 1, f (x), 19, , 10. Let f : N → R be the function defined by, 2x − 1, and g : Q → R be another function, 2, 3, defined by g (x) = x + 2. Then (gof), is, 2, , f (x) =, , (a) 1 , (c), , 7, , 2, , (b) – 1, (d) 3, , Competency Based Questions n C–1, , BASED, , (a) {(1, 3)}, (b) {(2, 4)}, (c) {(1, 8), (2, 4), (1, 4)}, (d) {(1, 3) (2, 4)}, 2. The maximum number of equivalence relations, on the set A = {2, 3, 4} are, (a) 1 , (b) 27, (c) 3 , (d) 5, 3. If a relation R on the set {1, 2, 3} be defined by, R = {(1, 2)}, then R is, (a) reflexive, (b) transitive, (c) symmetric, (d) none of these, 4. If the set A contains 7 elements and the set B, contains 8 elements, then number of one-one, and onto mappings from A to B is, (a) 24, (b) 120, (c) 0, (d) none of these, , COMPETENCY, , Competency Based Questions

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C–2 n Shiv Das Chapterwise Question Bank (MATHEMATICS XII), , Relations and Functions, , HINTS AND ANSWERS, 1. (d); Equivalence class of [(1, 3)] is given by set of ordered pair (a, b) ∈ A × A such that, (1, 3) R (a, b), ⇒ 1 + b = 3 + a , , \ [(1, 3)] = {(1, 3), (2, 4)}, 2. (d); A = {2, 3, 4}, Number of equivalence relations is as follows :, R1 = {(2, 2), (3, 3), (4, 4)}, R2 = {(2, 2), (3, 3), (4, 4), (2, 3), (3, 2)}, R3 = {(2, 2), (3, 3), (4, 4), (2, 4), (4, 2)}, R4 = {(2, 2), (3, 4), (4, 4), (2, 4), (4, 2)}, , \ Maximum number of equivalence relations on the set A = {2, 3, 4} = 5, 3. (b); R on the set {1, 2, 3} be defined by R = {(1, 2)}, , \ R is transitive., ∵ If A and B are two non-empty finite sets containing m and n elements, 4. (c); Here m = 7, n = 8, respectively, then the number of one-one and onto mappings from, ..., , \ m≠n, A to B is 0 if m ≠ n, and, , \ Number of mappings = 0, n! if m = n, 5. (d); f (x) = 2 ,∀ x ∈ R, x, , s, a, iv D, , 2, , When x = 0, f (0) =, ( Not defined) , 0, 6. (b); f (x) = x + 2, For one-one, f (x1) = f (x2) ⇒ x1 + 2 = x2 + 2 ⇒ x1 = x2, For onto,, Let y = x + 2 , ⇒ y – 2 = x , Now, f (y – 2) = y – 2 + 2 = y , Hence f (x) = x + 2 is one-one and onto and hence bijective., 7. (b); Let y = f (x), , Sh, , y = x3 + 5 , , ⇒ y – 5 = x3 , ⇒ x = (y – 5)1/3, , \ f is not defined., \ f is one-one, , ⇒ f (x) = x + 2, \ f is onto, , \ f–1(x) = (x – 5)1/3, , 8. (a); Given f : A → B and g : B → C be the bijective functions., (gof)–1 = f–1og–1, 9. (a); Let y = f (x) , y =, , 3x + 2, , 5x − 3, , ⇒ 5xy – 3y = 3x + 2, ⇒ x(5y – 3) = 3y + 2 , , , ⇒ 5xy – 3x = 3y + 2, , ⇒ , x=, , 3y + 2, , 5y − 3, , , \ f–1(x) = f (x) , , ⇒ f–1(x) =, … [∵ f ( x ) =, , 3x + 2, 5x − 3, , 3x + 2, 5x − 3, ,   3, , 3,  2  2  − 1 ,  3 − 1  = g  2  = g(1) = 1 + 2 = 3, 10. (d) ; (gof), = g,  ,  = g , , 2, 2, 2 , 2, , , , 

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Competency Based Questions n C–3, ,  Case Based Questions, , , Explanation. Clearly, (1, 1), (2, 2), (3, 3), ∈ R. So, R is, reflexive on A., Since, (1, 2) ∈ R but (2, 1) ∉ R. So, R is not symmetric, on A., Since, (2, 3), ∈ R and (3, 1) ∈ R but (2, 1) ∉ R., So, R is not transitive on A., , , Explanation. We have, R = {(x, y) : 3x – y = 0}, where, x, y ∈ A = {1, 2, ...., 14}, , ∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}, Clearly, (1, 1) ∉ R. So R is not reflexive on A., Since (1, 3) ∈ R but (3, 1) ∉ R., So, R is not symmetric on A., Since, (1, 3) ∈ R and (3, 9) ∈ R but (1, 9) ∉ R., So, R is not transitive on A., , (v) If the relation R on the set A = {1, 2, 3} defined, as R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3),, (3, 1), (3, 2), (3, 3)}, then R is, (a) reflexive only, (b) symmetric only, (c) transitive only, (d) equivalence, Sol. (d) equivalence, , Explanation. Clearly, (1, 1), (2, 2), (3, 3) ∈ R., So, R is reflexive on A., We find that the ordered pairs obtained by interchanging the components of ordered pairs in R are, also in R. So, R is symmetric on A., For 1, 2, 3 ∈ A such that (1, 2) and (2, 3) are in R, implies that (1, 3) is also, in R. So R is transitive on A., Thus, R is an equivalence relation., , s, a, iv D, , (ii) If the relation R = {(1, 2), (2, 1), (1, 3), (3, 1)}, defined on the set A = {1, 2, 3}, then R is, (a) reflexive, (b) symmetric, (c) transitive, (d) equivalence, Sol. (b) Symmetric, , Sh, , (iii) If the relation R on the set N of all natural, numbers defined as R = {(x, y) : y = x + 5 and, (x < 4), then R is, (a) reflexive, (b) symmetric, (c) transitive, (d) equivalence, Sol. (c) transitive, Explanation. We have, R = {(x, y) : y = x + 5 and x < 4},, where x, y ∈ N., ∴ R = {(1, 6), (2, 7), (3, 8)}, Clearly, (1, 1), (2, 2) etc. are not in R., So, R is not reflexive., Since, (1, 6) ∈ R but (6, 1) ∉ R., So, R is not symmetric., Since, (1, 6) ∈ R and there is not order pair in R which, has 6 as the first element. Same is the case for (2, 7) and, (3, 8). So, R is transitive., , x−1, , = x − 2 such that f is a bijection., , Based on the above information, answer the, following questions:, (i) Domain of f is, (a) R – {2}, (b) R, (c) R – {1, 2}, (d) R – {0}, Sol. (a) R – {2}, , Explanation. For f(x) to be defined x – 2 ≠ 0 i.e. x ≠ 2, ∴ Domain of f = R – {2}, , (ii) Range of f is, (a) R, (c) R – {0}, , (b) R – {1}, (d) R – {1, 2}, , Sol. (b) R – {1}, , x−1, Explanation. Let y = f(x), then y = x − 2, , 2y − 1,  xy – 2y = x – 1 ⇒ xy – x = 2y – 1 ⇒ x =, y−1, Since, x ∈ R – {2}, therefore y ≠ 1, Hence, range of f = R – {1}, , (iii) If g : R – {2}  R – {1} is defined by, , g(x) = 2f(x) – 1, then g(x) in terms of x is, x+2, x, x−2, (c) x, x, Sol. (d) x − 2, , (a), , x+1, , (b) x − 2, x, , (d) x − 2, , BASED, , , Explanation. Since, (1, 1), (2, 2) and (3, 3) are not in R., So, R is not reflexive on A., Now, (1, 2) ∈ R ⇒ (2, 1) ∈ R, and (1, 3) ∈ R ⇒ (3, 1) ∈ R., So, R is symmetric, Clearly, (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∉ R., So, R is not transitive., , 2. Consider the mapping f : A  B is defined by f(x), , COMPETENCY, , 1. A relation R on a set A is said to be an equivalence, relation on A if it is, , • Reflexive i.e., (a, a) ∈ R  a ∈ A., , • Symmetric i.e., (a, b) ∈ R ⇒ (b, a) ∈ R , a, b ∈ A., , • Transitive i.e., (a, b) ∈ R and (b, c) ∈ R, ⇒ (a, c) ∈ R ∀ a, b, c ∈A., , Based on the above information, answer the, following questions:, (i) If the relation R = {(1, 1), (1, 2), (1, 3), (2, 2),, (2, 3), (3, 1), (3, 2), (3, 3)} defined on the set A =, {1, 2, 3}, then R is, (a) reflexive, (b) symmetric, (c) transitive, (d) equivalence, Sol. (a) reflexive, , (iv) If the relation R on the set A = {1, 2, 3, ... 13, 14}, defined as R = {(x, y) : 3x – y = 0}, then R is, (a) reflexive, (b) symmetric, (c) transitive, (d) equivalence, Sol. (d) equivalence

