Page 1 :
www.tntextbooks.in, , GOVERNMENT OF TAMIL NADU, , HIGHER SECONDARY FIRST YEAR, , BUSINESS MATHEMATICS, AND, STATISTICS, , A publication under Free Textbook Programme of Government of Tamil Nadu, , Department of School Education, Untouchability is Inhuman and a Crime, , 00_11th_BM-STAT_FM_EM.indd 1, , 21-04-2020 12:06:39 PM

Page 2 :
www.tntextbooks.in, , Government of Tamil Nadu, First Edition, , Revised Edition -, , 2018, 2019, 2020, , (Published under New Syllabus), , NOT FOR SALE, , Content Creation, , The wise, possess all, , State Council of Educational, Research and Training, © SCERT 2018, , Printing & Publishing, , Tamil NaduTextbook and Educational, Services Corporation, www.textbooksonline.tn.nic.in, ii, , 00_11th_BM-STAT_FM_EM.indd 2, , 21-04-2020 12:06:41 PM

Page 3 :
www.tntextbooks.in, , HOW TO USE, THE BOOK, Career Options, , Learning Objectives:, Note, , List of Further Studies & Professions., , Learning objectives are brief statements that describe what, students will be expected to learn by the end of school year,, course, unit, lesson or class period., , Additional information about the concept., , Amazing facts, Rhetorical questions to lead students, to Mathematical inquiry, , Exercise, ICT, Web links, Quick Response, Code, Miscellaneous, Problems, Glossary, References, , Assess students’ critical thinking and their understanding, , To enhance digital skills among students, , List of digital resources, , To motivate the students to further explore the content, digitally and take them in to virtual world, , Additional problems for the students, , Tamil translation of Mathematical terms, , List of related books for further studies of the topic, , Let’s use the QR code in the text books!, •, •, •, •, , Download DIKSHA app from the Google Play Store., Tap the QR code icon to scan QR codes in the textbook., Point the device and focus on the QR code., On successful scan, content linked to the QR code gets listed., , Note: For ICT corner, Digi Links QR codes use any other QR scanner., , iii, , 00_11th_BM-STAT_FM_EM.indd 3, , 21-04-2020 12:06:42 PM

Page 4 :
www.tntextbooks.in, , CAREER OPTIONS IN BUSINESS, MATHEMATICS AND STATISTICS, Higher Secondary students who have taken commerce with, Business mathematics and statistics can take up careers in BCA, B.Com.,, and B.Sc. Statistics. Students who have taken up commerce stream, have, a good future in banking and financial institutions., A lot of students choose to do B.Com with a specialization in computers., Higher Secondary Commerce students planning for further studies can take up careers, in professional fields such as Company Secretary , Chartered Accountant (CA), ICAI and, so on. Others can take up bachelor’s degree in commerce (B.Com), followed by M.Com,, Ph.D and M.Phil. There are wide range of career opportunities for B.Com graduates., After graduation in commerce, one can choose MBA, MA Economics, MA, Operational and Research Statistics at Postgraduate level. Apart from these, there are, several diploma, certificate and vocational courses which provide entry level jobs in the, field of commerce., Career chart for Higher Secondary students who have taken commerce, with Business Mathematics and statistics., Courses, B.Com., B.B.A., B.B.M., B.C.A.,, B.Com (Computer), B.A., , Institutions, •, •, •, •, , B.Sc Statistics, , •, •, •, •, •, •, , Scope for further, studies, , Government Arts & Science Colleges,, Aided Colleges, Self financing Colleges., Shri Ram College of Commerce (SRCC),, Delhi, Symbiosis Society’s College of Arts &, Commerce, Pune., St. Joseph’s College, Bangalore, , C.A., I.C.W.A, C.S., , Presidency College, Chepauk, Chennai., Dr. Ambedkar Govt. Arts College,, Vyasarpadi, Chennai - 39., Govt. Arts College, Tindivanam,, Villupram & Nagercoil., Madras Christian College, Tambaram, Loyola College, Chennai., D.R.B.C.C Hindu College, Pattabiram,, Chennai., , M.Sc., Statistics, , B.B.A., LLB, B.A., LLB,, B.Com., LL.B. (Five years, integrated Course), , •, •, , Government Law College., School of excellence, Affiliated to, Dr.Ambethkar Law University, , M.L., , M.A. Economics (Integrated Five, Year course) – Admission based on, All India Entrance Examination, , •, , Madras School of Economics,, Kotturpuram, Chennai., , Ph.D.,, , B.S.W., , •, , School of Social studies, Egmore, Chennai, , M.S.W, , iv, , 00_11th_BM-STAT_FM_EM.indd 4, , 21-04-2020 12:06:42 PM

Page 5 :
www.tntextbooks.in, , CONTENTS, Ch. No., 1, , TITLE, , Page No., , Matrices and Determinants, , 1-24, , 1.1, , Determinants, , 1, , 1.2, , Inverse of a Matrix, , 8, , 1.3, , Input –Output Analysis, , 15, , 2, , Algebra, Partial Fractions, , 25, , 2.2, , Permutations, , 29, , 2.3, , Combinations, , 36, , 2.4, , Mathematical Induction, , 39, , 2.5, , Binomial Theorem, , 41, , Analytical Geometry, Locus, , 51, , 3.2, , System of Straight Lines, , 53, , 3.3, , Pair of Straight Lines, , 57, , 3.4, , Circles, , 60, , 3.5, , Conics, , 66, , Trigonometry, , July, , July, , 76-95, , 4.1, , Trigonometric Ratios, , 78, , 4.2, , Trigonometric Ratios of Compound Angles, , 81, , 4.3, , Transformation Formulae, , 85, , 4.4, , Inverse Trigonometric Functions, , 89, , 5, , June, , 51-75, , 3.1, , 4, , June, , 25-50, , 2.1, , 3, , Month, , Differential Calculus, , August, , 96-127, , 5.1, , Functions and their Graphs, , 97, , 5.2, , Limits and Derivatives, , 106, , 5.3, , Differentiation Techniques, , 115, , August, September, , v, , 00_11th_BM-STAT_FM_EM.indd 5, , 21-04-2020 12:06:42 PM

Page 6 :
www.tntextbooks.in, , 6, , Applications of Differentiation, , 128-160, , 6.1, , Applications of Differentiation in Business, and Economics, , 128, , 6.2, , Maxima and Minima, , 139, , 6.3, , Applications of Maxima and Minima, , 144, , 6.4, , Partial Derivatives, , 149, , 6.5, , Applications of Partial Derivatives, , 152, , 7, , Financial Mathematics, , 161-177, , 7.1, , Annuities, , 161, , 7.2, , Stocks, Shares, Debentures and Brokerage, , 167, , 8, , Descriptive Statistics and Probability, Measures of Central Tendency, , 178, , 8.2, , Measures of Dispersion, , 188, , 8.3, , Probability, , 196, , Correlation and Regression Analysis, , October, , 178-209, , 8.1, , 9, , October, , November, , 210-235, , 9.1, , Correlation, , 210, , 9.2, , Rank Correlation, , 215, , 9.3, , Regression Analysis, , 218, , 10, , Operations Research, , November, , 236-255, , 10.1, , Linear Programming Problem, , 236, , 10.2, , Network Analysis, , 244, , Answers, , 256-274, , Tables, , 275-280, , Books for Reference, , December, , 281, , As the Statistics component of this text book involves problems based on numerical, calculations, Business Mathematics and Statistics students are advised to use calculator, , E- Book, , Assessment, , DIGI Links, , vi, , 00_11th_BM-STAT_FM_EM.indd 6, , 21-04-2020 12:06:42 PM

Page 7 :
www.tntextbooks.in, , Chapter, , 1, , MATRICES AND DETERMINANTS, , Learning Objectives, After studying this chapter, the students, will be able to understand, , •, , the definition of matrices and, determinants, , •, •, •, •, , the properties of determinants, , •, , the input-output analysis, , the concept of inverse matrix, the concept of adjoint matrix, the solving simultaneous linear, equations, , coefficients were represented by calculating, bamboos or sticks. Later the German, Mathematician Gottfried Wilhelm Von, Leibnitz formally developed determinants., The present vertical notation was given in, 1841 by Arthur Cayley., Determinant was invented, independently by Crammer, whose well known rule, for solving simultaneous, equations was published in, 1750., In class X, we have studied matrices, and algebra of matrices. We have also, learnt that a system of algebraic equations, can be expressed in the form of matrices., We know that the area of a triangle with, vertices (x1, y1) (x2, y2) and (x3, y3) is, 16, x (y - y3) + x2 (y3 - y1) + x3 (y1 - y2) @, 2 1 2, , Seki Kowa, , 1.1, , Determinants, , Introduction, The idea of a determinant was, believed to be originated from a Japanese, Mathematician Seki Kowa (1683) while, systematizing the old Chinese method of, solving simultaneous equations whose, , To minimize the difficulty in, remember­, ing this type of expression,, Mathematicians developed the idea of, representing the expression in determinant, form and the above expression can be, represented in the form, x1 y1 1, 1 x y 1, 2 2 2, x3 y3 1, Thus a determinant is a particular, type of expression written in a special, concise form. Note that the quantities are, arranged in the form of a square between, Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 1, , 1, , 21-04-2020 12:19:40 PM

Page 8 :
www.tntextbooks.in, , two vertical lines. This arrangement is, It is shortly written as 6aij@ mxn i = 1, 2,......m; j = 1,, 6aij@ mxn i = 1, 2,......m; j = 1, 2,.. ..n. Here aij is the element, called a determinant., in the ith row and jth column of the matrix., In this chapter, we study determinants, up to order three only with real entries., Also we shall study various properties, of determinant (without proof), minors,, cofactors, adjoint, inverse of a square matrix, and business applications of determinants., We have studied matrices in the, previous class. Let us recall the basic concepts, and operations on matrices., , Row matrix, A matrix having only one row is, called a row matrix., For example, A = 6 a11 a12 a13 ............a1i ..........a1n@1 ×n ;, B = 71 2A1 X 2, , Column matrix, , 1.1.1 Recall, , A matrix having only one column is, called a column matrix, , Matrix, Definition 1.1, A matrix is a rectangular, arrangement of numbers in horizontal, lines (rows) and vertical lines (columns)., Numbers are enclosed in square brackets, or a open brackets or pair of double bars., It is denoted by A, B, C, …, 1 4 2, G, For example , A = =, 5 8 6, Order of a matrix, If a matrix A has m rows and n, columns, then A is called a matrix of order, m × n., , G then, For example, If , A = =, 5 8 6, order of A is 2 × 3, 1 4 2, , General form of a Matrix, Matrix of order m × n is represented, RS, V, SS a11 a12 a13 g g a1i g g a1n WWW, SS a21 a22 a23 g g a2i g g a2n WW, S, W, as SSS g g g g g g g g g WWW, SS g g g g g g g g g WW, SS, W, Sam1 am2 am3 g g ami g g amnWW, T, X, 2, , Types of matrices, , For examples,, R V, S a11 W, S a21 W, S W, 3, ., A = SS WW , B = = G, - 5 2×1, ., S . W, S W, Sam1W, T Xm × 1, Zero matrix (or) Null matrix, If all the elements of a matrix are, zero, then it is called a zero matrix. It is, represented by an English alphabet ‘O’, For examples,, , RS0 0 0VW, SS, WW, 0 0, S, G, , O = S0 0 0WW, O = 60@ , O = =, 0 0, SS, W, S0 0 0WW, T, X, are all zero matrices., NOTE, Zero matrix can be of any order., Square matrix, If the number of rows and number, of columns of a matrix are equal then it is, called a square matrix., , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 2, , 21-04-2020 12:19:42 PM

Page 9 :
www.tntextbooks.in, , G is a square, For examples, A = =, 4 2, matrix of order 2, R, V, S 3 -1 0 W, B = S 2 0, 5 W is a square, S, W, 3, 4, S2, W, 5, T, X, matrix of order 3., 8 2, , NOTE, The sum of the diagonal elements of a, square matrix is called the trace of matrix., Triangular matrix, A square matrix, whose elements above, or below the main diagonal (leading diagonal), are all zero is called a triangular matrix., For examples,, 1 2 3, 1 0 0, A= >0 2 4H, B = >2 3 0H, 0 0 5, 3 4 7, , For examples, I2 = = G is a unit, 0 1, matrix of order 2., R, V, S1 0 0W, I3 = S0 1 0W is a unit matrix of, S, W, order 3. S0 0 1 W, T, X, 1 0, , Multiplication of a matrix by a scalar, If A = 6aij @ is a matrix of any, order and if k is a scalar, then the scalar, multiplication of A by the scalar k is, defined as kA = 6kaij@ for all i, j., , In other words to multiply a matrix, A by a scalar k, multiply every element of, A by k., , 1 2 4, G then,, For example, if A= =, 3 -2 8, , 2A = =, , 2 4 8, G, 6 - 4 16, , Diagonal matrix, , Negative of a matrix, , A square matrix in which all the, elements other than the main diagonal, elements are zero is called the diagonal matrix., , is defined as - A = 6- aij@ m # n for all i, j and, , For example,, , 5 0 0, A = >0 2 0 H is a diagonal matrix of, 0 0 1, order 3, , Scalar matrix, A diagonal matrix with all diagonal, elements are equal to K (a scalar) is called, a scalar matrix., , The negative of a matrix A = 6aij@ m # n, , is obtained by changing the sign of every, element., For example, if, -2 5 -7, G, then - A = =, 0 -5 -6, , 2 -5 7, G, A==, 0 5 6, , Equality of matrices, Two matrices A and B are said to be, equal if, , 5 0 0, For example, A = >0 5 0 H is a scalar, 0 0 5, matrix of order 3., , (i) they have the same order and, (ii) their corresponding elements are equal., , Unit matrix (or) Identity matrix, , Addition and subtraction of matrices, , A scalar matrix having each, diagonal element equal to one (unity) is, called a Unit matrix., , Two matrices A and B can be added,, provided both the matrices are of the, same order. Their sum A+B is obtained by, Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 3, , 3, , 21-04-2020 12:19:44 PM

Page 10 :
www.tntextbooks.in, , adding the corresponding entries of both, the matrices A and B., Symbolically if A = [aij]m× n and, B = [bij]m × n, then A + B = [aij+ bij]m × n, Similarly A − B = A + (−B) =, [aij]m× n+ [− bij]m × n= [aij− bij]m × n, Multiplication of matrices, Multiplication of two matrices is, possible only when the number of columns, of the first matrix is equal to the number, of rows of the second matrix. The, product of matrices A and B is obtained, by multiplying every row of matrix A, with the corresponding elements of every, column of matrix B element-wise and add, the results., Let A = 6aij @ be an m # p matrix, and B = 6bij @ be a p # n matrix, then the, product AB is a matrix C = 6cij@ of order, m # n., Transpose of a matrix, , Let A = 6aij @ be a matrix of order, m # n. The transpose of A, denoted by AT, of order n # m is obtained by interchanging, either rows into columns or columns into, rows of A., For example,, 1 3, 1 2 5, T, =, G then A, >2 4 H, if A = =, 3 4 6, 5 6, , NOTE, It is believed that the students, might be familiar with the above concepts, and our present syllabus continues from, the following., 4, , Definition 1.2, To every square matrix A of, order n with entries as real or complex, numbers, we can associate a number, called determinant of matrix A and it is, denoted by | A | or det (A) or Δ., Thus determinant can be formed by, the elements of the matrix A., If A = =, , a11 a12, G then its, a21 a22, , |A|=, , a11 a12, = a11 a22 - a21 a12, a21 a22, , Example 1.1, Evaluate:, , 2 4, -1 4, , Solution, 2 4, = (2)(4) - (- 1) 4, -1 4, , = 8+4 =12, To evaluate the determinant of order, 3 or more, we define minors and, cofactors., , 1.1.2 Minors:, Let | A | = |[aij]| be a determinant, of order n. The minor of an arbitrary, element aij is the determinant obtained, by deleting the ith row and jth column in, which the element aij stands. The minor of, aij is denoted by Mij., , 1.1.3 Cofactors:, The cofactor is a signed minor., The cofactor of aij is denoted by Aij and is, defined as, Aij = (- 1) i + j Mij, , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 4, , 21-04-2020 12:19:45 PM

Page 12 :
www.tntextbooks.in, , Cofactor of 4 is A31 = (- 1) 3 + 1 M31 = M31 = 1, Cofactor of 1 is A32, = (- 1) 3 + 2 M32 = - M32 = - 11, Cofactor of 0 is A33, = (- 1) 3 + 3 M33 = M33 = 4, a11, , NOTE, , Value of a determinant can be, obtained by using any row or column, For example, a11 a12 a13, If D = a21 a22 a23 , then, a31 a32 a33, Δ = a11A11+ a12A12+ a13A13 (or), a11M11−a12M12+ a13M13, (expanding along R1), Δ = a11A11+ a21A21+ a31A31 (or), a11M11−a21M21+ a31M31, (expanding along C1), Example 1.4, , Solution, , 1 2 4, Evaluate: - 1 3 0, 4 1 0, , 1 2 4, - 1 3 0 = (Minor of 1) –2 (Minor of 2) +, 4 (Minor of 4), 4 1 0, , =1, , 3 0, −1 0, −1 3, −2, +4, 1 0, 4 0, 4 1, , = 0 –0 –52 = –52., , 1.1.4 Properties of determinants, (without proof), 1. The value of a determinant is, unaltered when its rows and columns, are interchanged, 6, , 2. If any two rows (columns) of a, determinant are interchanged, then, the value of the determinant changes, only in sign., 3. If the determinant has two identical, rows (columns), then the value of the, determinant is zero., 4. If all the elements in a row (column), of a determinant are multiplied by, constant k, then the value of the, determinant is multiplied by k., 5. If any two rows (columns) of a, determinant are proportional, then, the value of the determinant is zero., 6. If each element in a row (column), of a determinant is expressed as the, sum of two or more terms, then the, determinant can be expressed as the, sum of two or more determinants of, the same order., 7. The value of the determinant is unaltered, when a constant multiple of the elements, of any row (column) is added to the, corresponding elements of a different, row (column) in a determinant., Example 1.5, Show that, , x, y, z, 2x + 2a 2y + 2b 2 z + 2 c = 0, c, a, b, , Solution, , x, y, z, 2x + 2a 2y + 2b 2 z + 2 c, c, a, b, x y z, x y z, = 2x 2y 2z + 2a 2b 2c, a b c, a b c, , =0+0, =0, , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 6, , 21-04-2020 12:19:48 PM

Page 13 :
www.tntextbooks.in, , Example 1.9, , e, , 3 2, 1 6, 4 8, o+e, o=e, o, 4 5, 7 8, 11 13, 3 2, 1 6, 4 8, +, !, 4 5, 7 8, 11 13, , Example 1.6, Evaluate, , x x+ 1, x- 1 x, , Solution, x x+ 1, = x2–(x–1)(x+1), x- 1 x, , , , = x2–(x2–1), , , , = x –x +1 = 1, 2, , 2, , 1 a a2, Evaluate 1 b b2 = (a–b) (b–c) (c–a), 1 c c2, , Solution, , 0 a - b a2 - b2, 1 a a2, - R2, 2, 2 R1 " R1, 1 b b2 = 0 b - c b - c, R2 " R2 - R3, 1 c, c2, 1 c c2, , 0 a - b ^ a - bh^ a + bh, = 0 b - c ^ b - ch^ b + ch, 1 c, c2, , 0 1 a+ b, = ^a - bh^ b - ch 0 1 b + c, 1 c c2, , = ^a - bh^ b - ch60 - 0 + " b + c - ^a + bh,@, , ^ a - bh^ b - ch60 - 0,   + " b + c - ^a + bh,@, , = ^a - bh^ b - ch^c - ah ., , NOTE, The value of the determinant of, triangular matrix is equal to the product, of the main diagonal elements., Example 1.7, , Solution, , x- 1 x x- 2, Solve 0 x - 2 x - 3 = 0, 0, 0, x- 3, , x- 1 x x- 2, 0 x- 2 x- 3 = 0, 0, 0, x- 3, , & (x–1)(x–2)(x–3)= 0, , x = 1, x = 2, x = 3, Example 1.8, , Solution, , 1 3 4, Evaluate 102 18 36, 17 3 6, , 1 3 4, 102 18 36, 17 3 6, , = 6 17 3 6, , 1 3 4, , , , =0, , 17 3 6, , (since R2 / R3), , Exercise 1.1, 1. Find the minors and cofactors of, all the elements of the following, determinants., 5 20, (i), 0 -1, , 1 -3 2, (ii) 4 - 1 2, 3 5 2, , 3 -2 4, 2. Evaluate 2 0 1, 1 2 3, 2 x 3, , 3. Solve: 4 1 6 = 0, 1 2 7, , ==, 4. Find ABABif ifA =A, , 33 00, 3 3- 1- 1, G and, G and, = = G G, B=, B=, 2, 1, 1 1- 2- 2, 2 1, , 7 4 11, 5. Solve: - 3 5 x = 0, -x 3 1, 1 a a2 - bc, 6. Evaluate: 1 b b2 - ca, 1 c c2 - ab, Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 7, , 7, , 21-04-2020 12:19:50 PM

Page 14 :
www.tntextbooks.in, , 1, bc b + c, a, 1, ca c + a =0, b, 1, ab a + b, c, , 7. Prove that, , - a2 ab ac, 8. Prove that ab - b2 bc = 4a2 b2 c2, ac bc - c2, , 1.2, , Inverse of a Matrix, , 1.2.1 Singular matrix:, A Square matrix A is said to be, singular, if A = 0, , 1.2.2 Non – singular matrix:, A square matrix A is said to be, non – singular, if A ! 0, , If A and B are non – singular, matrices of the same order then, AB and BA are also non – singular, matrices of the same order., , 1.2.3 Adjoint of a matrix, The adjoint of a square matrix A, is defined as the transpose of a cofactor, matrix. Adjoint of the matrix A is denoted, by adj A, i.e., adj A = 6Aij@T where 6Aij@ is the, cofactor matrix of A., Example 1.12, Find adj A for A = =, , 2 3, G, 1 4, , Example 1.10, Show that =, , 1 2, G is a singular matrix., 2 4, , Solution, A==, , 2 3, G, 1 4, , Solution, Let A = =, , 1 2, G, 2 4, , A =, , ` A is a singular matrix., Example 1.11, Show that =, , 8 2, G is non – singular., 4 3, , Solution, Let A = =, , 8 2, G, 4 3, , , , =, , 8 2, 4 3, , = 24 – 8 = 16 ! 0, , ` A is a non-singular matrix, 8, , = =, , 4 -3, G, -1 2, , 1 2, 2 4, , = 4 – 4 = 0, , A, , adj A = 6Aij@T, , NOTE, (i), (ii), , adjA = A, , n- 1, , matrix A, , n is the order of the, , kA = kn A n is the order of the, matrix A, , (iii) adj (kA) = k n - 1 adjA n is the order, of the matrix A, (iv) A (adj A) = (adj A) A = A lI, (v) AdjI = I, I is the unit matrix., (vi) adj (AB) = (adj B)(adj A), AB == AA BB, (vii) AB, , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 8, , 21-04-2020 12:19:51 PM

Page 15 :
www.tntextbooks.in, , Example 1.13, , V, RS, SS 1 - 2 - 3WWW, 1, 0WW, Find adjoint of A = SS 0, SS, W, 1, 0WW, S- 4, X, T, Solution, Aij = (–1)i+jMij, A11 = (–1), , 1+1, , 1.2.4 Inverse of a matrix, Let A be any non-singular matrix of, order n. If there exists a square matrix B of, order n such that AB = BA = I then, B is called, -1, the inverse of A and is denoted by A, , We have A (adj A) = (adj A)A, , 1 0, M11 =, =0, 1 0, , A12 = (–1)1+2M12 = A13 = (–1) M13, 1+3, , A22 = (–1)2+2M22 =, , 1 -3, -4 0, , = 0–12 = –12, , A23 = (–1)2+3M23 = , , = I, ( AB = BA = I where B = 1 adjA, A, , -2 -3, 1 0, , = -^0 - ]- 3gh = - 3, , , , ( A c 1A adj A m = c 1A adj A m A, , 0 0, =0, -4 0, , 0 1, =, = 0–(–4) = 4, -4 1, , A21 = (–1)2+1M21 = , , = A I, , A- 1 =, , NOTE, (i) If B is the inverse of A then, A is the, inverse of B, i.e., B = A- 1 & A = B- 1, , 1 -2, = - ]1 - 8g = 7, -4 1, (ii) AA-1 = A-1 A = I, , 1 -2, (iii), = - ]1 - 8g = 7, -4 1, -2 -3, = 0 - (- 3) = 3 (iv), A31 = (–1)3+1M31 =, 1 0, - 2 - 3, = 0 - (- 3) = 3, (v), 1 0, 1 -3, (vi), = 0-0 = 0, A32 = (–1)3+2M32 = 0 0, -, , , , 1, adjA, A, , 1 -3, = 0-0 = 0, 0 0, 1 -2, =, = 1-0 = 1, A33 = (–1)3+3M33 =, 0 1, 1 - 2, =, = 1-0 = 1, 0 1, 0, 0 4, [Aij] = >- 3 - 12 7H, 3, 0 1, , The inverse of a matrix, if it exists,, is unique., Order of inverse of A will be the, same as that of A., I-1 = I , where I is the identity matrix., 1, (AB) - 1 = B- 1 A- 1 (vii) A -1 = A, , -, , Adj A = 6Aij@T, 0 -3 3, = >0 -12 0H, 4, 7 1, , Let A be a non-singular matrix, then, A 2 = I + A = A -1, Example 1.14, If A= =, , 2 4, G then, find A- 1 ., -3 2, , Solution, , = =, , 2, -3, 2, A =, -3, , A, , 4, G, 2, 4, = 16 ≠ 0, 2, , Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 9, , 9, , 21-04-2020 12:19:54 PM

Page 16 :
www.tntextbooks.in, , Since A is a nonsingular matrix, A- 1, exists, Now, , adj A = =, , 2 -4, G, 3 2, , A- 1 =, , -5 8 -4, adj A = > 2 - 2 0H, 0 -2 2, A, , 1, adjA, A, , -1, , -5 8 -4, 1, 1, =, 2 - 2 0H, adjA =, -2 >, A, 0 -2 2, , RS, VW, SS 5 - 4, WW, 2, SS 2, WW, = S- 1, 1, 0, WW, SS, 1 - 1WW, S 0, T, X, , 1 =2 - 4G, = 16, 3 2, , Example 1.15, -2 6, G then, find A- 1, If A = =, 3 -9, , Solution, A =, , -2, 6, =0, 3 -9, , Since A is a singular matrix, A- 1, does not exist., Example 1.16, 2 4 4, If A = >2 5 4 H then find A 1, 2 5 3, , Solution, 2 4 4, A = 2 5 4, 2 5 3, , =2, , 5 4, 2 4, 2 5, -4, +4, 5 3, 2 3, 2 5, , = 2 (15 - 20) - 4 (6 - 8) + 4 (10 - 10),   , 2 (15 - 20, ) - 4 (6 - 8) + 4 (10 - 10), = 2 (- 5) - 4 (- 2) + 4 (0), = - 10 + 8 + 0 = - 2 ! 0, , Example 1.17, If A = =, , 2 3, G satisfies the equation, 1 2, A2 - kA + I2 = O then, find k and also A- 1 ., , Solution, , A ==, , 2 3, G, 1 2, , A2 = A.A, , = =, , 2 3 2 3, G = G, 1 2 1 2, , = =, , 7 12, G, 4 7, , Given A2 - kA + I2 = O, &=, , 7 12, 2 3, 1 0, 0 0, G - k=, G+ =, G= =, G, 4 7, 1 2, 0 1, 0 0, , &=, , 0 0, 7 - 2k + 1 12 - 3k + 0, G= =, G, 0 0, 4 - k + 0 7 - 2k + 1, , 0 0, 8 - 2k 12 - 3k, G== G, &=, 0 0, 4 k 8 2k, &4-k = 0, & k = 4., , Since A is a nonsingular matrix, A- 1, exists., , Also, A11 = - 5; A21 = 8; A31 = - 4, A12 = 2; A22 = - 2; A32 = 0, A13 = 0; A23 = - 2; A33 = 2, -5 2 0, 6Aij @ = > 8 - 2 - 2H, -4 0 2, , 10, , A =, , 2 3, 1 2, , = 4 – 3 = 1 ! 0, adj A = =, , 2 -3, G, -1 2, , A- 1 =, , 1, adjA, A, , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 10, , 21-04-2020 12:19:57 PM

Page 17 :
www.tntextbooks.in, , 2 -3, G, = 11 =, -1, , adj B = =, , 2 1, G, -1 0, , 2, , 1, B- 1 = B adjB, , 2 -3, G, = =, -1 2, , 2 1, 2 1, G==, G, = 11 =, , Example 1.18, 1 2, 0 -1, G then, show, If A= = G, B = =, 1 1, 1 2, that (AB) - 1 = B- 1 A- 1 ., , Solution, A =, , 1 2, = 1- 2 =- 1 ! 0, 1 1, , -1, , B, , 1 2 0 -1, 2 3, G==, G =, G, 1 1 1 2, 1 1, , AB = =, , 2 3, = -1 ! 0, 1 1, , (AB) - 1 exists., , -1 3, G, 1 -2, , Solution, 1 -3, =, G, -1 2, , R, S1 3, AB = S4 2, S, S1 2, T, , 1, adj (AB), AB, , (AB) - 1 =, , 1 -3, G, = -1 =, 1 -1, , 2, , = -1, =, , 3 , G, 1 -2, , adj A = =, , 1 -2, G, -1 1, , A- 1 =, , ==, , Show that the matrices, RS, SS - 4 11, SS 35 35, 1 3 7, A = >4 2 3 H and B = SSS - 1 - 6, SS 35 35, 1 2 1, 1, SS 6, S 35 35, T, are inverses of each other., , also exists., , adj(AB) =, , 2 1 -1 2, G, G =, B- 1 A- 1 = =, -1 0, 1 -1, , 1, adjA, A, , 1 -2, 1, G, = -1 =, -1, 1, -1 2, G, ==, 1 -1, , … (2), , Example 1.19, , 0 -1, = 0+1 = 1 ! 0, 1 2, , AB =, , -1 0, , From (1) and (2), (AB) - 1 = B- 1 A- 1 ., Hence proved., , ` A- 1 exists., B =, , -1 0, , … (1), , R, S1 3, = S4 2, S, S1 2, T, , V, - 5 WW, 35 WW, 25 WWW, 35 WW, - 10 WW, 35 WW, X, , V, R, - 4 11, -5 W, S, V, 35 W, 7W S 35 35, S, 1, 6, 25 W, 3W S, 35 W, W 35 35, S, W, 1, 6, 1 - 10 W, S, X S 35 35, 35 WW, X, T, V, R, V, 7W, S- 4 11 - 5W, 1 S- 3W, 1 6 25W, W, W 35 S, S 6 1 - 10W, 1W, X, T, X, , 35 0 0, 1, = 35 > 0 35 0 H, 0 0 35, 1 0 0, 35, = 35 >0 1 0 H, 0 0 1, , 1 0 0, = >0 1 0 H = I, 0 0 1, Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 11, , 11, , 21-04-2020 12:20:00 PM

Page 18 :
www.tntextbooks.in, , R, V, S - 4 11 - 5 W, S 35 35 35 W 1 3 7, - W 4 2 3, BA = SS 351 356 25, H, 35 W >, 1, 2, S 6, 1 - 10 W 1, S 35 35, 35 W, T, X, - 4 11 - 5 1 3 7, 1 - = 35 > 1 6 25H >4 2 3 H, 6 1 - 10 1 2 1, 35 0 0, 1, = 35 > 0 35 0H, 0 0 35, , 5., , (adj A)A = O, 6., , 8., , 1 0 0, = >0 1 0 H = I, 0 0 1, , Thus AB = BA = I, 9., , Find the adjoint of the matrix, 2 3, G, A==, 1 4, , 1 3 3, If A = >1 4 3 H then verify that, 1 3 4, , A(adj A) = |A| I and also find A- 1, 3., , Find the inverse of each of the, following matrices, 1 -1, G, 2 3, , 4., , (i) =, , (ii) =, , 1 2 3, (iii) >0 2 4 H, 0 0 5, , -3 -5 4, (iv) >- 2 3 - 1H, 1 -4 -6, , 3 1, G, -1 3, , -1 4, 2 3, G, then, G and B = =, If A = =, 1 -6, 1 -2, , verify adj (AB) = (adj B)(adj A), , 12, , If A = =, , 3 7, 6 8, G and B = =, G then,, 2 5, 7 9, , -1, -1 -1, verify that (AB) = B A, , Exercise 1.2, , 2., , 1 0 3, If A = >2 1 - 1H then, find A., 1 -1 1, 2 2 1, Show that the matrices A = >1 3 1 H, R, V, 1 2 2, S 4 -2 -1W, 5, 5W, S 5, 1, 3, 1, S, and B = S- 5 5 - 5 WW are inverses, S 1, 2, 4W, S- 5 - 5, 5W, T, X, -1, , of each other., , Therefore A and B are inverses of, each other., , 1., , -1 2 -2, If A = > 4 - 3 4H then, show that, 4 -4 5, , the inverse of A is A itself., 7., , 1 0 0, 35, = 35 >0 1 0 H, 0 0 1, , 2 -2 2, If A = >2 3 0H then, show that, 9 1 5, , 10., , 11., , 1 1 3, Find m if the matrix >2 m 4H has, 9 7 11, no inverse., 2 1 -1, 8 -1 -3, If X = >- 5 1 2H and Y = >0 2 1H, 5 p q, 10 - 1 - 4, -1, then, find p, q if Y = X, , 1.2.5 Solution of a system of linear, equations, Consider a system of n linear non –, homogeneous equations with n unknowns, x1, x2, ......xn as, a11 x1 + a12 x2 + .... + a1n xn = b1, a21 x1 + a22 x2 + .... + a2n xn = b2, , ………………………………, ………………………………, am1 x1 + am2 x2 + ... + amn xn = bm, , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 12, , 21-04-2020 12:20:02 PM

Page 19 :
www.tntextbooks.in, , This can be written in matrix form as, R, VR V R V, Sa11 a12 ....a1n W Sx1 W Sb1 W, Sa21 a22 ...a2n W Sx2W Sb2 W, S, WS W S W, S... ... .. .. ... ... W S. W = S. W, S... ... .. .. ... ... W S. W S. W, SSa a ...a WW SSx WW SSb WW, m1 m2, mn, n, m, T, XT X T X, Thus we get the matrix equation, AX = B, … (1), , R, V, Sa11 a12 .....a1n W, Sa21 a22 .....a2n W, S, W, where A = S.. ... .... .... ..... W ;, S.. ... .... .... ..... W, SSa a ....a WW, m1 m2, mn, T, X, Rb V, Rx V, S 1W, S 1W, Sb2 W, Sx2W, S W, S W, X = S. W and B = S. W, S W, S. W, S. W, S W, Sx W, SbmW, T nX, T X, From (1) solution is, X=A B, –1, , adj A = =, , 2 -5, G, -3 2, , A–1 = 1 adjA, A, , 2 -5, G, = -1 =, 11 - 3, , 2, , X = A–1B, 2 -5 1, G = G, = -1 =, 11 - 3, , 2, , 7, , 1 =, G== G, = - 11, -1, 11, - 33, , 3, , x, 3, = G== G, -1, y, , , x = 3 and y = –1., Example 1.21, Solve by using matrix inversion, method:, 3x - 2y + 3z = 8; 2x + y - z = 1; 4x - 3y + 2z = 4, , 3x - 2y + 3z = 8; 2x + y - z = 1; 4x - 3y + 2z = 4, , Example 1.20, , Solution, , Solve by using matrix inversion method:, 2x + 5y = 1, , 8, 3 -2 3 x, =, 2, 1, 1, y, 1, >, H>H > H, 4, 4 -3 2 z, , 3x + 2y = 7, Solution, The given system can be written as, 1, 2 5 x, = G = G= = G, 7, 3 2 y, , i.e.,, , AX = B, X = A–1B, , x, 2 5, 1, G, X = = G and B = = G, where A = =, y, 3 2, 7, A =, , The given system can be written as, , 2 5, 3 2, , = –11 ] 0, A–1 exists., , i.e., AX = B, X = A–1B, 3 -2 3, x, Here A = >2 1 1H, X = >y H and, 4 -3 2, z, 8, B = >1 H, 4, 3 -2 3, A = 2 1 -1, 4 -3 2, , = –17 ] 0, Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 13, , 13, , 21-04-2020 12:20:04 PM

Page 20 :
www.tntextbooks.in, , A–1 exists, , Then, 4x + 3y + 2z = 320, , A11 = –1, , A12 = –8, , A13 = –10, , A21 = –5, , A22 = –6, , A23 = 1, , A31 = –1, , A32 = 9, , A33 = 7, , - 1 - 8 - 10, [Aij] = >- 5 - 6, 1H, -1 9, 7, , adj A = [Aij], , T, , -1 -5 -1, = > - 8 - 6 9H, - 10 1 7, , 2x + 4y + 6z = 560, 6x + 2y + 3z = 380., It can be written as, 320, 4 3 2 x, >2 4 6H >yH = >560 H, 380, 6 2 3 z, , AX = B, 4 3 2, 320, x, where A = >2 4 6 H, X = >y H and B = >560 H, 6 2 3, 380, z, 4 3 2, A = 2 4 6, 6 2 3, , A–1 = 1 adjA, A, , -1 -5 -1, 1, = - 17 > - 8 - 6 9H, - 10 1 7, , X = A–1B, -1 -5 -1 8, 1, = - 17 > - 8 - 6 9H >1 H, - 10 1 7 4, - 17, 1, 1, = - 17 >- 34 H = >2 H, - 51, 3, 1, x, >y H = >2 H, 3, z, , x = 1, y = 2 and z = 3., Example 1.22, The cost of 4 kg onion, 3 kg wheat, and 2 kg rice is `320. The cost of 2kg onion,, 4 kg wheat and 6 kg rice is `560. The cost, of 6 kg onion, 2 kg wheat and 3 kg rice is, `380. Find the cost of each item per kg by, matrix inversion method., Solution, Let x, y and z be the cost of onion,, wheat and rice per kg respectively., 14, , = 50 ] 0, A–1 exist, A11 = 0, , A12 = 30, , A13 = –20, , A21 = –5, , A22 = 0, , A23 = 10, , A31 = 10, , A32 = –20, , A33 = 10, , 0 30 - 20, [Aij] = > 5, 0 10H, 10 20 10, , adj A = [Aij]T, 0 - 5 10, = > 30 0 - 20H, - 20 10 10, 1, A–1 = A adjA, 0 - 5 10, 1, = 50 > 30 0 - 20H, - 20 10 10, , X = A–1B, 0 - 5 10 320, 1, = 50 > 30 0 - 20H >560 H, - 20 10 10 380, 0 - 5 10 32, 1, = 5 > 30 0 - 20H >56H, - 20 10 10 38, , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 14, , 21-04-2020 12:20:06 PM

Page 21 :
www.tntextbooks.in, , 0 - 280 + 380, 1, = 5 > 960 + 0 - 760 H, - 640 + 560 + 380, 100, 1, = 5 >200 H, 300, 20, x, >y H = >40 H, 60, z, , x = 20, y = 40 and z = 60., Cost of 1 kg onion is `20, cost of 1 kg, wheat is `40 and cost of 1 kg rice is `60., Exercise 1.3, 1. Solve by matrix inversion method:, 2x + 3y – 5 = 0 ; x – 2y + 1 = 0, 2. Solve by matrix inversion method:, (i) 3x – y + 2z = 13 ; 2x + y – z = 3;, x + 3y – 5z = –8, (ii) x – y + 2z = 3 ; 2x + z = 1;, 3x + 2y + z = 4., (iii) 2x – z = 0; 5x + y = 4; y + 3z = 5, 3. A sales person Ravi has the following, record of sales for the month of, January, February and March 2009, for three products A, B and C He, has been paid a commission at fixed, rate per unit but at varying rates for, products A, B and C, Months, , Sales in Units, , Commission, , January, , A, 9, , B, 10, , C, 2, , 800, , February, , 15, , 5, , 4, , 900, , March, , 6, , 10, , 3, , 850, , Find the rate of commission payable on, A, B and C per unit sold using matrix, inversion method., , 4. The prices of three commodities A,, B and C are `x, `y and `z per unit, respectively. P purchases 4 units of C, and sells 3 units of A and 5 units of B., Q purchases 3 units of B and sells 2 units, of A and 1 unit of C. R purchases 1 unit, of A and sells 4 units of B and 6 units of, C. In the process P, Q and R earn `6,000,, `5,000 and `13,000 respectively. By, using matrix inversion method, find the, prices per unit of A, B and C., 5. The sum of three numbers is 20. If, we multiply the first by 2 and add the, second number and subtract the third, we get 23. If we multiply the first by 3, and add second and third to it, we get, 46. By using matrix inversion method, find the numbers., 6. Weekly expenditure in an office, for three weeks is given as follows., Assuming that the salary in all the, three weeks of different categories of, staff did not vary, calculate the salary, for each type of staff, using matrix, inversion method., Number of, employees, A, , B, , C, , Total weekly, salary, (in `), , 1st week, , 4, , 2, , 3, , 4900, , 2nd week, , 3, , 3, , 2, , 4500, , 3rd week, , 4, , 3, , 4, , 5800, , Week, , 1.3, , Input–Output Analysis, , Input–Output analysis is a technique, which was invented by Prof. Wassily, W. Leontief. Input–Output analysis is a, form of economic analysis based on the, interdependencies between economic, Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 15, , 15, , 21-04-2020 12:20:06 PM

Page 22 :
www.tntextbooks.in, , sectors. The method is most commonly, used for estimating the impacts of positive, or negative economic shocks and analyzing, the ripple effects throughout an economy., , the assumption that the structure of the, economy does not change., Let aij be the rupee value of the, output of Ai consumed by Aj , i, j = 1, 2, , Let x1 and x2 be the rupee value of the, The foundation of Input–Output, current outputs of A1 and A2 respectively., analysis involves input–output tables. Such, tables include a series of rows and columns, Let d1 and d2 be the rupee value of, of data that quantify the supply chain for, the final demands for the outputs of A1, sectors of the economy. Industries are listed, and A2 respectively., in the heads of each row and each column., The assumptions lead us to frame, The data in each column corresponds to, the two equations, the level of inputs used in that industry’s, a11+ a12+ d1 = x1;, production function. For example the, a21+ a22+ d2 = x2... (1), column for auto manufacturing shows the, aij, resources required for building automobiles, Let, bij = x i, j = 1, 2, j, (ie., requirement of steel, aluminum, plastic,, electronic etc.,). Input–Output models, b11 = ax11 ; b12 = ax12 ; b21 = ax21 ; b22 = ax22, 1, 2, 1, 2, typically includes separate tables showing, a11, a, a, a, ; b12 = 12 ; b21 = 21 ; b22 = 22, x1, x2, x1, x2, the amount of labour required per rupee, unit of investment or production., The equations (1) take the form, Consider a simple economic model, consisting of two industries A1 and A2, where each produces only one type of, product. Assume that each industry, consumes part of its own output and rest, from the other industry for its operation., The industries are thus interdependent., Further assume that whatever is produced, that is consumed. That is the total output, of each industry must be such as to meet, its own demand, the demand of the other, industry and the external demand (final, demand)., Our aim is to determine the output, levels of each of the two industries in, order to meet a change in final demand,, based on knowledge of the current outputs, of the two industries, of course under, 16, , b11x1+ b12x2+ d1 = x1, b21x1+ b22x2+ d2 = x2, The above, rearranged as, , equations, , ^1 - b11h x1 - b12 x2, , can, , be, , = d1, , - b21 x1 + ^1 - b22h x2 = d2, , The matrix form of the above, equations is, x1, 1 - b11 - b12, =, G = G, - b21 1 - b22 x2, , )e, , == G, d1, d2, , b11 b12, x1, d1, 1 0, o-e, o3 = G = = G, b21 b22, x2, d2, 0 1, , (1–B)X = D, where B = =, , b11 b12, 1 0, G, I == G, b21 b22, 0 1, , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 16, , 21-04-2020 12:20:07 PM

Page 23 :
www.tntextbooks.in, , X = = G and D = = G, x1, x2, , d1, d2, , By solving we get, X=(I–B)-1D, The matrix B is known as the, Technology matrix., , 1.3.1, , The Hawkins – Simon conditions, , Therefore, the given system is not, viable., Example 1.24, The following inter – industry, transactions table was constructed for an, economy of the year 2016., Final, Total, consumption output, , Industry, , 1, , 2, , Hawkins – Simon conditions ensure, the viability of the system., , 1, , 500, , 1,600, , If B is the technology matrix then, Hawkins – Simon conditions are, , Labours, , (i) the main diagonal elements in, I – B must be positive and, (ii), , I - B must be positive., , Example 1.23, The technology matrix of an, economic system of two industries is, 0.8 0.2, G. Test whether the system is viable, =, 0.9 0.7, , as per Hawkins – Simon conditions., Solution, B, , ==, , 0.8 0.2, G, 0.9 0.7, , I–B ==, , 1 0, 0.8 0.2, G– =, G, 0 1, 0.9 0.7, , = =, , 0.2 - 0.2, G, - 0.9 0.3, , I- B =, , 0.2 - 0.2, - 0.9 0.3, , 2, , 1,750 1,600, 250, , 4,800, , 2,500, , 4,650, , 8,000, , -, , -, , Construct technology co-efficient, matrix showing direct requirements., Does a solution exist for this system., Solution, a11 = 500, , a12 = 1600, , x1 = 2500, , a21 = 1750, , a22 = 1600, , x2 = 8000, , 500 =, b11 = ax11 = 2500, 0.20 ;, 1, = 0.20, b12 = ax12 = 1600, 8000, 2, 1750 =, b21 = ax21 = 2500, 0.70 ;, 1, = 0.20, b22 = ax22 = 1600, 8000, 2, , The technology matrix is = =, , 0.2 0.2, G, 0.7 0.2, , I–B= =, , 1 0, 0.2 0.2, G –=, G, 0 1, 0.7 0.2, , = =, , 0.8 - 0.2, G,, - 0.7 0.8, , elements, , = (0.2)(0.3)–(–0.2)(–0.9), , , , = 0.06 – 0.18, , Now, I - B =, , , , = –0.12 < 0, , , , , , 0.8 - 0.2, - 0.7 0.8, , = (0.8)(0.8)–(–0.7)(–0.2), = 0.64 – 0.14, = 0.50 >0, , Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 17, , of, , main diagonal are positive, , , , Since I - B is negative, Hawkins –, Simon conditions are not satisfied., , 400, , 17, , 21-04-2020 12:20:08 PM

Page 24 :
www.tntextbooks.in, , Since diagonal elements of I–B are, positive and I - B is positive,, Hawkins – Simon conditions are, satisfied, Hence this system has a solution., Example 1.25, In an economy there are two, industries P1 and P2 and the following, table gives the supply and the demand, position in crores of rupees., , P1, , Consumption, Final Gross, sector, demand output, P2, P1, 10, 25, 15, 50, , P2, , 20, , Production, sector, , 30, , 10, , 60, , Determine the outputs when the, final demand changes to 35 for P1 and 42, for P2., Solution, a11 = 10, , a12 = 25, , x1 = 50, , a21 = 20, , a22 = 30, , x2 = 60, , a, 10 1, b11 = x11 = 50 = 5 ;, 1, a, 25, 5, b12 = x12 = 60 = 12, 2, a, 20 2, b21 = x21 = 50 = 5 ;, 1, a, 30 1, b22 = x22 = 60 = 2, 2, , R, V, S1 5 W, 5 12 W, The technology matrix is B = S2, 1 W, SS, 5, 2 W, R, V, T, X, 1, 5, W, S, 1 0 S5 12 W, I–B= = G– 2 1, W, 0 1 SS, 5 2 W, T V, X, R, 4, 5, S, - W, 5, S, W , elements of, = S 2 12, 1, W, S- 5, 2W, T, X, , main diagonal are positive, , 18, , Now, I - B, , 4 -5, 5, 12, = 2, 1, 5, 2, , , , 7, = 30, 20, , Main diagonal elements of I - B, are positive and I - B is positive, , ∴ The problem has a solution, R, V, S1 5 W, 12 W, adj (I–B), = S2, SS2 4 WW, 5 5, T, X, (I–B)–1 = 1 adj (I - B) (Since I - B ! 0, (I - B) - 1 e, I- B, , 1, adj (I - B) (Since I - B ! 0, (I - B) - 1 exist), I- B, , =, , 1, , 7, 30, , adj (I - B), , R, V, S1 5 W, 25, S2 12 W = 1 >15 2 H, = 30, 7 S2 4 W 7 12 24, S5 5 W, T, X, 35, –1, X = (I–B) D, where D = = G, 42, 25 35, 150, = 17 >15 2 H = G = = G, 12 24 42, , 204, , The output of industry P1 should be, `150 crores and P2 should be `204 crores., Example 1.26, An economy produces only coal, and steel. These two commodities serve, as intermediate inputs in each other’s, production. 0.4 tonne of steel and 0.7, tonne of coal are needed to produce a, tonne of steel. Similarly 0.1 tonne of, steel and 0.6 tonne of coal are required to, produce a tonne of coal. No capital inputs, are needed. Do you think that the system, is viable? 2 and 5 labour days are required, to produce a tonnes of coal and steel, , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 18, , 21-04-2020 12:20:10 PM

Page 25 :
www.tntextbooks.in, , respectively. If economy needs 100 tonnes, of coal and 50 tonnes of steel, calculate the, gross output of the two commodities and, the total labour days required., Solution, Here the technology matrix is given, under, , 30, = 0.117 = G, 95, , 176.5, G, ==, 558.8, Steel output = 176.5 tonnes, Coal output = 558.8 tonnes, Total labour days required, , Steel, , Coal, , Final, demand, , Steel, , 0.4, , 0.1, , 50, , Coal, , 0.7, , 0.6, , 100, , Labour days, , 5, , 2, , -, , The technology matrix is B = =, , 0.4 0.1, G, 0.7 0.6, , I–B==, , 1 0, 0.4 0.1, G– =, G, 0 1, 0.7 0.6, 0.6 - 0.1, G, - 0.7 0.4, , 0.6 - 0.1, - 0.7 0.4, , , , = (0.6)(0.4)–(–0.7)(–0.1), , , , = 0.24 – 0.07 = 0.17, , Since the diagonal elements of I – B, are positive and value of I - B is positive,, the system is viable., adj (I–B) = =, , 0.4 0.1, G, 0.7 0.6, , (I–B)–1 =, , 1, adj (I - B), I- B, , 0.4 0.1, G, = 0.117 =, 0.7 0.6, , X = (I–B)–1D, where D = =, , 50, G, 100, , 0.4 0.1 50, G= G, = 0.117 =, 0.7 0.6 100, , = 5(176.5) + 2(558.8), = 882.5 + 1117.6 = 2000.1, - 2000 labour days., Exercise 1.4, 1. The technology matrix of an economic, 0.50 0.30, G., system of two industries is =, 0.41 0.33, Test whether the system is viable as per, Hawkins Simon conditions., , ==, I- B =, , = 5(steel output) + 2(coal output), , 2. The technology matrix of an economic, 0.6 0.9, G., system of two industries is =, 0.20 0.80, Test whether the system is viable as per, Hawkins-Simon conditions., 3. The technology matrix of an economic, 0.50 0.25, G., system of two industries is =, 0.40 0.67, Test whether the system is viable as per, Hawkins-Simon conditions., 4. Two commodities A and B are produced, such that 0.4 tonne of A and 0.7 tonne of, B are required to produce a tonne of A., Similarly 0.1 tonne of A and 0.7 tonne of B, are needed to produce a tonne of B. Write, down the technology matrix. If 68 tonnes, of A and 10.2 tonnes of B are required,, find the gross production of both of them., Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 19, , 19, , 21-04-2020 12:20:11 PM

Page 26 :
www.tntextbooks.in, , 5. Suppose the inter-industry flow of the, product of two industries are given as, under., , Find the gross output when the, domestic demand changes to 12 for X, and 18 for Y., , Production Consumption Domestic Total, sector, sector, demand output, X, , Y, , X, , 30, , 40, , 50, , 120, , Choose the correct answer, , Y, , 20, , 10, , 30, , 60, , 0 1 0, 1. The value of x if x 2 x = 0 is, 1 3 x, (a) 0, –1, (b) 0, 1, , Determine the technology matrix and, test Hawkin’s -Simon conditions for the, viability of the system. If the domestic, demand changes to 80 and 40 units, respectively, what should be the gross, output of each sector in order to meet, the new demands., 6. You are given the following transaction, matrix for a two sector economy., Sector, , Sales, , Final, demand, , Gross, output, , 1, , 2, , 1, , 4, , 3, , 13, , 20, , 2, , 5, , 4, , 3, , 12, , (i) Write the technology matrix, (ii) Determine the output when the, final demand for the output sector, 1 alone increases to 23 units., 7. Suppose the inter-industry flow of the, product of two sectors X and Y are, given as under., Production Consumption Domestic Gross, sector, Sector, demand output, , 20, , Exercise 1.5, , X, , Y, , X, , 15, , 10, , 10, , 35, , Y, , 20, , 30, , 15, , 65, , (d) –1, –1, , (c) –1, 1, , 2x + y x y, 2. The value of 2y + z y z is, 2z + x z x, (a) x y z, , (b) x + y + z, , (c) 2x + 2y + 2z, , (d) 0, , 3. The cofactor of –7, in the determinant, , 2 -3 5, 6 0 4 is, 1 5 -7, , (a) –18, , (b) 18, , (c) –7, , (d) 7, , 1 2 3, 3 1 2, 4. If D = 3 1 2 then 1 2 3 is, 2 3 1, 2 3 1, (a) D, , (b) –D, , (c) 3D, , (d) –3D, , 5. The, a 0, 0 b, 0 0, , value of, 0 2, 0 is, c, , the, , determinant, , (a) abc, , (b) 0, , (c) a2 b2 c2, , (d) –abc, , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 20, , 21-04-2020 12:20:12 PM

Page 27 :
www.tntextbooks.in, , 6. If A is square matrix of order 3 then, | kA |is, (a) k A, , (b) – k A, , (c) k3 A, , (d) – k3 A, , (a) e, , 7. adj ] ABg is equal to, (b) adjAT adjBT, , (a) adjA adjB, , T, , 8. The inverse matrix of, 7 2, (a) 30 f 2, 5, , 1, , 5, 12, 4, 5, , 1, , 5, 12, 4, 5, , 30 2, (c) 7 f 2, 5, , T, , (d) adjB adjA, , (c) adjB adjA, , 4, f –25, 5, , –5, 12, 1, 2, , p, , 1, , -5, 12, 1, 5, , p, , p, , 30 2, (d) 7 f -2, 5, , 1, , -5, 12, 4, 5, , p, , a b, o such that ad - bc ! 0, 9. If A = e, c d, then A -1 is, d b, 1, o, (a) ad - bc e, -c a, d b, 1, o, (b) ad - bc e, c a, d -b, 1, o, (c) ad - bc e, -c, a, d -b, 1, o, (d) ad - bc e, c, a, 10. The number of Hawkins­, -Simon, conditions for the viability of an, input-output analysis is, (a) 1 , (b) 3, (c) 4 , (d) 2, 11. The inventor of input - output analysis is, (b) Fisher, (c) Prof. Wassily W. Leontief, (d) Arthur Caylay, , (c) e, , -1, 1, o, 1 -4, , (b) e, , (d) e, , sin a sin a, o, - cos a cos a, 3 1, o is, 5 2, , 2 -1, o, -5, 3, , (b) e, , -2, 5, o, 1 -3, , 3 -1, o, -5 -3, , (d) e, , -3, 5, o, 1 -2, , (a) e, (c) e, , 2 -1, o, -4, 2, , cos a sin a, o, - sin a cos a, , 13. The inverse matrix of e, , p is, , 7 2, (b) 30 f -2, 5, , (a) Sir Francis Galton, , 12. Which of the following matrix has no, inverse, , 14. If A = e, , -1, 2, o then A ^adjA h is, 1 -4, , (a) e, , -4 -2, o, -1 -1, , (b) e, , 4 -2, o, -1, 1, , (c) e, , 2 0, o, 0 2, , (d) e, , 0 2, o, 2 0, , 15. If A and B non-singular matrix then,, which of the following is incorrect?, (a) A 2 = I implies A -1 = A, (b) I -1 = I, (c) If AX = B then X = B -1 A, (d) If A is square matrix of order 3 then, adj A = A 2, 5, 5, 5, 4y, 4z is, 16. The value of 4x, - 3x - 3y - 3z, (a) 5, , (b) 4, , (c) 0, , (d) –3, , 17. If A is an invertible matrix of order 2, then det ^ A -1h be equal to, 1, (a) det(A), (b), det ] Ag, (c) 1, (d) 0, Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 21, , 21, , 21-04-2020 12:20:18 PM

Page 28 :
www.tntextbooks.in, , 18. If A is 3 3 matrix and A = 4 then, A -1 is equal to, , 1, (a) 4, , 1, (b) 16, , (c) 2, , (d) 4, , 19. If A is a square matrix of order 3 and, A = 3 then adjA is equal to, , (a) 81, , (b) 27, , (c) 3, , (d) 9, , 2, , x x - yz 1, 2, 20. The value of y y - zx 1 is, z z 2 - xy 1, (a) 1 , , (b) 0, , (c) –1, , (d) - xyz, cos i sin i, G , then 2A is, - sin i cos i, , 21. If A = =, equal to, , (a) 4 cos 2i, , (b) 4, , (c) 2, , (d) 1, , a11 a12 a13, 22. If D = a21 a22 a23 and Aij is cofactor, a31 a32 a33, of aij, then value of D is given by, , (a) a11 A31 + a12 A32 + a13 A33, (b) a11 A11 + a12 A21 + a13 A31, (c) a21 A11 + a22 A12 + a23 A13, (d) a11 A11 + a21 A21 + a31 A31, 23. If, , x 2, = 0 then the value of x is, 8 5, , -5, (a) 6, - 16, (c) 5, 24. If, , 22, , 5, (b) 6, 16, (d) 5, , 4 3, 20 15, = –5 then the value of, is, 3 1, 15 5, , (a) –5, , (b) –125, , (c) –25, , (d) 0, , 25. If any three rows or columns of a, determinant are identical then the, value of the determinant is, (b) 2, , (a) 0, , (c) 1, , (d) 3, , Miscellaneous Problems, x 2 -1, x =0, 1. Solve: 2 5, -1 2, x, 10041 10042 10043, 2. Evaluate 10045 10046 10047, 10049 10050 10051, 3. Without actual expansion show that the, , 5 5 2 53, 2, 3, 4, value of the determinant 5 5 5 is, 5 4 55 5 6, zero., 0 ab 2 ac 2, 2, 2, 4. Show that a b 0 bc = 2a3 b3 c3 ., a2 c b2 c 0, RS, V, 1 1WW, SS1, W, 4 7WW verify that, 5. If A = SS3, SS, W, S1 - 1 1WW, T, X, A ^adjA h = ^adjA h] Ag = A I3 ., , V, RS, 1WW, SS 3 - 1, W, 6 - 5WW then, find the, 6. If A = SSS- 15, W, SS 5 - 2, 2WW, X, T, Inverse of A., , 1 -1, H show that, 2, 3, , 7. If A = >, , A 2 - 4A + 5I2 = O and also find A -1, 8. Solve by using matrix inversion method :, x - y + z = 2, 2x - y = 0, 2y - z = 1 ., 9. The cost of 2 Kg of Wheat and 1 Kg, of Sugar is `70. The cost of 1 Kg of, Wheat and 1 Kg of Rice is `70. The, , 11th Std. Business Mathematics and Statistics, , 01_11th_BM-STAT_Ch-1-EM.indd 22, , 21-04-2020 12:20:23 PM

Page 29 :
www.tntextbooks.in, , cost of 3 Kg of Wheat, 2 Kg of Sugar, and 1 Kg of rice is `170. Find the, cost of per kg each item using matrix, inversion method., 10. The data are about an economy of, two industries A and B. The values, are in crores of rupees., , User, A, B, , Producer, , Final, demand, , Total, output, , A, , 50, , 75, , 75, , 200, , B, , 100, , 50, , 50, , 200, , Find the output when the final, demand changes to 300 for A and 600 for B, , Summary, , zz, , Determinant is a number associated to a square matrix., , zz, , If A = , , zz, , Minor of an arbitrary element aij of the determinant of the matrix is the determinant, obtained by deleting the ith row and jth column in which the element aij stands., The minor of aij is denoted by Mij., , zz, , zz, ,  a11 a12 ,  then its A = a11 a22 - a21 a12,  a21 a22 , , The cofactor is a signed minor. The cofactor of aij is denoted by Aij and is defined as, , Aij = (- 1) i + j Mij ., , a11 a12 a13, Let ∆ = a21 a22 a23 , then, a31 a32 a33, Δ= a11A11+ a12A12+ a13A13 or a11M11−a12M12+ a13M13, , zz, zz, zz, , adj A = 7AijA where [Aij] is the cofactor matrix of the given matrix., T, , adjA = A n - 1 , where n is the order of the matrix A., , A ^adjA h = ^adjA h] Ag = A I ., , zz, , adjI = I, I is the unit Matrix., , zz, , adj(AB) = (adj B)(adj A)., , zz, , A square matrix A is said to be singular, if | A | = 0., , zz, , A square matrix A is said to be non-singular if | A | ! 0, , zz, , Let A be any square matrix of order n and I be the identity matrix of order n. If, there exists a square matrix B of order n such that AB = BA = I then, B is called the, inverse of A and is denoted by A–1, 1, Inverse of A ( if it exists ) is A-1 = A adjA ., Hawkins - Simon conditions ensure the viability of the system. If B is the technology, , zz, zz, , matrix then Hawkins-Simon conditions are, (i) the main diagonal elements in I - B must be positive and, (ii) I - B must be positive., Matrices and Determinants, , 01_11th_BM-STAT_Ch-1-EM.indd 23, , 23, , 21-04-2020 12:20:25 PM

Page 31 :
www.tntextbooks.in, , Chapter, , 2, , different cases of combination, such as word formation, , Algebra is built on experiences with numbers, and operations along with geometry, and data analysis. The word “algebra” is, derived from the Arabic word “al-Jabr”. The, Arabic mathematician Al-Khwarizmi has, traditionally been known, as the “Father of algebra”., Algebra is used to find the, perimeter and area of any, plane region, volume and, CSA of solid., , difference between permutation, and combination, , 2.1 Partial Fractions, , Learning Objectives, After studying this chapter, the students, will be able to understand, , •, •, , definition of partial fractions, , •, , different cases of permutation, such as word formation, , •, •, •, •, , ALGEBRA, , the various techniques to resolve, into partial fractions, , the concept and principle of, mathematical induction, Binomial expansion techniques, , Rational Expression, An expression of the form, , p ]xg, a0 xn + a1 xn - 1 + g + an ,[q(x)≠0], =, q ]xg b0 xm + b1 xm - 1 + g + bm, , is called a rational algebraic expression., , If the degree of the numerator p (x), is less than that of the denominator q (x), p ]xg, then ]xg is called a proper rational, q, expression. If the degree of p (x) is greater, p ]xg, than or equal to that of q (x) then ]xg, q, is called an improper rational expression., , Al-Khwarizmi, , Introduction, Algebra is a major component of mathematics, that is used to unify mathematics concepts., , An improper rational expression, can be expressed as the sum of an integral, function and a proper rational expression., p ]xg, r ]xg, That is, ]xg = f ]xg + ]xg, q, q, For example,, , x, x2 + x + 1, =1– 2, x + 2x + 1, x 2 + 2x + 1, Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 25, , 25, , 21-04-2020 12:17:53 PM

Page 32 :
www.tntextbooks.in, , Partial Fractions, We can express the sum or difference, 3, 2, of two rational expressions x - 1 and x - 2, as a single rational expression., i.e.,, 5x - 8, 3, 2, x - 1 + x - 2 = ]x - 1g]x - 2g, 5x - 8, 3, 2, ]x - 1g]x - 2g = x - 1 + x - 2, , 1, A, B, = x-1 + x+1, ^ x - 1h, 2, , Solution, Let, , 2.1.1 Denominator contains, non-repeated Linear factors, , p ]xg, In the rational expression ]xg , if, q, q ]xg is the product of non-repeated linear, factors of the form ]ax + bg]cx + dg , then, p ]xg, can be resolved into partial fraction, q ]xg, A, B, of the form ax + b + cx + d , where A and, , B are constants to be determined., , For example,, 3x + 7, A, B, ]x - 1g]x - 2g = ]x - 1g + ]x - 2g , the values, of A and B are to be determined., Example 2.1, Find the values of A and B if, 26, , A, B, = x-1 + x+1, , Multiplying both sides by ]x - 1g]x + 1g ,, we get, 1 = A ]x + 1g + B ]x - 1g , , ... (1), , Put x =1 in (1) we get, 1 = A ]2 g, , Process of writing a single rational, expression as a sum or difference of two or, more simple rational expressions is called, splitting up into partial fractions., Generally if p (x) and q (x) are two, rational integral algebraic functions of x, p ]xg, and the fraction ]xg be expressed as the, q, algebraic sum (or difference) of simpler, fractions according to certain specified, p ]xg, rules, then the rational expression ]xg, q, is said to be resolved into partial fractions., , 1, , ^ x2 - 1h, , 1, ` A= 2, , Put x = –1 in (1) we get, 1 = A(0)+B(–2), 1, ` B=–2, Example 2.2, Resolve into partial fractions:, 7x - 1, x 2 - 5x + 6, Solution, Write the denominator into the, product of linear factors., Here x 2 - 5x + 6 = ]x - 2g]x - 3g, , Let, , A, B, 7x - 1, = ]x - 2g + ]x - 3g …(1), x2 - 5x + 6, , Multiplying both the sides of (1) by, (x–2)(x–3), we get, 7x - 1 = A ]x - 3g + B ]x - 2g , , …(2), , Put x = 3 in (2),we get, 21 – 1 = B (1), &, , B = 20, , Put x = 2 in (2),we get, 14 - 1 = A ]- 1g, &, , A = - 13, , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 26, , 21-04-2020 12:17:58 PM

Page 33 :
www.tntextbooks.in, , Substituting the values of A and B, in (1), we get, - 13, 20, 7x - 1, +, =, x 2 - 5x + 6 ]x - 2g ]x - 3g, , 7x - 1, =, x - 5x + 6, 1 < ]7x - 1g F, 1 < ]7x - 1g F, +, x, 3g x = 2 ] x 3g ] x - 2g x = 3, ] x 2g ], 2, , =, , - 13, 20, ] x - 2g + ] x - 3g, , Example 2.3, Resolve into partial fraction:, x+4, ^ x 2 - 4h]x + 1g, , Solution, , Put x = –1 in (2), we get, , - 1 + 4 = A ]0 g + B ]0 g + C ]- 3g]1 g, , `, , C = -1, , Substituting the values of A, B and, C in (1), we get, 1, 1, x+4, = x1- 2 +, - x+1, +, 2, (, x, 2, ), ], g, 2, +, x, 1, ^ x - 4h], g, 2, , 2.1.2 Denominator contains Linear, factors, repeated n times, p (x), In the rational fraction q (x) , if, q (x) is of the form ]ax + bgn [the linear, , factor is the product of (ax+ b), n times],, p (x), then q (x) can be resolved into partial, fraction of the form., A3, A, A1, A2, +, + g + ax +n b n, +, ]ax + bg ]ax + bg2 ]ax + bg3, ], g, , Write the denominator into the, product of linear factors, , For example,, , Here ^ x 2 - 4h]x + 1g = ]x - 2g]x + 2g]x + 1g, , 9x + 7, A, B, C, = x+, 4g + ]x + 1g + ]x + 1g2, ], ]x + 4g]x + 1g2, , `, , x+4, , ^ x 2 - 4h]x + 1g, , =, , A, B, C, ]x - 2g + ]x + 2g + ]x + 1g ...(1), , Multiplying both the sides by (x–2), (x+2)(x+1), we get, , Example 2.4, Find the values of A, B and C if, x, A, B, C, 2 = x-1 + x+1 +, ]x - 1g]x + 1g, ]x + 1g2, , Solution, x + 4 = A ]x + 2g]x + 1g + B ]x - 2g]x + 1g + C ]x - 2g]x x+ 2g 2 = A + B + C 2, x - 1 x + 1 ] x + 1g, ]x - 1g]x + 1g, ]x + 1g + B ]x - 2 , g]x + 1g + C ]x - 2g]x + 2g , ... (2), x = A ]x + 1g2 + B ]x - 1g]x + 1g + C ]x - 1g ... (1), Put x = –2 in (2), we get, Put x = 1 in (1), - 2 + 4 = A ]0 g + B ]- 4g]- 1g + C ]0 g, 1, B= 2, Put x = 2 in (2), we get, , `, , 2 + 4 = A ]4g]3 g + B ]0 g + C ]0 g, , `, , 1, A= 2, , 1 = A ]1 + 1g2 + B ]0 g + C ]0 g, 1, `A = 4, Put x = –1 in (1), , - 1 = A ]0 g + B ]0 g + C ]- 1 - 1g, 1, `C = 2, Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 27, , 27, , 21-04-2020 12:18:06 PM

Page 34 :
www.tntextbooks.in, , Equating the constant term on both, sides of (1),we get, A-B-C = 0, & B = A-C, 1, ` B = –4, Example 2.5, , Example 2.6, Resolve into partial fraction, 9, ]x - 1g]x + 2g2, , Solution, , 9, ]x - 1g]x + 2g2, A, B, C, , = ] x - 1g + ] x + 2g +, … (1), ]x + 2g2, , Resolve into partial fraction, x+1, ]x - 2g2 ]x + 3g, , Multiplying both sides by ]x - 1g]x + 2g2, , ` 9 = A ]x + 2g2 + B ]x - 1g]x + 2g + C ]x - 1g, Solution, x+1, … (2), A ]x + 2g2 + B ]x - 1g]x + 2g + C ]x - 1g , Let, 2, ] x - 2g ] x + 3g, Put x = –2 in (2), A, B, C, = ] x - 2g +, ...(1), +, ]x - 2g2 ]x + 3g, 9 = A ]0 g + B ]0 g + C ]- 2 - 1g, Multiplying both the sides by, ]x - 2g2 ]x + 3g , we get, , ` C = –3, , x + 1 = A ]x - 2g]x + 3g + B ]x + 3g + C ]x - 2g, , 2, , …(2), Put x = 2 in (2), we have, 2+1, , = A ]0 g + B ]5 g + C ]0 g, , 3, ` B = 5, , Put x = –3 in (2), we have, , –3+1 = A ]0 g + B ]0 g + C ]- 3 - 2g2, -2, ` C = 25, Equating the coefficient of x 2 on, both the sides of (2), we get, 0, , = A+C, , A, , = -C, 2, ` A = 25, Substituting the values of A, B and, C in (1),, x+1, =, ]x - 2g2 ]x + 3g, 2, 3, 2, +, 25 ]x - 2g 5 ]x - 2g2 25 ]x + 3g, 28, , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 28, , Put x = 1 in (2), , 9 = A ]3 g2 + B ]0 g + C ]0 g, , ` A= 1, , Equating the coefficient of x 2 on, both the sides of (2), we get, 0 = A+B, B = -A, ` B = –1, Substituting the values of A, B and, C in (1), we get, 9, 1, 1, 3, =, ]x - 1g]x + 2g2 x - 1 x + 2 ]x + 2g2, , 2.1.3 Denominator contains quadratic, factor, which cannot be factorized, into linear factors, In the rational fraction, , p ]xg, , if one, q ]xg, , of the quadratic factor q ]xg is of the form, ax 2 + bx + c which cannot be factorized, p ]xg, into linear factors, then ]xg can be, q, , 21-04-2020 12:18:13 PM

Page 35 :
www.tntextbooks.in, , resolved into partial fractions by taking, the numarator of ax 2 + bx + c of the form, Ax+B., Example 2.7, Resolve into partial fractions:, 2x + 1, ]x - 1g^ x 2 + 1h, Solution, Here x 2 + 1 cannot be factorized, , Exercise 2.1, Resolve into partial fractions for, the following:, 1., , 3x + 7, x 2 - 3x + 2, , 3., , 1, ]x - 1g]x + 2g2, , 5., , into linear factors., Let, , A, Bx + c, 2x + 1, = x-1 + 2, ... (1), 2, ]x - 1g^ x + 1h, x +1, , Multiplying both sides by ]x - 1g^ x2 + 1h, , 2x+1 = A ^ x2 + 1h + ]Bx + Cg]x - 1g ... (2), , Put x = 1 in (2), 2+1 = A ]1 + 1g + 0, , 3, ` A= 2, , Put x = 0 in (2), , 0+1 = A ]0 + 1g + ]0 + C g]- 1g, , 1 = A–C, 1, ` C= 2, Equating the coefficient of x 2 on, both the sides of (2), we get, A+B = 0, B = -A, -3, 2, Substituting the values of A, B and, C in (1), we get, -3, 3, 1, x+ 2, 2x + 1, 2, 2, = x-1 +, ]x - 1g^ x 2 + 1h, x2 + 1, 3, 3x - 1, , = ]x - 1g 2, 2 ^ x2 + 1h, ` B =, , x-2, ]x + 2g]x - 1g2, , 4x + 1, 2. ]x - 2g]x + 1g, 4., , 1, x2 - 1, , 6., , 2x 2 - 5x - 7, ]x - 2g3, , 7., , x 2 - 6x + 2, x 2 ]x + 2g, , 8., , x2 - 3, ]x + 2g^ x 2 + 1h, , 9., , x+2, x, 1, ], g]x + 3g2, , 10., , 1, ^ x + 4h]x + 1g, , 2.2, , 2, , Permutations, , 2.2.1 Factorial, For any natural number n, ‘n, factorial’ is defined as the product of the, first n natural numbers and is denoted by, n! or n ., For any natural number n,, n! = n ]n - 1g]n - 2g … 3 # 2 # 1, For examples,, 5! = 5 # 4 # 3 # 2 # 1 =120, 4! = 4 # 3 # 2 # 1=24, 3! = 3 # 2 # 1=6, 2! = 2 # 1=2, , NOTE, 0! = 1, , 1! = 1, , For any natural number n, , n! = n ]n - 1g]n - 2g … 3 # 2 # 1, = n ]n - 1g !, , = n ]n - 1g]n - 2g !, Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 29, , 29, , 21-04-2020 12:18:18 PM

Page 36 :
www.tntextbooks.in, , In particular,, 8! = 8×(7×6×5×4×3×2×1), = 8×7!, Example 2.8, Evaluate the following:, 7!, (i) 6!, , 8!, (ii) 5!, , 9!, (iii) 6!3!, , Solution, (i), , 7!, 7 # 6!, = 7, 6! =, 6!, , (ii), , 8!, 8 # 7 # 6 # 5!, = 336, 5! =, 5!, , # 8 # 7 = 84, (iii) 69!3! ! = 9 # 68! ## 73!# 6! = 39 #, 2#1, , Example 2.9, Rewrite 7! in terms of 5!, Solution, , 2.2.2 Fundamental principle of, counting, Multiplication principle of counting:, Consider the following situation in, an auditorium which has three entrance, doors and two exit doors. Our objective is, to find the number of ways a person can, enter the hall and then come out., Assume that, P1, P2 and P3 are the, three entrance doors and S1 and S2 are, the two exit doors. A person can enter the, hall through any one of the doors P1, P2 or, P3 in 3 ways. After entering the hall, the, person can come out through any of the, two exit doors S1 or S2 in 2 ways., Hence the total number of ways of, entering the hall and coming out is, 3 # 2 = 6 ways., , 7! = 7×6!, = 7×6×5! = 42×5!, , These possibilities are explained in, the given flow chart (Fig. 2.1)., , Example 2.10, 1, 1, n, Find n, if 9! + 10! = 11!, Solution, 1, 1, n, 9! + 10! = 11!, n, 1, 1, 9! + 10 # 9! = 11!, 1 :1 + 1 D, n, 10 = 11!, 9!, 1 11, n, 9! # 10 = 11!, n =, , 11! # 11, 9! # 10, , 11! # 11, 10!, 11 # 10! # 11, =, 10!, n = 121, =, , 30, , P1, , P2, , S1, , P1 S1, , S2, , P1 S2, , S1, , P2 S1, , S2, , P2 S2, S1, , P3 S1, , S2, , P3 S2, , P3, , Fig. 2.1, , The above problem can be solved, by applying multiplication principle of, counting., , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 30, , 21-04-2020 12:18:21 PM

Page 37 :
www.tntextbooks.in, , Solution, , Definition 2.1, There are two jobs. Of which one, job can be completed in m ways, when it, has completed in any one of these m ways,, second job can be completed in n ways,, then the two jobs in succession can be, completed in m # n ways. This is called, , multiplication principle of counting., Example 2.11, , Find the number of 4 letter words,, with or without meaning, which can, be formed out of the letters of the word, “NOTE”, where the repetition of the letters, is not allowed., , Since each question can be answered, in 4 ways, the total number of ways of, answering 4 questions, = 4 # 4 # 4 # 4 = 256 ways., Example 2.13, How many 3 digits numbers can, be formed if the repetition of digits is not, allowed?, Solution, Let us allot 3 boxes to represent the, digits of the 3 digit number, 100th place, , 10th place, , 1, , 2, , Solution, Let us allot four vacant places, (boxes) to represent the four letters., N,O,T,E of the given word NOTE., 1, , 2, , 3, , 4, , Fig. 2.2, , The first place can be filled by, any one of the 4 letters in 4 ways. Since, repetition is not allowed, the number of, ways of filling the second vacant place by, any of the remaining 3 letters in 3 ways., In the similar way the third box can be, filled in 2 ways and that of last box can be, filled in 1 way. Hence by multiplication, principle of counting, total number of, words formed is 4 # 3 # 2 # 1 = 24 words., Example 2.12, If each objective type questions, having 4 choices, then find the total number, of ways of answering the 4 questions., , unit place, 3, , Fig. 2.3, , Three digit number never begins, with “0”. ` 100th place can be filled with, any one of the digits from 1 to 9 in 9 ways., Since repetition is not allowed,, place can be filled by the remaining 9, numbers (including 0) in 9 ways and the, unit place can be filled by the remaining 8, numbers in 8 ways. Then by multiplication, principle of counting, number of 3 digit, numbers = 9 # 9 # 8 = 648., , 10th, , 2.2.3 Addition principle of counting, Let us consider the following, situation in a class, there are 10 boys and, 8 girls. The class teacher wants to select, either a boy or a girl to represent the class, in a function. Our objective is to find, the number of ways that the teacher can, select the student. Here the teacher has to, perform either of the following two jobs., Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 31, , 31, , 21-04-2020 12:18:21 PM

Page 38 :
www.tntextbooks.in, , One student can be selected in 10, ways among 10 boys., One student can be selected in 8, ways among 8 girls, , 3., , If (n+2)! = 60[(n–1)!], find n, , 4., , How many five digits telephone, numbers can be constructed using, the digits 0 to 9 if each number, starts with 67 with no digit appears, more than once?, , 5., , How many numbers lesser than, 1000 can be formed using the digits, 5, 6, 7, 8 and 9 if no digit is repeated?, , Hence the jobs can be performed in, 10 + 8 = 18 ways, This problem can be solved by, applying Addition principle of counting., Definition 2.2, If there are two jobs, each of which, can be performed independently in m, and n ways respectively, then either of, the two jobs can be performed in (m+n), ways. This is called addition principle of, counting., Example 2.14, There are 6 books on commerce and, 5 books on accountancy in a book shop., In how many ways can a student purchase, either a book on commerce or a book on, accountancy?, Solution, Out of 6 commerce books, a book, can be purchased in 6 ways., Out of 5 accountancy books, a book, can be purchased in 5 ways., Then by addition principle of, counting the total number of ways of, purchasing any one book is 5+6=11 ways., Exercise 2.2, 1., 2., 32, , 1, 1, x, Find x if 6! + 7! = 8! ., n!, Evaluate r! (n - r) ! when n = 5, and r = 2., , 2.2.4 Permutation, Suppose we have a fruit salad with, combination of “APPLES, GRAPES &, BANANAS”. We don’t care what order the, fruits are in. They could also be “bananas,, grapes and apples” or “grapes, apples and, bananas”. Here the order of mixing is not, important. Any how, we will have the, same fruit salad., But, consider a number lock with, the number code 395. If we make more, than three attempts wrongly, then it, locked permanently. In that situation we, do care about the order 395. The lock will, not work if we input 3-5-9 or 9-5-3. The, number lock will work if it is exactly 3-9-5., We have so many such situations in our, practical life. So we should know the, order of arrangement, called Permutation., Definition 2.3, The number of arrangements that can be, made out of n things taking r at a time, is called the number of permutation of n, things taking r at a time., For example, the number of three, digit numbers formed using the digits, 1, 2, 3 taking all at a time is 6., , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 32, , 21-04-2020 12:18:21 PM

Page 39 :
www.tntextbooks.in, , 6 three digit numbers are 123, 132,, 231, 213, 312, 321., Notations: If n > 1 and 0 < r < n, then, the number of all permutations of, n distinct things taken r at a time is, denoted by, P(n,r ) (or) nPr, We have the following theorem, without proof., Theorem:, , n!, npr = (n - r) !, , Example 2.15, Evaluate: 5P3 and P(8, 5), Solution, , 5!, 5!, 5P3 = ]5 - 3g = 2! = 60., !, P(8, 5) = 8P5, 8!, = ]8 - 5g, !, 8!, = 3!, 8×7×6×5×4×3!, =, = 6720, 3!, , Results, (i) 0!, , = 1, , n!, n!, ]n - 0g ! = n! = 1, n ]n - 1g !, n!, (iii) np1 = ]n - 1g = ]n - 1g = n, !, !, n!, n!, (iv) npn = ]n - ng = 0! = n!, !, (v) npr = n(n–1) (n–2) ... [n–(r–1)], (ii) np0 =, , In how many ways 7 pictures can be, hung from 5 picture nails on a wall ?, Solution, The number of ways in which 7, pictures can be hung from 5 picture nails, on a wall is nothing but the number of, permutation of 7 things taken 5 at a time, 7!, = 7P5 = ]7 - 5g = 2520 ways., !, , Example 2.18, , Find how many four letter words, can be formed from the letters of the word, “LOGARITHMS” (words are with or, without meanings), Solution, There are 10 letters in the word, LOGARITHMS, Therefore n = 10, Since we have to find four letter, words,, r =4, Hence the required number of four, letter words, = npr = 10 p4, = 10 × 9 × 8 × 7, = 5040., Example 2.19, If nPr = 360, find n and r., Solution, , Example 2.16, Evaluate: (i) 8P3, , Example 2.17, , (ii) 5P4, , Solution, (i) 8P3 = 8 # 7 # 6 = 336., (ii) 5P4 = 5 # 4 # 3 # 2 = 120., , nPr = 360 = 36 × 10, = 3×3×4×5×2, = 6 × 5 × 4 × 3 = 6P4, , Therefore n = 6 and r = 4, , Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 33, , 33, , 21-04-2020 12:18:23 PM

Page 40 :
www.tntextbooks.in, , Permutation of repeated things:, , S, , occurs 4 times, , The number of permutation of n, different things taken r at a time, when, repetition is allowed is nr., , P, , occurs twice., , Example 2.20, Using 9 digits from 1, 2, 3, … , 9 taking, 3 digits at a time, how many 3 digits numbers, can be formed when repetition is allowed?, , Therefore, , required, 11!, permutation = 4! 4! 2!, (ii), , ` Number of 3 digit numbers = nr, , , = 93 = 729 numbers, , Permutations when all the objects are, not distinct:, The number of permutations of, n things taken all at a time, of which p, things are of one kind and q things are of, n!, another kind, such that p + q = n is p! q! ., In general, the number of, permutation of n objects taken all at a, time, p1 are of one kind, p2 are of another, kind, p3 are of third kind, … pk are of, kth kind such that p1 + p2 + …+pk = n is, n!, ., p1 !p2 ! ... pk !, , In this word M occurs once, I occurs 4 times, 34, , , , T, , occurs twice, , , , A, , occurs twice, , and the rest are all different. Therefore, 11!, number of permutations = 2! 2! 2!, , 2.2.5, , Circular permutation, In the last section, we have studied, , permutation of n different things taken, all together is n!. Each permutation is, a different arrangement of n things in a, row or on a straight line. These are called, Linear permutation. Now we consider the, permutation of n things along a circle,, called circular permutation., D, , C, , D, , How many distinct words can be, formed using all the letters of the following, words., (i) MISSISSIPPI (ii) MATHEMATICS., (i) There are 11 letters in the word, MISSISSIPPI, , In the word MATHEMATICS, Here M occurs twice, , Example 2.21, , Solution, , of, , There are 11 letters, , Solution, Here, n = 9 and r = 3, , number, , B, , C(–g, –f), , B, , A, , C(–g, –f), , A, , B, , A, , B, , C(–g, –f), , D, , C, , C, , C(–g, –f), , C, , A, , D, Fig. 2.4, , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 34, , 21-04-2020 12:18:24 PM

Page 41 :
www.tntextbooks.in, , Consider the four letters A, B, C, D, and the number of row arrangement of these, 4 letters can be done in 4! Ways. Of those, 4! arrangements, the arrangements ABCD,, BCDA, CDAB, DABC are one and the same, when they are arranged along a circle., So, the number of permutation of, 4!, ‘4’ things along a circle is 4 = 3!, In general, circular permutation of n, different things taken all at the time = (n–1)!, NOTE, , The number of permutation is, ]n - 1g ! ]10 - 1g ! 9!, =, = 2, 2, 2, Example 2.24, Find the rank of the word ‘RANK’, in dictionary., Solutions, Here maximum number of the word, in dictionary formed by using the letters, of the word ‘RANK’ is 4!, The letters of the word RANK in, alphabetical order are A, K, N, R, , If clock wise and anti-clockwise, circular permutations are considered, to be same (identical), the number of, circular permutation of n objects taken, ]n - 1g !, all at a time is, 2, , Rank of word in dictionary, The rank of a word in a dictionary is, to arrange the words in alphabetical, order. In this dictionary the word may, or may not be meaningful., , Example 2.22, In how many ways 8 students can be, arranged, (i) in a line   (ii) along a circle, Solution, (i) Number of permutations of 8, students along a line = 8P8 = 8!, (ii) When the students are arranged, along a circle, then the number, of permutation is (8–1)! = 7!, Example 2.23, In how many ways 10 identical keys, can be arranged in a ring?, , Number of words starting with A, , =3! =6, , Number of words starting with K, , =3! =6, , Number of words starting with N, , =3! =6, , Number of words starting with RAK =1! = 1, Number of words starting with RANK = 0! = 1, ` Rank of the word RANK is, , 6 + 6 + 6 + 1+1= 20, Exercise 2.3, 1., , If nP4 = 12(nP2), find n., , 2., , In how many ways 5 boys and 3, girls can be seated in a row, so that, no two girls are together?, , 3., , How many 6-digit telephone numbers, can be constructed with the digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each numbers, , Solution, Since keys are identical, both, clock wise and anti-clockwise circular, permutation are same., , Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 35, , 35, , 21-04-2020 12:18:24 PM

Page 42 :
www.tntextbooks.in, , starts with 35 and no digit appear, more than once?, 4., , 5., , 6., , 2.3, , Find the number of arrangements, that can be made out of the letters, of the word “ASSASSINATION”, (a) In how many ways can 8 identical, beads be strung on a necklace?, (b) In how many ways can 8 boys, form a ring?, Find the rank of the word ‘CHAT’, in dictionary., , Combinations, , A combination is a selection of items, from a collection such that the order of, selection does not matter. That is the act of, combining the elements irrespective of their, order. Let us suppose that in an interview, two office assistants are to be selected out, of 4 persons namely A, B, C and D. The, selection may be of AB, AC, AD, BC, BD,, CD (we cannot write BA, since AB and BA, are of the same selection). Number of ways, of selecting 2 persons out of 4 persons is 6., 4×3, It is represented as 4C2 = 1×2 = 6., The process of different selection, without repetition is called Combination., Definition 2.4, Combination is the selection of n things, taken r at a time without repetition. Out of n, things, r things can be selected in ncr ways., n!, nCr = ]n - r g , n > 1, 0 < r < n, r!, !, Here n ≠ 0 but r may be 0., For example, out of 5 balls 3 balls, can be selected in 5C3 ways., 5×4×3, 5!, 5!, 5C3 = ]5 - 3g = 3! 2! = b 1 × 2 × 3 l, 3!, !, = 10 ways., 36, , Example 2.25, Find 8C2, Solution, , 8×7, 8C2 = 2 × 1 = 28, , Permutations are for lists (order, matters) and combinations are for, groups (order doesn’t matter), Properties, (i), , nC0 = ncn = 1, , (ii), , nC1 = n, n ]n - 1g, nC2 =, 2!, , (iii), , (iv) nCx = nCy , then either, x = y or x + y = n, (v), , nCr = nCn–r., , nCr+ nCr–1 = (n + 1)Cr, npr, (vii) nCr = r!, (vi), , Example 2.26, If nC4 = nC6, find 12Cn ., Solution, If nCx = nC y , then x + y = n., Here nC4 = nC6, ` n = 4 + 6 = 10, 12Cn = 12C10, , , = 12C2, , 12 × 11, = 1 × 2 = 66, Example 2.27, If nPr = 720 ; nCr = 120, find r, , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 36, , 21-04-2020 12:18:26 PM

Page 43 :
www.tntextbooks.in, , Solution, , Solution, , nP, We know that nCr = r! r, 720, 120 = r!, 720, r! = 120 = 6 = 3 !, (r = 3, , In part A, out of 10 questions 8 can, be selected in 10C8 ways. In part B out of, 10 questions 5 can be selected in 10C5., Therefore by multiplication principle the, total number of selection is, 10C8 × 10C5 = 10C2 × 10C5, =, , Example 2.28, If 15C3r = 15Cr+3 , find r, Solution, 15C3r = 15Cr+3, Then by the property,, nCx = nCy ( x + y = n, we have, 3r + r + 3 = 15, (, , r =3, , Example 2.29, From a class of 32 students, 4 students, are to be chosen for a competition. In how, many ways can this be done?, Solution, , =, Example 2.30, , = 11340, Example 2.31, A Cricket team of 11 players is to be, formed from 16 players including 4 bowlers, and 2 wicket-keepers. In how many different, ways can a team be formed so that the team, contains at least 3 bowlers and at least one, wicket-keeper?, Solution, A Cricket team of 11 players can be, formed in the following ways:, (i) 3 bowlers, 1 wicket keeper and 7 other, players can be selected in, 4C3 × 2C1 ×10C7 ways, , The number of combination = 32C4, =, , 10 × 9 10 × 9 × 8 × 7 × 6, 2 × 1 × 5 × 4 × 3 × 2 ×1, , 32!, - 4g !, 32, ], 4!, , 32!, 4! ]28g !, , A question paper has two parts, namely Part A and Part B. Each part, contains 10 questions. If the student has, to choose 8 from part A and 5 from part, B, in how many ways can he choose the, questions?, , 4C3 ×2C1 ×10C7 = 4C1 × 2C1 × 10C3, = 960 ways, (ii) 3 bowlers, 2 wicket keepers and 6, other players can be selected in, 4C1 × 2C2 × 10C6 ways, 4C1 × 2C2 × 10C6 = 4C1 × 2C2 × 10C4, = 840 ways, (iii) 4 bowlers and 1 wicket keeper and, 6 other players 4C4 # 2C1 # 10C6, 4C4 # 2C1 # 10C4 = 420 ways, Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 37, , 37, , 21-04-2020 12:18:27 PM

Page 44 :
www.tntextbooks.in, , 4 bowlers, 2 wicket keepers and 5, other players can be selected in, 4C4 # 2C2 # 10C5 ways, , Exercise 2.4, 1., , If nPr = 1680 and nCr = 70, find n, and r ., , 2., , Verify that 8C4 + 8C3 = 9C4 ., , 4C4 # 2C2 # 10C5 = 252 ways, By addition principle of counting., Total number of ways, = 960 + 840 + 420 + 252, , 3., , How many chords can be drawn, through 21 points on a circle?, , 4., , How many triangles can be formed, by joining the vertices of a hexagon?, , 5., , Out of 7 consonants and 4 vowels,, how many words of 3 consonants, and 2 vowels can be formed?, , 6., , If four dice are rolled, find the, number of possible outcomes in, which atleast one die shows 2., , 7., , There are 18 guests at a dinner party., They have to sit 9 guests on either, side of a long table, three particular, persons decide to sit on one particular, side and two others on the other side., In how many ways can the guests to, be seated?, , 8., , If a polygon has 44 diagonals, find, the number of its sides., , 9., , In how many ways can a cricket, team of 11 players be chosen out of, a batch of 15 players?, , = 2472, Example 2.32, , If 4 ]nC2g = ]n + 2g C3 , find n, , Solution, , 4 (nC2) = ]n + 2g C3, , n (n - 1), ]n + 2g]n + 1g]ng, =, 4 1#2, 1#2#3, 12 ]n - 1g = (n+2) (n+1), 12 ]n - 1g = ^n 2 + 3n + 2h, , n 2 - 9n + 14 = 0, , ]n - 2g]n - 7g = 0 & n = 2, n = 7, Example 2.33, , If ]n + 2g Cn = 45, find n, , Solution, , ]n + 2g Cn = 45, , (i) There is no restriction on the, selection., (ii) A particular player is always, chosen., (iii) A particular player is never, chosen., , ]n + 2g Cn + 2 - n = 45, ]n + 2g C2 = 45, , (n + 2) ]n + 1g, = 45, 2, n 2 + 3n - 88 = 0, , ]n + 11g]n - 8g = 0, , n = –11, 8, , n = - 11 is not possible, `n = 8, 38, , 10., , A Committee of 5 is to be formed, out of 6 gents and 4 ladies. In how, many ways this can be done when, (i) atleast two ladies are included, (ii) atmost two ladies are included, , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 38, , 21-04-2020 12:18:31 PM