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C–4 n Shiv Das Chapterwise Question Bank (MATHEMATICS XII), , Explanation. We have, g(x) = 2f(x) – 1,  x − 1  − 1 = 2x − 2 − x + 2 = x, = 2,  x − 2, x−2, x−2, , (iv) The function g defined above, is, (a) One-one, (b) Many-one, (c) into, (d) None of these, Sol. (a) One-one, x, x−2, x1, x2, Let g(x1) = g(x2) ⇒, =, x1 − 2 x 2 − 2, , , Explanation. We have, g(x) =, , Relations and Functions, , ⇒ x1x2 – 2x1 = x1x2 – 2x2 ⇒ 2x1 = 2x2 ⇒ x1 = x2, Thus, g(x1) = g(x2) ⇒ x1 = x2, Hence, g(x) is one-one., , (v) A function f(x) is said to be one-one if, (a) f(x1) = f(x2), ⇒ –x1 = x2, , (b) f(–x1) = f(–x2), ⇒ –x1 = x2, (c) f(x1) = f(x2), ⇒ x1 = x2, (d) None of these, Sol. (c) f(x1) = f(x2) ⇒ x1 = x2, 3. A general election of Lok Sabha is a gigantic, exercise. About 911 million people were eligible, to vote and voter turnout was about 67%, the, highest ever, , ONE – NATION, ONE – ELECTION, FESTIVAL OF, DEMOCRACY, GENERAL ELECTION, — 2019, , (iii) Three friends F1, F2 and F3 exercised their, voting right in general election — 2019, then, which of the following is true?, (a) (F1, F2) ∈ R, (F2, F3) ∈ R and (F1, F3) ∈ R, (b) (F1, F2) ∈ R, (F2, F3) ∈ R and (F1, F3) ∉ R, (c) (F1, F2) ∈ R, (F2, F3) ∈ R but (F1, F3) ∉ R, (d) (F1, F2) ∉ R, (F2, F3) ∉ R and (F1, F3) ∉ R, Sol. (a) (F1, F2) ∈ R, (F2, F3) ∈ R and (F1, F3) ∈ R, (iv) The above defined relation R is ............. ., , (a) Symmetric and transitive but not reflexive, , (b) Universal relation, (c) Equivalence relation, (d) Reflexive but not symmetric and transitive, Sol. (c) Equivalence relation, (v) Mr. Shyam exercised his voting right in General, Election — 2019, then Mr. Shyam is related to, which of the following?, , (a) All those eligible voters who cast their votes, , (b) Family members of Mr. Shyam, (c) All citizens of India, (d) Eligible voters of India, Sol. (a) All those eligible voters who cast their votes, 4. Sherlin and Danju are playing Ludo at home, during Covid-19. While rolling the dice, Sherlin’s, sister Raji observed and noted the possible, outcomes of the throw every time belongs to set, {1, 2, 3, 4, 5, 6}. Let A be the set of players while B, be set of all possible outcomes., A = {S, D}, B = {1, 2, 3, 4, 5, 6}, , s, a, iv D, , Sh, , Let I be the set of all citizens of India who were, eligible to exercise their voting right in general, election held in 2019. A relation ‘R’ is defined on, I as follows:, R = {(V1, V2)} : V1, V2 ∈ I and both use their voting, right in general election — 2019}, , Based on the above information answer the following:, (i) Two neighbours X and Y ∈ I. X exercised his, voting right while Y did not cast her vote in, general election — 2019. Which of the following, is true?, (a) (X, Y) ∈ R, (b) (Y, X) ∈ R, (c) (X, X) ∉ R, (d) (X, Y) ∉ R, Sol. (d) (X, Y) ∉ R, (ii) Mr. ‘X’ and his wife ‘W’ both exercised their, voting right in general election — 2019, which, of the following is true?, (a) Both (X, W) and (W, X) ∈ R, (b) (X, W) ∈ R but (W, X) ∉ R, (c) Both (X, W) and (W, X) ∉ R, (d) (W, X) ∈ R but (X, W) ∉ R, Sol. (a) Both (X, W) and (W, X) ∈ R, , Based on the above information answer the, following:, (i) Let R : B  B be defined by R = {(x, y) : y is, divisible by x} is, , (a) Reflexive and transitive but not symmetric, , (b) Reflexive and symmetric but not transitive, , (c) Not reflexive but symmetric and transitive, , (d) Equivalence, Sol. (a) Reflexive and transitive but not symmetric, (ii) Raji wants to know the number of functions, from A to B. How many number of functions, are possible?, (a) 62, (b) 26, (c) 6!, (d) 212, Sol. (a) 62

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Competency Based Questions n C–5, , (iii) Ravi wants to know among those relations, how, many functions can be formed from B to G?, (a) 22, (b) 212, (c) 32, (d) 23, Sol. (d) 23, (iv) Let R : B  G be defined by R = {(b1, g1),, (b2, g2), (b3, g1)}, then R is ______ ., , (a) Injective, , (b) Surjective, (c) Neither Surjective nor Injective, (d) Surjective and Injective, Sol. (b) Surjective, (v) Ravi wants to find the number of injective, functions from B to G. How many numbers of, injective functions are possible?, , (a) 0, (b) 2!, (c) 3!, (d) 0!, Sol. (a) 0, 6. Students of Grade 9, planned to plant saplings along, straight lines, parallel to each other to one side of, the playground ensuring that they had enough, play area. Let us assume that they planted one of, the rows of the saplings along the line y = x – 4., Let L be the set of all lines which are parallel on the, ground and R be a relation on L., , s, a, iv D, , Sh, , E, , BASED, , Ravi decides to explore these sets for various, types of relations and functions., Based on the above information answer the, following:, (i) Ravi wishes to form all the relations possible, from B to G. How many such relations are, possible?, , (a) 26, (b) 25, , (c) 0, (d) 23, Sol. (a) 26, (ii) Let R : B  B be defined by R = {(x, y) : x and y, are students of same sex}, then this relation R is, ............. ., , (a) Equivalence, , (b) Reflexive only, , (c) Reflexive and symmetric but not transitive, , (d) Reflexive and transitive but not symmetric, Sol. (a) Equivalence, , Answer the following using the above information:, (i) Let relation R be defined by R = {(L1, L2) : L1 ||, L2 where L1, L2 ∈ L}, then R is ____ relation., , (a) Equivalence, (b) Only reflexive, , (c) Not reflexive, , (d) Symmetric but not transitive, Sol. (a) Equivalence, (ii) Let R = {(L1, L2) : L1 ^ L2 where L1, L2 ∈ L} which, of the following is true?, , (a) R is Symmetric but neither reflexive nor, transitive., , (b) R is Reflexive and transitive but not symmetric, , (c) R is Reflexive but neither symmetric nor, transitive., , (d) R is an Equivalence relation., Sol. (a) R is Symmetric but neither reflexive nor, transitive., (iii) The function f : R  R defined by f(x) = x – 4 is, _______ ., (a) Bijective, , COMPETENCY, , (iii) Let R be a relation on B defined by R = {(1, 2),, (2, 2), (1, 3), (3, 4), (3, 1), (4, 3), (5, 5)}. Then R is, (a) Symmetric, (b) Reflexive, (c) Transitive, (d) None of these three, Sol. (d) None of these three, (iv) Raji wants to know the number of relations, possible from A to B. How many numbers of, relations are possible?, , (a) 62, (b) 26, (c) 6!, (d) 212, Sol. (d) 212, (v) Let R : B  B be defined by R = {(1, 1), (1, 2),, (2, 2)(3, 3), (4, 4), (5, 5), (6, 6)}, then R is, , (a) Symmetric, , (b) Reflexive and Transitive, (c) Transitive and symmetric, (d) Equivalence, Sol. (b) Reflexive and Transitive, 5. An organization conducted bike race under 2, different categories—boys and girls. Totally, there was 250 participants. Among all of them, finally three from Category 1 and two from, Category 2 were selected for the final race. Ravi, forms two sets B and G with these participants for, this college project., Let B = {b1, b2, b3} and G = {g1, g2} where B, represents the set of boys selected and G the set of, girls who were selected for the final race.

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Relations and Functions, , C–6 n Shiv Das Chapterwise Question Bank (MATHEMATICS XII), , , (b) Surjective but not injective, (c) Injective but not Surjective, (d) Neither Surjective nor Injective, Sol. (a) Bijective, (iv) Let f : R  R be defined by f(x) = x – 4. Then, the range of f(x) is ______ ., , (a) R, (b) Z, (c) W, (d) Q, Sol. (a) R, (v) Let R = {(L1, L2) : L1 is parallel to L2 and, L1 : y = x – 4} then which of the following can, be taken as L2?, , (a) 2x – 2y + 5 = 0, , (b) 2x + y = 5, (c) 2x + 2y + 7 = 0, , (d) x + y = 7, Sol. (a) 2x – 2y + 5 = 0, 7. Raji visited the Exhibition along with her family., The Exhibition had a huge swing, which attracted, many children. Raji found that the swing traced, the path of a Parabola as given by y = x2., , Answer the following questions using the above, information:, (i) Let f : R  R be defined by f(x) = x2 is ____, , (a) Neither Surjective nor Injective, , (b) Surjective, (c) Injective, , (d) Bijective, Sol. (a) Neither Surjective nor Injective, (ii) Let f : N  N be defined by f(x) = x2 is _____, , (a) Surjective but not Injective, , (b) Surjective, (c) Injective, , (d) Bijective, Sol. (c) Injective, (iii) Let f: {1, 2, 3,...}  {1, 4, 9, ...} be defined by, f(x) = x2 is _______ ., (a) Bijective, , (b) Surjective but not injective, (c) Injective but Surjective, (d) Neither Surjective nor Injective, Sol. (a) Bijective, (iv) Let : N  R be defined by f(x) = x2. Range of the, function among the following is ______, , (a) {1, 4, 9, 16, ...}, (b) {1, 4, 8, 9, 10,...}, (c) {1, 4, 9, 15, 16,...} (d) {1, 4, 8, 16,...}, Sol. (a) {1, 4, 9, 16, ...}, (v) The function f : Z  Z defined by f(x) = x2 is, ________, , (a) Neither Injective nor Surjective, (b) Injective, (c) Surjective, , (d) Bijective, Sol. (a) Neither Injective nor Surjective, , s, a, iv D, , Sh, , vvvv

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Chapter: Two, , Inverse Trigonometric Functions, ,  Multiple Choice Questions (MCQs), 1. The value of tan–1, , , , , , (d) − π, 3, −1 1  , , π, (b), 4, , π, , 3, , (d) 2, , 3. If tan–1 4 + cot–1 x =, (a) 4 , (c) 8 , , π, , π, , then value of x is, 2, , (b) – 4, , (d) 2, , 4. If 4 sin–1 x + cos –1 x = p, then value of x is, (a) 4 , , (b) 2, , 1, (c) , 2, , (d) 3, , ,  − 3 , 5. Value of sin  2 cos −1 , is,  5  , , , (a), , −6, , 5, , (b) –6, , (c), , 24, , 25, , (d), , − 24, 25, , −5, 12, , (c) 169 , , (d), , 5, 12, , 7. The principle value of the expression, cos–1[cos(–680°)] is, (a), , − 2π, 9, π, (d), 9, , 2π, , 9, , (b), , (c) 34 π , 9, , , ,  33π  , , 8. The value of sin–1  cos  5   is, − 2π, 5, , (a), , 3π, , 5, , (b), , (c), , π, , 10, , (d) 10, , −π, , 2, 9. If cos  sin −1 + cos −1 x  = 0, then x is, , , , , 5, , (a), , −2, , 5, , (b), , 2, 5, , (c), , 1, , 5, , (d), , −1, 5, , 4π, , 10. If tan–1 x + tan–1 y =, , then cot–1 x + cot–1 y, 5, equals, (a) π, 5, , (b), , 2π, 5, , (c), , 3π, 5, , (d) p, , Competency Based Questions n C–7, , BASED, , (a) π , 6, , (b), , s, a, iv D, , Sh, ,  is equal to, 2  , , − 5 , is equal to, , 12  , , − 120, , 169, , (a), , π, (b), 2, , , , (c), , , , 3 – sec–1 (–2) is, , (a) π , 6, π, (c), , 3, 2. tan–1  2 cos  2 sin, , , , , 6. sin  2 cot −1 , , COMPETENCY, , Competency Based Questions

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C–8 n Shiv Das Chapterwise Question Bank (MATHEMATICS XII), , HINTS AND ANSWERS, 1. (d); tan–1, , , 3 – sec–1(–2), , π, π, , = tan–1  tan π  – [p – sec–1 2 ] = − π + sec −1  sec  = π − π + π = − π, 3, 3, , , 3, , 3, , 3, , 3, , 2. (b); tan–1  2 cos  2 sin −1 1  , , , Inverse Trigonometric Functions, , , , 2 , , , π , π , , , , , tan–1  2 cos  2 sin −1  sin    = tan–1  2 cos  2 ×   = tan–1  2 cos  π  , , , 6, 6,  , , , , , , , , , , , 3 , , , , π, , 1 , , = tan–1  2    = tan–1 (1) =, 4,   2 , 3. (a); tan–1 4 + cot–1 x = π, , , 2, π, = tan–1 4 =, – cot–1 x ⇒ tan–1 4 = tan–1 x, 2, , \ x=4, , 4. (c); 4 sin–1 x + cos–1 x = p ⇒ 4 sin–1 x + π – sin–1 x = p , , 2, π, π, π, = 3 sin–1 x = p –, ⇒ 3 sin–1 x =, ⇒ sin–1 x =, 6, 2, 2,  5 , , , , 3 , , , , 6, , Sh, , = sin  2  π − cos−1   … ∵ cos−1 ( −θ) = π − cos−1 θ, , 5 ,  , , , , 3, = sin  2π − 2 cos−1  = – sin  2 cos−1 3  , , , , = – sin(2A), , , , = – 2 sinA cos A, , , , 24, = – 2  4   3  = −, 25,  5  5, , 5, , , , , , 5, ,  − 5 ,  , 12  , , 6. (a); sin  2 cot −1 , , , , , =, , =, , , , π, − sin −1 x, 2, , 2, , , , , , =, , \ x = sin π = 1, , s, a, iv D, , ,  − 3 , , 5. (d); sin  2 cos−1 , , , , … ∵ cos−1 x, , … [∵ sin(2π − θ) =, , − sin θ, , , −1 3,  Let A = cos 5, , 3, …  cos A = 5, , ∴ sin A = 4, , 5, , 5, , 4, , A, 3, , , ,  − 5,  =y, 12 , , …  Let cot −1 , , , −5, ,  ⇒ 12 = cot y, sin(2y) , ,  ⇒ tan y = − 12, , 5, , 2 tan y, , … ∵ sin 2A = 2 tan A, 1 + tan 2 y, 1 + tan2 A, , ,  − 12 , 2,  5 , , 24, − 24, − 120, 25, 5, =, ×, =, =, =, , 2, 25, 144, +, 5, 169, 169,  − 12 , 1+, 25,  5 , −, , 7. (a); cos–1(cos(–680°)), = cos–1(cos 680°) , , … [∵ cos( −A) = cos A