Page 45 :
www.tntextbooks.in, , 2.4, , Mathematical Induction, , Mathematical induction is one of, the techniques which can be used to prove, variety of mathematical statements which, are formulated in terms of n, where n is a, positive integer., Mathematical Induction is used in, the branches of Algebra, Geometry and, Analysis where it turns out to be necessary, to prove some truths of propositions., , Step 2: Let us assume that the statement is, true for n = k., i.e., P(k) is true, , k ]k + 1g, is true, 2, Step 3: To prove that P(k + 1) is true, , , , 1 + 2 + 3+… + k =, , P(k + 1) = 1 + 2 + 3 + … + k + (k +1), = P(k) + k +1, =, =, , The principle of mathematical induction:, Let P(n) be a given statement for n ! N ., (i) Initial step: Let the statement is true, for n = 1 i.e., P(1) is true and, (ii) Inductive step: If the statement is, true for n = k (where k is a particular, but arbitrary natural number) then, the statement is true for n = k + 1., i.e., truth of P(k) implies the truth, of P(k+1). Then P(n) is true for all, natural numbers n., , k ]k + 1g + 2 ]k + 1g, 2, , ]k + 1g]k + 2g, 2, , ` P(k +1) is true, Thus if P(k) is true, then P(k +1) is also true., ` P(n) is true for all n ! N, Hence 1+ 2 + 3 +… +n =, , n ]n + 1g, ,n ! N, 2, , Example 2.35, By the principle of Mathematical, Induction, prove that, 1 + 3 + 5 + … + (2n–1) = n2, for all n ! N ., , Example 2.34, Using mathematical induction method,, Prove that, 1+2+3+…+n=, Solution, , =, , k ]k + 1g, + k +1, 2, , n ]n + 1g, ,n!N., 2, , Let the given statement P(n) be, defined as, n ]n + 1g, 1+2+3+ … + n =, for n ! N, 2, Step 1:, Put n = 1, LHS P(1) = 1, 1 ]1 + 1g, RHS P(1) =, =1, 2, LHS = RHS for n = 1, ` P(1) is true, , Solution, Let P(n) denote the statement, 1 + 3+ 5 +…+ (2n–1) = n2, Put n = 1, LHS = 1, RHS = 12, = 1, LHS = RHS, ` P(1) is true., Assume that P(k) is true., i.e., 1 + 3 + 5 + ... + (2k–1) = k2, Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 39, , 39, , 21-04-2020 12:18:33 PM

Page 46 :
www.tntextbooks.in, , To prove: P(k + 1) is true., ` 1 + 3 + 5 + … + (2k–1) + (2k+1), = P ]k g + ]2k + 1g, , = k 2 + 2k + 1, = ]k + 1g2, , ` P(n) is true for all n ! N ., , P(k + 1) is true whenever P(k) is true., , Example 2.37, , ` P(n) is true for all n ! N ., , Show by the principle of mathematical, induction that 23n–1 is a divisible by 7, for, all n ! N ., , Example 2.36, By Mathematical Induction, prove that, n ]n + 1g]2n + 1g, , for, 12 + 22 + 32 +…+ n2 =, 6, all n ! N ., Solution, , ` LHS = 12, =1, , 1 ]1 + 1g]2 + 1g, 6, =1, , RHS =, , LHS = RHS, ` P(1) is true., Assume that P(k) is true., P(k), , Solution, Let the given statement P(n) be, defined as P(n) = 23n–1, Step 1: Put n = 1, , Let P(n) denote the statement:, n ]n + 1g]2n + 1g, 12 + 22 + 32 + … +n2 =, 6, Put n = 1, , = 12 + 22 + 32 + … +k2, k ]k + 1g]2k + 1g, =, 6, , Now,, , 12 + 22+ 32+… + k2 + (k+1)2 = p ]k g + ]k + 1g2, k ]k + 1g]2k + 1g ]k + 1g2, +, 6, k ]k + 1g]2k + 1g + 6 ]k + 1g2, =, 6, ]k + 1g6k ]2k + 1g + 6 ]k + 1g@, =, 6, =, , 40, , ]k + 1g^2k 2 + 7k + 6h, 6, ]k + 1g]k + 2g]2k + 3g, =, 6, ` P(k + 1) is true whenever P(k) is true, =, , ` P(1) = 23 –1, = 7 is divisible by 7, i.e., P(1) is true., Step 2: Let us assume that the statement, is true for n = k i.e., P(k) is true., We assume 23k –1 is divisible by 7, & 2 3k - 1 = 7 m, Step 3: To prove that P(k + 1) is true, P(k +1) = 23]k + 1g –1, , = 2]3k + 3g - 1, = 23k $ 23 - 1, = 23k $ 8 - 1, , = 23k $ ]7 + 1g - 1, , = 23k $ 7 + 23k - 1, = 23k $ 7 + 7m = 7 (23k + m), which is divisible by 7, P(k +1) is true whenever p(k) is true, ` P(n) is true for all n ! N ., Hence the proof., , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 40, , 21-04-2020 12:18:37 PM

Page 47 :
www.tntextbooks.in, , Example 2.38, By the principle of mathematical, induction prove that n2 + n is an even, number, for all n ! N ., Solution, Let P(n) denote the statement that,, “n2+ n is an even number”., Put n = 1, ` 12 + 1 = 1 + 1 = 2, an even number., , 1 + 4 + 7 + … + (3n –2) =, for all n ! N ., , 5., , 32n –1 is a divisible by 8, for all, n!N., , 6., , an - bn is divisible by a – b, for all, n!N., , 7., , 52n –1 is divisible by 24, for all, n!N., , 8., , n(n + 1) (n+2) is divisible by 6, for, all n ! N ., , 9., , 2n > n, for all n ! N ., , Let us assume that P(k) is true., ` k2 + k is an even number is true, , n ]3n - 1g, ,, 2, , 4., , ` Take k2 + k = 2m…(1), , 2.5 Binomial Theorem, , To prove P(k + 1) is true, , An algebraic expression of sum, or the difference of two terms is called a, binomial. For example ^x + y h, ]5a - 2bg, c x + 1y m, c p + 5p m, , ` ]k + 1g2 + ]k + 1g, , = k 2 + 2k + 1 + k + 1, , 7 1, ^x + y h, ]5a - 2bg, c x + 1y m, c p + 5p m, d 4 + y2 n etc., are binomials., = k + k + 2k + 2, The Binomial theorem or Binomial, = 2m + 2(k + 1) by (1), Expression is a result of expanding the, = 2(m + k + 1) ( a multiple of 2), powers of binomials. It is possible to, expand (x + y)n into a sum involving, ` (k + 1)2 + (k + 1) is an even number, terms of the form axbyc, exponents b and, ` P(k + 1) is true whenever P(k) is true., c are non-negative integers with b + c = n,, P(n) is true for n ! N ., the coefficient ‘a’ of each term is a positive, integer called binomial coefficient., Exercise 2.5, Expansion of (x + a)2 was given, by Greek mathematician Euclid on 4th, By the principle of mathematical induction,, century and there is an evidence that the, prove the following, binomial theorem for cubes i.e., expansion, n 2 (n + 1) 2, 3, 3, 3, 3, 1. 1 + 2 + 3 +… +n =, of (x+a)3 was known by 6th century in, 4, for all n ! N ., India. The term Binomial coefficient was, first introduced by Michael Stifle in 1544., 2. 1.2 + 2.3 + 3.4 + … + n(n + 1), Blaise Pascal (19 June 1623 to 19 August, n ]n + 1g]n + 2g, ,, for, all, ., =, n, !, N, 3, 1662) a French Mathematician, Physicist,, inventor, writer and catholic theologian., 3. 4 + 8 + 12 + … + 4n = 2n(n+1), for, all n ! N ., In his Treatise on Arithmetical triangle, 2, , Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 41, , 41, , 21-04-2020 12:18:40 PM

Page 48 :
www.tntextbooks.in, , of 1653 described the convenient tabular, presentation for Binomial coefficient now, called Pascal’s triangle. Sir Issac Newton, generalized the Binomial theorem and, made it valid for any rational exponent., Now we study the Binomial theorem, for (x + a)n, Theorem(without proof ), If x and ‘a’ are real numbers, then for all, n!N, , NOTE, (i) Number of terms in the expansion, of (x + a)n is n+1, (ii) Sum of the indices of x and a in each, term in the expansion is n, (iii) nC0, nC1, nC2, nC3, ... nCr, ... nCn, are also represented as C0, C1, C2,, C3, ..., Cr, ..., Cn, are called Binomial, co-efficients., , General, (iv), n - r r term in the expansion of, ]x + agn = nC0 xn a0 + nC1 xn - 1 a1 + nC2 xn - 2 a 2 + ..., + nC, a, rx, n-r r, n is t, (x+a), a, r+1 = nCr x, nC0 xn a0 + nC1 xn - 1 a1 + nC2 xn - 2 a 2 + ... + nCr xn - r ar + ..., n, (v) Since nCr = nCn - r, for r =0, 1, 2, …, n, + nCn - 1 x1 an - 1 + nCn x0 an = / nCr xn - r ar, in the expansion of (x + a)n , binomial, r=0, co-efficients of terms equidistant, from the beginning and from the end, NOTE, of the expansion are equal., 0, When n = 0, (x+a) = 1, (vi) Sum of the co-efficients in the, When n = 1,, ]x + ag = 1C0 x + 1C1 a = x + a, , expansion of (1+x)n is equal of 2n, , (vii) In the expansion of (1+x) n, the sum, When n = 2,, of the co-efficients of odd terms =, 2, 2, a, 2, 2, 2, the sum of the co-efficients of the, ]x + ag = 2C0 x + 2C1 x + 2C2 a = x + 2xa + a, even terms =2n–1, = 2C0 x 2 + 2C1 xa + 2C2 a 2 = x 2 + 2xa + a 2, When n = 3 ,, , ]x + ag3 = 3C0 x3 + 3C1 x 2 a + 3C2 xa 2 + 3C3 a3, , = x3 + 3x 2 a + 3xa 2 + a3 and so on., , Given below is the Pascal’s Triangle, showing the co- efficient of various terms, in Binomial expansion of ]x + agn, n = 0 1 , n = 1 1 1, n = 2 , , 1 , , 2 , , 1, , n = 3 1 3 3 1, n = 4 1 4 6 4 1, n = 5 1 5 10 10 5 1, , Fig. 2.5, , 42, , Middle term of (x + a)n, Case (i), , If n is even, then the number of, terms n + 1 is odd, then there, is only one middle term, given, by tt, n, 2 +1, , Case (ii) If n is odd, then the number of, terms (n+1) is even. Therefore,, we have two middle terms, given by tt and tt, n+1, 2, , n+3, 2, , Sometimes we need a particular, term in the expansion of (x + a) n. For this, we first write the general term tr + 1 . The, , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 42, , 21-04-2020 12:18:44 PM

Page 50 :
www.tntextbooks.in, , Example 2.42, , Example 2.44, , Find the 5th term in the expansion, 3 10, of c x - x 2 m, Solution, , Find the middle term in the expansion, , 9, x, of b 3 + 9y l, , Solution, , General term in the expansion of, ]x + agn is tr + 1 = nCr xn - r ar ...(1), 3 10, term of c x - x 2 m, To find the, For this take r = 4, 5th, , Then,, 3 4, t4 + 1 = t5 = 10C4 ]x g 6 c - x 2 m, , (Here n = 10 , x = x , a = –, = 10C4 ]x g 6, =, , 34, x8, , 3, ), x2, , 9, x, Compare b 3 + 9y l with (x + a)n, , Since n = 9 ,we have 10 terms(even), ` There are two middle terms, t, t, t, t, namely n 2+ 1 , n2+ 3 i.e., 92+ 1 , 92+ 3, General term in the expansion of, (x + a)n is, tr + 1 = nCr xn - r ar , , Here t5 and t6 are middle terms., put r = 4 in (1),, , 17010, x2, , x 5, 4, t4 + 1 = t5 = 9C4 b 3 l $ ^9y h, , Example 2.43, , x5 4 4, $9 y, 35, 9 × 8 × 7 × 6 x5, = 4 × 3 ×2 × 1 $ 5 $ 9 4 y 4, 3, = 9C4, , Find the middle term in the expansion, 2 10, of b x 2 - x l, , = 3402 x5 y 4, , Solution, 2 10, Compare b x 2 - x l with ]x + agn, 2, n = 10, x = x 2, a = - x, Since n = 10, we have 11 terms (odd), , put r = 5 in (1),, x 4, 5, t5 + 1 = t6 = 9C5 b 3 l $ ^9y h, = 9C5, , ` 6th term is the middle term., The general term is, tr + 1 = nCr x, , a , , ...(1), , 5, 5 -2, t5 + 1 = t6 = 10C5 ^ x 2h b x l, , = 10C5 x, , 10, , ]- 2g 5, x5, , = – 8064x5, 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 44, , x4 5 5, $9 y, 34, , = 91854 x 4 y5, Example 2.45, , n-r r, , To get t6 , put r = 5, , 44, , ... (1), , Find the Coefficient of x10 in the, 3 11, binomial expansion of b 2x 2 - x l, Solution, General term of (x + a)n is, tr + 1 = nCr xn - r ar (1), 3 11, Compare b 2x 2 - x l with (x + a)n, , 21-04-2020 12:18:59 PM

Page 51 :
www.tntextbooks.in, , tr + 1 = 11Cr ^2x2 h, , 11 - r, , -3 r, b x l, , = 11Cr 211 - r x2]11, , - rg, , Exercise 2.6, , r, ]- 3g r b 1x l, , = 11Cr 211 - r ]- 1gr $ 3r x22 - 3r, , To find the co-efficient of x10, take, 22–3r = 10, r = 4. Put r = 4 in (1), t5 = 11C4 211 - 4 3 4 x10 = 11C4 $ 27 3 4 x10, ` Co-efficient of, , x10, , 7 4, , is 11C4 2 3 ., , Example 2.46, Find the term independent of x in, 1 9, the expansion of c 2x + 2 m ., 3x, Solution, General term in the expansion of, (x + a)n is tr + 1 = nCr xn - r ar, Compare c 2x +, ` tr + 1, , 1 r, 9-r, 2, x, c, m, ], g, = 9Cr, 3x 2, = 9Cr 2, , tr + 1, , 1 9, m with (x + a)n, 3x 2, , = 9Cr, , 9-r 9-r, , x, , 1 -2r, x, 3r, , 2 9 - r 9 - 3r, , $x, 3r, , ... (1), , To get the term independent of x,, equating the power of x as 0, , 1. Expand the following by using binomial, theorem, 1 7, (ii) c x + y m, (i) ]2a - 3bg4, 1 6, (iii) c x + x 2 m, 2. Evaluate the following using binomial, theorem:, (i) (101)4, , 3. Find the 5th term in the expansion of, (x–2y)13., 4. Find the middle terms in the expansion of, 2 8, 1 11, (ii) b3x + x2 l, (i) b x + x l, 3 10, 2, (iii) c 2x - x3 m, , 5. Find the term independent of x in the, expansion of, 2 15, 2 9, (i) b x 2 - 3x l (ii) c x - x 2 m, 1 12, (iii) b 2x 2 + x l, 6. Prove that the term independent of, 1 2n, x in the expansion of b x + x l is, 1 $ 3 $ 5... ]2n -1g 2n ., n!, 7. Show that the middle term in the expansion, 1.3.5… (2n - 1) 2n xn, 2n, +, 1, x, ], g, of, is, n!, , 9 – 3r = 0, `, , Exercise 2.7, , r = 3, , Choose the correct answer, , Put r = 3 in (1), 29 - 3 0, $x, 35, 26, = 9C3 5, 3, , t3 + 1 = 9C3, , = 1792, , (ii) (999)5, , 1., , If nC3 = nC2 then the value of nc4 is, (a) 2  (b) 3  (c) 4  (d) 5, , 2., , The value of n, when np2 = 20 is, (a) 3, , (b) 6, , (c) 5, , (d) 4, , Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 45, , 45, , 21-04-2020 12:19:05 PM

Page 52 :
www.tntextbooks.in, , 3., , The number of ways selecting 4, players out of 5 is, , 12., , (a) 4!   (b) 20   (c) 25   (d) 5, 4., , If nPr = 720 ]nCr g , then r is equal to, (a) 4   (b) 5   (c) 6   (d) 7, , 5., , 6., , 7., , 8., , 9., , The possible out comes when a coin, is tossed five times, 5, (a) 25    (b) 52  (c) 10  (d) 2, The number of diagonals in a, polygon of n sides is equal to, (a) nC2, , (b) nC2 - 2, , (c) nC2 - n, , (d) nC2 - 1, , (c) 162, 13., , 14., , (a) 81, , (b) 16, , (c) 8 2, , (d) 27 3, , kx, 4, 1, If ]x + 4g]2x - 1g = x + 4 + 2x - 1, then k is equal to, , The greatest positive integer which, divide n ]n + 1g]n + 2g]n + 3g for all, n ! N is, (a) 2   (b) 6   (c) 20   (d) 24, , (a) 206, , If n is a positive integer, then the, number of terms in the expansion, of ]x + agn is, , (a) n, , (b) n + 1, , (c) n – 1, , (d) 2n, , For all n > 0, nC1+ nC2+ nC3+…, + nCn is equal to, , 15., , 16., , (b) 133, , (c) 216, , (d) 300, , The number of parallelograms that, can be formed from a set of four, parallel lines intersecting another, set of three parallel lines is, (a) 18   (b) 12   (c) 9   (d) 6, , 17., , (b) 2n –1, (d) n2–1, , There are 10 true or false questions in, an examination. Then these questions, can be answered in, (a) 240 ways, (b) 120 ways, (c) 1024 ways, (d) 100 ways, , The term containing x3 in the, 18. The value of ]5C0 + 5C1g + ]5C1 + 5C2g + ]5C2 + 5C3g + ]5C3, 7, expansion of ^x - 2y h ]is, 5C0 + 5C1g + ]5C1 + 5C2g + ]5C2 + 5C3g + ]5C3 + 5C4g + ]5C4 + 5C5g, The middle term in the expansion, 1 10, of b x + x l is, 1, (b) 10C5, (a) 10C4 b x l, (c) 10C6, , 46, , The last term in the expansion of, 8, ^3 + 2 h is, , (a) 9   (b) 11   (c) 5   (d) 7, , is, , (a) 3rd   (b) 4th   (c) 5th   (d) 6th, , 11., , (d) 160, , The number of 3 letter words that, can be formed from the letters of the, word number when the repetition is, allowed are, , (a) 2n, (c) n2, 10., , The constant term in the expansion, 2 6, of b x + x l is, (a) 156, (b) 165, , (d) 10C7 x 4, , 19., , (a) 26 –2, , (b) 25 –1, , (c) 28, , (d) 27, , The total number of 9 digit number, which have all different digit is, (a) 10!, (c) 9 × 9!, , (b) 9!, (d) 10 × 10!, , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 46, , 21-04-2020 12:19:09 PM

Page 53 :
www.tntextbooks.in, , 20., , The number of ways to arrange the, letters of the word “ CHEESE ”, (a) 120   (b) 240   (c) 720   (d) 6, , 21., , 22., , Thirteen guests has participated in a, dinner. The number of handshakes, happened in the dinner is, (a) 715, , (b) 78, , (c) 286, , (d) 13, , Number of words with or without, meaning that can be formed using, letters of the word “EQUATION” ,, with no repetition of letters is, , 3., , 4., , 5., , 24., , Sum of Binomial co-efficient in a, particular expansion is 256, then, number of terms in the expansion is, (a) 8, (b) 7, (c) 6, (d) 9, , 25., , n!, (c) ]n - r g, !, , How many code symbols can be, formed using 5 out of 6 letters A, B,, C, D, E, F so that the letters a) cannot, be repeated b) can be repeated, c) cannot be repeated but must, begin with E d) cannot be repeated, but end with CAB., , 7., , From 20 raffle tickets in a hat, four, tickets are to be selected in order., The holder of the first ticket wins, a car, the second a motor cycle,, the third a bicycle and the fourth a, skateboard. In how many different, ways can these prizes be awarded?, , 8., , In how many different ways, 2, Mathematics, 2 Economics and 2, History books can be selected from, 9 Mathematics, 8 Economics and 7, History books?, , 9., , Let there be 3 red, 2 yellow and, 2 green signal flags. How many, different signals are possible if we, wish to make signals by arranging, all of them vertically on a staff?, , 10., , Find the Co-efficient of x11 in the, 2 17, +, x, c, m, expansion of, x2, , (b) nr, , n!, (d) ]n + r g, !, , Sum of the binomial coefficients is, (a) 2n, (c) 2n, , (b) n 2, (d) n+17, , Miscellaneous Problems, 1., 2., , Resolve into Partial Fractions :, 5x + 7, ]x - 1g]x + 3g, Resolve into Partial Fractions:, x-4, 2, x - 3x + 2, , (3!)! × 2!, 5!, , 6., , The number of permutation of n, different things taken r at a time,, when the repetition is allowed is, (a) r n, , Evaluate the following., (ii) 3! +2 1!, (i) 3! × 0! + 0!, 2!, (2 )!, (iii), , (a) 7!   (b) 3!   (c) 8!   (d) 5!, 23., , Decompose into Partial Fractions:, 6x 2 - 14x - 27, ]x + 2g]x - 3g2, Decompose into Partial Fractions:, 5x 2 - 8x + 5, ]x - 2g^ x 2 - x + 1h, , Algebra, , 02_11th_BM-STAT_Ch-2-EM.indd 47, , 47, , 21-04-2020 12:19:10 PM

Page 54 :
www.tntextbooks.in, , Summary, , zz For any natural number n, n factorial is the product of the first n, natural numbers and is denoted by n! or n ., , zz For any natural number n, n! = n(n–1) (n–2)…3×2×1, zz 0!=1, zz If there are two jobs, each of which can be performed independently in m and n ways, respectively, then either of the two jobs can be performed in (m+n) ways., , zz The number of arrangements that can be made out of n things taking r at a time is, called the number of permutation of n things taking r at a time., , n!, zz npr = ]n rg!, , n!, n!, zz npq = ]n 0g ! = n! = 1, , n ^n - 1h !, , n!, zz np1 = ]n 1g ! = ]n - 1g ! = n, n!, n!, zz npn = ]n ng ! = 0! = n!, , zz npr = n (n - 1) (n - 2) … [n - (r - 1)], zz The number of permutation of n different things taken r at a time, when repetition, is allowed is nr., , zz The number of permutations of n things taken all at a time, of which p things are of, n!, one kind and q things are of another kind, such that p+q = n is p! q!, zz Circular permutation of n different things taken all at a time = (n–1)!, , zz The number of circular permutation of n identical objects taken all at a time is, ]n - 1g !, 2, zz Combination is the selection of n things taken r at a time without repetition., , zz Out of n things, r things can be selected in ncr ways., zz nCr = r! ^nn-! r h ! , 0 # r # n, zz nC0 = nCn = 1., zz nC1 = n., , n ^n - 1h, 2!, zz nCx = nC y , then either x = y or x + y = n., , zz nC2 =, , 48, , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 48, , 21-04-2020 12:19:13 PM

Page 56 :
www.tntextbooks.in, , ICT Corner, Expected final outcomes, Step – 1, Open the Browser and type the URL given (or) Scan, the QR Code., GeoGebra Work book called “11th BUSINESS, MATHEMATICS and STATISTICS” will appear. In this several work sheets for Business Maths, are given, Open the worksheet named “Pascal’s Triangle”, Step - 2, Pascal’s triangle page will open. Click on the check boxes “View Pascal Triangle” and “View, Pascal’s Numbers” to see the values. Now you can move the sliders n and r to move the red point, to respective nCr. Now click on view Combination to see the nCr calculation., Browse in the link, 11th Business Mathematics and Statistics: https://ggbm.at/qKj9gSTG (or), scan the QR Code, , ICT Corner, Expected final outcomes, Step – 1, Open the Browser and type the URL given (or) Scan, the QR Code., GeoGebra Work book called “11th BUSINESS, MATHEMATICS and STATISTICS” will appear. In this several work sheets for Business Maths, are given, Open the worksheet named “Combination Exercise”, Step - 2, Combination practice exercise will open. You can generate as many questions you like by clicking, “New question” Solve the problem yourself and check your answer by clicking on the “Show, solution” box. Also click “Short method” and follow this method., Browse in the link, 11th Business Mathematics and Statistics: https://ggbm.at/qKj9gSTG (or), scan the QR Code, , 50, , 11th Std. Business Mathematics and Statistics, , 02_11th_BM-STAT_Ch-2-EM.indd 50, , 21-04-2020 12:19:15 PM

Page 57 :
www.tntextbooks.in, , Chapter, , 3, , ANALYTICAL GEOMETRY, , Learning Objectives, After studying this, chapter, the students will, be able to understand, , •, •, •, •, •, •, •, •, •, •, , the locus, the angle between two lines., the concept of concurrent lines., the pair of straight lines, the general equation and, parametric equation of a circle., the centre and radius of the, general equation of a circle., the equation of a circle when the, extremities of a diameter are given., the equation of a tangent to the, circle at a given point., identification of conics., the standard equation of a parabola,, its focus, directrix, latus rectum, and commercial applications, , Introduction, The word “Geometry” is derived, from the word “geo” meaning “earth” and, “metron” meaning “measuring”. The need, of measuring land is the origin of geometry., “Geometry” is the study of points,, lines, curves, surface etc., and their properties., The importance of analytical geometry is, that it establishes a correspondence between, geometric curves and algebraic equations., A systematic study of geometry, by the use of algebra was first carried out, by celebrated French Philosopher and, mathematician Rene Descartes (1596-1650),, in his book’ La Geometry’, published, in 1637. The resulting combination of, analysis and geometry is referred now as, analytical geometry. He is known as the, father of Analytical geometry, Analytical geometry is extremely, useful in the aircraft industry, especially, when dealing with the shape of an airplane, fuselage., , 3.1, , Locus, , Definition 3.1, The path traced by a moving, point under some specified geometrical, condition is called its locus., , 3.1.1 Equation of a locus, Rene Descartes, (1596-1650), , Any relation in x and y which is, satisfied by every point on the locus is, called the equation of the locus., Analytical Geometry, , 03_11th_BM-STAT_Ch-3-EM.indd 51, , 51, , 21-04-2020 12:14:34 PM

Page 58 :
www.tntextbooks.in, , Solution, , a, C, , a, , P2, , Let P ^ x1, y1h be any point on the, locus and A be the foot of the perpendicular, from P ^ x1, y1h to the y-axis., Given that OP = 3 AP, , a, , OP2 = 9AP2, , P1, , 2, ^ x1 - 0 h2 + ^ y1 - 0 h2 = 9x1, , Fig. 3.1, , x12 + y12 = 9x12, , For example,, (i) The locus of a point P ^ x1, y1h whose, distance from a fixed point C(h, k) is, constant, is a circle. The fixed point, ‘C’ is called the centre., (ii) The locus of a point whose distances, from two points A and B are equal is, the perpendicular bisector of the line, segment AB., , 8x12 - y12 = 0, , ` The locus of P ^ x1, y1h is, 8x 2 - y 2 = 0, , Example 3.2, Find the locus of the point which is, equidistant from (2, –3) and (3, –4)., Solution, Let A(2, –3) and B (3, –4) be the, given points, locus., , A, , Let P ^ x1, y1h be any point on the, , Given that, , B, , PA = PB., PA2 = PB2, , ^ x1 - 2h2 + ^ y1 + 3 h2 =, ^ x1 - 3 h2 + ^ y1 + 4 h2, x12 - 4x1 + 4 + y12 + 6y1 + 9 =, , Fig. 3.2, , x12 - 6x1 + 9 + y12 + 8y1 + 16, , , , i.e., 2x1 - 2y1 - 12 = 0, Straight line is the locus of a point, which moves in the same direction., Example 3.1, A point in the plane moves so that, its distance from the origin is thrice its, distance from the y- axis. Find its locus., 52, , i.e.,, , x1 - y1 - 6 = 0, , The locus of P ^ x1, y1h is x - y - 6 = 0., , Example 3.3, Find the locus of a point, so that the, join of (–5, 1) and (3, 2) subtends a right, angle at the moving point., , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 52, , 21-04-2020 12:14:39 PM

Page 59 :
www.tntextbooks.in, , Solution, , P so that the area of triangle APB =, 8 sq.units., , Let A(–5, 1) and B(3, 2) be the given, points, locus., , 3.2, , Let P ^ x1, y1h be any point on the, , System of Straight Lines, , 3.2.1 Recall, In lower classes, we studied the, basic concept of coordinate geometry, like, distance formula, section - formula, area, of triangle and slope of a straight lines etc., , Given that +APB = 90o., Triangle APB is a right angle, triangle., BA2 = PA2+PB2, , We also studied various form of, 2 h2 in X std. Let us recall, (3+5) + (2–1) = ^ x1 + 5 h + ^ y1 - 1 h + ^ xequations, 1 - 3 h + ^ yof, 1 - lines, the equations of straight lines. Which, ^ x1 + 5 h2 + ^ y1 - 1 h2 + ^ x1 - 3 h2 + ^ y1 - 2 h2, will help us for better understanding the, 2, 2, 2, + y12 - 4y1and, + 4 definitions of XI std, 65 = x1 + 10x1 + 25 + y1 - 2y1 + 1 + x1 - 6xnew, 1 + 9concept, co-ordinate geometry., x12 + 10x1 + 25 + y12 - 2y1 + 1 + x12 - 6x1 + 9 + y12 - 4y1 + 4, 2, , 2, , 2, , i.e., 2x12 + 2y12 + 4x1 - 6y1 + 39 - 65 = 0, x12 + y12 + 2x1 - 3y1 - 13 = 0, , i.e.,, , The locus of P ^ x1, y1h is, , x 2 + y 2 + 2x - 3y - 13 = 0, Exercise 3.1, , 1., , Find the locus of a point which is, equidistant from (1, 3) and x axis., , 2., , A point moves so that it is always at, a distance of 4 units from the point, (3, –2), , 3., , 4., , 5., , If the distance of a point from the, points (2, 1) and (1, 2) are in the, ratio 2 :1, then find the locus of the, point., Find a point on x axis which is, equidistant from the points (7, –6), and (3, 4)., If A(–1, 1) and B(2, 3) are two fixed, points, then find the locus of a point, , 2, , 2, , Various forms of straight lines:, (i), , Slope-intercept form, , Equation of straight line having, slope m and y-intercept ‘c’ is y = mx+c, (ii), , Point- slope form, , Equation of Straight line passing, through the given point P ^ x1, y1h and, having a slope m is, y - y1 = m ^ x - x1h, , (iii), , Two-Point form, , Equation of a straight line joining, , the given points A ^ x1, y1h, B ^ x2, y2h is, y - y1, x - x1, =, y2 - y1 x2 - x1, , In determinant form, equation of, straight line joining two given points, A ^ x1, y1h and B ^ x2, y2h is, x y 1, x1 y1 1 = 0, x2 y2 1, Analytical Geometry, , 03_11th_BM-STAT_Ch-3-EM.indd 53, , 53, , 21-04-2020 12:14:42 PM

Page 60 :
www.tntextbooks.in, , (iv), , Intercept form, , Equation of a straight line whose x, x y, and y intercepts are a and b, is a + = 1., b, (v), General form, Equation of straight line in general, form is ax + by + c = 0 where a, b and c are, constants and a, b are not simultaneously, zero., , 3.2.2 Angle between two straight, lines, Let l1 and l2 be two straight lines, represented by the equations l1: y = m1 x + c1, and l2: y = m2 x + c2 intersecting at P., , NOTE, , m -m, (i) If 1 +1 m m2 is positive, then i ,, 1 2, the angle between l1 and l2 is acute, and if it is negative, then, i is the, obtuse., , (ii) We know that two straight lines are, parallel if and only if their slopes, are equal., (iii) We know that two lines are, perpendicular if and only if the, product of their slopes is –1. (Here, the slopes m1 and m2 are finite.), , If i1 and i 2 are two angles made, by l1 and l2 with x-axis then slope of the, lines are m1 = tan i1 and m2 = tan i 2 ., y, l1, , l2, , The straight lines x-axis and, y-axis are perpendicular to each, other. But, the condition m1 m2 = - 1, is not true because the slope of the, x-axis is zero and the slope of the, y-axis is not defined., , P, i, i2, , O, , Example 3.4, , i1, x, , Fig. 3.3, , From fig 3.3, if i is angle between, the lines l1 & l2 then, i = i1 - i 2, , ` tan i = tan ^i1 - i 2h, , =, , tan i1 - tan i 2, 1 + tan i1 tan i 2, , m -m, tanθ = 1 +1 m m2, 1 2, θ = tan, 54, , -1, , m1 - m2, 1 + m1 m2, , Find the acute angle between the, lines 2x - y + 3 = 0 and x + y + 2 = 0, Solution, Let m1 and m2 be the slopes of, 2x - y + 3 = 0 and x + y + 2 = 0, Now m1 = 2, m2 = –1, Let i be the angle between the, given lines, m -m, tan i = 1 +1 m m2, 1 2, =, , 2+1, =3, 1 + 2 ^- 1 h, , i = tan -1 ^3 h, , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 54, , 21-04-2020 12:14:47 PM

Page 61 :
www.tntextbooks.in, , 3.2.3 Distance of a point from a line, (i), , The length of the perpendicular, , from a point P(l, m) to the line ax + by + c = 0, al + bm + c, is d =, a2 + b2, (ii), , The length of the perpendicular, , form the origin (0,0) to the line ax + by + c = 0, c, is d =, 2, a + b2, , r, Given m1 = 3 and i = 4, 3-m, r, ` tan 4 = 1 + 3m2, 2, 3-m, 1 = d 1 + 3m2 n, 2, 1 + 3m2 = 3 - m2, 1, m2 = 2, 1, Hence the slope of the other line is 2 ., &, , Example 3.5, , 3.2.4 Concurrence of three lines, , Show that perpendicular distances, of the line x - y + 5 = 0 from origin and, from the point P(2, 2) are equal., , If two lines l1 and l2 meet at a, common point P, then that P is called point, of intersection of l1 and l2 . This point of, intersection is obtained by solving the, equations of l1 and l2 ., , Solution, Given line is x - y + 5 = 0, Perpendicular distance of the given, 2-2+5, line from P(2, 2) is =, 12 + 12, =, , =, , 5, 5, =, 2, 2, , Distance of (0,0) from the given line, 5, 5, 5, =, 2, 2 =, 2, 2, 1 +1, , The given line is equidistance from, origin and (2, 2), Example 3.6, If the angle between the two lines is, r, 4 and slope of one of the lines is 3, then, find the slope of the other line., Solution, We know that the acute angle i, between two lines with slopes m1 and m2, is given by, m -m, tan i = 1 +1 m m2, 1 2, , If three or more straight lines will, have a point in common then they are said, to be concurrent., The lines passing through the common, point are called concurrent lines and the, common point is called concurrent point., Conditions for three given straight lines, to be concurrent, Let a1 x + b1 y + c1 = 0, , " (1), , a2 x + b2 y + c2 = 0, , " (2), , a3 x + b3 y + c3 = 0 , , " (3), , be the equations of three straight, lines, then the condition that these lines, to be concurrent is, a1 b1 c1, a2 b2 c2 = 0, a3 b3 c3, Example 3.7, Show that the given lines 3x − 4y − 13 = 0,, 8x − 11y = 33 and 2x − 3y − 7 = 0 are concurrent, and find the concurrent point., Analytical Geometry, , 03_11th_BM-STAT_Ch-3-EM.indd 55, , 55, , 21-04-2020 12:14:51 PM

Page 62 :
www.tntextbooks.in, , Solution, Given lines 3x - 4y - 13 = 0 ... (1), , , , 8x - 11y = 33 ... (2), , 1(35+33) – k(–21+55) = 0, , 2x - 3y - 7 = 0  ... (3), a1, a2, a3, , Conditon for concurrent lines is, b1 c1, b2 c2 =0, b3 c3, , 3 - 4 - 13, i.e., 8 - 11 - 33 =, 2 -3 -7, 3(77–99) + 4(–56+66) – 13(–24+22), = –66 + 40 + 26 = 0, &, , Given lines are concurrent., , To get the point of concurrency, solve the equations (1) and (3), , & 34k = 68. ` k = 2., Example 3.9, A private company appointed a clerk, in the year 2012, his salary was fixed as, `20,000. In 2017 his salary raised to `25,000., (i) Express the above information as a, linear function in x and y where y, represent the salary of the clerk and, x-represent the year, (ii) What will be his salary in 2020?, Solution, Let y represent the salary (in Rs), and x represent the year, , Equation (1) # 2, , & 6x–8y =26, , year (x), , salary (y), , Equation (3) # 3, , , &, , 6x–9y =21, _________, , 2012(x1), , 20,000(y1), , y=5, _________, , 2020, , ?, , , , , When y=5 from (2) 8x = 88, x = 11, Point of concurrency is (11, 5), Example 3.8, If the lines 3x − 5y − 11 = 0,, 5x + 3y − 7 = 0 and x + ky = 0 are concurrent,, find the value of k., Solution, Given the lines are concurrent., Therefore, 56, , 3 - 5 - 11, 5 3 -7 = 0, 1 k, 0, , a1 b1 c1, a2 b2 c2 =0, a3 b3 c3, , 2017(x2), , 25,000(y2), , The equation of straight line expressing, the given information as a linear equation in, x and y is, y - y1, x - x1, =, y2 - y1, x2 - x1, y - 20, 000, x - 2012, 25, 000 - 20, 000 = 2017 - 2012, y - 20, 000, x - 2012, =, 5000, 5, y = 1000x – 2012000+20,000, y = 1000x – 19,92,000, In 2020 his salary will be, y = 1000(2020)–19,92,000, y = `28000, , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 56, , 21-04-2020 12:14:53 PM

Page 64 :
www.tntextbooks.in, , Then, , Example 3.10, , m -m, tan i = 1 +1 m m2, 1 2, , ±2 h 2 - ab, a+b, Let us take ‘ i ’ as acute angle, , , ` , , =, , i = tan -1 <, , 2 h - ab, F, a+b, 2, , Find the combined equation of the, given straight lines whose separate equations, are 2x + y - 1 = 0 and x + 2y - 5 = 0 ., Solution, The combined equation of the given, straight lines is, , ^2x + y - 1 h ^x + 2y - 5 h = 0, , i.e., 2x 2 + xy - x + 4xy + 2y 2 - 2y - 10x - 5y + 5 = 0, , NOTE, , 2, 2, (i) If i is the angle between the pair of 2x + xy - x + 4xy + 2y - 2y - 10x - 5y + 5 = 0, straight lines, i.e.,, 2x 2 + 5xy + 2y 2 - 11x - 7y + 5 = 0, , ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 ,, then i = tan -1 <, , 2 h - ab, F, a+b, 2, , (ii) If the straight lines are parallel,, then h 2 = ab ., (iii) If the straight lines are perpendicular,, then a + b = 0, i.e., cqefficient qf x 2 + cqefficient qf y 2 = 0, , 3.3.4 The condition for general second, degree equation to represent the, pair of straight lines, The condition for a general second, degree equation in x, y namely, ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0, to represent a pair of straight lines is, abc + 2fgh–af 2 –bg 2 –ch 2 = 0 ., , Example 3.11, Show that the equation 2x2 + 5xy +, 3y2 + 6x + 7y + 4 = 0 represents a pair of, straight lines. Also find the angle between, them., Solution, Compare the equation, 2x 2 + 5xy + 3y 2 + 6x + 7y + 4 = 0, with, a x 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 ,, we get, 5, 7, a = 2, b = 3, h = 2 , g = 3, f = 2 and, c=4, Condition for the given equation, to represent a pair of straight lines is, abc + 2fgh–af 2 –bg 2 –ch 2 = 0 ., , 105 49, abc + 2fgh–af 2 –bg 2 –ch 2 = 24 + 2 - 2 - 27 - 25 =, The condition in determinant, is–af 2 –bg 2 –ch 2 = 24 + 105 - 49 - 27 - 25 = 0, + 2fgh, abcform, 2, 2, a h g, Hence the given equation represents, h b f =0, a pair of straight lines., g f c, Let θ be the angle between the lines., , NOTE, , 58, , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 58, , 21-04-2020 12:15:04 PM

Page 66 :
www.tntextbooks.in, , Hence the given equation represents, a pair of parallel straight lines., Now 4x2 - 12xy + 9y2 = ^2x - 3y h2, Consider,, , 4x2 - 12xy + 9y2 + 18x - 27y + 8 = 0, &, Put, , ^2x - 3y h2 + 9 ^2x - 3y h + 8 = 0, , 2x–3y = z, , The given line represents a pair of, straight lines if,, abc + 2fgh - af 2 - bg 2 - ch 2 = 0, 675, 225 25, i.e., 4k + 2 - 162 - 2 - 4 k = 0, &, , 16k + 1350 - 648 - 450 - 25k = 0, , &   9k = 252, , z 2 + 9z + 8 = 0, , Exercise 3.3, , (z+1)(z+8) = 0, z+1 = 0  , 2x–3y+1 = 0, , ` k = 28, , z+8 = 0, , 1., , 2x–3y+8 =0, , Hence the separate equations are, 2x–3y+1 = 0 and 2x–3y+8 =0, , If the equation ax2 + 5xy - 6y2 +, 12x + 5y + c = 0 represents a pair, of perpendicular straight lines, find, a and c., , 2. Show that the equation 12x2 - 10xy + 2y2 + 14x - 5y +, Example 3.15, 12x2 - 10xy + 2y2 + 14x - 5y + 2 = 0 represents a, Find the angle between the straight, pair of straight lines and also find, 2, 2, lines x + 4xy + y = 0, the separate equations of the straight, lines., Solution, The given equation is x2 + 4xy + y2 = 0, , 3., , Here a = 1, b = 1 and h = 2., If i is the angle between the given, straight lines, then, θ = tan -1 <, = tan -1 <, , 2 h2 - ab, F, a+b, 2 4-1, F, 2, , = tan -1 ^ 3 h, r, θ= 3, , Find the angle between the pair of, straight lines 3x2 - 5xy - 2y2 + 17x + y + 10, 3x2 - 5xy - 2y2 + 17x + y + 10 = 0, , Example 3.16, For what value of k does 2x2 + 5xy +, 2y2 + 15x + 18y + k = 0 represent a pair of, straight lines., Solution, , 5, 15, Here a = 2, b = 2, h = 2 , g = 2 ,, f = 9, c = k., 60, , Show that the pair of straight lines, 4x2 + 12xy + 9y2 - 6x - 9y + 2 = 0, represents two parallel straight lines, and also find the separate equations, of the straight lines., , 4., , 3.4, , Circles, , Definition 3.2, A circle is the locus of a point, which moves in such a way that its, distance from a fixed point is always, constant. The fixed point is called the, centre of the circle and the constant, distance is the radius of the circle., , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 60, , 21-04-2020 12:15:16 PM

Page 67 :
www.tntextbooks.in, , 3.4.1 The equation of a circle when, the centre and radius are given, y, C, , Equation of circle is x2 + y2 = r2, Here r = 3, , P(x, y), , r, , Solution, , i.e equation of circle is x2 + y2 = 9, , R, , 3.4.2 Equation of a circle when the end, points of a diameter are given, P(x, y), , L, , O, , M, , x, , Fig. 3.4, , Let C(h, k) be the centre and ‘r’ be, the radius of the circle, , A(x1, y1), , Let P(x, y) be any point on the circle, CP = r, CP2 = r2, (x–h)2 + (y–k)2 = r2, is the equation of the circle., In particular, if the centre is at the, origin, the equation of circle is x2 + y2 = r2, Example 3.17, Find the equation of the circle with, centre at (3, –1) and radius is 4 units., Solution, Equation of circle is, , C, , B(x2, y2), , Fig. 3.5, , Let A ^x1, y1 h and B ^x2, y2 h be the, end points of a diameter of a circle and, P(x, y) be any point on the circle., We know that angle in the semi, circle is 90o, ` APB = 90c, ` (Slope of AP) (Slope of BP) = –1, d, , y - y1, y - y2, x - x1 n # d x - x2 n = –1, , ^ y - y1 h^ y - y2 h = -^ x - x1h^ x - x2h, , ^ x - h h2 + ^ y - k h2 = r2, , & ^ x - x1 h^ x - x2 h + ^ y - y1 h^ y - y2 h = 0 is, the required equation of circle., , Equation of circle is, , Example 3.19, , Here (h, k) = (3, –1) and r = 4, ^ x - 3 h2 + ^ y + 1 h2 = 16, , x2 - 6x + 9 + y2 + 2y + 1 = 16, x 2 + y 2 - 6x + 2y - 6 = 0, Example 3.18, Find the equation of the circle with, centre at origin and radius is 3 units., , Find the equation of the circle when, the end points of the diameter are (2, 4), and (3, –2)., Solution, Equation of a circle when the end, points of the diameter are given is, ^ x - x1 h^ x - x2 h + ^ y - y1 h^ y - y2 h = 0, Analytical Geometry, , 03_11th_BM-STAT_Ch-3-EM.indd 61, , 61, , 21-04-2020 12:15:19 PM