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Competency Based Questions n C–9, , , , = cos–1(cos(2 × 360° – 40°)), , , , = cos–1(cos 40°) = 40° , , , , = 40 ×, , … [∵ cos(2 × 360° − A) = cos A, , π, 2π, =, 180, 9, 33 π  ,  5  , , , , , 8. (d); sin–1  cos , , = sin–1  cos, , , 30 π + 3 π , , 5, , 3π , , = sin–1  cos  6π +   = sin–1  cos 3 π , , , , 5 , 5 , , , , … [∵ cos(2nπ + θ) = cos θ, , 2π  , 5π − 2π , , –1 , = sin–1  cos,  = sin  cos  π −, , , 5  , 5, 2π , , , , , , 2π  , , = sin–1  − cos  = –sin–1  cos , 5, 5, = –sin–1  sin  π − 2π  , 2, , 5 , = –sin–1  sin, , , π, , − θ, 2, , , , , … ∵ cos θ = sin , , , , , , s, a, iv D, , , , 9. (b); cos  sin −1 + cos−1 x  = 0, , , 5, , ⇒  sin −1 2 + cos−1 x  = cos–1 0, , , − sin −1 A, , π , π,  = − 10, 10 , , 2, , , , … [∵ sin−1 ( −A) =, , 5, , , , ⇒ sin −1, , Sh, , 2, π, + cos−1 x =, 5, 2, , 2, π, 2, 2, , ⇒ sin −1 = − cos−1 x , ⇒ sin −1 = sin −1 x, \ x=, 5, 2, 5, 5, 4π, 10. (a); tan–1 x + tan–1 y =, 5, 4π, π, π, −, 1, −, − cot x + − cot 1 y =, , 2, 2, 5, π π 4π, + −, 2 2, 5, , ⇒ cot–1 x + cot–1 y =, ,  Case Based Questions, 1. The Government of India is planning to fix a, hoarding board at the face of a building on the, road of a busy market for awareness on COVID-19, protocol. Ram, Robert and Rahim are the three, engineers who are working on this project. ‘A’ is, considered to be a person viewing the hoarding, board 20 metres away from the building, standing, at the edge of a pathway nearby, Ram Robert and, Rahim suggested to the film to place the hoarding, board at three different locations namely C, D, and E. ‘C’ is at the height of 10 metres from the, ground level. For the viewer ‘A’, the angle of, elevation of ‘D’ is double the angle of elevation, of ‘C’. The angle of elevation of ‘E’ is triple the, angle of elevation of ‘C’ for the same viewer., , , , =, , π, , − cot −1 A , 2, , , π, 5π + 5π − 8 π, \ cot–1 x + cot–1 y =, 5, 10, , , Look at the figure given and based on the above, information answer the following:, E, D, , A′, , 5m, , b, , A, , a, , C, 10 m, B, , 20 m, , (i) Measure of ∠CAB =, 1, , (a) tan–1(2), , (b) tan–1  , 2, , (c) tan–1(1), , (d) tan–1(3), , 1, , Sol. (b) tan–1  , 2, , BASED, , cot–1 x + cot–1 y =, , …  tan−1 A, , COMPETENCY

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C–10 n Shiv Das Chapterwise Question Bank (MATHEMATICS XII), BC 10 1, 1, =, =, \ α = tan–1  , AB 20 2,  2, , (ii) Measure of ∠DAB =, , (a) R + ,  − π , π ,  2 2,  π π, , (a) tan–1  3 , , (b) tan–1(3), , , (b) R − ,  − 2 , 2 , , (c) tan–1  4 , , (d) tan–1(4), , π π, (c) R,  − , , 2 2, ,  4,  3, , Inverse Trigonometric Functions, , (v) Domain and Range of tan–1 x =, , Sol. (c) tan–1  4 , , , (d) R,  0 ,, ,  3, , 2 tan α, , 1 − tan 2 α, , , Explanation. tan 2α =, ⇒ tan 2a =, , 1, 2 ,  2, 1, 1− ,  2, , 2, , 1 4, =, 3 3, 4, , =, , , , (iii) Measure of ∠EAB, (a) tan–1 (11), , 1, ... ∵ tan α =, 2, , ,  11 , , 2, , (d) tan–1  2 , ,  11 , , Sol. (d) tan–1  2 , , Explanation. tan 3a =, , , tan 3a =, , 1, 1, 3  −  ,  2  2, , 1, 1 − 3 ,  2, 11, ⇒ tan 3a =, 2,  11 , ∴ 3a = tan–1  , 2, , 2, , 3, , s, a, iv D, , 3 tan α − tan 3 α, 1 − 3 tan 2 α, , 3 1 12 − 1, −, 8, = 2 8 =, 4−3, 1, 4, 4, , Sh, , (iv) A’ is another viewer standing on the same line, of observation across the road. If the width of, the road is 5 meters, then the difference between, ∠CAB and ∠CA’B is,  1, ,  1, , (b) tan–1  8 , , 2, , (d) tan–1  21 , , (a) tan–1  12 , ,  11 , , (c) tan–1  , 5,  1, , Sol. (a) tan–1  12 , , 10, CB, , Explanation. CA′B = tan β =, =, 5 + 20, A ′B, 10, 2, tan β = 25 or  5 ,  2, \ β = tan–1  5 , , Now, a – β = tan–1 1 − tan −1 2 , 2, , =, , tan–1, ,  1−2 ,  2 5 , = tan–1, , 1 2, +, ×, 1, , 2 5 , , π π, Sol. (c) R,  − , , 2 2, , 2. Two men on either side of a temple of 30, meters high observe its top at the angles of, elevation α and β respectively (as shown in the, given figure). The distance between the two, men is 40 3 meters and the distance between, the first person A and the temple is 30 3, meters., , Based on the above information answer the, following:, ,  4, , \ 2a = tan–1  3 , , (b) tan–1 (3), , (c) tan–1  11 , , π, 2, , ...[From part (i), 5,  1 ,  10  = tan −1 1,  12 , 12,  10 , , A, , 30m, , Explanation. tan α =, , α, , b, , E, , d, , C, , D, E, , 30 3 m, , 40 3 m, , (i) ∠CAB = α =, 1, ,  2 ,  3 , , (a) sin–1 , , (b) sin–1  , 2, , (c) sin–1(2), , (d) sin–1  3 ,  2 , , , , , , 1, , Sol. (b) sin–1  , 2, , Explanation. In DABD, using pythagoras theorem, AB =, , (, , ( 30 )2 + 30 3, , ), , 2, , =, , ∠CAB = α ⇒ sin α =, ∴ α = sin–1  1 ,  2, , 3600 = 60 m, BE, = 30 = 1, AB, 60 2