Page 68 :
www.tntextbooks.in, , Centre = C ^- g, - f h = C ^4, - 3h and, , Here ^x1, y1 h = (2, 4) and, ^x2, y2 h = (3, –2), , Radius: r =, , ` ]x - 2g]x - 3g + _ y - 4 i_ y + 2 i = 0, 2, , x + y - 5x - 2y - 2 = 0, , 3.4.3 General equation of a circle, The general equation of circle is, x + y + 2gx + 2fy + c = 0 where g, f and, c are constants., 2, , 2, , x2 + y2 + 2gx + 2fy = –c, , i.e, , x2 + 2gx + g2 - g2 + y2 + 2fy + f 2 - f 2 = –c, , ^x + g h2 - g2 + ^ y + f h2 - f 2 = –c, , ^x + g h + ^ y + f h = g2 + f 2 - c, 2, , 2, , 7x - ^- g hA + 7 y - ^- f hA = 7 g 2 + f 2 - c A, 2, , 2, , 16 + 9 + 24 = 7 unit., , Example 3.21, For what values of a and b does the, equation, , ]a - 2g x2 + by2 + ]b - 2g xy + 4x + 4y - 1 = 0, represents a circle? Write down the resulting, equation of the circle., , Solution, The given equation is, , ]a - 2g x2 + by2 + ]b - 2g xy + 4x + 4y - 1 = 0, , As per conditions noted above,, , 2, , Comparing this with the circle, ^ x - h h2 + ^ y - k h2 = r2, , We get, centre is ^- g, - f h and radius, , g2 + f 2 - c, , is, , =, , 2, , g2 + f 2 - c, , NOTE, , (i) coefficient of xy = 0 & b - 2 = 0, ` b=2, , , , (ii) coefficient of x2 = Coefficient of y2, & a–2=b, a–2=2 & a=4, ` Resulting equation of circle is, 2x + 2y2 + 4x + 4y - 1 = 0, 2, , The general second degree equation, Example 3.22, ax2 + by2 + 2hxy + 2gx + 2fy + c = 0, If the equation of a circle x2 + y2 + ax + by, represents a circle if, 2, 2, (i) a = b i.e., co-efficient of x2 = x + y + ax + by = 0 passing through the points (1, 2), and (1, 1), find the values of a and b, co-efficient of y2., (ii) h = 0 i.e., no xy term., Example 3.20, Find the centre and radius of the, circle x2 + y2 - 8x + 6y - 24 = 0, Solution, Equation of circle is, x2 + y2 - 8x + 6y - 24 = 0, Here g = –4 , f = 3 and c = –24, 62, , Solution, The circle x2 + y2 + ax + by = 0, passing through (1, 2) and (1, 1), ` We have 1 + 4 + a + 2b = 0 and, 1 + 1+ a + b = 0, &, , a + 2b = –5, , (1), , and, , a + b = –2, , (2), , Solving (1) and (2), we get a = 1, b, = –3., , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 62, , 21-04-2020 12:15:25 PM

Page 69 :
www.tntextbooks.in, , Example 3.23, , Example 3.25, , If the centre of the circle x2 + y2 + 2x - 6y + 1 Find the equation of the circle passing, x2 + y2 + 2x - 6y + 1 = 0 lies on a straight line, through the points (0,0), (1, 2) and (2,0)., ax + 2y + 2 = 0, then find the value of ‘a’, Solution, Solution, , Centre C(–1, 3), , Let the equation of the circle be, x + y + 2gx + 2fy + c = 0, , It lies on ax + 2y + 2 = 0, , The circle passes through the point (0, 0), , 2, , 2, , c = 0, , –a + 6 + 2 = 0, , ... (1), , The circle passes through the point (1, 2), , a =8, , 1 2 + 2 2 + 2g (1) + 2f (2) + c = 0, , Example 3.24, Show that the point (7, –5) lies on the, circle x2 + y2 - 6x + 4y –12 = 0 and find the, coordinates of the other end of the diameter, through this point., Solution, , 2g + 4f + c = – 5, , ... (2), , The circle passes through the point (2, 0), 22 + 0 + 2g (2) + 2f (0) + c = 0, 4g + c = –4 , , ... (3), , Solving (1), (2) and (3), we get, , Let A(7, –5), , 3, g =- 1, f =- 4 and c = 0, , Equation of circle is, Substitute (7, –5) for (x, y) , we get, , ` The equation of the circle is, -3, x 2 + y 2 + 2 ( - 1) x + 2 b 4 l y + 0 = 0, , x2 + y2 - 6x + 4y - 12 =, , ie.,, , x2 + y2 - 6x + 4y - 12 = 0, , 72 + ]- 5g2 - 6 ]7 g + 4 ]- 5g - 12, = 49 + 25 – 42 – 20 – 12 = 0, ` (7, –5) lies on the circle, Here g = –3 and f = 2, `, , Centre = C(3, –2), , Let the other end of the diameter be, B (x, y), Midpoint of AB =, y-5, c x + 7,, m = C ^3, - 2h, 2, 2, x+7, 2 =3, x = –1 , , y-5, 2 =-2, , 2x2 + 2y2 - 4x - 3y = 0, , 3.4.4 Parametric form of a circle, Consider a circle with radius r and, centre at the origin. Let P (x, y) be any point, on the circle. Assume that OP makes an, angle i with the positive direction of x-axis., Draw PM perpendicular to x-axis., y, P(x, y), r, i, O, , M, , x, , y=1, , Other end of the diameter is (–1 , 1)., , Fig. 3.6, Analytical Geometry, , 03_11th_BM-STAT_Ch-3-EM.indd 63, , 63, , 21-04-2020 12:15:29 PM

Page 70 :
www.tntextbooks.in, , From the figure,, x, cos i = r & x = r cos i, y, sin i = r & y = r sin i, The equations x = r cos i , y = r sin i, are called the parametric equations of the, circle x2 + y2 = r2 . Here ‘ i ’ is called the, parameter and 0 # i # 2r ., Example 3.26, Find the parametric equations of, the circle x2 + y2 = 25, Solution, Here r2 = 25 & r = 5, Parametric equations are x = r cos i ,, y = r sin i, & x = 5 cos i, y = 5 sin i, 0 # i # 2r, Exercise 3.4, , 5. Find the equation of the circle passing, through the points (0, 1),(4, 3) and, (1, –1)., 6. Find the equation of the circle on the, line joining the points (1,0), (0,1), and having its centre on the line, x+y=1, 7. If the lines x + y = 6 and x + 2y = 4 are, diameters of the circle, and the circle, passes through the point (2, 6) then, find its equation., 8. Find the equation of the circle having, (4, 7) and (–2, 5) as the extremities of, a diameter., 9. Find the Cartesian equation of the, circle whose parametric equations are, x = 3cos i ,y =3 sin i , 0 # i # 2r ., , 3.4.5 Tangents, , 1. Find the equation of the following, circles having, , C(–g, –f), , (i) the centre (3,5) and radius 5 units, (ii) the centre (0,0) and radius 2 units, P(x1, y1), , 2. Find the centre and radius of the circle, (ii) x2 + y2 - 22x - 4y + 25 = 0, (iii) 5x2 + 5y2 + 4x - 8y - 16 = 0, , (iv) ]x + 2g]x - 5g + ^ y - 2 h^ y - 1 h = 0, , 3. Find the equation of the circle, whose centre is (–3, –2) and having, circumference 16r, 4. Find the equation of the circle whose, centre is (2,3) and which passes, through (1,4), 64, , Fig. 3.7, , x2 + y2 = 16, , (i), , T, , The equation of the tangent to the, circle, x2 + y2 + 2gx + 2fy + c = 0 at, ^x1, y1 h is, xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0., Corollary:, The equation of the tangent at, ^x1, y1 h to the circle, x2 + y2 = a2 is xx1 + yy1 = a2, , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 64, , 21-04-2020 12:15:33 PM

Page 72 :
www.tntextbooks.in, , At S(2, 2), ST2 = 4 + 4 – 8 + 8 – 8 = 0, ` The point P lies inside the circle., The points Q and R lie outside the circle, and the point S lies on the circle., Condition for the straight line, y = mx + c to be a tangent, Result, to the circle x2 + y2 = a2 is, c 2 = a 2 (1 + m 2 ), , 3.5, , Definition 3.3, If a point moves in a plane such, that its distance from a fixed point bears, a constant ratio to its perpendicular, distance from a fixed straight line, then, the path described by the moving point, is called a conic., l, , Find the value of k so that the line, 3x + 4y - k = 0 is a tangent to x2 + y2 - 64 = 0, Solution, The given equations are x2 + y2 - 64 = 0, and 3x + 4y - k = 0, , , , , k2 = 64 × 25, k = ± 40, Exercise 3.5, , 1. Find the equation of the tangent to the, circle x2 + y2 - 4x + 4y - 8 = 0 at (–2, – 2), 2. Determine whether the points P(1, 0),, Q(2, 1) and R(2, 3) lie outside the, circle, on the circle or inside the circle, x 2 + y 2 - 4x - 6y + 9 = 0, 3. Find the length of the tangent from (1, 2) to, the circle x2 + y2 - 2x + 4y + 9 = 0, , P(Moving point), , M, , Example 3.30, , The condition for the tangency is, 2, 2, c = a ( 1 + m 2), -3, k, Here a2 = 64 , m = 4 and c = 4, 9, k2, c2 = a2 (1 + m2) & 16 = 64 b1 + 16 l, , Conics, , Fixed line (directrix), , At R(–2, 3), RT2 = 4 + 9 + 8 + 12 – 8 = 25 > 0, , F(Fixed point), Fig. 3.9, , In figure, the fixed point F is called, focus, the fixed straight line l is called, directrix and P is the moving point such, FP, that PM = e, a constant. Here the locus of, P is called a conic and the constant ‘e’ is, called the eccentricity of the conic., Based on the value of eccentricity, we can classify the conics namely,, a) If e = 1, then, the conic is called a, parabola, b) If e < 1, then, the conic is called an ellipse, c) If e > 1, then, the conic is called a, hyperbola., The general second degree equation, ax + 2hxy + by2 + 2gx + 2fy + c = 0, represents,, 2, , (i) a pair of straight lines if abc + 2fgh - af 2 - bg 2 - ch 2 =, 4. Find the value of P if the line, abc + 2fgh - af 2 - bg 2 - ch 2 = 0, 3x + 4y - P = 0 is a tangent to the, (ii) a circle if a = b and h = 0, circle x2 + y2 = 16, 66, , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 66, , 21-04-2020 12:15:41 PM

Page 73 :
www.tntextbooks.in, , If the above two conditions are not, Axis, 2, 2, satisfied, then ax + 2hxy + by + 2gx + 2fy + c = 0, 2, xy + by + 2gx + 2fy + c = 0 represents,, (i), , a parabola if h2 –ab = 0, , (ii) an ellipse if h2 –ab < 0, (iii) a hyperbola if h2 –ab > 0, In this chapter, we study about, parabola only., , 3.5.1 Parabola, , Vertex, , Definition 3.4, The locus of a point whose distance, from a fixed point is equal to its distance, from a fixed line is called a parabola., y2 = 4ax is the standard equation of the, parabola. It is open rightward., , 3.5.2 Definitions regarding a, parabola: y2 = 4ax, y, P(x, y), M(–a, y), , Z(–a, 0), , V, , F(a, 0), , x, , Fig. 3.10, , Focus, , The fixed point used to draw, the parabola is called the, focus (F)., , Here, the focus is F(a, 0)., Directrix The fixed line used to draw, a parabola is called the, directrix of the parabola., Here, the equation of the, directrix is x = − a., , Focal, distance, Focal, chord, Latus, rectum, , The axis of the parabola is the, axis of symmetry. The curve, y2 = 4ax is symmetrical, about x-axis and hence x-axis, or y = 0 is the axis of the, parabola y2 = 4ax . Note, that the axis of the parabola, passes through the focus and, perpendicular to the directrix., The point of intersection of, the parabola with its axis is, called its vertex. Here, the, vertex is V(0, 0)., The distance between a point, on the parabola and its focus, is called a focal distance, A chord which passes, through the focus of the, parabola is called the focal, chord of the parabola., It is a focal chord, perpendicular to the axis, of the parabola. Here, the, equation of the latus rectum, is x = a . Length of the latus, rectrum is 4a., , 3.5.3 Other standard parabolas, , 1. Open leftward : y2 = - 4ax 6a 2 0@, If x > 0, then y become imaginary., i.e., the curve exist for x ≤ 0., y, y2=–4ax, , F(–a, 0), , x=a, , V, , Fig. 3.11, Analytical Geometry, , 03_11th_BM-STAT_Ch-3-EM.indd 67, , x, , 67, , 21-04-2020 12:15:43 PM

Page 74 :
www.tntextbooks.in, , 2. Open upward : x2 = 4ay 6a 2 0@, , If y < 0, then x becomes imaginary., i.e., the curve exist for y ≥ 0., , The process of shifting the origin or, translation of axes., Y, , y, , (X, Y), , x = 4ay, 2, , y, , P(x, y), Y, , O′, , F(0, a), , X, , V, , h, O (0, 0), , x, , k, M, , L, , x, , y = –a, Fig. 3.14, Fig. 3.12, , 3. Open downward : x2 = - 4ay 6a 2 0@, , If y > 0, then x becomes imaginary., i.e., the curve exist for y ≤ 0., y, , Let the co-ordinates of Ol with, respect to xoy system be (h, k), , y=a, , The co-ordinate of P with respect to, xoy system:, , V, x, x2=–4ay, , Consider the xoy system. Draw a, line parallel to x -axis (say X axis) and, draw a line parallel to y – axis (say Y axis)., Let P ^ x, y h be a point with respect to xoy, system and P ^ X, Y h be the same point, with respect to XOl Y system., , F(0, –a), , , OL, , = OM + ML, , , , =h+X, , i.e) x = X + h, similarly, , Fig. 3.13, Equations y2 = 4ax y2 = –4ax x2 = 4ay x2 = –4ay, , y=0, , y=0, , x=0, , x=0, , Vertex, , V(0, 0), , V(0, 0), , V(0, 0), , V(0, 0), , Focus, , F(a, 0), , F(–a, 0), , F(0, a), , F(0, –a), , Equation, of, directrix, , x = –a, , x=a, , y = –a, , y=a, , Length, of Latus, rectum, , 4a, , 4a, , 4a, , 4a, , Equation, of Latus, rectum, , x=a, , x = –a, , y=a, , y = –a, , Axis, , 68, , y =Y+k, , ` The new co-ordinates of P with, respect to XOl Y system., X = x–h and Y = y–k, , 3.5.4 General form of the standard, equation of a parabola, which, is open rightward (i.e., the, vertex other than origin):, Consider a parabola with vertex V, whose co-ordinates with respect to XOY, system is (0, 0) and with respect to xoy, system is (h, k). Since it is open rightward,, , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 68, , 21-04-2020 12:15:45 PM

Page 76 :
www.tntextbooks.in, , The parabola is open left, its focus is, F ^- a, qh, F ^- 3, 0h, , Its vertex is V ^h, kh = V ^0, 0h, , X2 = 4Y, Comparing with X2 = 4aY, 4a = 4, a = 1, , The equation of the directrix is x = a, , Referred, Referred to (x, y), to (X, Y) x = X - 3, y = Y + 3, x = X - 3, y = Y + 3, , i.e., x = 3, Its axis is x – axis whose equation, is y = 0., Length of the latus rectum = 4a = 12, , Axis y = 0, , Y=0, , y=3, , Example 3.33, , Vertex V(0,0), , V(0,0), , V(–3,3), , Show that the demand function, x = 10p - 20 - p2 is a parabola and price, is maximum at its vertex., , Focus F(0, a), , F(0, 1), , F(–3, 4), , Equation of, directrix (y=–a), , Y = –1, , y=2, , Length of Latus, rectum (4a), , 4(1) = 4, , 4, , Solution, x = 10p - 20 - p2, = - p2 - 20 + 10p + 5 - 5, x - 5 = - p2 + 10p - 25, , Example 3.35, , = -_ p2 - 10p + 25 i, , The supply of a commodity is related, to the price by the relation x = 5P - 15, . Show that the supply curve is a parabola., , Put X = x - 5 and P = p - 5, , Solution, , = -^ p - 5h2, , X =- P2 i.e P2 =- X, It is a parabola open downward., ` At the vertex p = 5 when x = 5., i.e., price is maximum at the vertex., Example 3.34, , The supply price relation is given by, x2 = 5p – 15, = 5(p – 3), & X2 = 4 aP where X = x and, P = p-3, , Find the axis, vertex, focus, equation, of directrix and the length of latus rectum, for the parabola x2 + 6x - 4y + 21 = 0, , ` the supply curve is a parabola, whose vertex is (X = 0, P = 0), , Solution, , i.e., The supply curve is a parabola, whose vertex is (0, 3), , 4y = x2 + 6x + 21, , 4y = ^ x2 + 6x + 9h + 12, , 4y–12 = ^ x + 3h2, , ^ x + 3 h2 = 4(y–3), , X = x + 3, Y = y - 3, , x = X - 3 and y = Y + 3, 70, , Exercise 3.6, 1. Find the equation of the parabola, whose focus is the point F ^- 1, - 2h and, the directrix is the line 4x–3y + 2 = 0 ., , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 70, , 21-04-2020 12:15:57 PM

Page 77 :
www.tntextbooks.in, , 2. The parabola y2 = kx passes through, the point (4,–2). Find its latus rectum, and focus., 3. Find the vertex, focus, axis, directrix, and the length of latus rectum of the, parabola y2 –8y–8x + 24 = 0, 4. Find the co-ordinates of the focus,, vertex, equation of the directrix, axis, and the length of latus rectum of the, parabola, , 2. The angle between the pair of straight, lines x2 –7xy + 4y2 = 0 is, 1, (a) tan -1 b 3 l, , 1, (b) tan -1 b 2 l, , 33, (c) tan -1 c 5 m, , 5, (d) tan -1 d 33 n, , 3. If the lines 2x - 3y - 5 = 0 and 3x - 4y - 7 = 0, 3x - 4y - 7 = 0 are the diameters of a circle,, then its centre is, (a) ( –1, 1), (b) (1,1), (c) (1, –1), , (a) y2 = 20x, (b) x2 = 8y, (c) x2 =- 16y, 5. The average variable cost of a monthly, output of x tonnes of a firm producing, 1, a valuable metal is ` 5 x2 - 6x + 100 ., Show that the average variable cost, curve is a parabola. Also find the, output and the average cost at the, vertex of the parabola., , (d) (–1, –1), , 4. The x–intercept of the straight line, 3x + 2y - 1 = 0 is, 1, 1, (a) 3, (b) 2, (c) 3, (d) 2, 5. The slope of the line 7x + 5y - 8 = 0 is, 7, 7, 5, 5, (a) 5, (b) – 5 (c) 7, (d) – 7, , 6. The locus of the point P which moves, such that P is at equidistance from, their coordinate axes is, 1, (a) y = x, (b) y =- x, 6. The profit `y accumulated in thousand, -1, (c) y = x, (d) y = x, 2, in x months is given by y = - x + 10x - 15, 2, y = - x + 10x - 15 . Find the best time to end, 7. The locus of the point P which moves, the project., such that P is always at equidistance, from the line x + 2y + 7 = 0 is, Exercise 3.7, Choose the correct answer, 1. If m1 and m2 are the slopes of the pair, of lines given by ax 2 + 2hxy + by2 = 0,, then the value of m1 + m2 is, 2h, (a) b, , 2h, (b) - b, , 2h, (c) a, , 2h, (d) – a, , (a) x + 2y + 2 = 0 (b) x - 2y + 1 = 0, (c) 2x - y + 2 = 0 (d) 3x + y + 1 = 0, 8. If kx2 + 3xy - 2y2 = 0 represent a, pair of lines which are perpendicular, then k is equal to, 1, 1, (b) – 2 (c) 2, (d) –2, (a) 2, 9. (1, –2) is the centre of the circle x2 + y2 + ax + by - 4, x2 + y2 + ax + by - 4 = 0 , then its radius, (a) 3, , (b) 2, , (c) 4, , (d) 1, , Analytical Geometry, , 03_11th_BM-STAT_Ch-3-EM.indd 71, , 71, , 21-04-2020 12:16:04 PM

Page 78 :
www.tntextbooks.in, , 10. The length of the tangent from (4,5), to the circle x2 + y2 = 16 is, (b) 5, , (a) 4, , (c) 16, , (d) 25, , 11. The focus of the parabola x2 = 16y is, (a) (4 ,0), , (b) (–4 , 0), , (c) (0, 4), , (d) (0, –4), , 12. Length of the latus rectum of the, parabola y2 =- 25x ., (a) 25, , (b) –5, , (c) 5, , (d) –25, , 17. ax2 + 4xy + 2y2 = 0 represents a pair, of parallel lines then ‘a’ is, (a) 2, , (b) –2, , (c) 4, , (d) –4, , 18. In the equation of the circle x2 + y2 = 16, then y intercept is (are), (a) 4, , (b)16, , (c) ±4, , (d) ±16, , 19. If the perimeter of the circle is 8π, units and centre is (2,2) then the, equation of the circle is, , (a) ^ x - 2h2 + ^ y - 2h2 = 4, 13. The centre of the circle x2 + y2 - 2x + 2y - 9 = 0, (b) ^ x - 2h2 + ^ y - 2h2 = 16, 2, 2, x + y - 2x + 2y - 9 = 0 is, (c) ^ x - 4h2 + ^ y - 4h2 = 2, (b) (–1, –1), (a) (1 ,1), (d) x2 + y2 = 4, (c) (–1,1), (d) (1, –1), 14. The equation of the circle with centre, on the x axis and passing through the, origin is, , (c) ^ x - 3h2 + ^ y - 4h2 = 16, , (b) y2 - 2ay + x2 = 0, , (d) x2 + y2 = 16, , (c) x2 + y2 = a2, (d) x2 - 2ay + y2 = 0, 15. If the centre of the circle is (–a, –b) and, radius is a2 - b2 , then the equation, of circle is, (a) x2 + y2 + 2ax + 2by + 2b2 = 0, (b) x2 + y2 + 2ax + 2by - 2b2 = 0, (c) x2 + y2 - 2ax - 2by - 2b2 = 0, (d) x2 + y2 - 2ax - 2by + 2b2 = 0, 16. Combined equation of co-ordinate, axes is, , 72, , 2, , (a) ^ x - 3h2 + ^ y - 4h2 = 4, , (b) ^ x - 3h2 + ^ y + 4h2 = 16, , (a) x2 - 2ax + y2 = 0, , 2, , 20. The equation of the circle with centre, (3, –4) and touches the x – axis is, , 2, , 2, , (a) x - y = 0, , (b) x + y = 0, , (c) xy = c, , (d) xy = 0, , 21. If the circle touches x axis, y axis and, the line x = 6 then the length of the, diameter of the circle is, (a) 6, , (b)3, , (c) 12, , (d) 4, , 22. The eccentricity of the parabola is, (a) 3, , (b) 2, , (c) 0, , (d) 1, , 23. The double ordinate passing through, the focus is, (a) focal chord, (c) directrix, , (b) latus rectum, (d) axis, , 24. The distance between directrix and, focus of a parabola y2 = 4ax is, (a) a, , (b) 2a, , (c) 4a, , (d) 3a, , 11th Std. Business Mathematics and Statistics, , 03_11th_BM-STAT_Ch-3-EM.indd 72, , 21-04-2020 12:16:11 PM

Page 81 :
www.tntextbooks.in, , ICT Corner, Expected final outcomes, Step – 1, Open the Browser and type the URL given (or) Scan, the QR Code., GeoGebra Work book called “11th BUSINESS, MATHEMATICS and STATISTICS” will appear., In this several work sheets for Business Maths are given, Open the worksheet named “LocusParabola”, Step - 2, Locus-Parabola page will open. Click on the check boxes on the Left-hand side to see the, respective parabola types with Directrix and Focus. On right-hand side Locus for parabola is, given. You can play/pause for the path of the locus and observe the condition for the locus., Step 1, , Step 2, Step 4, , Browse in the link, 11th Business Mathematics and Statistics: https://ggbm.at/qKj9gSTG (or), scan the QR Code, Analytical Geometry, , 03_11th_BM-STAT_Ch-3-EM.indd 75, , 75, , 21-04-2020 12:16:19 PM

Page 82 :
www.tntextbooks.in, , Chapter, , 4, , TRIGONOMETRY, , Learning Objectives, After studying this chapter, the students, will be able to understand, , •, •, , trigonometric ratio of angles, , •, , transformation of sums into products, and vice versa, , •, , basic concepts of inverse trigonometric, functions, , •, , properties of inverse trigonometric, functions, , addition formulae, multiple and, sub-multiple angles, , trigonometric functions in terms of right, angles. The study of trigonometry was, first started in India. The ancient Indian, Mathematician, Aryabhatta, Brahmagupta,, Bhaskara I and Bhaskara II, obtained important results., Bhaskara I gave formulae, to find the values of sine, functions for angles more, than 90 degrees.The earliest, applications of trigonometry were in the fields, of navigation, surveying and astronomy., , Currently, trigonometry is used in, many areas such as electric circuits,, describing the state of an atom,, predicting the heights of tides in the, ocean, analyzing a musical tone., Recall, , Brahmagupta, , Introduction, The word trigonometry is derived, from the Greek word ‘tri’ (meaning three),, ‘gon’ (meaning sides) and ‘metron’ (meaning, measure). In fact, trigonometry is the study, of relationships between the sides and angles, of a triangle. Around second century A.D., George Rheticus was the first to define the, 76, , 1., , sinθ =, , side opposite to angle i, Hypotenuse, , 2., , cosθ =, , Adjacent side to angle i, Hypotenuse, , 3., , tanθ =, , side opposite to angle i, Adjacent side to angle i, , 4., , cotθ =, , Adjacent side to angle i, side opposite to angel i, , 5., , secθ =, , Hypotenuse, Adjacent side to angle i, , 6. cosecθ =, , Hypotenuse, side opposite to angle i, , 11th Std. Business Mathematics and Statistics, , 04_11th_BM-STAT_Ch-4-EM.indd 76, , 21-04-2020 12:09:49 PM

Page 83 :
www.tntextbooks.in, , Relations between trigonometric ratios, 1. sinθ =, , 1, 1, or cosecθ =, sin i, cosec i, , 2. cosθ =, , 1, sec i, , or, , secθ =, , 1, cos i, , 3. tanθ =, , sin i, cos i, , or, , cotθ =, , cos i, sin i, , 4. tanθ =, , 1, cot i, , or, , cotθ =, , 1, tan i, , Trigonometric Identities, 1., , sin 2 i + cos 2 i = 1, , 2., , 1 + tan 2 i = sec2θ, , 3., , 1 + cot 2 i = cosec2θ., , rm, Te, , O, , Degree measure, If a rotation from the initial position, 1 th, to the terminal position b 360 l of the, revolution, the angle is said to have a, measure of one degree and written as 1o., A degree is divided into minutes and minute, is divided into seconds., , Radian measure, The angle subtended at the centre of, the circle by an arc equal to the length of, the radius of the circle is called a radian,, and it is denoted by 1c, , B, sid, , Two types of units of measurement, of an angle which are most commonly used, namely degree measure and radian measure., , One degree = 60 minutes (60l ), One minute = 60 seconds (60m ), , Angle, , l, ina, , Measurement of an angle, , e, , B, r, 1c, , Initial side, Angle ∠AOB, , r, , r, , O, , A, , A, , Fig. 4.1, , Angle is a measure of rotation of a, given ray about its initial point. The ray, ‘OA’ is called the initial side and the final, position of the ray ‘OB’ after rotation is, called the terminal side of the angle. The, point ‘O’ of rotation is called the vertex., If the direction of rotation is, anticlockwise, then angle is said to be, positive and if the direction of rotation is, clockwise, then angle is said to be negative., , Fig. 4.2, , NOTE, The number of radians in an angle, subtended by an arc of a circle at the, centre of a circle is, length of the arc, =, radius, s, i.e., i = r , where θ is the angle, subtended at the centre of a circle of radius, r by an arc of length s., Trigonometry, , 04_11th_BM-STAT_Ch-4-EM.indd 77, , 77, , 21-04-2020 12:09:52 PM

Page 84 :
www.tntextbooks.in, , Relation between degrees and radians, We know that the circumference of, a circle of radius 1unit is 2π. One complete, revolution of the radius of unit circle subtends, 2π radians. A circle subtends at the centre an, angle whose degree measure is 360c., 2π radians = 360c, , π radians = 180c, , 1radian =, , 180c, r, , Find the quadrants in which the, terminal sides of the following angles lie., (i) - 70c, , 4.1, , (ii) - 320c, , (iii) 1325c, , Solution (i), (i) The terminal side of - 70c lies in IV, quadrant., , 4r, 4, 180c, 5 radians = 5 r # r = 144c, , 1, iii) 4 radian =, =, , The parts XOY, YOX' , X'OY l and, Y l OX are known as the first, second, third, and the fourth quadrant respectively., Example 4.2, , Example 4.1, Convert: (i) 160c into radians, 4r, (ii) 5 radians into degree, 1, (iii) 4 radians into degree, Solution, r, 8, i) 160c = 160 # 180 radians = 9 r, ii), , Let X l OX and Y l OY be two lines, at right angles to each other as in the, figure. We call X l OX and Y l OY as X axis, and Y axis respectively. Clearly these axes, divided the entire plane into four equal, parts called “Quadrants”., , Y, , 1 180, 4 r, 1 #, 7, 4 180 # 22 =14c19l 5ll, O, , X′, , Trigonometric Ratios, , X, , 4.1.1 Quadrants:, , Fig. 4.4, , P, Y, , II, , Y′, , (ii) The terminal side - 320c of lies in I, quadrant., , I, , Y, O, , X′, III, , P, , X, IV, X′, , X, , O, , Fig. 4.5, , Y′, Fig. 4.3, , 78, , Y′, , 11th Std. Business Mathematics and Statistics, , 04_11th_BM-STAT_Ch-4-EM.indd 78, , 21-04-2020 12:09:56 PM

Page 85 :
www.tntextbooks.in, , (iii) 1325c = ]3 # 360g + 180c + 65c, terminal side lies in III quadrant., , the, , Y, , X′, , X, , Fig. 4.6, , P, Y′, , In the first quadrant both x and y, are positive. So all trigonometric ratios, are positive. In the second quadrant, ]90c 1 i 1 180cg x is negative and y is, positive. So trigonometric ratios sin θ, and cosec θ are positive. In the third, quadrant ]180c 1 i 1 270cg both x and y, are negative. So trigonometric ratios tan, θ and cot θ are positive. In the fourth, quadrant ]270c < i < 360cg x is positive, and y is negative. So trigonometric ratios, cos θ and sec θ are positive., , 4.1.2 , Signs of the trigonometric, A function f (x) is said to be, ratios of an angle θ as it varies, odd function if f (- x) = – f (x) . sin i, tan i, cot i and co, from 0º to 360º, sin i, tan i, cot i and cosec i are all odd function., Y, , A function f (x) is said to be even function, if f (- x) = f (x) . cos i and sec i are even, function., , I Quadrant, All, , II Quadrant, sin & cosec, O, , X′, , X, IV Quadrant, cos & sec, , III Quadrant, tan & cot, , Y′, ASTC : All Sin Tan Cos, Fig. 4.7, , 4.1.3 Trigonometric ratios of allied, angles:, Two angles are said to be allied angles, when their sum or difference is either, zero or a multiple of 90c. The angles - i ,, 90c ! i, 180c ! i, 360c ! i etc .,are angles, allied to the angle θ. Using trigonometric, ratios of the allied angles we can find the, trigonometric ratios of any angle., , 90c - i 90c + i 180c - i 180c + i 270c - i 270c + i 360c - i 360c + i, -i, or, or, or, or, or, or, or, or, r, r, 3r, 3r, 2 -i 2 +i r-i r+i, 2 - i 2 + i 2r - i 2r + i, Sine, - sin i cos i, cos i, sin i - sin i - cos i - cos i - sin i sin i, Cosine, cos i, sin i - sin i - cos i - cos i - sin i sin i, cos i, cos i, tangent, - tan i cot i - cot i - tan i tan i, cot i - cot i - tan i tan i, cotangent - cot i tan i - tan i - cot i cot i, tan i - tan i - cot i cot i, secant, sec i cosec i - cosec i - sec i - sec i - cosec i cosec i sec i, sec i, cosecant - cosec i sec i, sec i cosec i - cosec i - sec i - sec i - cosec i cosec i, Angle/, Function, , Table : 4.1, Trigonometry, , 04_11th_BM-STAT_Ch-4-EM.indd 79, , 79, , 21-04-2020 12:10:03 PM

Page 87 :
www.tntextbooks.in, , 3. Determine the quadrants in which, the following degree lie., (i) 380o, , (ii) –140o, , (iii) 1195o, , 4. Find the values of each of the following, trigonometric ratios., (i) sin300o, (ii) cos(–210o), (iv) tan(–855o), (iii) sec390o, (v) cosec1125o, , 4.2 Trigonometric Ratios of, Compound Angles, 4.2.1 Compound angles, When we add or subtract angles,, the result is called a compound angle. i.e.,, the algebraic sum of two or more angles, are called compound angles, For example, If A, B, C are three, angles then A ± B, A + B + C, A – B + C etc.,, are compound angles., , 5. Prove that:, (i) tan(–225o) cot(–405o) –, tan(–765o) cot(675o) = 0, 4.2.2 Sum and difference formulae, r, r 3, 7r, (ii) 2 sin2 6 + cosec 2 6 cos2 3 = 2, of sine, cosine and tangent, 3r, 5r, 5r, 5rBlg, A+, l tansin, b i]= 1sinA cosB + cosA sinB, (iii) sec b 2 - i l sec b i - 2 l + tan b 2 + i(i), 2 =3r, 5r, 5r, 5r, (ii) sin ] A - Bg = sinA cosB – cosA sinB, ec b 2 - i l sec b i - 2 l + tan b 2 + i l tan b i - 2 l =- 1, (iii) cos ] A + Bg = cosA cosB – sinA sinB, 6. If A, B, C, D are angles of a cyclic, quadrilateral, prove that :, , cosA + cosB + cosC + cosD = 0, 7. Prove that:, (i), , sin ]180c - ig cos ]90c + ig tan ]270 - ig cot ]360 - ig, sin ]360 - ig cos ]360 + ig sin ]270 - ig cosec ]- ig, , = –1, , (iv) cos ] A - Bg = cosA cosB + sinA sinB, , (v) tan(A+B), (vi) tan(A–B), , tan A + tan B, = 1 - tan A tan B, tan A - tan B, = 1 + tan A tan B, , Example 4.6, , 4, 12, r, r, = 5 and cosB = 13 ,, 1, (ii) sin i $ cos i 'sin b 2 - i l $ cosec i + cos b 2 - i l $ secIficosA, 3r, r, r, 2 < (A, B) < 2r , find the value of, sin b 2 - i l $ cosec i + cos b 2 - i l $ sec i 1 = 1, (i) sin(A–B), 8. Prove that : cos510o cos330o + sin390o, cos120o = –1, , (ii) cos(A+B)., , Solution, 9. Prove that:, 3r, Since 2 < (A, B) < 2r , both A and B, +, +, r, x, x, r, 2, r, x, 2, r, x, ], g, ], g, ], g, ], g, - cos, (i) tan, cot, cos, lie in the fourth quadrant,, 2, x - rg - cos ]2r - xg cos ]2r + xg = sin x, tan ]r + xg cot ]   , , ]180c + Ag cos ]90c - Ag tan ]270c - Ag, 2 ` sinA and sinB are negative., (ii) sin, 540, c - Ag cos ]360c + Ag cosec ]270c + Ag =- sin A cos A, ], sec, n, 4, 12, c - Ag tan ]270c - Ag, 2, Given cosA = 5 and cosB = 13 ,, c + Ag cosec ]270c + Ag =- sin A cos A, 3, 1, r, 3rTherefore,, 10. If sin i = 5 , tan { = 2 and 2 < i < r < { < 2, sinA = - 1 - cos 2 A, r, 3r , then find the value of, 2 <i<r<{< 2, 16, = - 1 - 25, 8 tan i - 5 sec {, Trigonometry, , 04_11th_BM-STAT_Ch-4-EM.indd 81, , 81, , 21-04-2020 12:10:10 PM

Page 89 :
www.tntextbooks.in, , Applying componendo and dividendo, rule, we get, m+1, sin A cos B + cos A sin B, sin A cos B - cos A sin B = m - 1, , sin ] A + Bg, m+1, = m-1 ,, A, B, ], g, sin, , which completes the proof., , 4.2.3 Trigonometric ratios of, multiple angles, Identities involving sin2A, cos2A, tan2A,, sin3A etc., are called multiple angle, identities., , 5. , , cos3A = 4 cos3 A - 3 cos A, , 6. , , tan3A =, , 3 tan A - tan3 A, 1 - 3 tan2 A, , NOTE, , 1, (i) cos2 A = 2 ]1 + cos 2Ag (or), 1 + cos 2A, cos A = !, 2, 1, (ii) sin2 A = 2 ]1 - cos 2Ag (or), 1 - cos 2A, sin A = !, 2, Example 4.9, , sin 2i, (i) sin 2A = sin ^ A + Ah = sin A cos A + cos A sin A = Show, 2 sin Athat, cos A1. + cos 2i = tan i, sin ^ A + Ah = sin A cos A + cos A sin A = 2 sin A cos A., Solution, , 2, 2, (ii) cos 2A = cos (A + A) = cos A cos A - sin A sinsin, A=, 2icos A2– sin iAcos i sin i, =, =, = tan i, 1 + cos 2i, cos i, 2 cos2 i, cos (A + A) = cos A cos A - sin A sin A = cos 2 A– sin 2 A, , (iii) sin 3A = sin(2A+A), , Example 4.10, , = sin 2A cos A + cos 2A sin A, , = ^2 sin A cos Ah cos A + ^1 - 2 sin Ah sin A, 2, , = 2 sin A cos 2 A + sin A - 2 sin3 A, , = 2 sin A ^1 - sin 2 Ah + sin A - 2 sin3 A, = 3sinA− 4sin3A, , Thus we have the following multiple, angles formulae, 1. (i) sin2A = 2sinA cosA, 2 tan A, (ii) sin2A =, 1 + tan 2 A, 2. (i) cos2A = cos2A – sin2A, (ii) cos2A = 2cos A–1, 2, , (iii) cos2A = 1– 2 sin2A, , Using multiple angle identity, find tan60o, Solution, , 2 tan A, 1 - tan 2 A, Put A = 30o in the above identity, we get, 2 tan 30c, tan60o =, 1 - tan2 30c, tan 2A =, , 1, 3, =, 1 =, 1- 3, Example 4.11, 1, If tanA = 7, that cos2A = sin4B, 2, , 2, 3, 2, 3, 2 = 3 #2 =, 3, , 1, and tanB = 3 , show, , Solution, 1 - tan 2 A, (iv) cos2A =, 2, 11 + tan 2 A, cos2A = 1 - tan2 A =, 1 + tan A 1 +, 2 tan A, 3., tan2A =, 2, 1 - tan A, 1, 1 - 49, 1 - tan2 A3, 48 49 24, = 49 # 50 = 25 , 4. , sin3A = 3 sin A - 4 sin, 2 A=, 1, 1 + tan A 1 +, 49, , 1, 49 = 48 # 49 = 24, 1, 49 50 25, 49, , Trigonometry, , 04_11th_BM-STAT_Ch-4-EM.indd 83, , 3, , … (1), 83, , 21-04-2020 12:10:19 PM

Page 90 :
www.tntextbooks.in, , Now, sin4B = 2sin2B cos2B, 2 tan B # 1 - tan 2 B 24, =, =2, 1 + tan 2 B 1 + tan 2 B 25, , 1, 1,    4 × 3 1 − 9 24 , ×, =, =, 1, 1 25, 1+, 1+, 9, 9, , … (2), , From(1) and (2) we get, cos2A = sin4B., Example 4.12, If tanA =, tan2A = tanB, , 1 - cos B, sin B then prove that, , Solution, Consider, , 1 - cos B, sin B, , B, 2 sin 2 2, B, B, 2 sin 2 cos 2, tanA, tanA, A, 2A, ` tan2A, , 1, 1, If tan a = 3 and tan b = 7 then, r, prove that ^2a + bh = 4 ., tan ^2a + bh, tan 2a, tan ^2a + bh, , 2a + b, 84, , 1. Find the values of the following:, (i) cosec15º (ii) sin(–105º) (iii) cot75º, 2. Find the values of the following, (i) sin76º cos16º + cos76º sin16º, r, r, r, r, (ii) sin 4 cos 12 + cos 4 sin 12, (iii) cos70º cos10º – sin70º sin10º, , (iv) cos2 15º –sin2 15º, 3, r, 12, 3r, 3. If sin A = 5 , 0 < A < 2 and cos B = -13, ,r< B < 2, - 12, 3r, cos B = 13 , r < B < 2 find the values of the, B, B, 2 sin 2 2, sin 2, B, following:, =, B, B =, B = tan 2, 2 sin 2 cos 2, cos 2, (i) cos(A+B), (ii) sin(A–B), B, (iii) tan(A–B), sin 2, B, =, =, tan, B, 2, 1, cos 2, 4. If cosA = 13, 14 and cosB = 7 where A, B, r, 1 - cos B, are acute angles prove that A–B = 3, = sin B, 5. Prove that 2tan80º = tan85º –tan 5º, B, = tan 2, -5, 1, 6. If cot α = 2 , sec β = 3 , where, B, = 2, r, 3r, π < α < 2 and 2 < b < r , find the, =B, value of tan(α + β). State the quadrant, = tanB, in which α + β terminates., , Example 4.13, , Solution, , Exercise 4.2, , 7. If A+B = 45º, prove that, (1+tanA)(1+tanB) = 2 and hence, 1c, deduce the value of tan 22 2, , 8. Prove that, (i) sin ] A + 60cg + sin ] A - 60cg = sin A, tan 2a + tan b, (ii) tan 4A tan 3A tan A + tan 3A + tan A - tan 4A = 0, =, 1 - tan 2a $ tan, b, tan 4A tan 3A tan A + tan 3A + tan A - tan 4A = 0, 1, 2# 3, 9. (i) If tanθ = 3 find tan3θ, 2 tan a, 3, =, =, =, 2, 12, 4, 1 - tan a 1 - 1, (ii) If sin A = 13 , find sin3A, 9, 3 1, 25, 3, +7, 10. If sinA = 5 , find the values of cos3A, 4, 28, =, =, 3 1, 25, and tan3A, 1- 4 # 7, 28, sin ]B - C g sin ]C - Ag sin ] A - Bg, r, 11. Prove that cos B cos C + cos C cos A + cos A cos B, = 1 = tan 4, r, sin ]B - C g sin ]C - Ag sin ] A - Bg = 0., = 4, cos B cos C + cos C cos A + cos A cos B, , 11th Std. Business Mathematics and Statistics, , 04_11th_BM-STAT_Ch-4-EM.indd 84, , 21-04-2020 12:10:24 PM

Page 94 :
b, , www.tntextbooks.in, , Example 4.21, , Exercise 4.3, , Find sin105o + cos105o, 1., , Solution, sin105o + cos105o, = sin(90 + 15 )+cos105, o, , o, , A, 3A, (i) sin 8 sin 8, , o, , (ii) cos ]60c + Ag sin ]120c + Ag, , = cos15o + cos105o, , 7A, 5A, (iii) cos 3 sin 3, , = cos105o + cos15o, 105c + 15c l b 105c - 15c l, = 2 cos b, cos, 2, 2, 120c, 90c, = 2 cos b 2 l cos b 2 l, 1, = 2cos60o cos45o =, 2, , (iv) cos 7i sin 3i, 2., , Express each of the following as the, product of sine and cosine, (i) sinA + sin2A, , Example 4.22, , (ii) cos2A + cos4A, , Prove that, ^cos a + cos bh2 + ^sin a + sin bh2 = 4 cos2 b, , (iii) sin 6i - sin 2i, , a-b, 2 l, , Solution, , (iv) cos 2i - cos i, 3., , cos a + cos b = , , 2 cos b, , Express each of the following as the, sum or difference of sine or cosine:, , a+b, a-b, 2 l $ cos b 2 l , , ….(1), , sin a + sin b =, a+b, a-b, 2 sin b 2 l $ cos b 2 l  ….(2), , Prove that, , 1, (i) cos 20c cos 40c cos 80c = 8, (ii) tan 20c tan 40c tan 80c = 3, , 4., , Prove that, , (i) ^cos a - cos bh + ^sin a - sin bh = 4 sin2 b, 2, , 2, , a-b, Squaring and adding, a -and, cos(2), bh2 + ^sin a - sin bh2 = 4 sin2 b, ^cos(1), 2 l, ^cos a + cos bh2 + ^sin a + sin bh2, 1, (ii) sin A sin (60c + A) sin (60c - A) = 4 sin 3A, a+b, a-b, 2 a+b, 2 a, 1- b l, l, b, l, b, +, = 4 cos2 b 2 l cos2 bsin A, sin, cos, 4, =, +, sin, (, 60, c, A, ), sin, (, 60, c, A, ), 2, 2, 4 2sin 3A, , a+b, 2 a-b, 2 a+b, 2 a-b, 2 l cos b 2 l + 4 sin b 2 l cos b 2 l, , 5., , Prove that, , a-b, 2 l, , (i) sin ] A - Bg sin C + sin ]B - C g sin A + sin ]C - Ag si, a-b, 2 a+b, 2 a+b, l, b, l, b, l, E, ;, +, cos, sin, 2, 2, 2, sin ] A - Bg sin C + sin ]B - C g sin A + sin ]C - Ag sin B = 0, a-b, a+b, a+b, os2 b 2 l;cos2 b 2 l + sin2 b 2 lE, r, 9r, 3r, 5r, (ii) 2 cos 13 cos 13 + cos 13 + cos 13 = 0, = 4 cos2 b, , = 4 cos2 b, , 88, , a-b, 2 l, , r, 9r, 3r, 5r, 2 cos 13 cos 13 + cos 13 + cos 13 = 0, , 11th Std. Business Mathematics and Statistics, , 04_11th_BM-STAT_Ch-4-EM.indd 88, , 21-04-2020 12:10:49 PM