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Competency Based Questions n C–11, , 3. If f (x) = sin–1 x, on the above function answer the following, question:, , B, , , Based, , 30 m, , (i) f(x) is equal x to, π, , b, , α, , A, , E, , 30 3 m, , (a) 2 + cos–1 x, (b) π – sin–1 x, , C, , 2, , 10 3 m, , (c) π – cos–1x, , 40 3 m, , 2, , (ii) ∠CAB = α =, (b), ,  5,  3, , (d) Not defined, , cos–1  2 , , π, ,  5, , Sol. (c) 2 – cos–1 x, , 4, , (d) cos–1  , 5, , (c) cos–1  2 , , , , , Explanation. Given. f(x) = sin–1x, , ∴ sin–1x =, , Explanation. We have, ∠CAB = α, ,  3, , ∴ a = cos–1  ,  2 , , s, a, D, hiv, , (iii) ∠BCA = β =,  1, , (b) tan–1 (2), ,  1 , (c) tan–1  , 3, , (d) tan–1, , , f(x) = sin–1x, , S, , ( 3), , ( 3), , BE, , Explanation. In DBCE, tan β =, EC, 30, 3, =, = 3, tan β =, 10 3, 3, , , Put x = –x, f(–x) = sin–1(–x), , = –sin­–1x = –f(x), ∴ f(x) is an odd function., , (iii) If function is f(–x) then equal to, (a) –f(x), (b) f(x), (c) cos f(–x), (d) None of these, Explanation. f(x) = sin–1x, , (iv) ∠ABC =, 4, π, Sol. (c), 2, , π, , (b) 6, , (c), , π, 2, , (d), , π, 3, , , Explanation. In DABC, ∠ABC = π – α – β, , , ∴ ∠ABC = π – cos–1  3  – tan–1 ( 3 ),  2 , , , , BASED, , Sol. (a) –f(x), , \ β = tan–1 ( 3 ), (a) π, , π, – cos–1 x = f(x), 2, , (ii) f(x) is, (a) Odd, , (b) Even, (c) Neither odd nor even, , (d) not defined, Sol. (a) Odd, Explanation., , AE, 30 3, 3, ⇒ In DBCE, cos α = AB = 60 = 2, , Sol. (d) tan–1, , π, 2, , As we know, sin–1x + cos–1x =, ,  3, Sol. (c) cos–1  2 , , , , (a) tan–1  2 , , COMPETENCY, , (a), ,  1, cos–1, , =π– π−π=π, 6, , 3, , Sol. (c) [–1, 1], [0, p], , (iv) If λ(x) =, (a) tan–1x, , 2x, , then f [λ(x)] is equal to, 1 + x2, , (b) 2 tan–1x, , (c) π – tan–1x, 2, , Sol. (b) 2, , (d) π – cot–1x, 2, , tan–1x, , Explanation. We have, λ(x) =, , 2, , (v) Domain and Range of cos–1 x =, (a) (1, 1), (0, p) , (b) [–1, –1], [0, p], (c) [–1, 1], [0, p] , , ⇒ f(–x) = sin–1(–x) ∴ –sin­–1x = –f(x), ,  π π, (d) (–1, 1),  − 2 , 2 , , , , , f (x) = sin–1x, f [λ(x)] = sin–1, , 2x, 1 + x2, ...[Given, , 2x, = 2 tan–1x, 1 + x2

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C–12 n Shiv Das Chapterwise Question Bank (MATHEMATICS XII),  2x , ,  2x , , , , 4. If f(x) = tan–1  1 − x 2 , , Explanation. We know, tan–1  1 − x 2 , , , Based, , on the above function answer the following, question:, (i) f(x) is equal to, (a) tan–1 x, (b) 2 tan–1 x, , Inverse Trigonometric Functions, , (c) cot–1 x, Sol. (b) 2 tan–1 x, , 1−, , x2 , , (d) cos–1  1 + x 2 , , (a) π + cot −1 x, 2, , (c), , π, + sin −1 x, 2, ,  1 + x2 ,  2 x , , , (d) cos −1 , , (b) π − cot −1 x, 2, ,  1 − x2 , , Sol. (c) cos–1  1 + x 2 ,  2x , Explanation. We know, tan–1  1 − x 2 , , (d) Not defined, , , 2, 1, −, x, , = 2 tan–1x = cos–1 ,  1 + x 2 , , π, ... ∵ tan −1 x + cot −1 x =, 2, , , ∴ f(x) = π – 2 cot–1x, , (v) Maximum value of f(x), (a) [–1, 1], (b) [0, ∞), (c) (–∞, 0], (d) not maxima, , v, i, h, S, , 2x , (b) sin −1 , 1 − x2 , ,  2x , (c) sin −1  1 + x 2 , , (d) Not defined, , 2x , sin −1 ,  1 + x2 , , , , s, a, D, , Sol. (d) not maxima, , x , (a) sin −1 , 1 + x2 , , Sol. (c), , 2, ,  1 − x2 , , , Explanation. From part (i), f(x) = 2 tan–1x, , (iii) f(x) is equal to, , 1+ x , (a) cos −1 , 1 − x 2 , , (c) cos–1  1 + x 2 , , Sol. (d) Not defined, π, , ⇒ f(x) = 2  − cot −1 x  , 2, , , (iv) f(x) is equal to, ,  1 − x2 , , (b) cos −1  2 x , , 2x , , Explanation. Given. f(x) = tan −1 ,  1 − x 2 , ∴ f(x) = 2 tan–1 x, , (ii) f(x) is equal to, , 2x , = 2 tan–1x = sin–1 ,  1 + x 2 , , Explanation., , −, , ∞, , –∞, −, , vvvv, , π, 2, , π, 2

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OMR Based, Sample Question Paper, Time Allowed : 90 minutes, , Maximum Marks : 40, , GENERAL INSTRUCTIONS:, (i) Question paper comprises two Sections—A and B. There are 40 questions in the question paper. All questions are, compulsory., (ii) Section A – Questions no. 1 to 16 are Multiple Choice Questions (MCQs)., (iii) Section B – Questions no. 17 to 40 are Case Study Based MCQs., (iv) The Answer sheet is inside this Test Booklet. When you are directed to open the Test Booklet, take out the, Answer Sheet and fill in the particulars on SIDE-2 carefully with Blue/Black ball point pen only. In no case, pencil is to be used. Read “General Instructions for Candidates” on SIDE-1 carefully., (v) The candidate should ensure that the Answer Sheet is not folded. Do not make any stary marks on the Answer, Sheet. Do not write your Roll Number anywhere else except in the specified space in the Test Booklet/Answer, Sheet., , s, a, iv D, , SECTION — A (Multiple Choice Questions), , Sh, , 1. The equation of tangent drawn to the curve y = sin x, at the point (0, 0) is, , (a) y + x = 1, (b) y – x = 1, (c) y + x = 0 (d) y – x = 0, 2. Differentiate cos2(x3) w.r.t. x3 is equal to, (b) –sin(2x3), , , (a) – cos(2x3), , (c) sin(2x3), , (d) cos (2x3), , 3. The function f (x) = [x], where [x] denotes the greatest integer function, is continuous at, (b) 6, (c) – 2.4, (d) 376, 4. Let A = {1, 2, 3, 4}. Let R be the equivalence relation on A × A defined by (a, b) R (c, d) if a + d = b + c., Then the equivalence class [(1, 3)] is, , , (a) – 5, , , (a) {(1,, , 3)} , , (c) {(1, 8), (2, 4), (1, 4)} , , (b) {(2, 4)}, (d) {(1, 3) (2, 4)}, , 5. Which of the following functions form Z into Z bijections?, (a) f (x) = x3, , (b) f (x) = x + 2, , (c) f (x) = 2x + 1, , (d) f (x) = x2 + 1, , (c) – cot x, , (d) – tan x, , π, 6. The point on the curve y2 = x, where the tangent makes an angle of, with x-axis is, 4,  1 1,  1 1, (a)  , , (b)  , , (c) (4, 2), (d) (1, 1), 2 4, 4 2, , 7. The derivative of cos x w.r.t. sin x is, , (a) cot x, (b) tan x, , 8. Z = 7x + y, subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at, (a) (3, 0), , 1 5, (b)  , , 2 2, , (c) (7, 0), , OMR–1, , (d) (0, 5)