Page 95 :
www.tntextbooks.in, , 6., , Prove that, 2A - cos 3A, A, (i) cos, =, tan, 2, sin 2A + sin 3A, cos 7A + cos 5A, (ii) sin 7A - sin 5A = cot A, , 7., , Prove that, 1, cos20, 20occos, cos 40occos, cos 60co cos 80, 80co = 16, cos, , 8., , Evaluate, (i) cos 20° + cos 100° + cos 140°, °, , (ii) sin 50 - sin 70c + sin 10, , °, , -r r, : 2 , 2 D, y ! 0, , R–(–1, 1), , cosec -1 x, , -r r, r, R–(–1, 1) : [0,, 2 , r2],D, y ! 2, , sec -1 x, , R, , cot -1 x, , (0, r ), , 4.4.2 Properties of Inverse, Trigonometric Functions, Property (1), , (i) sin -1 ]sin xg = x, , 1, 1, If cosA, cos A++cosB, cos B = 2 and sinA, sin A++sinB, sin B = 4 ,, A+B, 1, prove that tan b 2 l = 2, , (ii) cos -1 ]cos xg = x, , 10., , If sin(y+z−x), sin(z+x−y), sin(x+y−z),, are in A.P, then prove that tan x, tan y, and tan z are in A.P, , (v) sec -1 ]sec xg = x, , 11., , If cosecA + secA = cosecB + secB, A+B, prove that cot b 2 l = tan A tan B, , 9., , 4.4, 4.4.1, , Inverse Trigonometric Functions, I nverse Trigonometric Functions, , The quantities such as sin-1x, cos-1x,, tan-1x, etc., are known as inverse trigonometric, functions., i.e., if sin i = x , then i = sin -1 x ., The domains and ranges (Principal value, branches) of trigonometric functions are, given below, Function, , Domain, x, , Range, y, , sin -1 x, , [–1, 1], , -r r, : 2 , 2D, , cos -1 x, , [–1, 1], , [0, r ], , tan -1 x, , R = ^- 3, 3h, , -r r, b 2 , 2l, , (iii) tan -1 ]tan xg = x, , (iv) cot -1 ]cot xg = x, (vi) cosec -1 ]cosec xg = x, , Property (2), (i), (ii), , 1, sin -1 b x l = cosec -1 (x), 1, cos -1 b x l = sec -1 (x), , 1, (iii) tan -1 b x l = cot-1 (x), , 1, (iv) cosec -1 b x l = sin -1 (x), (v), , 1, sec -1 b x l = cos -1 (x), , 1, (vi) cot-1 b x l = tan -1 (x), Property (3), (i), , sin -1 (- x) =- sin -1 (x), , (ii), , cos -1 (- x) = r - cos -1 (x), , (iii) tan -1 (- x) = - tan -1 (x), (iv) cot-1 (- x) =- cot-1 (x), (v), , sec -1 (- x) = r - sec -1 (x), , (vi) cosec -1 (- x) =- cosec -1 (x), Trigonometry, , 04_11th_BM-STAT_Ch-4-EM.indd 89, , 89, , 21-04-2020 12:10:56 PM

Page 96 :
www.tntextbooks.in, , Property (4), (i), (ii), , r, sin -1 (x) + cos -1 (x) = 2, r, tan -1 (x) + cot-1 (x) = 2, , Solution, , 5, (i) Let b sin -1 13 l = i , , 5, sin i = 13, , r, (iii) sec -1 (x) + cosec -1 (x) = 2, Property (5), If xy<1, then, x+y, (i) tan -1 (x) + tan -1 (y) = tan -1 d 1 - xy n, x-y, (ii) tan -1 (x) - tan -1 (y) = tan -1 d 1 + xy n, Property (6), , cos i =, , 1 - sin 2 i, , =, , 25, 1 - 169, , 12, = 13 , ... (2), From (1) and (2), we get, 5, 12, Now cos b sin -1 13 l = cos i = 13, 8, (ii) Let b cos -1 17 l = i, , sin -1 (x) + sin -1 ^ y h = sin -1 _ x 1 - y 2 + y 1 - x 2 i, sin -1 (x) + sin -1 ^ y h = sin -1 _ x 1 - y 2 + y 1 - x 2 i, Example 4.23, Find the principal value of, , (i) sin -1 _1 2 i (ii) tan -1 ^- 3 h, , Solution, , r, r, (i) Let sin -1 _1 2 i = y where - 2 # y # 2, r, ` sin y = _1 2 i = sin b 6 l, r,    y = 6, r, The principal value of sin -1 _1 2 i is 6, , r, r, (ii) Let tan-1 ^- 3 h = y where - 2 # y # 2, r, ` tan y = – 3 = tan b - 3 l, r,   , y= -3, r, The principal value of tan -1 ^- 3 h is - 3, , Example 4.24, Evaluate the following, 5, 8, (i) cos b sin -1 13 l (ii) tan b cos -1 17 l, 90, , ... (1), , cos i, , 8, = 17, , sin i, , =, , 1 - cos 2 i, , 64, 1 - 289, 15, = 17 , ... (2), From (1) and (2), we get, =, , 8, sin i, Now tan b cos -1 17 l = tan i =, cos i, =, Example 4.25, , 15, 17, 8, 17, , 15, = 8, , Prove that, 1, 1, 2, (i) tan -1 b 7 l + tan -1 b 13 l = tan -1 b 9 l, 4, 3, 27, (ii) cos -1 b 5 l + tan -1 b 5 l = tan -1 b 11 l, Solution, , 1, 1, (i) tan -1 b 7 l + tan -1 b 13 l, = tan >, -1, , 1, 7, , + 131, 1 H, 1, 1 - _ 7 i_ 13 i, , 20, = tan -1 b 90 l, 2, = tan -1 b 9 l, , 11th Std. Business Mathematics and Statistics, , 04_11th_BM-STAT_Ch-4-EM.indd 90, , 21-04-2020 12:11:02 PM

Page 97 :
www.tntextbooks.in, , 4, 4, (ii) Let cos -1 b 5 l = i . Then cos i = 5, 3, & sin i = 5, 3, 3, ` tan i = 4 & i = tan -1 b 4 l, 4, 3, ` cos -1 b 5 l = tan -1 b 4 l, , 4, 3, Now, cos -1 b 5 l + tan -1 b 5 l, , 3, 3, = tan -1 b 4 l + tan -1 b 5 l, = tan e, -1, , 3, 4, , 1-, , 3, 5, , +, 3, 4, , #, , 3, 5, , o, , 27, = tan -1 b 11 l, 1, r, Find the value of tan ; 4 - tan -1 b 8 lE, , Solution, , 1, Let tan b 8 l = i, 1, Then tan i = 8, 1, r, r, ` tan ; 4 - tan -1 b 8 lE = tan b 4 - i l, -1, , r, 4, , =, , tan - tan i, 1 + tan r4 tan i, , =, , 11+, , 1, 8, 1, 8, , 7, =9, , Example 4.27, , x-1, x+1, r, Solve tan -1 b x - 2 l + tan -1 b x + 2 l = 4, , Solution, x-1, x+1, tan -1 b x - 2 l + tan -1 b x + 2 l, = tan -1 >, , 1, , i, , ` tan -1 b, , &x =!, Example 4.28, , 1, 2, Simplify: sin -1 ` 3 j + sin -1 ` 3 j, , Solution, We know that, , sin -1 ]xg + sin -1 (y) =, , sin -1 7x 1 - y 2 + y 1 - x 2 A, , 1, 2, ` sin -1 b 3 l + sin -1 b 3 l =, 1, sin -1 : 3, , , , 4 2, 1- 9 + 3, , 1, = sin -1 : 3, , 5 2, 9 +3, , = sin -1 c, , 5 +4 2 m, 9, , 8D, 9, , Solve tan -1 ]x + 2g + tan -1 ]2 - xg =, 2, tan -1 b 3 l, , Solution, tan -1 <, , ^ x + 2 h + ^2 - x h, F = tan -1 ` 32 j, 1 - ^ x + 2h^2 - x h, , & 2x 2 – 6 = 12, , 2, , &, `, , x2 = 9, x = +3, , Example 4.30, If tan(x + y) = 42 and x = tan–1(2),, then find y, Trigonometry, , 04_11th_BM-STAT_Ch-4-EM.indd 91, , 1, 1- 9D, , Example 4.29, , 4, H = tan -1 b 2x- 3 l, , r, 2x - 4 l, = 4, -3, , 1, 2, , 4, 2, & tan -1 c 1 - ^4 - x 2h m = tan -1 ` 3 j, , Given that, x-1, x+1, r, tan -1 b x - 2 l + tan -1 b x + 2 l = 4, 2, , r, = tan 4 = 1, , 2x 2 - 4 = –3, & 2x2 –1 = 0, , , , Example 4.26, , x-1, x+1, x-2 + x+2, x+1, x-1, - _ x - 2 i_ x + 2, , 2x 2 - 4, -3, , 91, , 21-04-2020 12:11:08 PM

Page 98 :
www.tntextbooks.in, , Solution, , 10., tan(x+y) = 42, x + y = tan–1(42), , tan–1(2)+y = tan–1(42), , Express, , cos x, r, 3r, tan -1 :1 - sin x D, - 2 < x < 2 , in, the simplest form., Exercise 4.5, , y = tan–1(42) – tan–1(2), 42 - 2, = tan -1 ;1 + ]42 # 2g E, 40, = tan -1 b 85 l, 8, y = tan -1 :17 D, , Exercise 4.4, 1., , 2., , Find the pricipal value of the following, 1, (ii) tan -1 ]- 1g, (i) sin -1 b - 2 l, (iii) cosec -1 ^2 h, , Prove that, , (iv) sec -1 ^- 2 h, , 2x, (i) 2 tan ]x g = sin -1 c 1 + x 2 m, -1, , (ii) tan -1 b 34 l - tan -1 b 17 l = r, 4, 3., , 4., 5., , 6., , 7., 8., 9., , 92, , Show that, 1, 2, 3, tan -1 b 2 l + tan -1 b 11 l = tan -1 b 4 l, r, Solve tan -1 2x + tan -1 3x = 4, Solve, 4, tan -1 ]x + 1g + tan -1 ]x - 1g = tan -1 b 7 l, 3, Evaluate (i) cos 8tan ` 4 jB, 1, 4, (ii) sin 8 2 cos -1 ` 5 jB, -1, , 4, 12, Evaluate: cos `sin -1 ` 5 j + sin -1 ` 13 jj, Prove that, m, m-n, r, tan -1 b n l - tan -1 b m + n l = 4, Show that, 3, 8, 84, sin -1 b - 5 l - sin -1 b -17l = cos -1 85, , Choose the correct answer, r, 1. The degree measure of 8 is, (a) 20c60l, (b) 22c30l, (c) 22c60l, (d) 20c30l, 2. The radian measure of 37c30l is, 3r, 5r, (b) 24, (a) 24, 7r, 9r, (d) 24, (c) 24, 1, and i lies in the first, 5, quadrant then cos i is, , 3. If tan i =, (a), (c), , 1, 6, 5, 6, , -1, 6, - 5, (d), 6, , (b), , 4. The value of sin 15o is, (a), (c), , 3 +1, 2 2, 3, 2, , 3 -1, 2 2, 3, (d), 2 2, , (b), , 5. The value of sin ]- 420cg is, 3, (a) 2, 1, (c) 2, , 3, (b) - 2, -1, (d) 2, , 6. The value of cos ]- 480cg is, (a), , 3, 1, (c) 2, , 3, (b) - 2, -1, (d) 2, , 11th Std. Business Mathematics and Statistics, , 04_11th_BM-STAT_Ch-4-EM.indd 92, , 21-04-2020 12:11:15 PM

Page 99 :
www.tntextbooks.in, , 7. The value of sin28o cos17o + cos28o, sin17o is, 1, (b) 1, (a), 2, -1, (d) 0, (c), 2, 8. The value of sin15º cos15º is, 1, (a) 1, (b) 2, 3, 1, (d) 4, (c) 2, 9. The value of secA sin(270o + A) is, (a) –1, (b) cos2A, (d) 1, (c) sec2A, 10. If sinA + cosA =1 then sin2A is, equal to, (a) 1, (b) 2, 1, (c) 0, (d) 2, 11. The value of cos245º – sin245º is, 3, 1, (b) 2, (a) 2, 1, (c) 0, (d), 2, 12. The value of 1–2 sin 45º is, 1, (a) 1, (b) 2, 1, (c) 4, (d) 0, 2, , 13. The value of 4cos340º –3cos40º is, 3, 1, (b) – 2, (a) 2, 1, 1, (d), (c) 2, 2, 2 tan 30c, 14. The value of, is, 1 + tan 2 30c, 1, 1, (b), (a) 2, 3, 3, (c) 2, (d) 3, 1, 15. If sin A= 2 then 4 cos3 A - 3 cos A is, (a)1, (b) 0, 3, (c) 2, , (d), , 1, 2, , 3 tan 10c - tan3 10c, is, 1 - 3 tan 2 10c, 1, 1, (b) 2, (a), 3, 1, (c) 3, (d), 2, 2, 2, 17. The value of cosec -1 d 3 n is, r, r, (b) 2, (a) 4, r, r, (c) 3, (d) 6, 16. The value of, , 2, 2, 18. sec -1 3 + cosec -1 3 =, -r, r, (b) 2, (a) 2, (c) r, (d) – r, r, 19. If a and b be between 0 and 2 and if, 12, 3, cos ^a + bh = 13 and sin ^a - bh = 5, then sin 2 a is, 16, (b) 0, (a) 15, 56, 64, (d) 65, (c) 65, 1, 1, 20. If tan A = 2 and tan B = 3, tan ^2A + B h is equal to, (a) 1, (c) 3, , (b) 2, (d) 4, , r, 21. tan b 4 - x l is, 1 + tan x, (a) b 1 - tan x l, (c) 1 - tan x, , 1 - tan x, (b) b 1 + tan x l, (d) 1 + tan x, , 3, 22. sin b cos -1 5 l is, 3, (a) 5, 4, (c) 5, , 5, (b) 3, 5, (d) 4, , 23. The value of, -1, 2, (c) 2, , (a), , 1, is, cosec ]- 45cg, 1, (b), 2, (d) - 2, Trigonometry, , 04_11th_BM-STAT_Ch-4-EM.indd 93, , then, , 93, , 21-04-2020 12:11:22 PM

Page 100 :
www.tntextbooks.in, , 5., , 24. If p sec 50c = tan 50c then p is, (a) cos 50c, , (b) sin 50c, , (c) tan 50c, , (d) sec 50c, , 6., , cos x, 25. b cosec x l - 1 - sin 2 x 1 - cos 2 x is, 2, , 2, , (a) cos x - sin x, (c) 1, , 2, , 2, , (b) sin x - cos x, (d) 0, , Miscellaneous Problems, 1., , Prove that, cos 4x + cos 3x + cos 2x, sin 4x + sin 3x + sin 2x = cot 3x, , Prove that, 3r, 3r, r, 2 sin 2 4 + 2 cos 2 4 + 4 sec2 4 = 10, Find the value of (i) sin 75c, (ii) tan 15c, , 7., , If sin A = 13 , sin B = 14 then find the, value of sin ] A + Bg , where A and B, are acute angles., , 8., , Show that, 12, 3, 56, cos -1 b 13 l + sin -1 b 5 l = sin -1 b 65 l, , 5, If cos (a + b) = 54 and sin (a - b) = 13, where, 3xg a + b and a - b are acute,, 3., Prove that cot 4x ]sin 5x + sin 3xg = cot x ]sin 5x - sin, then find tan 2a, cot 4x ]sin 5x + sin 3xg = cot x ]sin 5x - sin 3xg ., 2., , Prove that, , 4., , If tan x = 34 , r < x < 32r then find, x, the value of sin 2 and cos 2x, , 3 cosec 20c - sin 20c = 4, , 9., , 10., , cos x - sin x, Express tan -1 b cos x + sin x l, 0 < x < r, , in the simplest form., , Summary, , zz If in a circle of radius r, an arc of length, , l subtends an angle of θ radians, then, l=rθ, , zz sin2 x + cos2 x = 1, zz tan x + 1 = sec x, 2, , 2, , zz tan 2a = 2 tan a2, , 1 - tan a, , zz sin 3a = 3 sin a - 4 sin3 a, zz cos 3a = 4 cos3 a - 3 cos a, 3, a, zz tan 3a = 3 tan a - tan, 2, , 1 - 3 tan a, x+y, x-y, zz sin(α + β) = sin α cos β + cos α sin β zz sin x + sin y = 2 sin 2 cos 2, x+y, x-y, zz sin(α − β) = sin α cos β – cos α sin β, zz sin x - sin y = 2 cos 2 sin 2, zz cos(α + β) = cos α cos β – sin α sin β, x+y, x-y, zz cos x + cos y = 2 cos 2 cos 2, zz cos ^a - bh = cos a $ cos b + sin a sin b, x+y, x-y, =z, z, cos, x, cos, y, 2, sin, sin, tan a + tan b, 2, 2, zz tan ^a + bh = 1 - tan a tan b, zz sin x cos y = 12 6sin ^ x + y h + sin ^ x - y h@, tan a - tan b, zz tan ^a - bh = 1 + tan a tan b, zz cos x sin y = 12 6sin ^ x + y h - sin ^ x - y h@, zz sin 2a = 2 sin a cos a, zz 2 cos x cos y = 12 6cos ^ x + yh + cos ^ x - y h@, 2, 2, 2, zz cos 2a = cos a – sin a = 1 - 2 sin a = 2 cos a - 1, zz sin x sin y = 12 6cos ^ x - yh - cos ^ x + y h@, cos 2a = cos 2 a – sin 2 a = 1 - 2 sin 2 a = 2 cos 2 a - 1, , zz cot2 x + 1 = cosec2 x, , 94, , 11th Std. Business Mathematics and Statistics, , 04_11th_BM-STAT_Ch-4-EM.indd 94, , 21-04-2020 12:11:30 PM

Page 102 :
www.tntextbooks.in, , Chapter, , 5, , DIFFERENTIAL CALCULUS, , Learning Objectives, After studying this, chapter, the students will, be able to understand, � the concept of limit and, continuity, , •, , formula of limit and definition of, continuity, , •, , basic concept of derivative, , •, , how to find derivative of a, function by first principle, , •, , derivative of a function by, certain direct formula and its, respective application, , •, , how to find higher order, derivatives, , Isaac Newton, , 96, , G.W. Leibnitz, , Introduction, Calculus is a Latin word which, means that Pebble or a small stone used, for calculation. The word calculation is, also derived from the same Latin word., Calculus is primary mathematical tool, for dealing with change. The concept of, derivative is the basic tool in science of, calculus. Calculus is essentially concerned, with the rate of change of dependent, variable with respect to an independent, variable. Sir Issac Newton (1642 – 1727, CE) and the German mathematician G.W., Leibnitz (1646 – 1716 CE) invented and, developed the subject independently and, almost simultaneously., In this chapter we will study, about functions and their graphs, limits,, derivatives and differentiation technique., , 11th Std. Business Mathematics and Statistics, , 05_11th_BM-STAT_Ch-5-EM.indd 96, , 21-04-2020 12:12:04 PM

Page 103 :
www.tntextbooks.in, , 5.1, , Functions and their Graphs, , Some basic concepts, , from point to point. a and b are intercept, values on the axes which are arbitrary., There are two kinds of variables, , 5.1.1 Quantity, Anything which can be performed, on basic mathematical operations like, addition, subtraction, multiplication and, division is called a quantity., , 5.1.2 Constant, A quantity which retains the, same value throughout a mathematical, investigation is called a constant., , (i), , A variable is said to be an independent, variable when it can have any arbitrary, value., , (ii) A variable is said to be a dependent, variable when its value depend on the, value assumed by some other variable., E xample: In the equation, y = 5x2 - 2x + 3 ,, , Basically constant quantities are of, two types, , “x” is the independent variable,, , (i), , “3” is the constant., , Absolute constants are those which, do not change their values in any, mathematical investigation. In other, words, they are fixed for ever., Examples: 3, 3 , p, ..., , (ii) Arbitrary constants are those which, retain the same value throughout a, problem, but we may assign different, values to get different solutions., The arbitrary constants are usually, denoted by the letters a, b, c, ..., Example: In an equation y=mx+4,, m is called arbitrary constant., , 5.1.3 Variable, A variable is a quantity which can, assume different values in a particular, problem. Variables are generally denoted, by the letters x, y, z, ..., Example: In an equation of the, straight line x + y = 1,, a, , “y” is the dependent variable and, , 5.1.4 Intervals, The real numbers can be represented, geometrically as points on a number line, called real line. The symbol R denotes, either the real number system or the real, line. A subset of the real line is an interval., It contains atleast two numbers and all the, real numbers lying between them., , −∞, , B, , a, , b, , ∞, , Fig. 5.1, , (i), , Open interval, , The set {x : a < x < b} is an open, interval, denoted by (a, b)., In this interval, the boundary points, a and b are not included., , b, , x and y are variables because they, assumes the co-ordinates of a moving point, in a straight line and thus changes its value, , A, , −∞, , A, , B, , (, , ), , a, , b, , ∞, , Fig. 5.2, Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 97, , 97, , 21-04-2020 12:12:05 PM

Page 104 :
www.tntextbooks.in, , For example, in an open interval, (4, 7), 4 and 7 are not elements of this, interval., But, 4.001 and 6.99 are elements of, this interval., (ii) Closed interval, The set {x : a < x < b} is a closed, interval and is denoted by [a, b]., In the, interval [a, b],, the boundary, points a and b −∞, are included., , A, , B, , [, , ], , a, , b, , ∞, , Fig. 5.3, , For example, in an interval [4, 7],, 4 and 7 are also elements of this interval., Also we can make a mention about, semi closed or semi open intervals., (a, b]= {x : a < x < b} is called left open, interval and, [a, b) = {x : a < x < b} is called right open, interval., In all the above cases, b - a = h is, called the length of the interval., , 5.1.5 Neighbourhood of a point, Let ‘a’ be any real number. Let e > 0, be arbitrarily small real number., Then ( a - e , a + e ) is called an, e - neighbourhood of the point ‘a’ and, denoted by Na, e, For examples,, N5, 14, , N2, 1, , 7, , 98, , 1, 1, = b5 - 4 , 5 + 4 l, 19, 21, = & x: 4 < x < 4 0, 1, 1, = b2 - 7 , 2 + 7 l, 13, 15, = & x: 7 < x < 7 0, , 5.1.6 Function, Let X and Y be two non-empty, sets of real numbers. If there exists a rule, f which associates to every element x ! X,, a unique element y ! Y, then such a rule, f is called a function (or mapping) from, the set X to the set Y. We write f : X $ Y., The function notation y = f(x), was first used by, Leonhard Euler in 1734 - 1735, , The set X is called the domain of f,, Y is called the co-domain of f and the, range of f is defined as f(X) = {f(x) / x ! X}., Clearly f(X) 3 Y., A function of x is generally denoted, by the symbol f(x), and read as “function, of x” or “f of x”., , 5.1.7 Classification of functions, Functions can be classified into two, groups., (i), , Algebraic functions, , Algebraic functions are algebraic, expressions using a finite number of terms,, involving only the algebraic operations, addition, subtraction, multiplication, division, and raising to a fractional power., Examples : y = 3x4 - x3 + 5x2 - 7,, 2x3 + 7x - 3, y= 3, , y = 3x 2 + 6x - 1 are, x + x2 + 1, algebraic functions., (a) y = 3x4- x3 + 5x2- 7 is a polynomial, or rational integral function., 2 x3 + 7 x − 3, (b) y= 3 2, is a rational, x + x +1, , function., , (c) y= 3x 2 + 6 x − 1 is an irrational, function., , 11th Std. Business Mathematics and Statistics, , 05_11th_BM-STAT_Ch-5-EM.indd 98, , 21-04-2020 12:12:06 PM

Page 105 :
www.tntextbooks.in, , (ii) Transcendental functions, A function which is not algebraic is, called transcendental function., Examples:, sin x, sin-1 x, e x, loga x, are transcendental functions., (a) sin x, tan 2x, ... are trigonometric, functions., (b) sin-1 x, cos-1 x, ... are inverse, trigonometric functions., , 5.1.8 Even and odd functions, A function f(x) is said to be an even, function of x, if f( - x ) = f( x )., A function f(x) is said to be an odd, function of x, if f( - x ) = - f( x )., Examples:, f(x) = x2 and f(x) = cos x are even, functions., f(x) = x3 and f(x) = sin x are odd, functions., NOTE, f(x) = x3 + 5 is neither even nor odd, function, , 5.1.9 Explicit and implicit functions, A function in which the dependent, variable is expressed explicitly in terms of, some independent variables is known as, explicit function., Examples: y = + 3 and y =, are explicit functions of x., , +, , 5.1.10 Constant function, , Examples: y = 3, f(x) = –5 are, constant functions., , (d) loga x, loge (sin x), ... are, logarithmic functions., , ex, , Example: x3 + y3-xy = 0 is an implicit, function., If k is a fixed real number then the, function f(x) given by f(x) = k for all x ! R, is called a constant function., , (c) ex, 2x, xx, ... are exponential, functions., , x2, , If two variables x and y are connected, by the relation or function f(x, y) = 0 and, none of the variable is directly expressed, in terms of the other, then the function is, called an implicit function., , e-x, , Graph of a constant function y = f(x), is a straight line parallel to x-axis, , 5.1.11 Identity function, A function that associates each real, number to itself is called the identity, function and is denoted by I., i.e. A function defined on R by, f(x) = x for all x ∈ R is an identity function., Example: The set of ordered pairs, {(1, 1), (2, 2), (3, 3)} defined by f : A " A, where A={1, 2, 3} is an identity function., Graph of an identity function on, R is a straight line passing through the, origin and makes an angle of 45c with, positive direction of x-axis, , 5.1.12 Modulus function, A function f(x) defined by f(x) = |x|,, , where |x| = ), , x if x $ 0, is called the, - x if x 1 0, , modulus function., Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 99, , 99, , 21-04-2020 12:12:08 PM

Page 106 :
www.tntextbooks.in, , It is also called an absolute value, function., Remark Domain is R and range set is [0, ∞), , NOTE, 4.7 = 5, –7.2 = –7, 5 = 5, 0.75 = 1., , 5.1.15 Rational function, , NOTE, |5| = 5, |–5| = –(–5) = 5, , 5.1.13 Signum function, A function f(x) is defined by, |x |, f(x) = * x if x ! 0 is called the, 0 if x = 0, , A function f(x) is defined by the, p (x), , rule f (x) = q (x) , q (x) ! 0 is called rational, function., x2 + 1, Example: f (x) = x - 3 , x ! 3 is a, rational function., , 5.1.16 Polynomial function, , For the real numbers a0, a1, a2,..., an ; a0 ! 0, Domain is R and range seta0is, a1, a2,..., an ; a0 ! 0 and n is a non-negative integer,, Remark, a function f(x) given by f ]xg = a0 xn + a1 xn - 1 + a2 xn - 2, {–1, 0, 1}, f ]x g = a0 xn + a1 xn - 1 + a2 xn - 2 + ... + an is called as a, 5.1.14 Step function, polynomial function of degree n., (i) Greatest integer function, Example: f (x) = 2x3 + 3x2 + 2x - 7, , Signum function., , The function whose value at any, real number x is the greatest integer less, than or equal to x is called the greatest, integer function. It is denoted by x., i.e. f : R $ R defined by f(x) = x, is called the greatest integer function., NOTE, 2.5 = 2, –2.1 = –3,, 0.74 = 0, –0.3 = –1, 4 = 4., , (ii) Least integer function, The function whose value at any real, number x is the smallest integer greater, than or equal to x is called the least integer, function. It is denoted by x., i.e. f : R $ R defined by f(x) = x, is called the least integer function., Remark, 100, , Function’s domain is R and, range is Z [set of integers], , is a polynomial function of degree 3., , 5.1.17 Linear function, For the real numbers a and b with, a ! 0, a function f(x) = ax + b is called a, linear function., Example: y = 2x + 3 is a linear, function., , 5.1.18 Quadratic function, For the real numbers a, b and c with, a ! 0, a function f(x) = ax2 + bx + c is, called a quadratic function., Example: f (x) = 3x2 + 2x - 7 is a, quadratic function., , 5.1.19 Exponential function, A function f(x) = ax, a ! 1 and, a > 0, for all x ∈ R is called an exponential, function, Remark, , Domain is R and range is (0, ∞), and (0, 1) is a point on the graph, , 11th Std. Business Mathematics and Statistics, , 05_11th_BM-STAT_Ch-5-EM.indd 100, , 21-04-2020 12:12:10 PM

Page 107 :
www.tntextbooks.in, , Examples: e2x, e x, exponential functions., , 2+, , 1, , and 2 x are, , Solution, Given that f(x) = g(x), &   2x2 – 1 = 1–3x, , 5.1.20 Logarithmic function, , 2x2 +3x – 2 = 0, , For x > 0, a > 0 and a ! 1, a function, f(x) defined by f(x) = logax is called the, logarithmic function., Remark, , (x+2) (2x–1) = 0, 1, ` x = –2, x =, 2, 1, Domain is $- 2, 2 ., , Domain is (0, ∞), range is R and, (1, 0) is a point on the graph, , Example: f(x) = loge(x+2) and, f(x) = loge (sinx) are logarithmic functions., , 5.1.21 Sum, difference, product and, quotient of two functions, If f(x) and g(x) are two functions, having same domain and co-domain, then, (i) (f + g) (x) = f(x) + g(x), , Example 5.2, f = {(1, 1), (2, 3)} be a function, described by the formula f(x) = ax+b., Determine a and b ?, Solution, Given that f(x) = ax + b , f(1) = 1 and, f(2) = 3, a + b = 1 and 2a + b = 3, `     , , (ii) (fg)(x) = f(x) g(x), f, f (x), (iii) c g m (x) = g (x) , g (x) ! 0, (iv) (cf)(x) = cf(x), c is a constant., , 5.1.22 Graph of a function, The graph of a function is the set, of points _ x, f ]x gi where x belongs to the, domain of the function and f(x) is the, value of the function at x., To draw the graph of a function, we, find a sufficient number of ordered pairs, _ x, f ]x gi belonging to the function and, join them by a smooth curve., NOTE, It is enough to draw a diagram for, graph of a function in white sheet itself., Example 5.1, Find the domain for which the, functions f(x) = 2x2–1 and g(x) = 1–3x are, equal., , &      a = 2 and b = –1, , Example 5.3, If f(x) =x + 1x , x > 0, then show that, , [f(x)]3 = f(x3) + 3 f ` 1x j, Solution, , f(x) = x + 1x , f(x3) = x3 + x13, , and f ` 1x j = f(x), , Now, LHS = [f(x)]3, , 1 3, = 8x + x B, 1, 1, 3, = x + 3 +3 ` x + x j, x, ,  , , = f (x3) + 3f (x), 1,       , = f (x3) + 3f ` j, x, = RHS, ,  , , Example 5.4, Let f be defined by f(x) = x – 5 and, g be defined by, Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 101, , 101, , 21-04-2020 12:12:11 PM

Page 108 :
www.tntextbooks.in, , x2 - 25, if x ! - 5, g(x) = * x + 5, Find l, m if x =- 5, such that f(x) = g(x) for all x ∈ R, Solution, `, , f(–5) = g(–5), , `, , –5 –5 = l, l = –10, , (1  x) 2, 1 x  2x, , log,  log, (1  x) 2, 1  x2  2 x, 2, ,  2 log, , 1 x,  2 f ( x)., 1 x, , Example 5.8, , Example 5.5, , If f(x) = x and g(x) = |x|, then find, , If f(x) = 2x, then show that, , (i) (f+g)(x), , f^ x h $ f^ y h = f^ x + yh, , f(x) = 2x (Given), f(x + y) = 2x+y, = 2x . 2y, , `, , = f^ xh $ f^ yh, , Example 5.6, , If f ( x) , , (ii) (f–g)(x), , (iii) (fg)(x), , Solution, , Solution, , x 1, , x > 0, then show that, x 1, , 1, f [ f ( x)]  , x, , (i) (f + g) (x) = f(x) + g(x) = x + |x|, =), , 2x if x $ 0, x + x if x $ 0, =), 0 if x < 0, x - x if x < 0, , (ii) (f – g) (x) = f(x) – g(x) = x – |x|, =), , 0 if x $ 0, x - x if x $ 0, =), 2x if x < 0, x ( x) if x < 0, , (iii) (f g) (x) = f(x) g(x) = x |x|, = ), , x2 if x $ 0, - x2 if x < 0, , Example 5.9, , Solution, f ( x) , , x 1, (Given), x 1, , f ]xg - 1, ` f 7 f ]x gA = ]xg + =, 1, f, , x-1, x+1, x-1, x+1, , -1, +1, , 1, 2, x 1 x 1, , , x 1  x 1, 2x, x, , Example 5.7, 1 x, , 0 < x < 1, then, 1 x,  2x ,  2 f ( x)., show that f , 2 ,  1 x , , If f ( x)  log, , 102, , 1 x, (Given), 1 x, 1 + 1 +2xx2, 2x, ` f c 1 + x2 m = log, 1 - 2x 2, f ( x)  log, , 1+x, , Given that f(x) = g(x) for all x ∈ R, , , , Solution, , A group of students wish to charter, a bus for an educational tour which can, accommodate atmost 50 students. The, bus company requires at least 35 students, for that trip. The bus company charges, ` 200 per student up to the strength of, 45. For more than 45 students, company, charges each student ` 200 less 15 times, the number more than 45. Consider the, number of students who participates the, tour as a function, find the total cost and, its domain., , 11th Std. Business Mathematics and Statistics, , 05_11th_BM-STAT_Ch-5-EM.indd 102, , 21-04-2020 12:12:15 PM

Page 109 :
00 -, , www.tntextbooks.in, , Solution, Let ‘x’ be the number of students, who participate the tour., , Plot the points (-2, 2), (-1, 1), (0, 0),, (1, 1), (2, 2) and draw a smooth curve., The graph is as shown in the figure, y, , Then 35 < x < 50 and x is a positive, integer., Formula :, , Total cost = (cost per student), × (number of students), , (i) If the number of students are, between 35 and 45, then, the cost per student is ` 200, , 2, , (-2, 2), , 1, , (-1, 1), , xl, , –2, , (2, 2), (1, 1), , (0, 0) 1, , –1, , x, , 2, , yl, , Fig. 5.4, , Example 5.11, , ` The total cost is y = 200x., , (ii) If the number of students are, between 46 and 50, then, , Draw the graph of the function, f(x) = x2- 5, , Solution, x, 1, the cost per student is ` $200 - 5 (x - 45) . = 209 - 5, Let y = f(x) = x2- 5, x, , 1 (x 45) . = 209 5, 5, , ` The total cost is y = `209 -, , x2, 209x - 5, , xj, x=, 5, , Choose suitable values for x and, determine y., Thus we get the following table., , 200x, ; 35 # x # 45, 2, and, ` y=*, x, 209x ; 46 # x # 50, 5, , the domain is {35, 36, 37, ..., 50}., Example 5.10, , x, , -3, , -2, , -1, , 0, , 1, , 2, , 3, , y, , 4, , -1, , -4, , -5, , -4, , -1, , 4, , Table : 5.2, Plot the points (-3, 4), (-2, -1),, (-1, -4), (0, -5), ( 1, -4), (2, -1), (3, 4), and draw a smooth curve., , Draw the graph of the function, f(x) = |x|, , The graph is as shown in the figure, , Solution, y = f(x) = |x|, , y, , =), , x if x $ 0, - x if x < 0, , (-3, 4), , Choose suitable values for x and, determine y., Thus we get the following table., x, y, , -2, 2, , -1, 1, , 0, 0, , Table : 5.1, , 1, 1, , 2, 2, , xl, , (3, 4), , (-2, -1), (-1, -4), (0, -5), , (2, -1), , x, , (1, -4), , yl, , Fig 5.5, Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 103, , 103, , 21-04-2020 12:12:16 PM

Page 110 :
www.tntextbooks.in, , Example 5.12, , y=ax, 0<a<1, , Draw the graph of f(x) = ax, a ! 1, and a > 0, Solution, We know that, domain set is R,, range set is (0, ∞) and the curve passing, through the point (0, 1), , (0, l), , xl, , y = f(x) =, , x, , Fig 5.7, , < 1 if x < 0, = *= 1 if x = 0, > 1 if x > 0, , NOTE, , We noticed that as x increases,, y is also increases and y > 0., ` The graph is as shown in the, , figure., y, , O, , yl, , Case (i) when a > 1, , ax, , y, , (i) The value of e lies between, 2 and 3. i.e. 2 < e < 3, (ii) The graph of f(x) = ex is identical to, that of f(x) = ax if a > 1 and, the graph of f(x) = e–x is identical to, that of f(x) = ax if 0 < a < 1, , y=ax, a>1, , Example 5.13, (0, l), o, , xl, , x, , Draw the graph of f(x) = logax;, x > 0, a > 0 and a ! 1., Solution, , yl, Fig 5.6, , We know that, domain set is (0, ∞),, range set is R and the curve passing the, point (1, 0), , Case (ii) when 0 < a < 1, , Case (i) when a > 1 and x > 0, , > 1 if x < 0, x, y = f(x) = a = *= 1 if x = 0, < 1 if x > 0, , < 0 if 0 < x < 1, f (x) = loga x = *= 0, if x = 1, >0, if x > 1, , We noticed that as x increases,, y decreases and y > 0 ., , We noticed that as x increases, f(x), is also increases., , ` The graph is as shown in the, , ` The graph is as shown in the, , figure., 104, , figure., 11th Std. Business Mathematics and Statistics, , 05_11th_BM-STAT_Ch-5-EM.indd 104, , 21-04-2020 12:12:17 PM

Page 111 :
www.tntextbooks.in, , y, , (i), , f(x)= logax, a>1, , (ii), O, , xl, , (1, 0), , x, , yl, , > 0 if 0 < x < 1, f (x) = loga x = *= 0, if x = 1, <0, if x > 1, We noticed that x increases, the, value of f(x) decreases., ` The graph is as shown in the, , f(x)= logax, 0<a<1, , xl, , O, , (1, 0), , x, , yl, Fig 5.9, , Exercise 5.1, 1., , Determine whether the following, functions are odd or even?, , (iii), , f(x) = sin x + cos x, , (iv), , f(x) = x2- |x|, , (v), , f(x) = x + x2, , Let f be defined by f(x) = x3- kx2 + 2x,, x ∈ R. Find k, if ‘f ’ is an odd function., , 3., , If f (x) = x3 -, , 1, , x ≠ 0, then show, x3, , that f (x) + f` 1x j = 0, 4., , +, If f (x) = x - 1 , x ≠ 1, then prove that, x 1, , 5., , x-1, For f(x), f (x)== 3x + 1 , x > 1, write the, , f^ f (x)h = x, , expressions of f` 1x j and f 1(x), , figure., y, , x2 + 1 i, , 2., , Fig 5.8, , Case (ii) when 0 < a < 1 and x > 0, ,  a x −1 , f(x) =  x ,  a +1 , f ]xg = log _ x2 +, , 6., , 7., , If f (x) = e x and g (x) = loge x , then, find, (i) (f+g)(1) , , (ii) (fg)(1), , (iii) (3f)(1) , , (iv) (5g)(1), , Draw the graph of the following, functions:, (i), , f (x) = 16 - x2, , (ii), , f (x) = | x - 2 |, , (iii), , f (x) = x | x |, , (iv), , f (x) = e2x, , (v), , f (x) = e-2x, , (vi), , f (x) =, , |x |, x, Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 105, , 105, , 21-04-2020 12:12:20 PM

Page 112 :
www.tntextbooks.in, , 5.2 Limits and Derivatives, In this section, we will discuss about, limits, continuity of a function, differentiability, and differentiation from first principle. Limits, are used when we have to find the value of a, function near to some value., , 5.2.2 Algorithm of left hand limit:, L[f(x)]x=a, , (i), , Write lim- f(x)., x"a, , (ii) Put x=a – h, h > 0 and replace, x " a– by h " 0., (iii) Obtain lim f(a–h)., h"0, , Definition 5.1, Let f be a real valued function of x., Let l and a be two fixed numbers. If f(x), approaches the value l as x approaches, to a, then we say that l is the limit of the, function f(x) as x tends to a, and written, as, lim f(x) = l, , x"a, , Notations:, (i) L[f(x)]x=a = lim- f(x) is known as, x"a, , left hand limit of f(x) at x=a., (ii) R[f(x)]x=a = lim+ f(x) is known as, x"a, , right hand limit of f(x) at x=a., , 5.2.1 Existence of limit, lim f(x) exist if and only if both, , x"a, , L[f(x)]x=a and R[f(x)]x=a exist and are, equal., i.e., lim f(x) exist if and only if, x"a, , L[f(x)]x=a = R[f(x)]x=a ., , (iv) The value obtained in step (iii) is, called as the value of left hand limit, of the function f(x) at x=a., , 5.2.3 Algorithm of right hand limit:, R[f(x)]x=a, , (i), , Write lim+ f(x)., x"a, , (ii) Put x = a+h, h>0 and replace x " a+, by h " 0., (iii), , Obtain lim f(a+h)., h"0, , (iv) The value obtained in step (iii) is, called as the value of right hand, limit of the function f(x) at x=a., Example 5.14, Evaluate the left hand and right, hand limits of the function, f(x) =, , *, , x- 3, if x ! 3, x- 3, 0 if x = 3, , at x = 3, , Solution, L[f(x)]x=3 = lim f(x), x"3, , = lim f(3 – h), x = 3 – h, h"0, , NOTE, L[f(x)]x=a and R[f(x)]x=a are shortly, written as L[f(a)] and R[f(a)], , (3 - h) - 3, = hlim, " 0 (3 - h) - 3, , h, = lim h"0, h, -, , = lim -h, h"0, h, 106, , 11th Std. Business Mathematics and Statistics, , 05_11th_BM-STAT_Ch-5-EM.indd 106, , 21-04-2020 12:12:26 PM

Page 113 :
www.tntextbooks.in, , = lim –1 = –1, h"0, , NOTE, , R[f(x)]x=3 = lim+ f(x), x"3, , = lim f(3 + h), h"0, , (3 + h) - 3, = hlim, " 0 (3 + h) - 3, , Let a be a point and f(x) be a function,, then the following stages may happen, (i) , lim f(x) exists but f(a) does not, x"a, , exist., , h, = hlim, "0 h, , (ii) The value of f(a) exists but lim f(x), x"a, does not exist., , = hlim, 1=1, "0, , (iii) Both lim f(x) and f(a) exist but are, x"a, , unequal., , NOTE, , (iv) Both lim f(x) and f(a) exist and are, x"a, , Here,, , equal., , L[f(3)] ! R[f(3)], , 5.2.4 Some results of limits, , ` lim f(x) does not exist., x"3, , If lim f(x) and lim g(x) exists, then, x"a, , Example 5.15, Verify the existence of the function, 5x - 4 if 0 1 x # 1, f(x) = ) 3, 4x - 3x if 1 1 x 1 2, , at x = 1., , Solution, L[f(x)]x=1 = lim- f(x), , lim f (x), , (i), (ii), , x"a, , lim (f+g)(x) = lim f(x) + lim g(x), , x"a, , x"a, , x"a, , lim (fg)(x) = lim f(x) lim g(x), , x"a, , (iii) , lim c, x"a, , x"a, , x"a, , lim f (x), f, m^ x h = x " a, whenever lim g (x) ! 0, g, lim g (x) , whenever, x"a, x"a, , x"a, , g (x) ! 0, lim g (x) , whenever xlim, "a, x"a, = lim f(1 – h), x = 1 – h, h"0, (iv) , lim k f(x) = k lim f(x), where k is, x"a, x"a, = lim [5(1–h)–4], a constant., x"1, , h"0, , = lim (1–5h) = 1, h"0, , R[f(x)]x=1 = lim+ f(x), x"1, , = lim f(1 + h), x = 1 + h, h"0, , = lim [4(1 + h)3 – 3(1 + h)], h"0, , = 4(1)3 – 3(1) = 1, Clearly, L[f(1)] = R[f(1)], ` lim f(x) exists and equal to 1, x"1, , 5.2.5 Indeterminate forms and, evaluation of limits, Let f(x) and g(x) are two functions, in which the limits exist., f^ a h, , If lim f(x) = lim g(x) = 0, then g^ a h takes 00, x"a, x"a, , f^ a h, 0, takes, takes form which is meaningless but does, 0, g^ a h, f_ xi, not imply that lim, is meaningless., x " a g_ xi, Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 107, , 107, , 21-04-2020 12:12:31 PM

Page 114 :
www.tntextbooks.in, , In many cases, this limit exists and, Example 5.17, have a finite value. Finding the solution, x2 - 4x + 6, Evaluate: lim x + 2, of such limits are called evaluation of the, x"2, indeterminate form., Solution, 0 3 #, lim ^ x2 - 4x + 6h, The indeterminate forms are 0 , 3 , 0 3, 3 - 32, 00, 3+0 and 13, x 4x 6, 0 3 #, = x"2, lim x + 2, 00, 3 00 and 1 3, 3, ,, ,, 0, 3, ,, 3, 3, ,, 0, ., lim ^ x + 2h, x"2, 0 3, x"2, Among all these indeterminate forms,, ^2 h2 - 4 ^2 h + 6 1, 0, =2, =, is the fundamental one., 2+2, 0, , 5.2.6 Methods of evaluation of, algebraic limits, (i), , 5 sin 2x - 2 cos 2x, Evaluate : lim 3 cos 2x + 2 sin 2x, r, , Direct substitution, , (ii), , Factorization, , (iii), , Rationalisation, , (iv), , By using some standard limits, , 5.2.7 Some standard limits, n, , n, , x" 4, , Solution, 5 sin 2x - 2 cos 2x, lim 3 cos 2x + 2 sin 2x, , r, x" 4, , lim ^5 sin 2x - 2 cos 2x h, , r, x" 4, , x -a, lim x - a = nan - 1 if n ! Q, x"a, , =, , sin i, tan i, = lim, = 1, q is in, i"0 i, i"0 i, radian., ax - 1, lim x = logea if a > 0, x"0, , r, r, 5 sin 2 - 2 cos 2, =, r, r, 3 cos 2 + 2 sin 2, , (i), , (ii) lim, (iii), , Example 5.18, , log (1 + x), =1, (iv) lim, x, x"0, (v), , lim ^3 cos 2x + 2 sin 2x h, , r, x" 4, , =, , 5 ^1 h - 2 ^0 h 5, =, 3 ^0 h + 2 ^1 h 2, , lim ^1 + x h x = e, , Example 5.19, , ex - 1, x =1, x"0, , Solution, , 1, , x"0, , 1 x, (vi) lim `1 + x j = e, x"3, (vii) lim, , Example 5.16, 2, , Evaluate: lim (3x + 4x - 5) ., x"1, , Solution, lim (3x2 + 4x - 5), , x3 - 1, Evaluate: lim x - 1, x"1, x3 - 1, lim x - 1 is of the type 00, x"1, ^ x - 1h^ x 2 + x + 1h, x3 - 1, lim x - 1 = lim, ^ x - 1h, x"1, x"1, , = lim ^ x2 + x + 1h, x"1, , x"1, , = (1)2 + 1 + 1, , = 3(1)2 + 4(1) – 5 = 2, , =3, , 108, , 11th Std. Business Mathematics and Statistics, , 05_11th_BM-STAT_Ch-5-EM.indd 108, , 21-04-2020 12:12:36 PM

Page 116 :
www.tntextbooks.in, , = lim ', x"0, , log ^1 + x3h, x3 1, #, x3, sin3 x, , log ^1 + x3h, 1, #, 3, x 3, sin, x, x"0, lim ` x j, , = lim, , x"0, , = 1 # 11 = 1, , Evaluate the following, x3 + 2, (i) lim x + 1, x"2, (ii) lim, , x"3, , (iii) lim, , x"3, , /n, , A function f(x) is continuous at, x = a if, (i) f(a) exists, , n2, , (ii) lim f(x) exists, , 1 +x - 1 -x, x, , (iii) lim f(x) = f(a), , x"0, 5, , x"a, , 5, , x8 - a8, (v) lim 2, 2, x"ax3 - a3, 2, , sin 3x, x2, x"0, , x9 + a9 =, lim, lim ^ x + 6h, then find, 2. If, x " –a x + a, x "3, the values of a., xn - 2n, 3. If lim x - 2 = 448, then find the, x"2, least positive integer n., 4. If f ^ x h =, , If a function is continuous at all the, point in an interval, then it is said to be a, continuous in that interval., , 5.2.8 Continuous function, , 2x + 5, 2, x + 3x + 9, , (iv) lim, , (vi) lim, , Before going into the topic, let us, first discuss about the continuity of a, function. A function f(x) is continues at, x = a if its graph has no break at x = a, If there is a break at the point, x = a then we say that the function is not, continuous at the point x = a., , Exercise 5.2, 1., , Derivative, , x7 - 128, , then find lim f ^ x h, x5 - 32, x"2, , x"a, , Observation, In the above statement, if atleast, any one condition is not satisfied at a, point x = a by the function f(x), then it, is said to be a discontinuous function at, x=a, , 5.2.9 Some properties of continuous, functions, If f(x) and g(x) are two real valued, continuous functions at x = a , then, (i) , f(x) + g(x) is also continuous at x = a., , ax + b, 5. Let f ^ x h = x + 1 , If lim f ^ x h = 2 and lim f ^(ii) , x h =f1(x), , g(x) is also continuous at x = a., x"0, x"3, (iii) kf(x) is also continuous at x = a,, , If lim f ^ x h = 2 and lim f ^ x h = 1, then show that f(–2) = 0., x"0, x"3, k be a real number., 110, , 11th Std. Business Mathematics and Statistics, , 05_11th_BM-STAT_Ch-5-EM.indd 110, , 21-04-2020 12:12:44 PM

Page 117 :
www.tntextbooks.in, , (iv) f^1x h is also continuous at x = a,, if f(a) ] 0., , f^ x h, (v) g^ x h is also continuous at x = a,, , Solution, L 7 f ]xgAx = 2 = lim- f ]xg, x"2, , = lim f ]2 - hg, x = 2 – h, h"0, , if g(a) ] 0., , = lim "2 - ]2 - hg,, , (vi)  f^ x h is also continuous at x = a., , h$0, , = lim h = 0, , Example 5.27, , h"0, , Show that, 5x - 4, if 0 1 x # 1, f(x) = ) 3, is, 4x - 3x, if 1 1 x < 2, continuous at x = 1, , R 7 f ]xgAx = 2 = lim+ f ]xg, x"2, , = lim f ]2 + hg , x = 2 + h, h"0, , = lim "2 + ]2 + hg,, , Solution, , h"0, , L 7 f ]xgAx = 1 = lim- f ]xg, , = lim ]4 + hg = 4, , x"1, , = lim f ]1 - hg, x =1–h, h"0, , h"0, , Now, f(2) = 2+2 = 4, , = lim 65 ]1 - hg - 4@, , Here lim - f ]x g ! lim + f ]x g, , = 5(1)–4 = 1, , Hence f(x) is not continuous at x = 2, , h"0, , R 7 f ]xgAx = 1 = lim+ f ]xg, , x$2, , x$2, , x"1, , Observations, , h"0, , (i) Constant function is everywhere, continuous., , = lim f ]1 + hg , x = 1+ h, , 74 ]1 + hg3 - 3 ]1 + hgA, = hlim, "0, , = 4(1)3 – 3(1), =4–3=1, Now, f(1) = 5(1) – 4 = 5–4 =1, , (ii) Identity function is everywhere, continuous., (iii) Polynomial function is everywhere, continuous., , ` lim- f ]xg = lim+ f ]xg = f(1),, , (iv) Modulus function is everywhere, continuous., , f(x) is continuous at x = 1, , (v) Exponential function ax, a>0 is, everywhere continuous., , x"1, , x"1, , Example 5.28, Verify the continuity of the function, f(x) given by, , f] x g = ), , 2 - x if x < 2, 2 + x if x $ 2, , at x = 2, , (vi) Logarithmic function is continuous, in its domain., (vii) Rational function is continuous at, every point in its domain., Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 111, , 111, , 21-04-2020 12:12:50 PM

Page 118 :
www.tntextbooks.in, , Z]a (x) if x < a, ]], ], If f(x) = [] k if x = a , then, ]], ] b (x) if x > a, \, L[f(x)]x=a = lim- f (x) = lim a (x), x"a, , x"a, , 5.2.11 L eft hand derivative and right, hand derivative, (i) , limx"c, , f ]c - hg - f ] c g, , is called the left, lim, -h, h"0, , hand derivative of f(x) at x = c and, it denoted by f l^c -h or L[ f l (c)]., , R[f(x)]x=a = lim+ f (x) = lim b (x), x"a, , x"a, , It is applicable only when the function, has different definition on both sides of, the given point x = a., , f ]xg - f ] c g, or, x-c, , (ii), , lim+, , x"c, , f ]xg - f ] c g, or, x-c, , f]c + hg - f] c g, is called the right, , lim, h, h"0, , Exercise 5.3, 1. Examine the following functions, for continuity at indicated points, x2 - 4, (a) f(x) = * x - 2 , if x ! 2 at x = 2, 0,, if x = 2, x2 - 9, (b) f(x) = * x - 3 , if x ! 3 at x = 3, 6,, if x = 3, 2. Show that f(x) = | x | is continuous, at x = 0, , hand derivative of f(x) at x = c and it, denoted by f l^c +h or R[ f l (c)]., , Result, , f(x) is differentiable at, x = c , L 7 f l] c gA = R 7 f l] c gA, , Remarks:, , (i) If L 7 f l] c gA ! R 7 f l] c gA , then we say, that f(x) is not differentiable at x=c., (ii) f(x) is differentiable at x = c ⇒ f(x), is continuous at x = c., , 5.2.10 Differentiability at a point, Let f(x) be a real valued function, defined on an open interval (a, b) and let, c ! (a, b). f(x) is said to be differentiable, or derivable at x = c if and only if, f ]x g - f ] c g, exists finitely., lim x - c, x"c, This limit is called the derivative (or), differential co-efficient of the function f(x), at x = c and is denoted by, d, (or) : dx _ f] x giD ., x=c, Thus f ′(c) = lim, x →c, , 112, , 11th, , 05_11th_BM-STAT_Ch-5-EM.indd 112, , (or) D f(c), , f ( x ) − f (c ), x−c, , A function may be continuous, at a point but may not be, differentiable at that point., Example 5.29, Show that the function f(x) = | x | is, not differentiable at x = 0., Solution, f(x) = | x | = ), , x if x $ 0, - x if x < 0, , L[ f l (0)] = lim- f] x g - f]0 g ,, x"0, x-0, , Std. Business Mathematics and Statistics, , 21-04-2020 12:12:54 PM

Page 119 :
www.tntextbooks.in, , f ]0 - hg - f ]0 g, 0 - h - 0 , x = 0–h, h"0, f ]- h g - f ] 0 g, = lim, -h, h"0, -h - 0, = lim, -h, h"0, , 1 + h 2 - 2h - 1, -h, h"0, , = lim, , = lim, , -h, = lim - h, h"0, , = lim ]- h + 2g = 2, h"0, , R[ f l (1)] = lim, +, x"1, , = lim, , h, = lim - h, h"0, , h"0, , h"0, , = lim, , h"0, , = lim, , 1 + h2 + 2h - 1, h, h"0, , = lim, , f]0 + hg - f]0 g, 0 + h - 0 , x = 0+h, , f ]hg - f ]0 g, h, , h - 0, = lim, h, h"0, h, = lim h, h"0, h, = lim h, h"0, = lim 1 = 1, , f]1 + hg - f]1 g, 1+h-1 ,x=1+h, , ]1 + hg2 - 1, h, h"0, , = lim ]- 1g = –1, h"0, f ] x g - f ]0 g, R[ f l (0)] = lim+ x - 0, x"0, = lim, , f ]x g - f ]1 g, x-1, , = lim ]h + 2g = 2, h"0, , Here, L[ f΄(1)] = R[ f΄(1)], ` f(x) is differentiable at x = 1 and f΄(1) = 2, , 5.2.12 Differentiation from first, principle, The process of finding the derivative, of a function by using the definition that, , Example 5.30, , f ]x + h g - f ] x g, is called, h, h$0, the differentiation from first principle. It, dy, is convenient to write f l (x) as dx ., , Show that f(x) =x2 differentiable at, x=1 and find f l (1)., , 5.2.13 Derivatives of some standard, functions using first principle, , Solution, , 1., , h"0, , Here, L[ f l (0)] ! R[ f l (0)], ` f(x) is not differentiable at x=0, , f(x) = x2, , f ]x g - f ]1 g, L[ f l (1)] = lim, x-1, x " 1f]1 - hg - f]1 g, = lim 1 - h - 1 , x=1–h, h"0, , ]1 - hg2 - 1, -h, h"0, , = lim, , l]x g = lim, ff΄(x), , d, For x ! R, dx ^ xnh = n xn - 1, Proof:, Let f(x) = xn, ` f(x+h) = (x+h)n, , f]x + hg - f] x g, d _ f] x gi, =, lim, dx, h, h"0, Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 113, , 113, , 21-04-2020 12:13:00 PM

Page 129 :
www.tntextbooks.in, , 1 1, $, t2 2at, 1, = - 3, 2at, (iii), y = x sin x, = -, , 5. If y = ^ x + 1 + x 2 h , then show, m, , that ^1 + x 2h y2 + xy1 - m 2 y = 0, , 6. If y = sin ^log x h , then show that, x 2 y2 + xy1 + y = 0 ., , dy, dx = x cos x + sin x, , Exercise 5.10, , 2, , d y, = –x sin x + cos x + cos x, dx2, = 2 cos x – x sin x, , Choose the correct answer, 1., , Example 5.47, If y = A sin x + B cos x, then prove, that y2 + y = 0, Solution, , y = A sin x + B cos x, , 2., , y2 = –A sin x –B cos x, y2 + y = 0, Exercise 5.9, 1., , 3., , 4., , (i) y = e3x + 2, 5., , (iii) x = a cos i, y = a sin i, , 4. If y = a cos mx + b sin mx , then show, that y2 + m 2 y = 0, , (c) 1, , (d) x 2 + x + 1, , If f ^ x h = *, , x2 - 4x if x $ 2, , then f(5) is, x + 2 if x 1 2, (b) 2, (d) 7, , x2 - 4x if x $ 2, , then f(0) is, If f (x) = *, x + 2 if x < 2, (b) 5, (d) 0, , 1-x, If f ^ x h = 1 + x , x > 1, then f(–x) is, equal to, 1, (a ) –f(x), (b), f^xh, 1, (c) (d) f(x), f^xh, The graph of the line y = 3 is, (a) Parallel to x-axis, (b) Parallel to y-axis, , 2. If y = 500e7x + 600e -7x , then show, that y2 - 49y = 0, 3. If y = 2 + log x , then show that, xy2 + y1 = 0, , (b) x, , (a) 2, (c) –1, , Find y2 for the following functions, , (ii) y = log x + a x, , (a) x2, , (a) –1, (c) 5, , y1 = A cos x – B sin x, y2 = –y, , If f ^ x h = x 2 - x + 1 , then f ^ x + 1h is, , (c) Passing through the origin, (d) Perpendicular to x-axis, 6., , The graph of y = 2x2 is passing, through, (a) (0,0), (c) (2,0), , (b) (2,1), (d) (0,2), Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 123, , 123, , 21-04-2020 12:13:48 PM

Page 130 :
www.tntextbooks.in, , 7., , The graph of y = e x intersect the, y-axis at, (a) (0, 0), (c) (0, 1), , 8., , (b) (1, 0), (d) (1, 1), , The minimum value of the function, f(x)=| x | is, (a) 0, (c) +1, , 9., , 10., , 11., , 2, , x +1, x, , 1, If f ^ x h = 2 x and g(x) = x , then, 2, (fg)(x) is, (a) 1, , (b) 0, , (c) 4x, , (d), , Which of the following function is, neither even nor odd?, (a) f ^ x h = x3 + 5, (c) f ^ x h = x10, , 12., , 1, 4x, , (b) f ^ x h = x5, , (d) f ^ x h = x2, , 14., , (d) constant function, 124, , The graph of f(x) = ex is identical to, that of, (b) f ^ x h = a x, a 1 1, , (c) f ^ x h = a x, 0 1 a 1 1, (d) y = ax + b, a ! 0, 15., , 16., , 17., , If f ^ x h = x2 and g(x) = 2x+1, then, (fg)(0) is, (a) 0, (c) 1, , (b) 2, (d) 4, , tan i, =, i"0 i, (a) 1, (c) –∞, , (b) ∞, (d) θ, , lim, , ex - 1, x, x"0, lim, , =, (b) nxn-1, (d) 0, , (a) e, (c) 1, 18., , x+2, For what value of x, f ^ x h = x - 1 is, not continuous?, (a) –2, (c) 2, , 19., , (b) 1, (d) –1, , If the function f(x) is continuous at, x = a, then lim f (x) is equal to, x"a, , (a) f(–a), (c) 2f(a), , (a) an identity function, , (c) exponential function, , (b) [0, ∞), (d) [1, ∞), , (a) f ^ x h = a x, a 2 1, , f ^ x h =- 5 , for all x ! R is a, (b) modulus function, , The range of f ^ x h = x , for all, x ! R is, (a) (0, ∞), (c) (–∞, ∞), , (b) –1, (d) –∞, , Which one of the following, functions, has, the, property, 1, f ^ x h = f ` x j , provided x ≠ 0, x2 - 1, (a) f ^ x h = x, 1 - x2, (b) f ^ x h = x, (c) f(x) = x, (d) f ^ x h =, , 13., , 20., , 1, (b) f ` a j, (d) f(a), , d 1, dx ` x j is equal to, 1, 1, (b) - x, (a) - 2, x, 1, (d) 2, (c) log x, x, , 11th Std. Business Mathematics and Statistics, , 05_11th_BM-STAT_Ch-5-EM.indd 124, , 21-04-2020 12:13:53 PM

Page 131 :
www.tntextbooks.in, , 21., , Miscellaneous Problems, , d ^ x, h, dx 5e - 2 log x is equal to, (a) 5e x - 2x, , If, , 2., , Draw the graph of y = 9–x2, , (b) 5e x - 2x, (c) 5e x - 1x, (d) 2 log x, 22., , (a) x2, (c), 23., , 1, (d) - 2, x, , –x2, , If y =, , e2x,then, , (c) 25., , 4., , d2 y, at x = 0 is, dx2, (b) 9, (d) 0, , (b) 2, x2, , Evaluate: lim, , ^2x - 3h^ x - 1h, , 5., 6., , Verify the continuity and differentiability, , x"1, , ]Z]1 - x, if x < 1, ]], of f(x) = [](1 - x)(2 - x) if 1 # x # 2 at, ]], if x > 2, ]3 - x, \, , x = 1 and x = 2, , 1, x2, , 7., , (d) e2, , d ^ xh, dx a =, 1, , if x ! 0, then show, 2, if x = 0, that lim f ^ x h does not exist., , 2x 2 + x - 3, Show that the function f(x) = 2x–| x |, is continuous at x = 0, , If y = log x,then y2 =, 1, (a) x, , If f(x)= *, , x- x, x, , x"0, , (b) 1, , (a) 4, (c) 2, 24., , 3., , dy, 1, If y = x and z = x ,then dz =, , 1, f ^ x h = 2x + 1 , x > - 12 , then, 2x + 1, show that f(f(x)) = 2x + 3 ., , 1., , 8., 9., , aa, , (a) x log a, e, , (b), , (c) x loge a, , (d) a x loge a, , 10., , If x y = y x , then prove that, dy, y x log y - y, d, n, =, x, dx, y log x - x, dy, If xy2 = 1 , then prove that 2 dx + y3 =0., If y =tanx, then prove that y2 - 2yy1 = 0., If y = 2 sin x + 3 cos x , then show, that y2 + y = 0., , Summary, , zz, zz, zz, zz, zz, zz, , Let A and B be two non empty sets, then a function f from A to B,, associates every element of A to an unique element of B., lim f ]xg exists + lim f ]xg = lim f (x) ., x"a, , x " a-, , x " a+, , For the function f(x) and a real number a, lim f ]xg and f(a) may not be the same., x"a, , A function f(x) is a continuous function at x = a if only if lim f ]xg = f(a)., x"a, A function f(x) is continuous, if it is continuous at every point of its domain., , If f(x) and g(x) are continuous on their common domain, then f ± g, f $ g, kf (k is, f, a constant) are continuous and if g ! 0 then g is also continuous., Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 125, , 125, , 21-04-2020 12:14:00 PM

Page 133 :
www.tntextbooks.in, , ICT Corner, Expected final outcomes, Step – 1, Open the Browser and type the URL given (or) Scan, the QR Code., GeoGebra Work book called “11th BUSINESS, MATHEMATICS and STATISTICS” will appear. In, this several work sheets for Business Maths are given, Open the worksheet named “Elementary, Differentiation”, Step – 2, Elementary Differentiation and Functions page will open. Click on the function check boxes on, the Right-hand side to see the respective graph. You can move the sliders a, b, c and d to change, the co-efficient and work out the differentiation. Then click the Derivatives check box to see the, answer and respective graphs, Step 1, , Step 2, Step 4, , Browse in the link, 11th Business Mathematics and Statistics: https://ggbm.at/qKj9gSTG (or), scan the QR Code, Differential Calculus, , 05_11th_BM-STAT_Ch-5-EM.indd 127, , 127, , 21-04-2020 12:14:02 PM

Page 134 :
Chapter, , www.tntextbooks.in, , 6, , APPLICATIONS OF DIFFERENTIATION, , Learning Objectives, After studying this chapter, the students, will be able to understand, , •, •, , the concepts of demand and supply, , •, •, •, , average and marginal concepts, , •, •, •, •, •, •, , meaning and uses of cost, revenue, and profit function, elasticity of demand and supply, relationship, among, average, revenue, marginal revenue and, elasticity of demand, application of increasing and, decreasing functions, application of maxima and minima, c onc ept of E c onom ic O rd e r, Quantity, concepts of partial differentiation, applications of partial derivatives, in Economics, partial elasticities of demand, , Introduction, Modern economic, theory is based on, both differential and, integral calculus. In, economics, differential, calculus is used to, Leonhard Euler, compute marginal cost,, marginal revenue, maxima and minima,, elasticities, partial elasticities and also, enabling economists to predict maximum, profit (or) minimum loss in a specific, condition. In this chapter, we will study, 128, , about some important, concepts and applications, of, differentiation, in, business and economics., , Euler’s theorem in Economics, If factors of production function, are paid as factors times of their, marginal productivities, then the total, factor payment is equal to the degree, of homogeneity times the production, function., , 6.1 Applications of Differentiation, in Business and Economics, , In an economic situation, consider, the variables are price and quantity. Let, p be the unit price in rupees and x be, the production (output / quantity) of a, commodity demanded by the consumer, (or) supplied by the producer., , 6.1.1 D,  emand, supply, cost, revenue, and profit functions, Demand function, In a market, the quantity of a, commodity demanded by the consumer, depends on its price. If the price of the, commodity increases, then the demand, decreases and if the price of the commodity, decreases, then the demand increases., The relationship between the quantity, and the unit price of a commodity demanded, by consumer is called as demand function, , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 128, , 21-04-2020 12:20:52 PM

Page 135 :
www.tntextbooks.in, , and is defined as x = f (p) or p = f (x ) ,, where x > 0 and p > 0 ., Graph of the demand function, x = f(p), x (demand), , x = f(p), , Demand curve, , o, , Obtuse, angle, Obtuse angle, \ slope is, Slope, is −𝑣𝑣𝑣𝑣, negative, , Supply function, In a market, the quantity of a, commodity supplied by producer depends, on its price. If the price of the commodity, increases, then quantity of supply increases, and if the price of the commodity decreases,, then quantity of supply decreases., The relationship between the quantity, and the unit price of a commodity supplied, by producer is called as supply function, and is defined as x = g(p) or p=g(x) where, x > 0 and p > 0., , The graph of the supply function, x= g(p), x (Supply), , p (Price), , Fig : 6.1, x = g(p), Supply curve, , The “demand - price relationship” curve illustrates the negative, relationship between price and, quantity demanded., , le, an g, angle, cute an, gle, AAcute, Acuteslope, 𝑣𝑣𝑣𝑣, s + is, pe i +𝑣𝑣𝑣𝑣, o, l, positive, SSlope is, , \, , o, , p (Price), , Fig : 6.2, , Observations, (i), , , Price and quantity of the demand, function are in inverse variation., , (ii) The graph of the demand function, lies only in first quadrant., (iii) Angle made by any tangent to the, demand curve with respect to the, positive direction of x – axis is always, an obtuse angle., (iv) Slope of the demand curve is, negative( –ve)., , The “supply - price relationship”, c ur ve i l lust r ate s t he p os it ive, relationship between price and, quantity supplied., , Observations, (i), , P,  rice and quantity of the supply, function are in direct variation., (ii) The graph of supply function lies, only in first quadrant., Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 129, , 129, , 21-04-2020 12:20:52 PM

Page 136 :
www.tntextbooks.in, , (iii) Angle made by any tangent to, the supply curve with respect to, positive direction of x – axis is, always an acute angle., (iv) , Slope of the supply curve is, positive (+ve)., , Diagrammatical explanation of, equilibrium price, equilibrium quantity, and equilibrium point, x (quantity), Demand, fuction, Equilibrium, quantity, , Supply, fuction, , Equilibrium, point, (pE , xE), , xE, , o, , The law of demand / supply tells, us the direction of change, but not the, rate at which the change takes place., , Equilibrium Price, The price at which the demand for a, commodity is equal to its supply is called as, Equilibrium Price and is denoted by pE., , Equilibrium Quantity, The quantity at which the demand, for a commodity is equal to its supply, is called as Equilibrium Quantity and is, denoted by xE., , Usually the demand and supply, functions are expressed as x in terms of, p, so the equilibrium quantity is obtained, either from the demand function (or), from the supply function by substituting, the equilibrium price., , Equilibrium Point, The point of intersection of the, demand and supply function (pE, xE) is, called as equilibrium point., 130, , p (Price), , Equilibrium, Price, , Fig: 6.3, , Average and Marginal concepts, Usually, the variation in the dependent, quantity ‘y’ with respect to the independent, quantity ‘x’ can be described in terms of two, concepts namely, (i) Average concept and, (ii) Marginal concept., , (i), , Average concept, , The average concept expressed as the, variation of y over a whole range of x and is, denoted by, , NOTE, , pE, , y, ., x, , (ii) Marginal concept, The marginal concept expressed as, the instantaneous rate of change of y with, respect to x and is denoted by, , dy, ., dx, , Remark:, If ∆x be the small change in x and, ∆y be the corresponding change in y of, the function y=f(x), then, f (x + ∆x ) − f (x ), ∆y, =, ∆x, ∆x, , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 130, , 21-04-2020 12:20:52 PM

Page 137 :
www.tntextbooks.in, , Fixed cost, , Instantaneous rate of change of y with, respect to x is defined as the limiting, case of ratio of the change in y to the, change in x., ∆y, , dy, , = lim, i.e. dx = ∆lim, x → 0 ∆x, ∆x → 0, , Fixed cost is the cost which does not, vary directly with the volume of production., If f(x) be the variable cost and k be, the fixed cost for production of x units, then, total cost is C(x) = f(x) + k, x > 0., , f (x + ∆x ) − f (x ), ∆x, , Y, , NOTE, Variable cost f(x) is a single valued, function., (ii) Fixed cost k is independent of the, level of output., (iii) f (x) does not contain constant term., (i), y, , y, , y = f(x), , {, , Some standard results, , {, , y + y, , x, , x x + x, , o, , X, , Fig : 6.4(a), , Y, y, , y, , o, , {, {, , y + y, , y = f(x), , x, , x x + x, , X, , Fig : 6.4(b), , Cost function, The amount spent for the production, of a commodity is called its cost function., Normally, total cost function [TC] consists, of two parts., (i) Variable cost; (ii) Fixed cost, , Variable cost, Variable cost is the cost which varies, almost in direct proportion to the volume of, production., , If C(x) = f(x) + k be the total cost, function, then, (i) Average cost:, f ]xg + k, Total Cost C ]xg, AC = Output = x =, x, (ii) Average variable cost:, f ]xg, Variable Cost, = x, AVC = Output, (iii) Average fixed cost:, Fixed Cost k, AFC = Output = x, (iv) Marginal cost:, dC, d, MC= dx = dx [C(x)]= C’(x), (v) Marginal average cost:, d, MAC = dx (AC), , (vi) Total cost:, TC = Average cost × output, , (vi) Average cost [AC] is minimum, when, MC = AC, , Remark:, The marginal cost [MC] is approximately, equal to the additional production cost, of (x+1)th unit, when the production, level is x units., Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 131, , 131, , 21-04-2020 12:20:54 PM

Page 138 :
www.tntextbooks.in, , Diagrammatical explanation of, marginal cost [MC], Marginal cost is the change in aggregate, cost when the volume of production is, increased or decreased by one unit., elgna e, , sutbO, Marginal, 𝑣𝑣𝑣𝑣−cost, si epol, S, , C(x+1), , t, , en, , ng, Ta, , dR, d, =, (R(x)) = R’ (x), dx, dx, , (iii) Marginal average revenue:, MAR =, , d, (AR) = AR’(x), dx, , Remarks:, C ¢ (x), , ost, arginal c, M, C(x+1)–C(x), , (i) A,  verage revenue [AR] and price [p], are the same. [i.e. AR=p], The marginal revenue [MR], (ii) , is approximately equal to the, additional revenue made on selling, of (x+1)th unit, when the sales level, is x units., , Diagrammatical explanation of, Marginal Revenue [MR], , y = C(x), , o, , MR =, , x, , x+1, , (output), , Fig: 6.5, , Revenue function, Revenue is the amount realised on a, commodity when it is produced and sold. If, x is the number of units produced and sold, and p is its unit price, then the total revenue, function R(x) is defined as R(x) =px, where, x and p are positive., , Marginal revenue is the change in, aggregate revenue when the volume of, selling unit is increased by one unit., elgna esutbO, Marginal, Revenue, 𝑣𝑣𝑣𝑣− si epolS, , (Revenue), , R(x+1), R(x), , e, in, tl, n, e, ng, Ta, , {, , C(x), , ne, , li, , {, , (cost), , (ii) Marginal revenue:, , R ¢ (x), , arginal cost, , M, R(x+1)–R(x), , y = R(x), , The revenue increases when the, Producer, supplies more quantity at a, higher price., , o, , x, , x+1, , (output), , Fig : 6.6, , Some standard results, If R(x) =px be the revenue function,, then, (i) Average revenue:, AR =, 132, , R (x ), Total Revenue, =, =p, Output, x, , Profit function, The excess of total revenue over the, total cost of production is called the profit. If, R(x) is the total revenue and C(x) is the total, cost, then profit function P(x) is defined as, P(x) = R(x) – C(x), , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 132, , 21-04-2020 12:20:55 PM

Page 139 :
www.tntextbooks.in, , Some standard results, , If P(x) = R(x) – C(x) be the profit, function, then, (i) Average profit:, Total Profit P (x), AP= Output = x, (ii) Marginal profit:, dP d, MP = dx = dx (P ]xg) = P'(x), (iii) Marginal average profit:, d, MAP = dx (AP) = AP ¢(x ), (iv) Profit [P(x)] is maximum when MR = MC, , 6.1.2, , (ii) Price elasticity of supply :, Price elasticity of supply is the degree, of responsiveness of quantity supplied to a, change in price., If x is supply and p is unit price of the, supply function x = g(p), then the elasticity, of supply with respect to the price is defined, p dx, as ηs = • ., x dp, NOTE, “price elasticity” is shortly called as, “elasticity”., , Elasticity, , Elasticity ‘η’ of the function y = f (x ), , at a point x is defined as the limiting case, of ratio of the relative change in y to the, relative change in x., \, , Þ, , ∆y, dy, E, y, y, η = y = lim, =, ∆, x, →, 0, Ex, ∆x, dx, x, x, x dy, η= ×, y dx, , Some important results on price, elasticity, (i), , If | η |>1, then the quantity demand or, supply is said to be elastic., , (ii) I f | η |=1, then the quantity demand or, supply is said to be unit elastic., (iii) I f | η |<1, then the quantity demand or, supply is said to be inelastic., , Remarks:, Marginal quantity of y, dy, with respect to x, η = dx =, Average quantity of y, y, with respect to x, x, , (i), , Price elasticity of demand, , Price elasticity of demand is the degree, of responsiveness of quantity demanded to a, change in price., If x is demand and p is unit price of, the demand function x = f(p), then the, elasticity of demand with respect to the price, is defined as η = − p • dx ., d, x dp, , (i) Elastic : A quantity demand or supply, is elastic when its quantity responds, greatly to changes in its price., Example: Cosumption of onion and, its price., (ii) I nelastic : A quantity demand or supply, is inelastic when its quantity responds, very little to changes in its price., Example: Consumption of rice and, its price., (iii) Unit elastic : A quantity demand or, supply is unit elastic when its quantity, responds as the same ratio as changes, in its price., Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 133, , 133, , 21-04-2020 12:20:56 PM

Page 140 :
www.tntextbooks.in, , Elasticity is predominantly used to, assess the change in consumer demand as, a result of a change in price., , Relationship among Marginal revenue, [MR], Average revenue [AR] and, Elasticity of demand[ηd]., We know that R(x)=px, i.e., R, , = px, , and ηd, , = − ⋅, , p dx, x dp, , d, Now, MR =, (R), dx, , =, =, =, =, , =, , d, (px ), dx, dp, p +x, dx, , x dp , p 1 + ⋅ , , p dx , , , 1 , p 1 +, , p dx , , ⋅, , x dp , , , 1, , p 1 −, , , p, dx, , − ⋅ , , x dp , , 1, p 1 − , hd , , , , , =, , i.e. MR, , =, , , 1, AR 1 −  (or), hd , , , ηd, , =, , AR, AR - MR, , Example 6.1, The total cost function for the, production of x units of an item is given by, C (x ) =, 134, , 1 3, x + 4x 2 − 25x + 7 ., 3, , Find, (i) Average cost function, (ii) Average variable cost function, (iii) Average fixed cost function, (iv) Marginal cost function and, (v) Marginal Average cost function, Solution:, C (x ) =, , 1 3, x + 4x 2 − 25x + 7, 3, , (i) Average cost:, AC =, , C, 1, 7, = x 2 + 4x − 25 +, x, 3, x, , (ii) Average variable cost:, AVC =, , f (x ), x, , =, , 1 2, x + 4x − 25, 3, , (iii) Average fixed cost:, AFC =, , k, 7, =, x, x, , (iv) Marginal cost:, , dC, d, (or) dx ^C ]xgh, dx, , d  1 3, x + 4x 2 − 25x + 7 , =, , dx  3, , 2, = x + 8x − 25, , MC =, , (v) Marginal Average cost:, MAC =, =, , d, d 1, 7, [AC] =  x 2 + 4x − 25 + , dx  3, x, dx, 2, 7, x +4− 2, 3, x, , Example 6.2, The total cost C in Rupees of making, x units of a product is C (x ) = 50 + 4x + 3 x ., Find the marginal cost of the product at 9, units of output., Solution:, C (x ) = 50 + 4x + 3 x, , Marginal cost (MC) =, , dC, d , =, C (x ), , dx, dx , , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 134, , 21-04-2020 12:20:58 PM

Page 141 :
www.tntextbooks.in, , =, , d, dx, , 50 + 4x + 3 x  = 4 + 3, , , 2 x, , When x = 9,, , =−, , dC, 1, 3, = 4 (or) ` 4.50, =4 +, 2, dx, 2 9, , p, , ,  20 , , , (p + 1), , ⋅, , −20, , (p + 1), , 2, , When p=3, ηd=, , =, , p, p +1, , 3, (or) 0.75, 4, , MC is ` 4.50 , when the level of output is, 9 units., , Here |ηd |<1, , Example 6.3, Find the equilibrium price and, equilibrium quantity for the following, demand and supply functions., , Example 6.5, Find the elasticity of supply for the supply, function x = 2 p 2 − 5 p + 1 , p> 3., , \, , Demand : x =, , 1, (5 − p ) and Supply : x = 2p–3., 2, , Solution:, At equilibrium, demand = supply, Þ, , 1, (5 − p ) = 2p–3, 2, , x = 2 p2 − 5 p + 1, dx, = 4p −5, dp, , =, , Þ p = 11, 5, 11 11, ∴ Equilibrium price: pE p=E `=, 5 5, 11, in x = 2 p − 3, Now, put p =, 5, 7,  11 , We get, x = 2   − 3 =, 5, 5, 7, ∴ Equilibrium quantity: xE = units., 5, , Example 6.4, 20, , p > 0,, For the demand function x =, p +1, , find the elasticity of demand with respect to, price at a point p = 3. Examine whether the, demand is elastic at p = 3., Solution:, dx, dp =, , Solution:, , Elasticity of supply: ηs=, , 5–p = 4 p − 6, , x=, , \ demand is inelastic., , p, 4 p2 − 5 p, ⋅, 4, −, 5, p, (, ), =, 2 p2 − 5 p + 1, 2 p2 − 5 p + 1, , Example 6.6, If y =, , 2x +1, , then obtain the elasticity at x = 1., 3x + 2, , Solution:, 2x +1, 3x + 2, dy ( 3 x + 2 )( 2 ) − ( 2 x + 1)( 3), 1, =, =, 2, 2, dx, ( 3x + 2 ), ( 3x + 2 ), y=, , x dy, x, 1, ⋅, =, ⋅, y dx  2 x + 1  ( 3 x + 2 )2, , ,  3x + 2 , x, =, ( 2 x + 1) ( 3x + 2 ), , Elasticity: η =, , , When x = 1, η =, , 20, p +1, −20, , (p + 1), , 2, , Elasticity of demand: ηd, , p dx, x dp, , =− ⋅, , p dx, ⋅, x dp, , 1, 15, , Example 6.7, A demand function is given by xp n = k, where n and k are constants. Prove that, elasticity of demand is always constant., Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 135, , 135, , 21-04-2020 12:21:15 PM

Page 142 :
www.tntextbooks.in, , Solution:, xp = k ⇒ x = kp, n, , When ηd = 1,, , −n, , 2(a − bx) = bx ⇒ output(x) =, , dx, = −nkp − n −1, dp, , p dx, x dp, , Elasticity of demand : ηd = − ⋅, = –-, , 2(a − bx), =1, bx, , p, (-nkp-n-1) = n , which is a constant., kp-n, , Example 6.8, For the given demand function p = 40–x,, find the output when ηd = 1, Solution:, p = 40–x ⇒ x = 40–p, , Example 6.10, Verify the relationship of elasticity of, demand, average revenue and marginal, revenue for the demand law p = 50 - 3x ., Solution:, p = 50 – 3x, dx, dp, 1, = –3 ⇒, =dx, dp, 3, p dx, x dp, , Elasticity of demand: ηd = − ⋅, , dx, = –1, dp, p dx, 40 - x, =, x dp, x, , 50 − 3 x  1  50 − 3 x, , −  =, 3x, x  3, , Elasticity of demand: ηd = − ⋅, , =−, , Given that ηd = 1, , Now, Revenue: R = px, , \, , 40 − x, = 1 ⇒ x = 20, x, , Average revenue: AR = p = 50 − 3x, , Example 6.9, Find the elasticity of demand in terms of x, , Marginal revenue: MR =, , 1, , for the demand law p = ( a − bx ) 2 . Also find, the output(x) when elasticity of demand is, unity., Solution:, , AR, AR − MR, , \, , 50 − 3 x, ( 50 − 3x ) − ( 50 − 6 x ), , =, , 50 − 3 x, , 3x, , From (1) and (2), we get, , Differentiating with respect to the price ‘p’ ,, , ηd =, , 1, , −1, 1, dx, ( a − bx ) 2 (−b) ⋅, 2, dp, 1, 2, , 2 (a − bx ), dx, =, −b, dp, , =−, , 136, , x, , ⋅, , 1, 2(a − bx) 2, , −b, , =, , … (2), , AR, , Hence verified., AR − MR, , Example 6.11, Find the elasticity of supply for the supply, p, when p = 20 and interpret, law x =, p+5, , your result., , Elasticity of η = − p ⋅ dx, d, demand:, x dp, 1, (a − bx) 2, , dR, = 50 − 6x, dx, , =, , p = ( a − bx ) 2 ., , \, , ... (1), , = ( 50 − 3x ) x = 50 x − 3 x 2, , , , \ output(x) = 20 units., , we get 1 =, , 2a, units., 3b, , Solution:, 2(a − bx), bx, , x=, , p, p+5, , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 136, , 21-04-2020 12:21:32 PM

Page 143 :
www.tntextbooks.in, , 5, ( p + 5) − p, dx, =, , 2, dp,  p  52, ( p + 5), Elasticity of supply: s, = ( p + 5), , 5, , ( p + 5), , 2, , =, , When p =20,  s =, , p dx, x dp, , 5, p+ 5, , Example 6.13, C = 0.05 x 2 + 16 +, , 100, is the manufacturer’s, x, , average cost function. What is the marginal, cost when 50 units are produced and, interpret your result., Solution:, , 5, = 0.2, 20 + 5, , Interpretation:, • I f the price increases by 1% from p = ` 20,, then the quantity of supply increases by, 0.2% approximately., , Total cost: C = AC × x = C ´ x, = 0.05 x3 + 16 x + 100, Marginal cost: MC =, , dC, = 0.15x2+16, dx, , 2,  dC , = 0.15 ( 50 ) + 16, , ,  dx  x =50, , • I f the price decreases by 1% from p = ` 20,, then the quantity of supply decreases by, 0.2% approximately., , = 375 + 16 = ` 391, , Example 6.12, , If the production level is increased by one, unit from x = 50, then the cost of additional, unit is approximately equal to ` 391., ,  x+5,  + 7 , prove,  x+2, , For the cost function C = 2 x , , that marginal cost (MC) falls continuously as, the output x increases., Solution:,  x+5, , C = 2x , +7 =,  x+2, , 2 x 2 + 10 x, +7, x+2, , Marginal cost:, MC =, , dC, , d  2 x + 10 x, =, +, 7, , , dx, dx  x + 2, , 2, , ( x + 2 ) ( 4 x + 10 ) − ( 2 x 2 + 10 x ), =, 2, ( x + 2), 2, 2 ( x 2 + 4 x + 10 ) 2 ( x + 2 ) + 6 , =, =, 2, 2, ( x + 2), ( x + 2), , 6 , = 2 1 +, , 2,  ( x + 2 ) , , ⇒ as x increases, MC decreases., Hence proved., , Interpretation:, , Example 6.14, For the function y = x3 + 19, find the values, of x when its marginal value is equal to 27., Solution:, y = x 3 + 19, dy, = 3x 2 , dx, , ... (1), , dy, = 27, dx, , ... (2) [Given], , From (1) and (2), we get, 3 x 2 = 27 ⇒ x = ±3, , Example 6.15, The demand, , function, , a, , Find the instantaneous rate of change of, demand with respect to price at p = 4. Also, interpret your result., Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 137, , for, , 4, commodity is p = ,where p is unit price., x, , 137, , 21-04-2020 12:21:44 PM

Page 144 :
www.tntextbooks.in, , Solution:, , Solution:, We know that average cost [AC] is minimum, when average cost [AC] = marginal cost, [MC]., , 4, 4, Þ x=, x, p, dx, 4, \, =− 2, dp, p, dx, 1, At p = 4,, = − = − 0.25, dp, 4, , p=, , Cost: C=, , 1 3, x − 3x 2 + 9x, 3, , \ Rate of change of demand with respect to, the price at p = ` 4 is − 0.25 ., , \ AC=, , Interpretation:, , Now, AC = MC Þ, , When the price increases by 1% from the, level of p = ` 4, the demand decreases (falls), by 0.25%., , Þ 2x 2 - 9x = 0 Þ x =, , Example 6.16, The demand and the cost function of a firm, are p = 497 – 0.2x and C = 25x + 10000, respectively. Find the output level and price, at which the profit is maximum., Solution:, We know that profit[P(x)] is maximum, when marginal revenue [MR] = marginal, cost [MC]., Revenue: R = px, , = (497–0.2x)x = 497x – 0.2x2, dR, MR = dx = 497 – 0 .4x, Cost: C = 25x+10000, \ MC = 25, , MR = MC Þ 497–0.4x = 25, , , Þ 472–0.4x = 0, , , , Þ x = 1180 units., , Now, p = 497–0.2x, at x = 1180, p = 497–0.2(1180) = ` 261., Example 6.17, 1, , The cost function of a firm is C = 3 x 3 − 3x 2 + 9x ., Find the level of output (x > 0) when average, cost is minimum., 138, , 1 2, x − 3x + 9 and MC = x 2 − 6x + 9, 3, 1 2, x − 3x + 9 = x 2 − 6x + 9, 3, , 9, units. [ x > 0], 2, , Exercise 6.1, 1. A firm produces x tonnes of output at a, total cost of C ( x) =, Find the, , 1 3, x − 4 x 2 − 20 x + 7 ., 10, , (i) average cost function, (ii), (iii), (iv), (v), , average variable cost function, average fixed cost function, marginal cost function and, marginal average cost function., , 2. The total cost of x units of output of a, 2, 3, , firm is given by C = x +, , 35, . Find the, 2, , (i) cost, when output is 4 units, , (ii) average cost, when output is 10 units, (iii) marginal cost, when output is 3 units,  evenue function ‘R’ and cost function, 3. R, ‘C’ are R = 14 x − x 2 and C = x ( x 2 − 2 ) ., Find the (i) average cost function (ii), marginal cost function, (iii) average, revenue function and (iv) marginal, revenue function., x, , 4. I f the demand law is given by p = 10e- 2 ,, then find the elasticity of demand., , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 138, , 21-04-2020 12:21:50 PM

Page 145 :
www.tntextbooks.in, , 5. F,  ind the elasticity of demand in terms of, x for the following demand laws and also, find the output (x), when the elasticity is, equal to unity, (ii) p = a − bx 2, , (i) p = (a − bx) 2, , 13. F,  ind the values of x, when the marginal, function of y= x3 + 10 x 2 − 48 x + 8 is twice, the x., 14. The total cost function y for x units is,  x+7,  + 5. Show that,  x+5, , given by y = 3x , , 6. F, ind the elasticity of supply for the, supply function x = 2 p 2 + 5 when p=3, 7. Th,  e demand curve of a commodity is, 50 − x, , given by p =, , find the marginal, 5, revenue for any output x and also find, marginal revenue at x = 0 and x = 25?, 8. Th,  e supply function of certain goods, is given by x = a p − b where p is unit, price, a and b are constants with p > b., Find the elasticity of supply at p = 2b., , , 1,  for the demand,  ηd , function p = 400 − 2 x − 3x 2 where p is unit, , 9. S how that MR = p 1 −, , price and x is quantity demand., 10. F,  or the demand function p = 550–3x–6x2, where x is quantity demand and p is unit, , 1, price. Show that MR = p 1 − ,  ηd , , the marginal cost [MC] decreases, continuously as the output (x) increases., 15. F,  ind the price elasticity of demand for, the demand function x = 10 – p where, x is the demand and p is the price., Examine whether the demand is elastic,, inelastic or unit elastic at p = 6., 16. F, , ind the equilibrium price and, equilibrium quantity for the following, functions., Demand: x =100 – 2p and supply: x = 3p –50, 17. Th,  e demand and cost functions of a firm, are x = 6000–30p and C = 72000+60x, respectively. Find the level of output and, price at which the profit is maximum., 18. T,  he co st f unc t i on of a f ir m is, C = x 3 − 12x 2 + 48x . Find the level of, output (x > 0) at which average cost is, , 25, 11. F,  or the demand function x = 4 , 1 ≤ p ≤5,, p, , determine the elasticity of demand., , 12. The demand function of a commodity is, p = 200 −, , x, and its cost is C=40x+120, 100, , where p is a unit price in rupees and x, is the number of units produced and, sold. Determine (i) profit function (ii), average profit at an output of 10 units, (iii) marginal profit at an output of 10, units and (iv) marginal average profit at, an output of 10 units., , minimum., , 6.2, , Maxima and Minima, , We are using maxima and minima in our, daily life as well as in every field such as, chemistry, physics, engineering and in, economics etc.,, In particular, we can use maxima and, minima, (i) T, o maximize the beneficial values like, profit, efficiency, output of a company etc.,, Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 139, , 139, , 21-04-2020 12:21:58 PM

Page 146 :
www.tntextbooks.in, , (ii) T,  o minimize the negative values like,, expenses, efforts etc.,, , Y, , (iii) U,  sed in the study of inventory control,, economic order quantity etc., , 6.2.1 I ncreasing and decreasing, functions, Before learning the concept of maxima and, minima, we will study the nature of the, curve of a given function using derivative., , f(x1), f(x2), o, , x1, , x 1 < x 2 ⇒ f (x 1 ) ≤ f (x 2 ) for all x 1, x 2 ∈ a,b , , A function f(x) is said to be strictly, increasing in [a,b] if, x 1 < x 2 ⇒ f (x 1 ) < f (x 2 ) for all x 1, x 2 ∈ a,b , Y, , y = f(x), , f(x2), , Strictly decreasing function, , Fig: 6.8, , o, x, x, , X, 2, , x 1 < x 2 ⇒ f (x 1 ) < f (x 2 ), Strictly increasing function, , Fig: 6.7, , (ii) Decreasing function, A function f(x) is said to be decreasing, function in [a, b] if, x 1 < x 2 ⇒ f (x 1 ) ≥ f (x 2 ), , for all x 1, x 2 ∈ a,b , , A function f(x) is said to be strictly, decreasing function in [a,b] if, x 1 < x 2 ⇒ f (x 1 ) > f (x 2 ) for all x 1, x 2 ∈ a,b , 140, , NOTE, A function is said to be monotonic function, if it is either an increasing function or a, decreasing function., , Derivative test for increasing and, decreasing function, Theorem: 6.1 (Without Proof), Let f(x) be a continuous function on [a,b], and differentiable on the open interval (a,b),, then, (i) f(x) is increasing in [a, b] if f ′ (x ) ≥ 0, , f(x1), 1, , X, , x 1 < x 2 ⇒ f (x 1 ) > f (x 2 ), , (i) Increasing function, A function f(x) is said to be increasing, function in the interval [a,b] if, , x2, , (ii) f(x) is decreasing in [a, b] if f ′ (x ) ≤ 0, , Remarks:, (i) f(x) is strictly increasing in (a,b) if, f ′ (x ) > 0 for every x ∈ (a,b ), (ii) f(x) is strictly decreasing in (a,b) if, f ′ (x ) < 0 for every x ∈ (a,b ), (iii) f(x) is said to be a constant function if, f ′ (x ) = 0, , 6.2.2 Stationary Value of a function, Let f(x) be a continuous function on [a, b], and differentiable in (a, b). f(x) is said to be, stationary at x = a if f ' (a)=0., , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 140, , 21-04-2020 12:21:59 PM