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OMR–2 n Shiv Das Chapterwise Question Bank (Mathematics XII), , 9. If y =, , sin x + y , then, , cos x, , , (a) 2 y − 1, , dy, is equal to, dx, cos x, (b) 1 − 2 y, , sin x, , (c) 1 − 2 y, , (d), , 4π, , then cot–1 x + cot–1 y equals, 5, 2π, 3π, (a) π, (b), (c), 5, 5, 5, 2, 11. If f : R → R be defined by f(x) = , x ∀ R, then f is, x, , sin x, 2y − 1, , 10. If tan–1 x + tan–1 y =, , (a) one-one, , (b) onto, , (d) p, , (d) f is not defined, , (c) bijective, ,  cos θ sin θ , 12. If A = ,  n ∈ N , then value of Det (An) is,  − sin θ cos θ , , (a) –1, , (c) n, , (b) 0, , 13. Find the point on the curve y =, , x2, , (d) 1, , – 2x + 3, when the tangent is parallel to x-axis., , , (a) (1, 2), (b) (2, 1), (c) (– 1, 2), 14. The value of |3I3| is …… when I3 is the identity matrix of order 3 , (a) 3, , (b) 9, , (d) (1, – 2), , (c) 27, , (d) 81, , 15. The line y = x + 1 is a tangent to the curve y2 = 4x at the point, , (a) (1, 2), , s, a, iv D, , (b) (2, 1), , (c) (1, – 2), , (d) (– 1, 2), , 16. The function f (x) = tan x – x, , (a) always increases, (b) never increases, , (c) always decreases , (d) sometimes increases and sometimes decreases, , Sh, , SECTION - B (Case Study Based Questions), Read the given para and answer the following MCQs by choosing the most appropriate option (Q. No. 17 to 20):, , q A trust fund has `35,000 that must be invested, in two different types of bonds, say X and Y. The first, bond pays 10% interest p.a. which will be given to an old, age home and second one pays 8% interest p.a. which, will be given to WWA (Women Welfare Assocation). Let, A be a 1 × 2 matrix and B be a 2 × 1 matrix, representing, the investment and interest rate on each bond respectively., 17. If `15,000 is invested in Bond X, then, Investment, , X, , Y, , (a) A = X  15, 000  ; B = 0.1 0.08 Interest rate, , , Y  20 , 000 , , X, , Y, , Interest Rate, , , (c) A = Investment 20 , 000 15, 000  ; B = X 0.08, , Y  0.1 , , X, , Y, , Interest rate, , (b) A = Investment 15, 000 20 , 000  ; B = X  0.1 , Y 0.08, (d) None of these, , 18. If `15,000 is invested in Bond X, then total amount of interest received on both bonds is, (a) `2,000, (b) `2,100, (c) `3,100, (d) `4,000, 19. If the trust fund obtains an annual total interest of `3,200, then the investment in two bonds is, (a) `15,000 in X, `20,000 in Y, (b) `17,000 in X, `18,000 in Y, (c) `20,000 in X, `15,000 in Y, (d) `18,000 in X, `17,000 in Y, 20. The total amount of interest on both bonds is given by, (a) AB, (b) A’B, (c) B′A, (b) None of these

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OMR Based Sample Question Paper n OMR–3, Read the given para and answer the following MCQs by choosing the most appropriate option (Q. No. 21 to 24):, , q If a real valued function f(x) is finitely derivable at any point of its domain. It is necessarily continuous at, that point. But its converse need not be true., For example, Every polynomial, constant function are both, continuous as well as differentiable and inverse trigonometric functions are continuous and differentiable, in its domains etc.,  x , for x ≤ 0, , 21. If f(x) = , , then at x = 0.,  0 , for x > 0, (a) f(x) is differentiable and continuous, (b) f(x) is neither continuous nor differentiable, (c) f(x) is continuous but not differentiable, (d) None of these, 22. If f(x) =|x – 1|, x ∈ R, then at x = 1., (a) f(x) is not continuous , (b) f(x) is continuous but not differentiable, (c) f(x) is continuous and differentiable, (d) none of these, 23. f(x) = x3 is, (a) continuous but not differentiable at x = 3, (b) continuous and differentiable at x = 3, (c) neither continuous nor differentiable at x = 3 (d) none of these, 24. If f(x) = |sin x|, then which of the following is true?, (a) f(x) is continuous and differentiable at x = 0 (b) f(x) is discontinuous at x = 0, (c) f(x) is continuous at x = 0 but not differentiable, (d) f(x) is differentiable but not continuous at x =, , π, 2, , Read the given para and answer the following MCQs by choosing the most appropriate option (Q. No. 25 to 28):, , q Shobhit′s father wants to construct a rectangular garden using a brick will on one side of the garden and, wire fencing for the other three sides as shown in figure. He has 200 ft. of wire fencing., 25. To construct a garden using 200 ft. of fencing, we need to maximise its, (a) volume, (b) area, (c) perimeter, (d) length of the side, 26. If x denotes the length of side of garden perpendicular to brick wall and y denotes the length of side, parallel to brick wall, then find the relation representing total amount of fencing wire., (a) x + 2y = 150, (b) x + 2y = 50, (c) y + 2x = 200, (d) y + 2x = 100, 27. Area of the garden as a function of x, say A(x), can be represented as, (a) 200 + 2x2, (b) x – 2x2, (c) 200x – 2x2, (d) 200 – x2, 28. Maximum value of A(x) occurs at x equals, (a) 50 ft., (b) 30 ft., (c) 26 ft., (d) 31 ft., , s, a, iv D, , Sh, , Read the given para and answer the following MCQs by choosing the most appropriate option (Q. No. 29 to 32):, , q The Government of India is planning to fix a hoarding board at the face, E, of a building on the road of a busy market for awareness on COVID-19, D, protocol. Ram, Robert and Rahim are the three engineers who are, working on this project. ‘A’ is considered to be a person viewing the, C, hoarding board 20 metres away from the building, standing at the edge, 10 m, of a pathway nearby, Ram Robert and Rahim suggested to the film to, a, B, A′ 5 m b, 20 m, A, place the hoarding board at three different locations namely C, D and, E. ‘C’ is at the height of 10 metres from the ground level. For the viewer ‘A’, the angle of elevation of, ‘D’ is double the angle of elevation of ‘C’. The angle of elevation of ‘E’ is triple the angle of elevation of, ‘C’ for the same viewer., 29. Measure of ∠CAB =, 1, (a) tan–1(2), (b) tan–1  , (c) tan–1(1), (d) tan–1(3), 30. Measure of ∠DAB =, , (a) tan–1  3 ,  4, , 2, , (b) tan–1(3), , (c) tan–1  4 , , (b) tan–1 (3), , (c) tan–1  11 , , 31. Measure of ∠EAB, (a) tan–1 (11), ,  3, 2, , (d) tan–1(4),  11 , , (d) tan–1  2 