Page 147 :
www.tntextbooks.in, , The stationary value of f(x) is f (a). The, point (a,f(a) ) is called stationary point., Y, , dy, = 0 Tangent, dx, , a, > sing, , dy, dx > 0, , Inc, re, , a, > sing, , dy, dx > 0, , Inc, re, o, , 0, , c, , re, > asin, g, , 0, , dy, dx, <, , ng, asi, cre >, , P, , dy <, dx, , De, , R, , De, , dy, = 0 Tangent, dx, , y = f(x), , Q, , Tangent, dy, =0, dx, a, , b, , Solution:, Given that f(x) = x2–4x+6, Differentiate with respect to x,, f ¢ (x) = 2x–4, , c, , X, , P, Q, R are called Stationary Points, , Fig: 6.9, , In figure 6.9 the function y = f(x) has, stationary at x = a,x = b and x = c., At these points,, , Example 6.19, Find the interval in which the function, f(x)=x2–4x+6 is strictly increasing and, strictly decreasing., , dy, = 0 . The tangents at, dx, , these points are parallel to x – axis., NOTE, , By drawing the graph of any function, related to economics data, we can study, the trend of the business related to the, function and therefore, we can predict or, forecast the business trend., , When f ¢ (x) = 0 Þ 2x–4=0 Þ x = 2., Then the real line is divided into two, intervals namely (–∞,2) and (2,∞), -∞, , ∞, , 2, , Fig :6.10, , [To choose the sign of f ¢ (x) choose any, values for x from the intervals ans substitute, in f ¢ (x) and get the sign.], Interval, , Sign of, , f ′ (x ) = 2x − 4 Nature of the function, , (-∞, 2), , <0, , f(x) is strictly decreasing in, (-∞, 2), , (2, ∞), , >0, , f(x) is strictly increasing in, (2, ∞), , Table: 6.1, , Example 6.18, Show that the function f(x) = x 3 − 3x 2 + 4x,x ∈ R, is strictly increasing function on R., , Example 6.20, Find the intervals in which the function f, given by f(x)=4x3–6x2–72x+30 is increasing, or decreasing., , Solution :, , Solution :, , f(x) = x 3 − 3x 2 + 4x , x ∈ R, f '(x) = 3x2–6x+4, = 3x2–6x+3+1, = 3(x–1)2+1 > 0, for all x Î R, Therefore, the function f is strictly increasing, on (-∞,∞)., , f(x) = 4x3–6x2–72x+30, f ¢ (x) = 12x2–12x–72, , = 12(x2–x–6), = 12(x–3)(x+2), f ¢ (x) = 0 Þ 12(x–3)(x+2)=0, , x = 3 (or) x = –2, f(x) has stationary at x =3 and at x = –2., Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 141, , 141, , 21-04-2020 12:21:59 PM

Page 148 :
www.tntextbooks.in, , These points divides the whole interval into, three intervals namely (–∞,–2),(–2,3) and, (3,∞)., -∞, , ∞, , 3, , -2, Fig : 6.11, , Interval, , Sign of, , f ¢ (x), , Intervals of increasing/, decreasing, , (–∞, –2), , (–) (–) > 0, , Increasing in (–∞, –2], , (–2, 3), , (–) (+) < 0, , Decreasing in [–2, 3], , (3, ∞), , (+) (+) > 0, , Increasing in [3, ∞), , Table: 6.2, Example 6.21, Find the stationary value and the stationary, points f(x)=x2+2x–5., Solution:, Given that f(x)= x2 + 2x – 5, , … (1), , f '(x) = 2x + 2, At stationary points, f ¢ (x) = 0, Þ 2x + 2 = 0 Þ x = –1, , f(x) has stationary value at x = –1, When x = –1, from (1), f(–1) = (–1)2+2(–1)–5 = – 6, Stationary value of f (x) is – 6, Hence stationary point is (–1,–6), Example 6.22, Find the stationary values and stationary points, for the function: f(x)=2x3+9x2+12x+1., Solution :, Given that f(x) = 2x3+9x2+12x+1., f '(x) = 6x2+18x+12, = 6(x2+3x+2) = 6(x+2)(x+1), f '(x) = 0 Þ 6 (x+2)(x+1) = 0, 142, , Þ x + 2 = 0 (or) x + 1 = 0., Þ x = –2 (or) x = –1, , f(x) has stationary points at x = – 2 and x = – 1, Stationary values are obtained by putting, x = – 2 and x = –1., When x = – 2, f(–2)=2(–8)+9(4)+12(–2)+1= –3, When x= –1, f(–1)= 2(–1)+9(1)+12(–1)+1=–4, The stationary points are (–2,–3) and (–1,–4)., Example 6.23, The profit function of a firm in, producing x units of a product is given by, x3, P(x)= 3 +x2+x. Check whether the firm is, running a profitable business or not., Solution:, x3, P(x) = 3 +x2+x., P'(x) = x2+2x+1=(x+1)2, It is clear that P'(x)>0 for all x., ` The firm is running a profitable business., IMPORTANT NOTE, Let R(x) and C(x) are revenue, function and cost function respectively, when x units of commodity is produced., If R(x) and C(x) are differentiable for all, x > 0, then P(x) = R(x) – C(x) is maximized, when Marginal Revenue = Marginal cost., i.e. when Rl (x)=C l (x) profit is maximum, at its stationary point., Example 6.24, , x2, Given C(x)= 6 +5x+200 and p(x) = 40–x, are the cost price and selling price when, x units of commodity are produced. Find, the level of the production that maximize, the profit., , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 142, , 21-04-2020 12:22:00 PM

Page 149 :
www.tntextbooks.in, , Solution:, , x2, Given C(x)= 6 +5x+200 , and p(x) = 40–x , , … (1), …(2), , Profit is maximized when,, marginal revenue = marginal cost., (i.e) Rl (x) = C l (x), x, = 3 + 5, R, = p ×x, , , = 40x– x2, , R'(x) = 40–2x, x, Hence 40 – 2x = 3 + 5, x = 15, At x = 15, the profit is maximum., , 6.2.3 L,  ocal and Global (Absolute), Maxima and Minima, , Definition 6.1, Local Maximum and local Minimum, A function f has a local maximum (or relative, maximum) at c if there is an open interval, (a,b) containing c such that f(c) ≥ f(x), for every x ! (a,b)., Similarly, f has a local minimum at c if there, is an open interval (a,b) containing c such, that f(c) ≤ f(x) for every x ! (a,b)., , Definition 6.2, , Absolute maximum and, absolute minimum, A function f has an absolute maximum at, c if f(c) ≥ f(x) for all x in domain of f. The, number f(c) is called maximum value of, f in the domain. Similarly f has an absolute, minimum at c if f(c) ≤ f(x) for all x in, domain of f and the number f(c) is called, the minimum value of f on the domain., , The maximum and minimum value of f are, called extreme values of f., NOTE, Absolute maximum and absolute, minimum values of a function f on an interval, (a,b) are also called the global maximum and, global minimum of f in (a,b)., Criteria for local maxima and, local minima, Let f be a differentiable function on an open, interval (a,b) containing c and suppose that, f ll (c) exists., (i) I f f l (c) = 0 and f ll (c) > 0, then f has a, local minimum at c., (ii) I f f l (c) = 0 and f ll (c) < 0,then f has a, local maximum at c., NOTE, In Economics, if y = f(x) represent cost, function or revenue function, then the, dy, point at which dx = 0, the cost or revenue, is maximum or minimum., Example 6.25, Find the extremum values of the function, f(x)=2x3+3x2–12x., Solution:, Given f(x)= 2x3+3x2–12x, f l (x) = 6x2+6x–12, f''(x) = 12x + 6, f l (x)= 0 Þ 6x2+6x–12 =0, Þ 6(x2+x–2)= 0, Þ 6(x+2)(x–1)= 0, Þ x = –2 ; x = 1, Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 143, , … (1), , 143, , 21-04-2020 12:22:00 PM

Page 150 :
www.tntextbooks.in, , When x = –2, f ll (–2) = 12(–2) + 6 = –18 < 0, ` f(x) attains local maximum at x = – 2 and, local maximum value is obtained from (1), by substituting the value x = – 2., f(–2) = 2 (–2)3+3(–2)2–12(–2), = –16+12+24 = 20., When x = 1, f ll (1)= 12(1) + 6 = 18, f(x) attains local minimum at x = 1 and, the local minimum value is obtained by, substituting x = 1 in (1)., f(1) = 2(1) + 3(1) – 12 (1) = –7, , Example 6.26, Find the absolute (global) maximum and, absolute minimum of the function, f(x)=3x5–25x3+60x+1 in the interval [–2,1], Solution :, … (1), , f l (x) = 15x4–75x2+60 = 15(x4–5x2+4), f l (x) = 0 Þ 15(x4–5x2+4)= 0, Þ (x2–4)(x2–1)= 0, Þ x = ±2 (or) x = ±1, , Of these four points −, –2, ±1 ∈ −2, 1 and, 2 ∉ −2, 1, , From (1), , f(–2) = 3 ^- 2h5 - 25 ^- 2h3 + 60 ^- 2h + 1 = –15, When x = 1, , f(1) = 3 ^1 h5 - 25 ^1 h3 + 60 ^1 h + 1 = 39, When x = – 1, , f(–1) =3 ^- 1h5 - 25 ^- 1h3 + 60 ^- 1h + 1 =–37, , Absolute maximum is 39 and, Absolute minimum is -37 ., 144, , 6.3.1 P,  roblems on profit maximization, and minimization of cost function:, Example 6.27, For a particular process, the cost function, is given by C = 56 - 8x + x2 , where C is, cost per unit and x, the number of unit’s, produced. Find the minimum value of the, cost and the corresponding number of units, to be produced., Solution:, C = 56 - 8x + x2, , Extremum values are –7 and 20., , f(x) = 3x5-25x3+60x+1, , 6.3 Applications of Maxima and, Minima, , Differentiate with respect to x,, dC, d2 C, =+, =2, and, 8, 2, x, dx, dx2, dC, dx = 0 &- 8 + 2x = 0 ⇒ x = 4, d2 C, =220, When x = 4,, dx2, ` C is minimum when x = 4, The minimum value of cost = 56–32+16 = 40, The corresponding, produced = 4., , number, , of, , units, , Example 6.28, The total cost function of a firm is, x3, C ^ x h = 3 - 5x2 + 28x + 10 , where x is the, output. A tax at the rate of ` 2 per unit of, output is imposed and the producer adds it, to his cost. If the market demand function, is given by p = 2530 – 5x, where p is the, price per unit of output, find the profit, maximizing the output and price., Solution:, Total revenue: R = p x, = (2530 – 5x)x = 2530x–5x2, Tax at the rate ` 2 per x unit = 2x., , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 144, , 21-04-2020 12:22:03 PM

Page 151 :
www.tntextbooks.in, , x3, ` C(x) + 2x = 3 - 5x2 + 28x + 10 + 2x, P = Total revenue – (Total cost + tax ), x3, = (2530–5x) x - ( 3 - 5x2 + 28x + 10 + 2x), x3, = - 3 + 2500x - 10, dP, d2 P, 2, +, =, and, = - 2x, x, 2500, dx, dx2, dP, 2, 2, dx = 0 & 2500 - x = 0 Þ x = 2500, ` x = 50 (–50 is not acceptable), d2 P, = -100 < 0, dx2, P is maximum when x = 50., , At x = 50, , ` P = 2530 – 5(50) = ` 2280., Example 6.29, The manufacturing cost of an item consists, of ` 1,600 as over head material cost ` 30 per, x2, item and the labour cost ` a 100 k for x items, produced. Find how many items be produced, to have the minimum average cost., Solution:, As per given information for producing x, units of certain item C(x) = labour cost +, material cost + overhead cost., x2, = 100 + 30x + 1600, x2, + 30x + 1600, x, C] g, AC = x = 100, x, x, 1600, = 100 + 30 + x, d2 ^ AC h 3200, d ^ AC h, 1, 1600, dx = 100 - x2 & dx2 = x3, , d ^ AC h = 0 &- 1600 + 1 =0, 100, x2, dx, 1600, 1, ⇒ x2 =160000, & 100 =, 2, x, x, =, 400, (–400, is not acceptable), `, , d2 ^ AC h, 3200, =, >0, When x = 400,, 2, dx, 4003, AC is minimum at x = 400, , Hence 400 items should be produced for, minimum average cost., , Exercise 6.2, 1. Th,  e average cost function associated with, producing and marketing x units of an, 50, item is given by AC = 2x - 11 + x ., Find the range of values of the output x,, for which AC is increasing., 2. A,  television manufacturer finds that, the total cost for the production and, marketing of x number of television sets is, C ^ x h = 300x2 + 4200x + 13, 500 . If each, product is sold for ` 8,400, show that the, profit of the company is increasing.,  monopolist has a demand curve, 3. A, x = 106 – 2p and average cost curve, x, AC = 5 + 50 , where p is the price per, unit output and x is the number of units, of output. If the total revenue is R = px,, determine the most profitable output and, the maximum profit., 4. A,  tour operator charges `136 per, passenger with a discount of 40 paisa, for each passenger in excess of 100. The, operator requires at least 100 passengers, to operate the tour. Determine the, number of passenger that will maximize, the amount of money the tour operator, receives., 5. F, ind the local minimum and local, maximum of y = 2x3 - 3x2 - 36x + 10., 6. , The total revenue function for a, x2, 1, commodity is R= 15x + 3 - 36 x 4 ., Show that at the highest point average, revenue is equal to the marginal revenue., Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 145, , 145, , 21-04-2020 12:22:08 PM

Page 152 :
www.tntextbooks.in, , 6.3.2 Inventory control, Inventory is any stored resource that is used, to satisfy a current or a future need. Raw, materials, finished goods are examples of, inventory. The inventory problem involves, placing and receiving orders of given, sizes periodically so that the total cost of, inventory is minimized., , An inventory decisions, 1. How much to order ?, , 2. When to order ?, , Costs involved in an, inventory problems,   (i) Holding cost (or) storage cost (or), inventory carrying cost (C1) :, , The cost associated with carrying or, holding the goods in stock is known as, holding cost per unit per unit time., , (ii) Shortage cost (C2) :, , , The penalty costs that are incurred, as a result of running out of stock are, known as shortage cost., , (iii) Setup cost (or) ordering cost (or), procurement cost (C3) :, , , This is the cost incurred with the, placement of order or with the initial, preparation of, , The derivation of this formula is given, for better understanding and is exempted, from examination., The formula is to determine the optimum, quantity ordered (or produced) and the, optimum interval between successive, orders, if the demand is known and uniform, with no shortages., Let us have the following assumptions., (i) Let R be the uniform demand per unit, time., (ii) Supply or production of items to the, inventory is instantaneous., (iii) Holding cost is ` C1 per unit time., Let there be 'n' orders (cycles) per, (iv) , year, each time 'q' units are ordered, (produced)., (v) Let ` C3 be the ordering (set up) cost, per order (cycle). Let 't' be the time, taken between each order., Diagrammatic representation of this model, is given below:, y, , production facility, , P, , production., , 6.3.3 E,  conomic Order Quantity (EOQ):, Economic order quantity is that size of, order which minimizes total annual cost of, carrying inventory and the cost of ordering, under the assumed conditions of certainty, with the annual demands known. Economic, order quantity (EOQ) is also called, Economic lot size formula., 146, , Q, , R, , q = Rt, , such as resetting the equipment for, , o, , →t ←A, , x, , Fig. 6.12, , If a production run is made at intervals t,, a quantity q = Rt must be produced in each, run. Since the stock in small time dt is Rt dt ,, the stock in period t is, , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 146, , 21-04-2020 12:22:09 PM

Page 153 :
www.tntextbooks.in, , (ii) Minimum inventory cost per unit time,, , t, , 1 2, ∫0 Rt dt = 2 Rt, , C0 = 2C1C3 R, , q0, R, × C1 , Ordering cost = × C3, q0, 2, q, R, Ordering cost, cost = × C3, = Area of the inventory triangle OAP (Fig.6.12) = 0 × C1 ,Ordering, 2, q0, , (iii) Carrying cost =, , 1, qt ( Rt = q ), , 2, =, , Cost of holding inventory per production, run =, , 1, C1 Rt 2, 2, , (v) Total optimum cost = Rp, C0 += 2C1C3 R, , Set up cost per production run = C3 ., 1, Total cost per production run = C1 Rt 2 + C3, 2, , Average total cost per unit time:, C(t) = 1 C1 Rt + C3 , t, C, d, 1, C (t ) = C1 R − 23 , dt, 2, t, 2, , d2 C ^ t h 2C3, = 3 , dt2, t, , C(t) is minimum if, , ... (1), ... (2), ... (3), , d, d2, C (t ) = 0 and 2 C (t ) > 0, dt, dt, , C, d, 1, C (t ) = 0 ⇒ C1 R − 23 = 0, 2, dt, t, 2C3, ⇒t =, C1 R, , When, , Example 6.30, A company uses 48000 units/year of a raw, material costing ` 2.5 per unit. Placing each, order costs `45 and the carrying cost is 10.8 %, per year of the unit price. Find the EOQ,, total number of orders per year and time, between each order. Also verify that at EOQ, carrying cost is equal to ordering cost., , Inventory cost: C1 =10.8% of, , d2 C ^ t h, 2C 3, ,, =, >0, 2, 3, dt, C 1R,  2C 2,  3 , C R ,  1 , , Thus C (t ) is minimum for optimum time, 2C3, C1 R, , 2.5 =, , 10.8, × 2.5 = 0.27, 100, , Ordering cost: C3 = 45, Economic order quantity: q0 =, =, , R, q0, , Time between orders: t0 =, , 48000, = 12, 4000, , q0 1, = = 0.083 year, R 12, , At EOQ, carrying cost:, , 2C3 R, C1, , This is known as the Optimal Lot-size, formula due to Wilson., (i) Optimum number of orders per year, C1, demand, =, n0 = R =, EOQ, 2C3 R, , 2C3 R, C1, , 2 × 45 × 48000, = 4000 units, 0.27, , Number of orders per year ==, , , Optimum quantity q0 to be produced, during each production run,, EOQ = =, q0 Rt, =, 0, , Here, we will discuss EOQ problems only, without shortage cost., , Solution:, Here demand rate: R = 48000, , 2C 3, , interval t0 =, , (iv) At EOQ, Ordering cost = Carrying cost, , RC1 1, =, 2C3 t0, , =, , q0, 4000, × 0.27 = ` 540, × C1 =, 2, 2, , Ordering cost =, , R, 48000, × C3 =, × 45 = ` 540, q0, 4000, , So at EOQ carrying cost is equal to ordering, cost., Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 147, , 147, , 21-04-2020 12:22:23 PM

Page 154 :
www.tntextbooks.in, , Example 6.31, A manufacturer has to supply 12,000 units, of a product per year to his customer. The, ordering cost (C3) is ` 100 per order and, carrying cost is ` 0.80 per item per month., Assuming there is no shortage cost and the, replacement is instantaneous, determine the, (i) economic order quantity, (ii) time between orders, (iii) number of orders per year, , Solution:, Demand : R, = 1000 per month, Setup cost : C3 = ` 200 per order, Carrying cost: C1= ` 20 per item per month., Set up cost, , Run, size, q, , R, ´C 3, q, , Ordering cost:, , C3 = ` 100/order, , Carrying cost:, , C1 = 0.80/item/month, , 500, 5400, 2 ×20 = 5000, 600 1000 ×200 = 333.3 600 ×20 = 6000 6333.3, 2, 600, 700 1000 ×200 = 285.7 700 ×20 = 7000 7285.7, 700, 2, 800, 1000, 800 800 ×200 = 250, 2 ×20 = 8000 8250, , Table : 6.3, , = 0.80×12 per year, = ` 9.6 per year, , , , 2C3 R, 2 ´ 100 ´ 12000, =, C1, 9.6, , = 500 units, , (ii) Number of order per year =, 12, 000, = 500 =24, , Demand, EOQ, , 1, 1, (iii) Optimal time per order = t = 24 year, 0, 12, 1, = 24 months = 2 month = 15 days, Example 6.32, A company has to supply 1000 item per, month at a uniform rate and for each time,, a production run is started with the cost of, ` 200. Cost of holding is ` 20 per item per, month. The number of items to be produced, per run has to be ascertained. Determine, the total of setup cost and inventory carrying, cost if the run size is 500, 600, 700, 800. Find, the optimal production run size using EOQ, formula., 148, , q, ´C 1, 2, , 500 1000 ×200 = 400, 500, , Solution:, Demand per year: R = 12,000 units, , (i) EOQ =, , Carrying cost, , Total, Inventory, cost, (Set up, cost +, Carrying, cost), , EOQ=, =, , 2RC3, C1, , 2×1000×200, 20, , =, , 20000 = 141 units (app.), , Example 6.33, A manufacturing company has a contract, to supply 4000 units of an item per year at, uniform rate. The storage cost per unit per, year amounts to ` 50 and the set up cost per, production run is ` 160. If the production, run can be started instantaneously and, shortages are not permitted, determine the, number of units which should be produced, per run to minimize the total inventory cost., Solution :, Annual demand: R = 4000, Storage cost: C1 = ` 50, Setup cost per production: C3= `160, EOQ =, , 2Rc3, c1 =, , 2 # 4000 # 160, = 160., 50, , ∴ To minimize the production cost, number, , of units produced per run is 160 units., , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 148, , 21-04-2020 12:22:26 PM

Page 155 :
www.tntextbooks.in, , Example 6.34, A company buys in lots of 500 boxes which, is a 3 month supply. The cost per box is `125, and the ordering cost in `150. The annual, inventory carrying cost is estimated at 20%, of unit price., (i) Determine the total inventory cost of, existing inventory policy., (ii) Determine EOQ in units, (iii) How much money could be saved by, applying the economic order quantity?, Solution:, Given, Ordering cost per order: C3= `150 per order., Number of units per order: q = 500 units, Annual demand = 500 × 4 = 2000 units, ∴ Demand rate: R = 2000 per year, , Carrying cost: C1= 20% of unit value, 20, , C1 = 100 # 125 = `25, (i) T,  otal inventory cost of the existing, inventory policy, q, R, 500, 2000, = q ×C3+ 2 C1= 500 ×150+ 2 ×25, = ` 6850, (ii) EOQ =, =, , 2RC3, C1, 2 # 2000 # 150, 25, , Exercise 6.3, 1. Th,  e following table gives the annual, demand and unit price of 3 items, Items, , Annual Demand (units), , Unit Price `, , A, , 800, , 0.02, , B, , 400, , 1.00, , C, , 13,800, , 0.20, , Ordering cost is Rs. 5 per order and annual, holding cost is 10% of unit price., Determine the following:, (i) EOQ in units, (ii) Minimum inventory cost, (iii) EOQ in rupees, (iv) EOQ in years of supply, (v) Number of orders per year., 2. A,  dealer has to supply his customer, with 400 units of a product per every, week. The dealer gets the product from, the manufacturer at a cost of ` 50 per, unit. The cost of ordering from the, manufacturers in ` 75 per order. The cost, of holding inventory is 7.5 % per year, of the product cost. Find (i) EOQ and, (ii) Total optimum cost., , 6.4, , Partial Derivatives, , = 12 # 2000, , Partial derivative of a function of several, , , , variables is its derivative with respect to one, , = 155 units (app.), , (iii) Minimum inventory cost =, , 2RC3 C1, , = 2 # 2000 # 150 # 25, , , = `3873., , By applying the economic order quantity,, money saved by a company = 6850–3873, , = `2977., , of those variables, keeping other variables as, constant. In this section, we will restrict our, study to functions of two variables and their, derivatives only., Let u = f ( x, y ) be a function of two, independent variables x and y., Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 149, , 149, , 21-04-2020 12:22:29 PM

Page 156 :
www.tntextbooks.in, , The derivative of u with respect to x when x, , third order partial derivatives, fourth order, , varies and y remains constant is called the, , partial derivatives etc., if they exist. The, , partial derivative of u with respect to x,, , process of finding such partial derivatives, , denoted by, , are called successive partial derivatives., , ∂u, (or) ux and is defined as, ∂x, , f (x + Tx, y) - f (x, y), 2u, =, , lim, Tx, 2x Tx " 0, , If we differentiate u = f(x,y) partially with, , change in x, , partially with respect to y,, , provided the limit exists. Here ∆x is a small, , The derivative of u with respect to y, when, y varies and x remains constant is called the, partial derivative of u with respect to y,, denoted by, , ∂u, (or) uy and is defined as, ∂y, , 2u, , 2y = Oylim, $0, , f (x, y + Ty) - f (x, y), Ty, , respect to x and again differentiating, ∂ 2u, 2 b 2u l, we obtain 2y 2x i.e.,, ∂y∂x, , Similarly, if we differentiate u = f(x,y), partially with respect to y and again, differentiating partially with respect to x,, we obtain, , ∂  ∂u , ∂ 2u,   i.e.,, ∂x  ∂y , ∂x∂y, , provided the limit exists. Here ∆y is a small, change in y., 2f, 2u, 2, is, also, written, as, f(x,y), (or), 2x ., 2x, 2x, Similarly, , ¶u, ¶, is also written as, f(x,y), ¶y, ¶y, , 2f, (or) 2y, , If u(x,y) is a continuous function of, x and y, then, , ∂ 2u, ∂ 2u, =, ., ∂y∂x, ∂x∂y, , Homogeneous functions, , The process of finding a partial derivative is, called partial differentiation., , 6.4.1 Successive partial derivatives, Consider the function u = f ( x, y ) . From this, 2u, 2u, 2u, 2u, we can find 2x and 2y . If 2x and 2y, are functions of x and y, then they may be, differentiated partially again with respect to, either of the independent variables, (x or y), 22 u 22 u 22 u 22 u, denoted by, ,, ,, ., 2y2 2x2 2y2x 2y2x, These derivatives are called second order, partial derivatives. Similarly, we can find the, 150, , NOTE, , A function f(x,y) of two independent, variables x and y is said to be homogeneous, in x and y of degree n if f (tx, ty ) = t n f ( x, y ), for t > 0., , 6.4.2 E,  uler’s theorem, and its applications, Euler’s theorem for two variables:, If u = f ( x, y ) is a homogeneous function of, degree n, then x, , ∂u, ∂u, +y, = nu, ∂x, ∂y, , Example 6.35, If u = x2(y–x) + y2 (x–y), then show that, ∂u ∂u, +, = –2 (x–y)2., ∂x ∂y, , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 150, , 21-04-2020 12:22:35 PM

Page 158 :
www.tntextbooks.in, , Example 6.39, x4 + y4, Let u = log x + y . By using Euler’s theorem, show that x ⋅, , ∂u, ∂u, + y⋅, = 3., ∂x, ∂y, , Solution:, x4 + y4, u = log x + y, x4 + y4, u, e = x + y = f(x,y), ... (1), x4 + y4, Consider f(x,y) = x + y, t4 x4 + t4 y4 3 x4 + y4, f ^tx, ty h= tx + ty = t d x + y n = t3 f ^ x, y h, ` f is a homogeneous function of degree 3., , Using Euler’s theorem we get,, x⋅, , ∂u, ∂u, +y ⋅, = 3f, ∂x, ∂y, , Consider f(x,y) = eu, x⋅, , `, , ∂e u, ∂e u, +y ⋅, ∂x, ∂y, , e ux ⋅, , = 3e u, , ∂u, ∂u, + e uy ⋅, = 3e u, ∂x, ∂y, , ∂u, ∂u, x⋅, +y ⋅, ∂x, ∂y, , =3, , Exercise 6.4, 2z, 1. I f z = ^ax + bh^cy + d h , then find 2x, 2z, and 2y ., , 22 u 22 u, +, =, 2. I f u = e xy , then show that, 2x2 2y2, 2, 2, u_ x + y i ., 3. L,  et u = x cos y + y cos x., 22 u, 22 u, Verify 2x2y = 2y2x ., , 4. V,  erify Euler’s theorem for the function, u= x3 + y3 + 3xy2 ., 152, , x, 5. Let u= x2 y3 cos a y k . By using Euler’s, 2u, 2u, theorem show that x. 2x + y. 2y =5u., , 6.5 Applications of Partial, Derivatives, In this section we solve problems on partial, derivatives which have direct impact on, Industrial areas., , 6.5.1 P,  roduction function and, marginal productivities of, two variables, (i) Production function:, Production P of a firm depends, upon several economic factors like, capital (K), labour (L), raw materials, (R), machinery (M) etc… Thus, P = f(K,L,R,M,…) is known as, production function. If P depends only, on labour (L) and capital (K), then we, write P=f(L,K)., , (ii) Marginal productivities:, Let P = f(L,K) be a production function., 2P, Then 2L is called the Marginal, 2P, productivity of labour and 2K is, called the Marginal productivity of, capital., , Euler’s, theorem, for, homogeneous production function P(L,K), 2P, 2P, of degree 1 states that L 2L + K 2K = P, , 6.5.2 Partial elasticity of demand, , Let q = f ^ p1, p2h be the demand for, commodity A, which depends upon the, prices., p1 and p2 of commodities A and B, respectively., , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 152, , 21-04-2020 12:22:46 PM

Page 159 :
www.tntextbooks.in, , The partial elasticity of demand q with, respect to p1 is defined to be, ηqp1 =, , Eq − p1 ∂q, =, Ep1, q ∂p1, , The partial elasticity of demand q with, respect to p2 is defined to be, ηqp2 =, , Eq − p2 ∂q, =, Ep2, q ∂p2, , Solution:, (i) Given the production is, P = 10L − 0.1L2 + 15K − 0.2K 2 + 2KL, , , ∂P, = 10 − 0.2 L + 2 K, ∂L, , , , ∂P, = 15 − 0.4 K + 2 L, ∂K, , When L = K = 10 units,, , Example 6.40, , Marginal productivity of labour:, , Solution:, ,  ∂P , = 15 − 4+20 = 31, , ,  ∂K (10,10 ), , ∂P, , , , Find the marginal productivities of capital, = 10 − 2+20 = 28, , , 2, 2 ∂L (10,10 ), (K) and labour (L) if P = 10L + 0.1L + 5K - 0.3K + 4KL, 2, Marginal productivity of capital :, 10L + 0.1L + 5K - 0.3K2 + 4KL when K = L = 10., , We have P =10L + 0.1L2 + 5K - 0.3K2 + 4KL, ∂P, ∂L, , = 10 + 0.2 L + 4 K, ∂P, ∂K, , (ii) Upper limit for use of labour when, 2P, K = 10 is given by a 2L k ≥ 0, 10 − 0.2L+20 ≥ 0, , = 5 − 0.6 K + 4 L, , 30 ≥ 0.2L, , Marginal productivity of labour:, , i.e., L ≤ 150, ,  ∂P , = 10+2+40 = 52, , ,  ∂L (10,10 ), , Marginal productivity of capital:,  ∂P , = 5 − 6+40 = 39, , ,  ∂K (10,10 ), , Example 6.41, The production function for a commodity is, P= 10L + 0.1L2 + 15K - 0.2K2 + 2KL where, , Hence the upper limit for the use of labour, will be 150 units., Example 6.42, 3, , 1, , For the production function, P = 4 L4 K 4, verify Euler’s theorem., Solution:, 3, , 1, , L is labour and K is Capital., , P = 4 L4 K 4 is a homogeneous function of, degree 1., , (i) Calculate the marginal products of two, , Marginal productivity of labour is, , inputs when 10 units of each of labour, and Capital are used, (ii) If 10 units of capital are used, what is, the upper limit for use of labour which a, rational producer will never exceed?, , 1, , ¶P, 3 −14 14,  K 4, = 4´ L K = 3 , ¶L, 4, L, , Marginal productivity of capital is, 3, , 3, 1 −43  L  4, ¶P, 4, = 4L × K =  , ¶K, 4, K, , Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 153, , 153, , 21-04-2020 12:22:57 PM

Page 161 :
www.tntextbooks.in, , 3. I f demand and the cost function of a, firm are p= 2–x and C=– 2x2 + 2x + 7 ,, then its profit function is, (b) x2 - 7, (a) x2 + 7, (c) - x2 + 7, , (d) - x2 - 7, , 4. If the demand function is said to be, elastic, then, (b) ηd =1, (a) ηd >1, (c) ηd <1, , (d) ηd =0, , 5. Th,  e elasticity of demand for the demand, 1, function x = p is, 1, (d) ∞, (a) 0, (b) 1, (c) - p, 6. Relationship among MR, AR and ηd is, AR, (a) ηd =, AR − MR, , (b) ηd = AR − MR, , (c) MR = AR = ηd, , (d) AR =, , MR, ηd, , 1, 7. F,  or the cost function C = 25 e5x , the, marginal cost is, 1, (a) 25, 1, (b) 5 e5x, 1, (c) 125 e5x, (d) 25e5x, 8. I, nstantaneous rate of change of, y = 2x2 + 5x with respect to x at x = 2 is, (a) 4, , (b) 5, , (c) 13, , (d) 9, , 9. If the average revenue of a certain firm, is ` 50 and its elasticity of demand is 2,, then their marginal revenue is, (a) ` 50 (b) ` 25 c) ` 100 (d) ` 75, 10. Profit P(x) is maximum when, (a) MR = MC, (c) MC = AC, , (b) MR = 0, (d) TR = AC, , 11. The maximum value of f(x)= sinx is, 3, 1, 1, (c), (d) (a) 1 (b) 2, 2, 2, 12. If f(x,y) is a homogeneous function of, 2f, 2f, degree n, then x 2x + y 2y is equal to, (a) (n–1)f, (b) n(n–1)f, (c) nf , (d) f, u = 4x2 + 4xy + y2 + 4x + 32y + 16 ., 22 u, then 2y2x is equal to, (a) 8x + 4y + 4, (b) 4, (c) 2y + 32, (d) 0, , 13. If, , 22 u, 14. If u = x3 + 3xy2 + y3 , then 2y2x is, (a) 3, , (b) 6y, , (c) 6x, , (d) 2, , 2, 2u, 15. If u = e x , then 2x is equal to, 2, 2, 2, (b) e x, (c) 2e x, (d) 0, (a) 2xe x, , 16. Average cost is minimum when, (a) Marginal cost = Marginal revenue, (b) Average cost = Marginal cost, (c) Average cost = Marginal revenue, (d) Average Revenue = Marginal cost, 17. A company begins to earn profit at, (a) Maximum point, (b) Breakeven point, (c) Stationary point, (d) Even point, 18. The demand function is always, (a) Increasing function, (b) Decreasing function, (c) Non-decreasing function, (d) Undefined function, 19. If q = 1000 + 8p1 - p2 , then, (a) –1 , (c) 1000, , ¶q, is, ¶p1, , (b) 8, (d) 1000 - p2, , Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 155, , 155, , 21-04-2020 12:23:12 PM

Page 162 :
www.tntextbooks.in, , 20. If R = 5000 units / year, C1= 20 paise,, C3= ` 20, then EOQ is, (a) 5000, , (b) 100, , (c) 1000, , The total cost function y for x units, 5. , ,  x+2,  + 6 . Prove,  x +1 , , is given by y = 4 x , , (d) 200, , that marginal cost[MC] decreases as x, increases., , Miscellaneous Problems, , 6. For the cost function C = 2000 + 1800x–75x2 + x3, 1. Th, e total cost function for the, production of x units of an item is given, 2000 + 1800x–75x2 + x3 find when the total cost (C) is, increasing and when it is decreasing., by C = 10 − 4 x3 + 3x 4 find the (i) average, cost function (ii) marginal cost function, 7. , A certain manufacturing concern has, (iii) marginal average cost fuction., total cost function C = 15 + 9x–6x2 + x3 ., Find x, when the total cost is minimum., , 2. F,  ind out the indicated elasticity for the, following functions, p dx, (i) p = xe x , x > 0; s, x dp, - 3x, η, (ii) p = 10e , x > 0; d, , 8. Let u = log, , 3. F,  ind the elasticity of supply when the, supply function is given by x = 2 p 2 + 5, at p=1., , 9. Verify, , 4. F,  or the demand function p = 100 − 6 x 2 ,, find the marginal revenue and also show, , 10. I f f ( x, y ) = 3x 2 + 4 y 3 + 6 xy − x 2 y 3 + 7, then, show that f yy (1,1) = 18., , , , that MR = p 1 −, , , x4 − y 4, . Using Euler’s theorem, x− y, ∂u, ∂u, show that x + y = 3 ., ∂x, ∂y, ∂ 2u, ∂ 2u, =, for u = x3 + 3x 2 y 2 + y 3, ∂x∂y ∂y∂x, , 1, , ηd , , Summary, zz Demand is the relationship between the quantity demanded and, the price of a commodity., , zz Supply is the relationship between the quantity supplied and the price of a, commodity., , zz C ost is the amount spent on the production of a commodity., zz Revenue is the amount realised by selling the output produced on commodity., zz Profit is the excess of total revenue over the cost of production., zz Elasticity of a function y = f(x) at a point x is the limiting case of ratio of the, relative change in y to the relative change in x, , zz Equilibrium price is the price at which the demand of a commodity is equal to, its supply., , zz Marginal cost is interpreted as the approximate change in production cost of, ^ x + 1hth unit, when the production level is x units., , 156, , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 156, , 21-04-2020 12:23:19 PM

Page 163 :
www.tntextbooks.in, , zz Marginal Revenue is interpreted as the approximate change in revenue made on, by selling of ^ x + 1hth unit, when the sale level is x units., , zz A function f(x) is said to be increasing function in the interval [a, b] if, x1 1 x2 & f ^ x1h # f ^ x2h for all x1, x2 e 6a, b@, , z z A, , function f(x) is said to be strictly, x1 1 x2 & f ^ x1h 1 f ^ x2h for all x1, x2 e 6a, b@, , increasing, , in, , [a,, , b], , if, , zz A function f(x) is said to be decreasing function in [a,b] if x1 1 x2 & f ^ x1h $ f ^ x2h, for all x1, x2 e 6a, b@, , zz A function f(x) is said to be strictly decreasing function in [a, b]if, x1 1 x2 & f ^ x1h > f ^ x2h for all x1, x2 e 6a, b@, , zz Let f be a differentiable function on an open interval (a,b) containing c and, suppose that f'’ (c) exists., , (i) If f '(c) = 0 and f "(c) > 0, then f has a local minimum at c., (ii) If f '(c) = 0 and f "(c) < 0,then f has a local maximum at c., , zz A function f(x, y) of two independent variables x and y is said to be homogeneous, in x and y of degree n if for t > 0 f (tx, ty ) = t n f ( x, y ), , zz If u = f ( x, y ) is a homogeneous function of degree n, then x, , ∂u, ∂u, +y, = nu, ∂x, ∂y, , zz The partial elasticity of demand q with respect to p1 is defined to be, ηqp1 =, , Eq − p1 ∂q, =, Ep1, q ∂p1, , zz The partial elasticity of demand q with respect to p2 is defined to be, ηqp2 =, , Eq − p2 ∂q, =, Ep2, q ∂p2, , Some standard results, , dC, 5. Marginal cost: MC = dx, , 1. Total cost: C ( x) = f ( x) + k, 2. Average cost: AC =, , 6. Marginal Average cost: MAC =, , f ( x) + k c( x), =, x, x, , 3. Average variable cost: AVC=, 4. Average fixed cost: AFC =, , k, x, , f ( x), x, , 7. Total cost: C (x ) = AC × x, 8. Revenue: R = px, 9. Average Revenue: AR=, , R, =, x, , p, , Applications of Differentiation, , 06_11th_BM-STAT_Ch-6-EM.indd 157, , d, ( AC ), dx, , 157, , 21-04-2020 12:23:27 PM

Page 164 :
www.tntextbooks.in, , 10. Marginal Revenue: MR =, , dR, dx, , 11. Profit: P( x) = R( x) − C ( x), 12. Elasticity: η =, , p dx, 13. Elasticity of demand: ηd = − ⋅, x dp, p dx, ⋅, x dp, , 15. Relationship between MR, AR and ηd :, , 1, AR, MR = AR 1 −  (or) ηd =, AR − MR,  ηd , , 16. M,  arginal function of y with respect to, x (or) Instantaneous rate of change of y, dy, with respect to x is, dx, , 17. A,  verage cost [AC] is minimum when, MC = AC, 18. T,  otal revenue [TR] is maximum when, MR = 0, 19. Profit [P(x)] is maximum when, MR = MC, , 158, , (a) If |η| > 1, then the function is elastic, (b) I f |η| = 1, then the function is unit, elastic, , x dy, ⋅, y dx, , 14. Elasticity of supply: η s =, , 20. In price elasticity of a function,, , (c) If |η| < 1, then the function is, inelastic., q0 Rt, =, 21. EOQ = =, 0, , 2C3 R, C1, , 22. Optimum number of orders per year, =, n0, , C1, demand, = R =, EOQ, 2C3 R, , RC1 1, =, 2C3 t0, , 23. Minimum inventory cost per unit time,, C0 = 2C1C3 R, , 24. Carrying cost =, ordering cost =, , q0, × C1 and, 2, R, × C3, q0, , 25. At EOQ, Ordering cost = Carrying cost, 26. I f u(x,y)is a continuous function of x, ∂ 2u, ∂ 2u, and y then,, =, ∂y∂x, ∂x∂y, , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 158, , 21-04-2020 12:23:33 PM

Page 166 :
www.tntextbooks.in, , ICT Corner, Application of Differentiation, Step – 1, Open the Browser type the URL Link, given below (or) Scan the QR Code., GeoGebra Workbook called “11th, BUSINESS MATHEMATICS and, STATISTICS” will appear. In that, there are several worksheets related, to your Text Book., , Expected Outcome ⇒, Step-2, Select the work sheet “Marginal function” Click on respective boxes to see the graph on left, side. Type the total cost function in the box seen on right side and proceed., Step 1, , Step 2, , Browse in the link, 11th Business Mathmatics and Statistics, https://ggbm.at/qKj9gSTG (or) scan the QR Code, , 160, , 11th Std. Business Mathematics and Statistics, , 06_11th_BM-STAT_Ch-6-EM.indd 160, , 21-04-2020 12:23:34 PM

Page 167 :
www.tntextbooks.in, , Chapter, , 7, , FINANCIAL MATHEMATICS, , Learning objectives, After studying this chapter, the, students will be able to understand, , •, •, •, , Annuities - types of annuities, Future and present values of annuities, ,  e concept of gain or loss in the sale, Th, and purchase of a stock., , • Brokerage in share transactions, • Effective rate of return, Introduction, , In our day-to-day, life we have lot of money, transactions., In many, transactions payment is, made in single payment or, in equal installments over a certain period, of time. The amounts of these installments, are determined in such a way that they, compensate for their waiting time. In other, cases, in order to meet future planned, expenses, a regular saving may be done. (i.e), at regular time intervals, a certain amount, may be kept aside, on which the person, gains interest. In such cases the concept of, annuity is used., , 7.1, , Annuities, , A sequence of equal payments made/, received at equal intervals of time is called, annuity. The amount of regular payment, of an annuity is called periodic payment., , The time interval between two successive, payments is called payment interval or, payment period. Note that, the payment, period may be annual, half yearly, quarterly,, monthly (or) any fixed duration of time., The time interval between the first payment, and the last payment of an annuity is called, term of an annuity., The sum of all payments made and, interest earned on them at the end of the, term of annuities is called future value of, an annuity. The present or capital value of, an annuity is the sum of the present values, of all the payments of the annuity at the, beginning of the annuity of purchase the, payments due in future. Here we note that, unless mentioned specifically, the payment, means yearly payment., , 7.1.1, , Types of annuities, , a)Based on the number of periods:, (i) Certain annuity: An annuity payable, for a fixed number of years is called, certain annuity., Installments of payment for a plot of, land, Bank security deposits, purchase, of domestic durables are examples, of certain annuity., Here the buyer, knows the specified dates on which, installments are to be made., , (ii) Annuity contingent: An annuity, payable at regular interval of time till the, happening of a specific event or the date, Financial Mathematics, , 07_11th_BM-STAT_Ch-7-EM.indd 161, , 161, , 21-04-2020 12:24:07 PM

Page 168 :
www.tntextbooks.in, , of which cannot be accurately foretold is, called annuity contingent.,  or example the endowment funds of, F, trust, where the interest earned is used, for welfare activities only. The principal, remains the same and activity continues, forever., , All the above types of annuities are, based on the number of their periods., An annuity can also be classified on the, basis of mode of payment as under., , b) Based on the mode of payment :, i) Ordinary annuity: An annuity in, which payments of installments are, made at the end of each period is called, ordinary annuity (or immediate annuity), , For example repayment of housing loan,, vehicle loan etc.,, , ii) Annuity due: An annuity in which, payments of installments are made in, the beginning of each period is called, annuity due., , In annuity due every payment is an, investment and earns interest. Next, payment will earn interest for one period, less and so on, the last payment will earn, interest of one period., , For example saving schemes, life, insurance payments, etc., , Deferred annuity:An annuity which is, payable after the lapse of a number of, periods is called deferred annuity., , The derivation of the following, formulae are given for better, understanding and are exempted, from examination., , (i) A,  mount of immediate annuity, (or) Ordinary annuity (or) Certain, annuity, Let ‘a’ be the ordinary annuity and, i percent be the rate of interest per period., In ordinary annuity, the first installment is, paid after the end of first period. Therefore, it earns interest for (n – 1) period, second, installment earns interest for (n – 2) periods, and so on. The last installment earns for, (n – n) periods. (i.e) earns no interest., For (n–1) periods,, The amount of first annuity = a(1+i)n-1, The amount of second annuity = a(1+i)n-2, The amount of third annuity = a(1+i)n-3, and so on., ` The total amount of annuity A for n period, at i percent rate of interest is, A = a(1+i)n-1+a(1+i)n-2+...a(1+i)+ a, = a[(1+i)n-1+(1+i)n-2+...+(1+i)+1], = a[1+(1+i)+(1+i)2+...+(1+i)n-1 ], = a[1+r+r2+...+rn-1], where 1+ =r, rn - 1, = a[ r - 1 ], G.P with common ratio r > 1, 1 + ign - 1, E, = a ;], 1+i-1, a, A = i [(1+i)n–1], , (ii) P,  resent Value of immediate annuity, (or ordinary annuity), Let ‘a’ be the annual payment of an ordinary, annuity, n be the number of years and, , 162, , 11th Std. Business Mathematics and Statistics, , 07_11th_BM-STAT_Ch-7-EM.indd 162, , 21-04-2020 12:24:08 PM