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OMR–4 n Shiv Das Chapterwise Question Bank (Mathematics XII), , 32. A’ is another viewer standing on the same line of observation across the road. If the width of the road, is 5 meters, then the difference between ∠CAB and ∠CA’B is,  1, , (a) tan–1  12 , ,  1, ,  11 , , 2, , (c) tan–1  , 5, , (b) tan–1  8 , , (d) tan–1  21 , , Read the given para and answer the following MCQs by choosing the most appropriate option (Q. No. 33 to 36):, , q Corner points of the feasible region for an LPP are (0, 3), (5, 0), (6, 8), (0, 8). Let Z = 4x – 6y be the objective, function. Based on the above information, answer the following questions:, 33. The minimum value of Z occurs at, (a) (6, 8), (b) (5, 0), (c) (0, 3), (d) (0, 8), 34. Maximum value of Z occurs at, (a) (5, 0), (b) (0, 8), (c) (0, 3), (d) (6, 8), 35. Maximum of Z - Minimum of Z =, (a) 58, (b) 68, (c) 78, (d) 88, 36. The corner points of the feasible region deter-mined by the system of linear inequalities are, x2, 7, , l1, , 6, 5, 4, 3, D(0, 3), , C(2, 3), , s, a, iv D, , 1, , (a) (0, 0), (–3, 0), (3, 2), (2, 3), (c) (0, 0), (3, 0), (3, 2), (2, 3), (0, 3), , A(3, 0), , Sh, O, , l2, , B(3, 2), , 2, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , x1, , l3, , (b) (3, 0), (3, 2), (2, 3), (0, –3), (d) None of these, , Read the given para and answer the following MCQs by choosing the most appropriate option (Q. No. 37 to 40):, , q If there is a statement involving the natural number n such that, , (i) The statement is true for n = 1, , (ii) When the statement is true for n = k (where k is some positive integer), then the statement is also true, for n = k + 1. Then, the statement is true for all natural numbers n., Also, if A is a square matrix of order n, then A2 is defined as AA. In general, Am = AA...A, (m times), where m is any positive integer., 3 −4 , , 37. If A = ,  , then for any positive integer n.,  1 −1,  3n −4n, − n , , (a) An = , n, , (b) An, , =, , 1 + 2 n −4n ,  3n −8n, n,  n, 1 − 2 n (c) A =  1 − n , , , 1 + 3n, , (d) An = ,  n, , −4n , 1 − 3n, , 1 2, , n, 38. If A = ,  , then |A |, where n ∈ N, is equal to, 0 1, , (a) 2n, , (b) 3n, , (c) n, , (d) 1, , 39. If A is a square matrix such that |A| = 2, then for any positive integer n, |An­| is equal to, (a) 0, , (b) 2n, , (c) 2n, , (d) n2, , a 0 0 , , 40. Let A =  0 a 0  and An = [aij]3×3 for some positive integer n, then the cofactor of a13 is, , ,  0 0 a, , (a) an, , (b) –an, , (c) 2an, , (d) 0

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OMR–6 n Shiv Das Chapterwise Question Bank (Mathematics XII), , osQoy uhys@dkys ckWy IokbaV isu dk iz;ksx djsa /, , i`"B&2 / SIDE-2, , iathdj.k la[;k @ Registration No., /, , mÙkj if=kdk la[;k/, Answer Sheet No., , USE BLUE/BLACK BALL POINT PEN ONLY, , fo|ky; dksM, , iz'u iqfLrdk la[;k, , School Code, , iz'u iqfLrdk dksM, , p;fur Hkk"kk, , Test Booklet Code, , Languages, attempted, , Test Booklet No., , A1, , /, , A 1 1, , 1, , 1 1 1 1 1, , 1 1 1 1, , 1 1 1 1 1, , 1 1 1 1 1 1 1, , B 2 2, , 2, , 2 2 2 2 2, , 2 2 2 2, , 2 2 2 2 2, , 2 2 2 2 2 2 2, , C 3, , 3, , 3 3 3 3 3, , 3 3 3 3, , 3 3 3 3 3, , 3 3 3 3 3 3 3, , D 4, , 4, , 4 4 4 4 4, , 4 4 4 4, , 4 4 4 4 4, , 4 4 4 4 4 4 4, , G 5, , 5, , 5 5 5 5 5, , 5 5 5 5, , 5 5 5 5 5, , 5 5 5 5 5 5 5, , M 6, , 6, , 6 6 6 6 6, , 6 6 6 6, , 6 6 6 6 6, , 6 6 6 6 6 6 6, , P 7, , 7, , 7 7 7 7 7, , 7 7 7 7, , 7 7 7 7 7, , 7 7 7 7 7 7 7, , R 8, , 8, , 8 8 8 8 8, , 8 8 8 8, , 8 8 8 8 8, , 8 8 8 8 8 8 8, , 9, , 9, , 9 9 9 9 9, , 9 9 9 9, , 9 9 9 9 9, , 9 9 9 9 9 9 9, , 0, , 0, , 0 0 0 0 0, , 0 0 0 0, , 0 0 0 0 0, , 0 0 0 0 0 0 0, , Q.No., , Response, , Hindi, English, , Q.No., , Response, , 01, , a, , b, , c, , d, , 21, , a, , b, , c, , d, , 02, , a, , b, , c, , d, , 22, , a, , b, , c, , d, , 03, , a, , b, , c, , d, , 23, , a, , b, , c, , d, , 04, , a, , b, , c, , d, , 24, , a, , b, , c, , d, , 05, , a, , b, , c, , d, , 25, , a, , b, , c, , d, , 06, , a, , b, , c, , d, , 26, , a, , b, , c, , d, , 07, , a, , b, , c, , d, , 27, , a, , b, , c, , d, , 08, , a, , b, , c, , d, , 28, , a, , b, , c, , d, , 09, , a, , b, , c, , d, , 29, , a, , b, , c, , d, , 10, , a, , b, , c, , d, , 30, , a, , b, , c, , d, , 11, , a, , b, , c, , d, , 31, , a, , b, , c, , d, , 12, , a, , b, , c, , d, , 32, , a, , b, , c, , d, , 13, , a, , b, , c, , d, , 33, , a, , b, , c, , d, , 14, , a, , b, , c, , d, , 34, , a, , b, , c, , d, , 15, , a, , b, , c, , d, , 35, , a, , b, , c, , d, , 16, , a, , b, , c, , d, , 36, , a, , b, , c, , d, , 17, , a, , b, , c, , d, , 37, , a, , b, , c, , d, , 18, , a, , b, , c, , d, , 38, , a, , b, , c, , d, , 19, , a, , b, , c, , d, , 39, , a, , b, , c, , d, , 20, , a, , b, , c, , d, , 40, , a, , b, , c, , d, , v, i, h, S, , s, a, D, , mÙkj if=kdk fujh{kd dks lkSaius ls igys vH;FkhZ ;g tk¡p dj ysa fd iathdj.k la[;k] fo|ky; dksM] iz'u iqfLrdk la[;k vkSj, p;fur Hkk"kk lgh <ax ls Hkj fn;s x, gSa vkSj fpfUgr dj fn;s x, gSaA, Before handling over the Answer Sheet to the invigilator, the candidate should check that Registration No., School Code,, Test Booklet No. & Language Attempted have been filled in and market correctly., , vH;FkhZ dk uke, , Candidate’s Name, , vH;FkhZ osQ gLrk{kj (pyrs gkFk ls fy[ksa), , vH;FkhZ osQ gLrk{kj@Invigilator’s Signature, , Candidate Signature (in running hand), , ----------

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OMR Based Sample Question Paper n OMR–7, , HINTS AND ANSWERS,        1. (d); y = sin x   , , \, , dy , , dy, = cos x, dx, , Now, dx , = cos 0 = 1 = m = slope of tangent, at(0 , 0), , Equation of tangent at (0, 0) is y – y1 = m(x – x1) ⇒ y – 0 = 1(x – 0) ⇒ y – x = 0, , 2. (b); Let y = cos2 (x3) , Let z = x3, , , , dy, dx, , = 2 cos (x3) (–sin x3). 3x2 = –3x2 sin (2x3) ...(i) [∵ 2 sin q cos q = sin 2q, , \, , 1, dy dy dx, =, ×, = –  3x2 sin (2x3) ×, = – sin(2x3), 3x2, dz dx dz, 3. (c); because the greatest integer function [x] is discontinuous at all integral values of x., , dz, = 3x2 , dx, , …(ii), , From (i) and (ii),, , So discontinuous at – 5, 6 and 376 and continuous at – 2.4., 4. (d); Equivalence class of [(1, 3)] is given by set of ordered pair (a, b) ∈ A × A such that, (1, 3) R (a, b), ⇒ 1 + b = 3 + a , \ [(1, 3)] = {(1, 3), (2, 4)}, 5. (b); f (x) = x + 2, For one-one, f (x1) = f (x2), ⇒ x1 + 2 = x2 + 2, ⇒ x1 = x2 , , For onto, Let y = x + 2, ⇒ y – 2 = x , ⇒ f (x) = x + 2, Now, f (y – 2) = y – 2 + 2 = y , \ f is onto, Hence f (x) = x + 2 is one-one and onto and hence bijective., , \ f is one-one, , 6. (b); y2 = x …(i), , \ 2y, , \, , dy, = 1 , dx, , s, a, iv D, , ⇒ Slope of tangent,, , Sh, , 1, = 1 , 2y, , ⇒ 2y = 1, , dy, 1, =, and tan π = 1, dx, 2y, 4, , π, , [∵ Slope of tangent = tan 4 (Given), , ⇒ y= 1, 2, , 2, 1,  1 1, Putting the value of y in (i), we get  1  = x ⇒ x = 4 ∴ Point (x, y) =  4 , 2 ,  2, , dy, ⇒, = – sin x …(i), dx, , 7. (d); Let y = cos x, From (i) and (ii),, , Let z = sin x, , ⇒, , dz, = cos x …(ii), dx, , dy, dy dx, 1, =, ×, = –sin x ×, = – tan x, dz, dx dz, cos x, , sin x + y, , ⇒ y2 = sin x + y…[squaring both sides, dy, dy, dy, cos x, Differentiating both sides w.r.t. x, 2y, = cos x +, ⇒, =, dx, dx, dx, 2y − 1, 4π, π, π, 4π, π, − cot −1 x + − cot −1 y =, 10. (a); tan–1 x + tan–1 y =, ⇒, … ∵ tan −1 A = − cot −1 A , 2, , , 5, 2, 2, 5, π, π, +, π, −, π, π, π, 4, π, 5, 5, 8, cot–1 x + cot–1 y = + −, ⇒ cot–1 x + cot–1 y =, \ cot–1 x + cot–1 y =, 5, 5, 10, 2 2, 2, 11. (d); f (x) =, ,∀ x ∈ R, x, 2, When x = 0, f (0) =, ( Not defined), \ f is not defined., 0, 9. (a); y =, ,  cos θ sin θ , 12. (d); A = , ,  − sin θ cos θ , 13. (a); y = x2 – 2x + 3, , ⇒, \, ,  cos nθ sin nθ , An = , ,  − sin nθ cos nθ , , ⇒, , |An | = cos2 nq + sin2 nq = 1, , dy, = 2x – 2 ⇒ 2x – 2 = 0 ⇒ x = 1 , dx, , Now, at x = 1, y = (1)2 – 2(1) + 3 ⇒ y = 4 – 2 = 2, , …[∵ tangent is || to x-axis, , \ Point (1, 2), , 14. (c); |3I3| = 33 |I3|= 27 (1) = 27… ∵| kA |= k n | A | and | I3 |=| I |= 1

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OMR–8 n Shiv Das Chapterwise Question Bank (Mathematics XII), 15. (a); Point (1, 2) satisfies both the equations., , y = x + 1  , , y2 = 4x    …[Putting x = 1, y = 2, , …[Putting x = 1, y = 2, , (2)2 = 4(1), , 2 = 1 + 1 ⇒ 2 = 2, true, , ⇒ 4 = 4, true, , , 16. (a); f (x) = tan x – x ⇒ f ′(x) = sec2 x – 1 For all x ∈ R, f ′(x) = sec2 x – 1 > 0 \ f (x) always increases., 17. (b); If `15,000 is invested in Bond X, then the amount invested in Bond Y = `(35,000 – 15,000) = `20,000, X, , A = Investment 15 , 000, , Interest rate, , Y, , Interest rate, , 20 , 000  and B = X 10% or X  0.1 , Y  8% , Y 0.08, , 18. (c); The amount of interest received on each bond is given by,  0.1 , AB = 15 , 000 20 , 000  ,  = [15,000 × 0.1 + 20,000 × 0.08] = [1,500 + 1,600] = `3,100, 0.08, 19. (c); Let `x be invested in Bond X and then `(35,000 – x) will be invested in Bond Y.,  0.1 , Now total amount of interest is given by [x 35,000 – x] ,  = [0.1x + (35,000 – x)0.08], 0.08, Total amount of interest = `3,200 ⇒ 0.1x + 2800 – 0.08x = 3,200 ⇒ 0.02x = 400 ⇒ x = 20,000, Thus, `20,000 invested in Bond X and `35,000 – `20,000 = `15,000 invested in Bond Y., 27. (c); Area of garden = A(x) = x . y = x(200 – 2x) = 200x – 2x2, ...[we know, y = 200 – 2x, 28. (a); A(x) = 200x – 2x2, ⇒ A′(x) = 200 – 4x, For the area to be maximum A′(x) = 0, 200 – 4x = 0, \ x = 50 ft., 29. (b); tan α =, , BC 10 1, 1, =, =, \ α = tan–1  , AB 20 2,  2, , 2 tan α, 30. (c); tan 2α =, 1 − tan 2 α, , ⇒ tan 2a =, , 3 tan α − tan 3 α, 31. (d); tan 3a =, 1 − 3 tan 2 α, , 1, 2 ,  2, , 2, , =, , s, a, iv D, , Sh, 1, 1− ,  2, , ⇒ tan 3a =, , 1 4, =, 3 3, 4, , 1, 1, 3  −  ,  2  2, 2, , 4, \ 2α = tan–1  , 3, , 3, , 1, 1 − 3 ,  2, 10, 10, 2, or, 32. (a); CA′B = β; tan β = CB =, =,    \, 5 + 20, 25, 5, A ′B, , =, , 1, ... ∵ tan α =, 2, , , 12 − 1, 8 = 11 ∴ tan 3a = tan–1  11 ,  2, 1, 2, 4,  2, β = tan–1  5 , ,  1 ,  1−2 , 1,  10 , , , 1, 2, 2, 5, Now a – β = tan–1 − tan −1 = tan–1 , = tan–1  12  = tan −1 12, 1 2, 2, 5,  10 ,  1 + 2 × 5 , 33. (d); Construct the following table of values of objective function., , Corner points, Value of Z = 4x – 6y, , (0, 3), –18, , (5, 0), 20 (Max.), , (6, 8), –24, , (0, 8), –48 (Min.), , , Minimum value of Z is –48 which occurs at (0, 8)., 35. (b); Maximum of Z – Minimum of Z = 20 – (–48) = 20 + 48 = 68,  3 −4 ,  3 −4   3 −4   5 −8, 2, 37. (b); We have, A = ,    ⇒ A =  1 −1  1 −1 =  2 −3 , which can be obtained from,  1 −1, , ,  , , 1, +, 2, n, −, 4, n, , , An = , for n = 2., 1 − 2 n ,  n, 38. (d); We have, |A| =, , 1 2, = 1 – 0 = 1   ∴ |An| = |A.A ... A (n times)| = |A|n = 1n = 1, 0 1, , 39. (c); We have |A| = 2 and |An| = |A.A ... A (n-times)| = |A| |A| ... |A|(n-times) = |A|n = 2n,  a2 0 0 ,  an 0 0 , 0 an, , , , 2, 40. (d); Here, A2 = A.A =  0 a 0  and similarly, An  0 an 0     ∴ Now, cofactor of a13 = (–1)1+3 0 0 = 0,  0 0 a 2 ,  0 0 an 