Page 169 :
www.tntextbooks.in, , i percent be the interest on one rupee per, year and P be the present value of the annuity., In the case of immediate annuity, payments, are made periodically at the end of specified, period. Since the first installment is paid at, a, the end of first year, its present value is 1 + i ,, the present value of second installment is, a, and so on. If the present value of, ]1 + ig2, a, , then we have, last installment is, ]1 + ign, a, a, a, a, +, + ..., P= 1+i +, 2, 3, +, ]1 ign, ]1 + ig, ]1 + ig, a, a, a, +, + ... + (1 + i), =, n, n, 1, ]1 + ig, ]1 + ig, a, = n 61 + r + r2 + ... + rn - 1@ , taking 1 + i = r, r, a n, = n ; rr --11 E , G.P with common ratio, r > 1, r, a < (1 + i) n - 1 F, =, (1 + i) n 1 + i - 1, 1, a, = i <1 - (1 + i) n F, a, , [, , P = i 1– 1 n, (1 + i), , ], , (iii) A,  mount of annuity due at the end of, n period, Annuity due is an annuity in which the, payments are made at the beginning of, each payment period. The first installment, will earn interest for n periods at the rate, of ‘i’ percent per period. Similarly second, installment will earn interest for (n – 1), periods, and so on the last interest for on, period., ` A = a(1+i)n+a(1+i)n-1+...+a(1+i)1, = a(1+i)[(1+i)n-1+(1+i)n-2)+...+1], = a(1+i)[1+(1+i)+(1+i)2+...+(1+i)n-1], = ar[1+r+r2+...+rn-1] , put 1+i = r, rn - 1, = ar[ r - 1 ], G.P with common ratio, r > 1, (1 + i) n - 1 F, = a(1+i) <, 1+i-1, , =, , a (1 + i ), [(1 + i) n - 1], i, , A=, , a (1 + i), , i, , [(1+i)n-1], , (iv) Present value of annuity due, Since the first installment is paid at the, beginning of the first period (year), its, present value will be the same as ‘a’, the, annual payment of annuity due. The second, installment is paid in the beginning of the, second year, hence its present value is given, a, by (1 + i) and so on. The last installment is, paid in the beginning of nth year, hence its, a, present value is given as, (1 + i) n - 1, If P denotes the present value of annuity, due, then, a, , a, , a, , 1, 1, 1, 1, = a <1 + 1 + i + (1 + i) 2 + (1 + i) 3 + ... + (1 + i) n - 1 F, , RS, nV, SS1 - b 1 l WWW, 1+i W, =a SSS, WW, SS 1 - 1, W, 1+i W, T, X, =a >, =, , ]1 + ign - 1, ]1 + ign, ]1 + ig - 1, 1+i, , H, , a (1 + i) (1 + i) n - 1, G, =, i, (1 + i) n, , , , 1 , a (1 + i) , 1 −, , P=, n, i, , (1 + i ) , , , (v) Perpetual Annuity, Perpetual annuity is an annuity whose, payment continuous for ever. As such the, amount of perpetuity is undefined as the, amount increases without any limit as time, passes on. We know that the present value, P of immediate annuity is given by, Financial Mathematics, , 07_11th_BM-STAT_Ch-7-EM.indd 163, , a, , P = a+ 1 + i + (1 + i) 2 + (1 + i) 3 + ... + (1 + i) n - 1, , 163, , 21-04-2020 12:24:13 PM

Page 170 :
www.tntextbooks.in, , 1, a, F, P = i <1 (1 + i) n, Now as per the definition of perpetual, 1, annuity as n → ∞, we know that, "0, (1 + i) n, since 1+i > 1., a, Here P = i [1-0], a, P= i, , Here A = 1,67,160, n = 16 i= 15 =0.15 a = ?, 100, To find : a, a, Now A = i [(1+i)n–1], a, 1,67,160 = 0.15 [(1+0.15)16–1], a, = 0.15 [(1.15)16–1], 1, 67, 160 # 0.15, ⇒a=, (1.15)16 - 1, =, , NOTE, In all the above formulae the period is of, one year. Now if the payment is made more, than once in a year then ‘i’ is replaced by, i, k and n is replaced by nk, where k is the, number of payments in a year., Example 7.1, A person pays ` 64,000 per annum, for 12 years at the rate of 10% per year., Find the amount of an ordinary annuity, [(1.1)12=3.3184]., Solution:, 10, Here a = 64,000, n = 12 and i = 100 =0.1, a, Amount of ordinary annuity (A) = i [(1+i)n-1], 64000, = 0.1 [(1+0.1)12–1], , = 6,40,000[(1.1)12 –1], , = 6,40,000[3.3184 – 1], = 6,40,000 [2.3184], = 64×23184, , ∴ A = ` 14,83,776, Example 7.2, What amount should be deposited, annually in an ordinary annuity scheme, so, that after 16 years, a person receives `1,67,160, if the interest rate is 15% [(1.15)16=9.358]., 164, , Solution:, , 1, 67, 160 # 0.15 1, 67, 160 # 0.15, =, = 3,000, 8.358, 9.358 - 1, , Therefore a = ` 3,000, , Example 7.3, The age of the girl is 2 years. Her father, wants to get `20,00,000 when his ward, becomes 22 years. He opens an account with, a bank at 10% rate of compound interest., What amount should he deposit at the end, of every month in this recurring account?, [(1.0083)240=6.194]., Solution:, 10, Here A = 20,00,000 ; i= 100 =0.1 n = 20 and, k = 12, a, i, [(1+ k )nk–1], A=, i k, a, .1 l20 # 12, 0.1, 20,00,000 = 12 ;b1 + 012, - 1E, 0.1 240, 12a, = 0.1 ;b1 + 12 l –1E, 12.1 240, = 120a ;b 12 l - 1E, = 120a [(1.0083)240–1], = 120a [6.194–1], = 120a (5.194), 20.00, 000, ⇒ a = 120 # 5.194 = 3208.83, ∴ a ≈ 3,209, , `3,209 is to be deposited at the end, of every month., , 11th Std. Business Mathematics and Statistics, , 07_11th_BM-STAT_Ch-7-EM.indd 164, , 21-04-2020 12:24:16 PM

Page 171 :
www.tntextbooks.in, , Example 7.4, A person deposits `4,000 in the beginning of, every year. If the rate of compound interest, is 14% then ,find the amount after 10 years ., [(1.14)10=3.707]., Solution :, Here a = 4,000 ; i = 0.14 and n = 10., a, A = (1+i) i [(1+i)n–1], 4000, = (1+0.14) 0.14 [(1+0.14)10–1], 4000, = (1.14) 0.14 [(1.14)10–1], 4000, = 1.14× 0.14 (3.707–1), 4000, = 1.14× 0.14 (2.707) =88,170.86, ∴ A ≈ `88,171, Example 7.5, A person purchases a machine on 1st January, 2009 and agrees to pay 10 installments each, of `12,000 at the end of every year inclusive, of compound rate of 15%. Find the present, value of the machine. [(1.15)10=4.016]., Solution:, Here n=10, a = 12,000 and i = 0.15, 1, a, Now P = i <1 - (1 + i) n F, 1, 12, 000, = 0.15 <1 - (1 + 0.15)10 F, 1, 12, 000, = 0.15 <1 - (1.15)10 F, 12, 00, 000, 1, :1 - 4.016 D, =, 15, 4.016 - 1, = 80,000 : 4.016 D, 3.016, = 80,000 : 4.016 D ≈ 60,080, ` P = ` 60,080, Example 7.6, A photographer purchases a camera on, installments. He has to pay 7 annual, , installments each of `36,000 right from the, date of purchase. If the rate of compound, interest is 16% then find the cost price, (present value) of the camera. [(1.16)7=2.828], Solution:, Here a = 36,000 ; n = 7 and i = 0.16, 1, a, Now P = (1+i) i <1 - (1 + i) n F, = (1+0.16), , 1, 36, 000 1 <, F, 0.16, (1 + 0.16) 7, , 1, 1.160, = 0.16 (36,000) <1 - (1 + 0.16) 7 F, 116 # 36000 :1 - 1 D, =, 2.828, 16, , 116 # 36000 : 2.828 - 1 D, 2.828, 16, 116 # 36, 000 # 1.828, =, 16 # 2.828, 116 # 36, 000 # 1828, =, ≈ 1,68,709., 16 # 2828, =, , ` P = ` 1,68,709, , Example 7.7, A person has taken a loan of ` 7,00,000, at 16% rate of interest from a finance, company. If the repayment period is of, 15 years then find the installment he has, to pay at the beginning of each month., [(1.0133)180=9.772]., Solution:, Here P = 7,00,000; n = 15; i = 0.16 and k = 12., 1, a, 1i, +, 1, e, o, b, l, i, > b1 + i lnk H, P=, k, k, k, .16 l, 700000= b1 + 012, d, , a, , 0.16, 12, , n >1 -, , 1, 15 # 12 H, 0, ., 16, b1 + 12 l, , 1, 12.16 12a 1 = b 12 lb 0.16 l > b 12.16 l180 H, 12, Financial Mathematics, , 07_11th_BM-STAT_Ch-7-EM.indd 165, , 165, , 21-04-2020 12:24:20 PM

Page 172 :
www.tntextbooks.in, , =, , 1, ]1216g a 1 16 ; ]1.0133g180 E, , 1, 1216, = 16 a :1 - 9.772 D, , 1216 9.772 - 1, = 16 a : 9.772 D, 1216 8.772, = 16 a : 9.772 D, ⇒a=, , 7, 00, 000 # 16 # 9772, ≈ ` 10,261, 1216 # 8772, , Example 7.8, The chairman of a society wishes to award a, gold medal to a student getting highest marks, in Business Mathematics and Statistics. If, this medal costs to ` 9,000 every year and, the rate of compound interest is 15%, then, what amount is to be deposited now., Solution:, Here a = 9,000 and i = 0.15, a, P= i, , 9, 00, 000, 15, , = 60,000., Therefore the amount to be deposited is, ` 60,000., Example 7.9, A limited company wants to create a, fund to help their employees in critical, circumstances. The estimated expenses, per month is `18,000. Find the amount to, be deposited by the company if the rate of, compound interest is 15%., Solution:, Here a = 18,000 ; i = 0.15 and k = 12., 166, , 18, 000, 0.15 12, , 18, 000, 0.15 # 12, 18, 00, 000 # 12, =, 15, = 14,40,000, , =, , Therefore the amount to be deposited is, `14,40,000., Example 7.10, Find the annual rate of interest, to get a, perpetuity of `675 for every half yearly from, the present value of `30,000., Solution :, Here P=30,000 ; a = 675 ; k = 2, i = ?, a, i k, , P== =, 30,000 =, , 18, 000, 0.15 12, , 675, i, 2, , 1350, i, 1350, 135, ⇒ i = 30, 000 = 3000 = 0.045, Rate of interest (i)= 0.045 × 100% = 4.5%, =, , 9000, = 0.15, =, , a, i k, , P== =, , Exercise 7.1, 1. F,  ind the amount of an ordinary annuity, of `3,200 per annum for 12 years at, the rate of interest of 10% per year., [(1.1)12 = 3.1384]., 2. I f the payment of `2,000 is made at the, end of every quarter for 10 years at the, rate of 8% per year, then find the amount, of annuity. [(1.02)40 =2.2080 ]., 3. Find the amount of an ordinary annuity of, 12 monthly payments of `1, 500 that earns, interest at 12% per annum compounded, monthly. [(1.01)12 = 1.1262 ]., , 11th Std. Business Mathematics and Statistics, , 07_11th_BM-STAT_Ch-7-EM.indd 166, , 21-04-2020 12:24:23 PM

Page 173 :
www.tntextbooks.in, , 4. A,  bank pays 8% per annum interest, compounded quarterly. Find the equal, deposits to be made at the end of each, quarter for 10 years to have `30,200?, [(1.02)40 =2.2080]., 5. A,  person deposits `2,000 at the end of, every month from his salary towards his, contributory pension scheme. The same, amount is credited by his employer also., If 8% rate of compound interest is paid,, then find the maturity amount at end of, 20 years of service., [(1.0067)240 = 4.9661]., 6. F,  ind the present value of `2,000 per, annum for 14 years at the rate of interest, of 10% per annum. If the payments are, made at the end of each payment period., [ (1.1)–14= 0.2632].,  ind the present value of an annuity of, 7. F, `900 payable at the end of 6th month for, 6 years. The money compounded at 8%, per annum. [(1.04)–12=0.6252]., 8. F,  ind the amount at the end of 12 years, of an annuity of `5,000 payable at the, beginning of each year, if the money, is compounded at 10% per annum., [(1.1)12 = 3.1384]., , possible for an individual to manage such, a large sum. Therefore the total sum of, money can be divided into equal parts, called shares. The holder of shares are, called shareholders., , 7.2.1 Types of shares:, There are two type shares namely common, (or equity) and preferred., Generally, the profit gained by the, company is distributed among their share, holders The preferred share holders have, a first claim on dividend. When they, have been paid, the remaining profit is, distributed among the common share, holders., , 7.2.2 Definitions, (i)   Capital stock is the total amount, invested to start a company., (ii) Th,  e share purchased by the individual is, also called stock., (iii) The persons who buy the shares are also, called stock holders, , 9. W,  hat is the present value of an annuity, due of `1,500 for 16 years at 8% per, annum? [(1.08)–16 = 0.2919]., , (iv)   Face value : The original value of a, share at which the company sells it, to investors is called a face value or, nominal value or par value. It is to be, noted that the original value of share, is printed on the share certificate., , 10. W,  hat is the amount of perpetual, annuity of `50 at 5% compound interest, per year?, , (v) M,  arket value : The price at which the, stock is bought or sold in the market is, called the market value (or cash value)., , 7.2 Stocks, Shares, Debentures, and Brokerage, To start any big business, a large sum, of money is needed. It is generally not, , The market value of a share keeps, on changing from time to time., Financial Mathematics, , 07_11th_BM-STAT_Ch-7-EM.indd 167, , 167, , 21-04-2020 12:24:23 PM

Page 174 :
www.tntextbooks.in, , Remarks:, (i), , , If the market value of a share is, greater than the face value then, the, share is said to be above par (or at, premium)., , (ii) If the market value of a share is the, same as its face value then, the share, is said to be at par., (iii) If the market value of a share is less, than the face value then, the share is, said to be below par (or at discount)., , Dividend :, The profit gained by a company,, distributed among their share holders is, called dividend. It is calculated on the, face value of the share., , Dividend is expressed as percentage., , Stock exchange:, The place where the shares are traded is, called the stock exchange (or) stock market., , Brokerage:, A broker who executes orders to buy and, sell shares through a stock market is called, Stock Broker. A fee or commission for their, service is called the brokerage., Brokerage is generally based on the face, value and expressed as a percentage., NOTE, (i) W,  hen the share is purchased, brokerage, is added to its market price., (ii) When the share is sold, brokerage is, subtracted from the market price., Example 7.11, Find the market value of 325 shares of face, value `100 at a premium of `18., Solution:, Face value of a share = `100, , Some useful results:, (i) Investment:, , Money invested = number of shares, , Premium per share = `18, , (ii) Income:, , Market value of 325 shares = number of, shares × M.V of a share, , × market value of a share., , Annual income = number of shares, × face value of a share × rate of, dividend., , (iii) Return percentage (or yield, percentage):, , Percentage of return, Income, = Investment # 100, , (iv) Number of shares:, , Number of shares purchased, Investment, = market value of a share, , 168, , M.V. of a share, , = `118, , = 325 × 118 = `38,350, Therefore market value of 325 shares is `38,350., Example 7.12, A man buys 500 shares of face value `100, at `14 below par. How much money does, he pay?, Solution:, Number of shares = 500, Face value of a share = `100, , 11th Std. Business Mathematics and Statistics, , 07_11th_BM-STAT_Ch-7-EM.indd 168, , 21-04-2020 12:24:23 PM

Page 175 :
www.tntextbooks.in, , Discount = `14, Market value of a share = 100 – 14 = `86, [face value – discount], Market value of 500 shares = Number of, shares × market value of a share, = 500 × 86 = 43,000, Market value of 500 shares = `43,000, Example 7.13, A person buys 20 shares of par value of `10, of a company which pays 9% dividend at, such a price that he gets 12% on his money., Find the market value of a share., Solution:, Face value of a share = `10, Face value of 20 shares = `200, 9, Dividend = 100 # 200 = `18, PNR, :S.I = 100 , N = 1D, 18 # 100, Investment = 1 # 12 = `150, [P = 100 × SI , N = 1], N×R, Market value of 20 shares = `150, The market value of one share = `, , 150, = `7.50, 20, , Example 7.14, If the dividend received from 10% of `25, shares is `2000. Find the number of shares., Solution:, Let x be the number of shares., Face value of ‘x’ shares = `25 x, 10, ×25x = `2,000, Now, 100, Þ x=, , 2000 ´ 100, = 800, 25 ´ 10, , Hence the number of shares = 800, , Example 7.15, Find the number of shares, which will give, an annual income of `3,600 from 12% stock, of face value `100., Solution:, Let ‘x’ be the number of shares., Face Value = `100, , Face value of ‘x’ shares = `100x, 12, ×100x = `3600, 100, , 12 x = 3600 & x = 300, Hence the number of shares = 300, Example 7.16, A man invest `96,000 on `100 shares at `80. If, the company pays him 18% as dividend, find, (i) the number of shares he bought, (ii) the dividend, (iii) percentage of return., Solution:, (i) Investment = `96,000, Face Value = `100, , Market Value = `80, The number of shares bought, Investments, = M.V of one share, 96, 000, = 80 = 1200 shares, (ii) Total dividend, = No. of shares × Rate of dividend, × Face value of a share, 18, = 1200× 100 ×100 = `21,600, (iii) Dividend on ` 96000 = `21600, 21, 600, Percentage of return = 96, 000 ×100, 45, = 2 = 22.5, Thus return on the shares = 22.5%, Financial Mathematics, , 07_11th_BM-STAT_Ch-7-EM.indd 169, , 169, , 21-04-2020 12:24:25 PM

Page 176 :
www.tntextbooks.in, , Example 7.17, A person bought 12% stock for `54,000 at a, discount of 17%. If he paid 1% brokerage,, find the percentage of his income., Solution:, Face value = `100, Market value = `(100 – 17 + 1) = `84, (12×100), ` percentage of his income =, 84, 100, 2, = 7 = 14 7, 2, ` % of income = 14 7 %, Example 7.18, Equal amounts are invested in 10% stock at, 89 and 7% stock at 90 (1% brokerage paid, in both transactions). If 10% stock bought, `100 more by way of dividend income than, the other, find the amount invested in each, stock., Solution:, Let x be the amount invested in each stock, Income on 10% stock at 89:, , F.V. = `100, M.V. = `90, x, investments, = 90, Number of shares =, M.V., x, 10, Annual income = 90 # 100 # 100, x, ... (1), , = 9 , Income on 7% stock at 90:, M.V. = `91, , x, investments, = 91, MV, x, 7, Annual income = 91 # 100 # 100, x, ... (2), , = 13 , x, x, Given : 9 - 13 = 100 ⇒ x = ` 2925, The amount invested in each stock ≈, `, `2,925., Number of shares =, , 170, , Example 7.19, A capital of a company is made up of, 1,00,000 preference shares with a dividend, rate of 16% and 50,000 ordinary shares.The, par value of each of preference and ordinary, shares is `10.The total profit of a company, is ` 3,20,000.If `40,000 were kept in reserve, and `20,000 were kept in depreciation fund,, what percent of dividend is paid to the, ordinary share holders., Solution:, F.V. = `10, Total face value of preference shares, = `1,00,000 × 10 = `10,00,000, Total face value of ordinary shares, = `50,000 × 10 = `5,00,000, Total dividend amount paid to shareholders, = `(3,20,000 – 40,000 – 20,000) = `2,60,000, Dividend for preference shares, 16, = 100 × 10,00,000 = `1,60,000, Dividend to ordinary shares, = 2,60,000 – 1,60,000 = `1,00,000, Dividend rate for ordinary share, Income, = Investment × 100%, 1, 00, 000, Dividend % = 5, 00, 000 ×100 = 20%, Example 7.20, A person sells a 20% stocks of face value, `10,000 at a premium of 42%. With the, money obtained he buys a 15% stock at a, discount of 22%. What is the change in his, income if the brokerage paid is 2%., Solution:, Step 1: For 20% stocks:, F.V. = `100, , 20, Income = 100 ×10000, = ` 2,000 , , ... (1), , 11th Std. Business Mathematics and Statistics, , 07_11th_BM-STAT_Ch-7-EM.indd 170, , 21-04-2020 12:24:27 PM

Page 177 :
www.tntextbooks.in, , Investment = `10,000, Face value = `100, , Market value = `100 + 42 – 2 = 140, investments, Number of shares =, FV, 10, 000, , = 100 = 100, = 100 × 140 = 14,000, Sales proceeds, Step 2: For 15% stocks:, M.V. = `100 – 22 + 2 = 80, investments, Number of shares =, FV, 14000, , = 80 = 175, 15, Income = 175 × 100 × 100 = `2625, Step 3: Change of income:, Change of income = ` 2625 – ` 2000 = ` 625, Example 7.21, Which is better investment: 12% `20 shares, at ` 16 (or) 15% ` 20 shares at `24 ., Solution:, Let the investment in each case be ` (16× 24), Step 1: Income on 12% shares:, , Income from 12 % `20 shares at ` 16, 12, = 16 ×(16× 24) = ` 288, Step 2: Income on 15% shares:, , Income from 15% `20 Shares at ` 24, 15, = 24 × (16× 24) = ` 240, Hence, the first investment is better., , Exercise 7.2, 1. F, ind the market value of 62 shares, available at `132 having the par value of, `100., 2. H,  ow much will be required to buy 125, of `25 shares at a discount of `7., , 3. I f the dividend received from 9% of `20, shares is `1,620, then find the number, of shares.,  ohan invested `29,040 in 15% of, 4. M, `100 shares of a company quoted at a, premium of 20%. Calculate,  (i) , the number of shares bought by, Mohan,  (ii) his annual income from shares, (iii) , the percentage return on his, investment,  man buys 400 of `10 shares at a, 5. A, premium of `2.50 on each share. If the, rate of dividend is 12%, then find,   (i) his investment, (ii) annual dividend received by him, (iii) rate of interest received by him on, his money, 6. S undar bought ` 4,500, 12% of `10 shares, at par. He sold them when the price rose, to `23 and invested the proceeds in `25, shares paying 10% per annum at `18., Find the change in his income., 7. A,  man invests `13,500 partly in 6% of, `100 shares at `140 and the remaining, in 5% of `100 shares at `125. If his, total income is `560, how much has he, invested in each?, 8. B, abu sold some `100 shares at 10%, discount and invested his sales proceeds, in 15% of `50 shares at `33. Had he sold, his shares at 10% premium instead of 10%, discount, he would have earned `450 more., Find the number of shares sold by him., 9. W,  hich is better investment? 7% of `100, shares at `120 (or) 8% of `100 shares, at `135., Financial Mathematics, , 07_11th_BM-STAT_Ch-7-EM.indd 171, , 171, , 21-04-2020 12:24:28 PM

Page 178 :
www.tntextbooks.in, , 10. W,  hich is better investment? 20% stock, at 140 (or) 10% stock at 70., , Exercise 7.3, Choose the correct answer, 1. Th,  e dividend received on 200 shares of, face value `100 at 8% is, (a) ` 1600, (b) ` 1000, (c) ` 1500, (d) ` 800, 2. W,  hat is the amount realised on selling, 8% stock of 200 shares of face value, `100 at `50., (a) ` 16,000, (b) ` 10,000, (c) ` 7,000, (d) ` 9,000, 3. A,  man purchases a stock of ` 20,000 of, face value ` 100 at a premium of 20%,, then investment is, (a) ` 20,000, (b) ` 25,000, (c) ` 24,000, (d) ` 30,000, 4. If a man received a total dividend of, ` 25,000 at 10% dividend rate on a stock, of face value `100, then the number of, shares purchased., (a) 3500, (b) 4500, (c) 2500, (d) 300, 5. Th,  e brokerage paid by a person on the, sale of 400 shares of face value `100 at, 1% brokerage, (a) ` 600, (b) ` 500, (c) ` 200, (d) ` 400, 6. P,  urchasing price of one share of face, value `100 available at a discount of, 1, 1, 9 2 % with brokerage 2 % is, 172, , (a) `89, (c) `91, , (b) `90, (d) `95, , 7. A,  person brought 100 shares of 9% stock, of face value `100 at a discount of 10%,, then the stock purchased is, (a) ` 9000, (b) ` 6000, (c) ` 5000, (d) ` 4000, 8. The % of income on 7 % stock at `80 is, (a) 9%, (b) 8.75%, (c) 8%, (d) 7%,  e annual income on 500 shares of face, 9. Th, value `100 at 15% is, (a) ` 7,500, (b) ` 5,000, (c) ` 8,000, (d) ` 8,500, 10. ` 5000 is paid as perpetual annuity every, year and the rate of C.I. 10 %. Then, present value P of immediate annuity is, (a) ` 60,000, (b) ` 50,000, (c) ` 10,000, (d) ` 80,000, 11. I f ‘a’ is the annual payment, ‘n’ is the, number of periods and ‘i’ is compound, interest for `1 then future amount of the, ordinary annuity is, a, (a) A= i ^1 + ih6^1 + ihn - 1@, a, (b) A = i 6^1 + ihn - 1@, a, (c) P = i, a, (d) P = i ^1 + ih61 - ^1 + i h-n@, 12. A,  invested some money in 10% stock at, `96. If B wants to invest in an equally, good 12% stock, he must purchase a, stock worth of, (a) ` 80, (b) ` 115.20, (c) ` 120, (d) ` 125.40, 13. A,  n annuity in which payments are, made at the beginning of each payment, period is called, , 11th Std. Business Mathematics and Statistics, , 07_11th_BM-STAT_Ch-7-EM.indd 172, , 21-04-2020 12:24:29 PM

Page 179 :
www.tntextbooks.in, , (a) Annuity due , (b) An immediate annuity, (c) perpetual annuity , (d) none of these, 14. , The present value of the perpetual, annuity of ` 2000 paid monthly at 10 %, compound interest is, (a) ` 2,40,000, (b) ` 6,00,000, (c) ` 20,40,000, (d) ` 2,00,400,  xample of contingent annuity is, 15. E, (a) Installments of payment for, a plot of land, (b) An endowment fund to give, scholarships to a student, (c) Personal loan from a bank, (d) All the above, , Miscellaneous Problems, 1. F,  ind the amount of annuity of `2000 payable, at the end of each year for 4 years of money is, worth 10% compounded annually., [(1.1)4 = 1.4641], 2. A, n equipment is purchased on an, installment basis such that `5000 on the, signing of the contract and four yearly, installments of `3000 each payable at the, end of first, second, third and the fourth, year. If the interest is charged at 5% p.a, find the cash down price. [(1.05)–4 = 0.8227], 3. (i) , Find the amount of an ordinary, annuity of `500 payable at the end, of each year for 7 years at 7% per, year compounded annually., [(1.07)7 = 1.6058], (ii) Calculate the amount of an ordinary, annuity of `10,000 payable at the, end of each half-year for 5 years at, 10% per year compounded halfyearly. [(1.05)10 = 1.6289], , (iii) , Find the amount of an ordinary, annuity of `600 is made at the end, of every quarter for 10 years at the, rate of 4% per year compounded, quarterly. [(1.01)40 = 1.4889], (iv) Find the amount of an annuity of, `2000 payable at the end of every, month for 5 years if money is, worth 6% per annum compounded, monthly. [(1.005)60 = 1.3489], 4. Naveen deposits `250 at the end of each, month in an account that pays an interest, of 6% per annum compounded monthly,, how many months will be required for, the deposit to amount to atleast `6390?, [log(1.1278) = 0.0523, log(1.005) = 0.0022], 5. A,  cash prize of `1,500 is given to the, student standing first in examination of, Business Mathematics by a person every, year. Find out the sum that the person, has to deposit to meet this expense. Rate, of interest is 12% p.a, 6. M,  achine A costs `15,000 and machine B, costs `20,000. The annual income, from A and B are `4,000 and `7,000, respectively. Machine A has a life of, 4 years and B has a life of 7 years. Find, which machine may be purchased., (Assume discount rate 8% p.a), [(1.08)–4 = 0.7350, (1.08)–7 = 0.5835], 7. V,  ijay wants to invest `27,000 in buying, shares. The shares of the following, companies are available to him. `100, shares of company A at par value; `100, shares of company B at a premium, of `25; `100 shares of company C at a, discount of `10; `50 shares of company, D at a premium of 20%. Find how many, Financial Mathematics, , 07_11th_BM-STAT_Ch-7-EM.indd 173, , 173, , 21-04-2020 12:24:29 PM

Page 180 :
www.tntextbooks.in, , shares will he get if he buys shares of, company (i) A (ii) B (iii) C (iv) D, 8. G,  opal invested `8,000 in 7% of `100, shares at `80. After a year he sold these, shares at `75 each and invested the, proceeds (including his dividend) in, 18% for `25 shares at `41. Find,  (i) his dividend for the first year,  (ii) , his annual income in the second, year, (iii) , The percentage increase in his, return on his original investment, 9. A,  man sells 2000 ordinary shares (par, value `10) of a tea company which pays, a dividend of 25% at `33 per share. He, , invests the proceeds in cotton textiles, (par value `25) ordinary shares at `44, per share which pays a dividend of 15%., Find (i) the number of cotton textiles, shares purchased and (ii) change in his, dividend income., 10. Th,  e capital of a company is made up of, 50,000 preferences shares with a dividend, of 16% and 25,000 ordinary shares. The, par value of each of preference and, ordinary shares is `10. The company, had a total profit of `1,60,000. If `20,000, were kept in reserve and `10,000 in, depreciation, what percent of dividend, is paid to the ordinary share holders, , Summary, Types of Annuities, , Certain Annuity, , Contingent Annuity, , Annuity Due (Payments are, made at the beginning of, each payment period), n, a, , , A = (1 + i ) (1 + i ) − 1, , , i, , Immediate Annuity (or), Ordinary Annuity (Payment, are made at the end of each, payment period), n, a, , A = (1 + i ) − 1, , i , , Present worth, −n , a, , P = (1 + i ) 1 − (1 + i ) , , , i, , Present worth, −n , a, P = 1 − (1 + i ) , , i , , 174, , Perpetual Annuity, , Present worth, a, P =, i, , 11th Std. Business Mathematics and Statistics, , 07_11th_BM-STAT_Ch-7-EM.indd 174, , 21-04-2020 12:24:29 PM

Page 181 :
www.tntextbooks.in, , zz, , Endowment or Scholarship fund, (i), (ii), , zz, , a, i, If the scholarship is awarded for a fixed period, say, n years, then, −n , a, P = 1 − (1 + i ) , , i , If the scholarship is awarded endlessly, then P =, , Face Value: The original Value of the share is called its nominal values or face, value or printed value., , zz, , Market Value: The price at which the share is sold (or) purchased in the capital, market through stock exchanges is called the market value., , zz, , A Share is called at par if the market value of the share is equal to its face (or), nominal value., , zz, , A share is said to be above par (or) at premium, if the market value of the share, is more than its nominal value., , zz, , A share is said to be below par (or) at discount, if the market value of the share, is less than its nominal value., , zz, , The part of the annual profit, which a share holder gets for his investment from, the company is called dividend., , zz, , Dividend is always declared on the face value of the share and the rate of dividend, is expressed as a percentage of the nominal value of a share per annum., , zz, , Annual income of a shareholders =, , n×r×F.V, 100, , Where n = number of shares with the shareholders, r = rate of dividend,, zz, , Annual income, Annual Return = investment in shares ×100%, , zz, , Number of Shares held =, , investment, M.V ^qr h F.V.qf qne share ^as the type qf investment h, (or), , Annual incqme, Tqtal F.V, = income from one share (or) = F.V.qf qne share, , Financial Mathematics, , 07_11th_BM-STAT_Ch-7-EM.indd 175, , 175, , 21-04-2020 12:24:30 PM

Page 183 :
www.tntextbooks.in, , ICT Corner, Financial Mathematics, Step – 1, Open the Browser type the URL Link given, below (or) Scan the QR Code., GeoGebra Workbook called “11th BUSINESS, MATHEMATICS and STATISTICS” will, appear. In that there are several worksheets, related to your Text Book., , Expected Outcome ⇒, Step-2, Select the work sheet “Dividend” Move the sliders to see the steps. Type the Amount, invested, Market value of one share and dividend rate in the box seen on right side and, proceed., Step 1, , Step 2, , Browse in the link, 11th Business Mathematics and Statistics, https://ggbm.at/qKj9gSTG (or) scan the QR Code, Financial Mathematics, , 07_11th_BM-STAT_Ch-7-EM.indd 177, , 177, , 21-04-2020 12:24:31 PM

Page 184 :
Chapter, , www.tntextbooks.in, , 8, , DESCRIPTIVE STATISTICS AND PROBABILITY, , Learning Objectives, After studying this chapter students will be, able to understand the following concepts., , •, •, •, •, •, •, •, , 8.1, , Measures of Central Tendency like, A.M, G.M & H.M, Relationship among the averages., Related positional measures like, Quartiles, Deciles and percentiles etc.,, Measures of Dispersion like Quartile, Deviation, mean Deviation, Relative measures like co-efficient, of Quartile Deviation, co-efficient of, Mean Deviation, Concept of conditional probability, and multiplication theorem., Baye’s theorem and its applications., , Measures of Central Tendency, , an Indian, average income,, etc,. Sir Ronald Fisher who, is known to be a father of, statistics and he made his, pioneering contributions, in the applications of, statistics in various disciplines., , 8.1.1 Average, - Recall, There are several measures of central, tendency for the data. They are, • Arithmetic Mean, • Median, • Mode, • Geometric Mean, • Harmonic Mean, , Arithmetic Mean (discrete case), Arithmetic, , Introduction:, One of the most, important, objectives, of Statistical analysis, is to get one single, value that describes the, characteristic of the, entire value for data., Sir Ronald Fisher, Such a value represent, the measure of central tendency for the, complete data. The word average is very, commonly used in day-to-day conversation., For example, we often talk of average boy, in a class, average height or average life of, 178, , mean, , of, , a, , set, , of, , observations is their sum divided by the, number of observations. The observation, are classified into a) Ungrouped data and, b) Grouped data., , a) Ungrouped data, (i) Direct Method:, X=, , X1 + X2 + X3 + ...Xn /X, = n, n, , where X is Arithmetic Mean, ∑X is, sum of all the values of the variable X and n, is number of observations., , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 178, , 21-04-2020 12:25:34 PM

Page 185 :
www.tntextbooks.in, , (ii) Short-cut method, The arithmetic mean can be calculated, by using any arbitrary value A as origin and, write d as the deviation of the variable X, then,, /d, X = A + n where d = X – A., , b) Grouped data, (i) Direct method, The formula for computing the mean, Rfx, is X = N, where f is frequency, X is the variable,, N = ∑ f i.e. total frequency., , (ii) Short-cut method, The arithmetic mean is computed by, applying the following formula:, Rfd, X = A+ N , where A is assumed mean,, d = X– A, N = ∑ f, , Arithmetic mean for Continuous case, The arithmetic mean may be computed, by applying any of the following methods:, (i) Direct method, (ii) Short-cut method, (iii) Step deviation method, (i) Direct method, When direct method is used arithmetic, mean is defined as, Rfm, X= N, Where m = midpoint of each of the class, interval,, f = the frequency of each class interval, N = ∑ f = total frequency, , (ii) Short-cut method, The arithmetic mean is computed by, applying the following formula., , Rfd, , X =A+ N, where A is assumed mean (or) arbitrary, value, d=m–A is deviations of mid-point from, assumed mean and N=∑f, , (iii) Step Deviation Method, , In case of grouped (or) continuous frequency, distribution, the arithmetic mean is, , Rfd, ]m - Ag, A, X = A + c N # c m , where d =, c, A is any arbitrary value (or) assumed mean, and c is the magnitude of class interval., , All the above three methods of, finding arithmetic mean in continuous, case gives us the same answer., , Mode:, , Mode is the value which repeats maximum, number of times among the given observations., , Median:, Median is exactly a middle value and it, exceeds and exceeded by the same number of, observations. Median is one of the positional, measure. Some other related positional, measures are also described below., NOTE, It is believed that the students might be, familiar with the above concepts and, our present syllabus continues, from the following., , 8.1.2 R, elated Positional Measures Quartiles, Deciles and Percentiles :, Besides median there are other measures, which divide a series into equal parts., Important amongest these are quartiles,, deciles and percentiles., Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 179, , 179, , 21-04-2020 12:25:35 PM

Page 186 :
www.tntextbooks.in, , (i) Quartiles:, , A measure which divides an array into four, equal parts is known as quartiles., Each portion contains equal number of, items. The first, second and third points, are termed as first quartile (Q1), second, quartile (Q2) (better named as median), and the third quartile(Q3). The first, quartile (Q1) or lower quartile, has 25%, of the items of the distribution below it, and 75% of the items are greater than it., Q2(median) the second quartile or median, has 50 % of the observations above it, and 50% of the observations below it., The upper quartile or third quartile(Q3), has 75% of the items of the distribution, below it and 25% of the items are above, it. Similarly the other two positional, measures can be defined., , (ii) Deciles:, , A measure which divides an array into ten, equal parts is known as deciles., That is Deciles are the values which divides, the series into ten equal parts. We get nine, dividing positions namely D1,D2,…,D9, which are called as deciles. Therefore there, are nine deciles. It is to be noted that D5 is, equal to median., , (iii) Percentiles:, A measure which divides an array into, hundred equal parts is known as Percentiles., That is percentiles are the values which, divides the series into hundred equal parts., We get 99 dividing positions P1 ,P2,…,P99, which are called as percentiles. Therefore, there are 99 percentiles. It is to be noted that, P50 is equal to median., 180, , 8.1.3 C,  omputations for Related, positional measure, The procedure for computing quartiles,, deciles and percentiles are the same as the, median., , (i) Ungrouped data:, Steps:, 1. Arrange the data either in ascending or, descending order of magnitude., 2. Apply the formula, n + 1 th, Q1= Size of b 4 l value, 3 ]n + 1g lth, value, Q3 = Size of b, 4, n+1, , th, , D1 = Size of b 10 l value, n + 1 th, D2 = Size of 2 b 10 l value, n + 1 th, P60 = Size of 60 b 100 l value, n + 1 th, P99 = Size of 99 b 100 l value, , Q2=D5= P50= Median, Example 8.1, Find D2 and D6 for the following series, 22, 4, 2, 12, 16, 6, 10, 18, 14, 20, 8, Solution:, Here n = 11 observations are arranged into, ascending order, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, n + 1 th, D2 = size of 2 b 10 l value, , n + 1 th, D6 = size of 6 b 10 l value, , D2 = size of 2.4th value ≈size of 2nd value = 4, D6 = size of 7.2th value ≈size of 7th value =14, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 180, , 21-04-2020 12:25:37 PM

Page 187 :
www.tntextbooks.in, , Example 8.2, Calculate the value of Q1, Q3, D6 and P50, from the following data, Roll No, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , Marks, , 20, , 28, , 40, , 12, , 30, , 15, , 50, , Solution:, Marks are arranged in ascending order, 12, , 15, , 20, , 28, , 30, , 40, , 50, , n = number of observations =7, n + 1 th, Q1 = Size of b 4 l value, 7 + 1 th, = Size of b 4 l value, , = Size of 2nd value = 15, , 3 ]n + 1g lth, value, Q3 = Size of b, 4, 3 # 8 th, = Size of b 4 l value, = Size of 6 value = 40, th, , 6 ]n + 1g th, D6 = Size of b 10 l value, 6 # 8 th, = Size of b 10 l value, = Size of 4.8th value, = Size of 5th value = 30, , 50 ]n + 1g, P50 = Size of b 100 l value, th, , = Size of 4th value = 28, Hence Q1=15, Q3=40, D6 =30 and P50=28, , (ii) Grouped data (discrete case):, Steps:, 1. A, rrange the data in ascending or, descending order of magnitude., 2. Find out cumulative frequencies.,  pply the formula:, 3. A, N + 1 th, Q1 = Size of b 4 l value, , 3, Q3 = Size of b, , ]N + 1g th, l value, 4, , Now look at the cumulative frequency (cf), column and find that total which is either, N+1, equal to b 4 l or next higher than that, and determine the value of the variable, corresponding to this. That gives the value, of Q1. Similarly Q3 is determined with, 3 ] N + 1g, reference of, value of the variable., 4, Example 8.3, Compute Q1, D2 and P90 from the following, data, Marks, , 10, , 20, , 30, , 40, , 50, , 60, , No. of, Students, , 4, , 7, , 15, , 8, , 7, , 2, , Solution:, Marks, X, , Frequency, f, , Cumulative Frequency, cf, , 10, , 4, , 4, , 20, , 7, , 11, , 30, , 15, , 26, , 40, , 8, , 34, , 50, , 7, , 41, , 60, , 2, , N= 43, , Table : 8.1, th, , N+1, Q1= Size of b 4 l, , value, , 43 + 1, th, 4 = 11 value = 20., 2 ]N + 1g lth, value, D2 = Size of b 10, 88, = Size of 10 = 8.8th value = 20., 90 ]N + 1g lth, value, P90 = Size of b 100, 3960, = Size of 100 = 39.6th value = 50., = Size of, , (iii) Grouped data (Continuous case):, In the case of continuous frequency, distribution, the classes are arranged either, in ascending or descending order and the, Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 181, , 181, , 21-04-2020 12:25:41 PM

Page 188 :
www.tntextbooks.in, , class corresponding to the cumulative, frequency (cf ) just equal or greater than, (N/4) is called Q1 class and the value of Q1is, obtained by the following formula:, , Solution:, CI, , f, , cf, , 10 − 20, , 12, , 12, , 20 − 30, , 19, , 31, , 30 − 40, , 5, , 36, , where L is the lower limit of the Q1 class,, , 40 − 50, , 10, , 46, , f is the frequency of the Q1 class,, , 50 − 60, , 9, , 55, , c is the magnitude of the Q1 class,, , 60 − 70, , 6, , 61, , pcf is the cumulative frequency of the pre Q1, class., , 70 – 80, , 6, , N= 67, , Q1 = L + f, , N, 4, , - pcf, p# c, f, , Similarly third quartile value can be obtained, by the same procedure with Q3class by the, following formula:, Q3 = L + f, , 3N, 4, , - pcf, p# c, f, , where L is the lower limit of the third, quartile class,, f is the frequency of the third quartile class,, c is the magnitude of the third quartile class,, pcf is the cumulative frequency of the pre, Q3 class., Similarly the same procedure is to be, followed for other positional measures such, as deciles and percentiles,, D4 = L + f, , 4N, 10, , P60 = L + f, , 60N, 100, , - pcf, p# c, f, , Frequency, , - pcf, p# c, f, , 182, , 19, , 5, , 10, , 9, , N th, 67, Q1= Size of b 4 l value = 4 = 16.75th value., Thus Q1 lies in the class (20 – 30) and its, corresponding values are L = 20;, N, 4 =16.75; pcf =12; f =19 ; c = 10, N, - pcf, f, p× c, Q1= L + 4, f, Q1 =20 + b, , 16.75 - 12 l, ×10 = 20 + 2.5 = 22.5, 19, 3N th, Q3 = Size of b 4 l value = 50.25th value, So Q3 lies in the class (50-60) corresponding, 3N, values are L = 50, b 4 l = 50.25; pcf = 46,, f = 9, c = 10, 3N, 4, , - pcf, p ×c, f, , Q3 =50+ b 50.259- 46 l × 10 = 54.72, , 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 – 80, 12, , Table : 8.2, , Q3 = L + f, , Example 8.4, Compute upper Quartiles, lower Quartiles,, D4 and P60, P75 from the following data., CI, , N = 67, , 6, , 6, , D4 =L + f, , 4N, 10, , - pcf, p ×c, f, , 4N th, D4 = Size of b 10 l value = 26.8th value., Thus D4 lies in the class (20 – 30) and its, corresponding values are, 4N, L = 20, 10 = 26.8; pcf = 12, f =19, c = 10., , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 182, , 21-04-2020 12:25:43 PM

Page 189 :
www.tntextbooks.in, , D4 = 20 + b, , 26.8 - 12 l, ×10 = 27.79, 19, , 75N th, P75 = Size of b 100 l value = 50.25th value., Thus P75 lies in the class (50 – 60) and its, corresponding values are, 75N, L = 50; 100 = 50.25; pcf = 46, f = 9,c = 10., P75= L + f, , 75N, 100, , - pcf, p ×c, f, , 50.25 - 46 l, = 50 + b, ×10 = 54.72, 9, , 8.1.4 Geometric mean, Geometric mean is defined as the nth root, of the product of n observations or values., If there are two observations, we take the, square root; if there are three observations, we have to take the cube root and so on, GM =, , n, , 1/n, X 1.X 2 . X 3 ..... X n = ^ X1, X2, X3,......Xn h, , where X 1 , X 2 , X 3 ,...., X n refer to the various, items of the series which are all greater than, zero and n refers number of observations., Thus the geometric mean of 3 values 2,3,4, would be, GM =, , 3, , (2) (3) (4) = 2.885, , When the number of items is three or more, the task of multiplying the numbers and, of extracting the root becomes excessively, difficult. To simplify calculations, logarithms, are used. Geometric mean is calculated as, follows:, log GM =, , (or ), , log X 1 + log X 2 + ........ + log X n, n, ,  Σ log X , log GM = ,  n , ,  Σ log X , ,, GM = Anti log ,  n , , where n is number of observation., , (i) In discrete observation,  Σ f log X , GM = Anti log ,  ; where N = Σf, , N, , (ii) In Continuous observation, GM = Anti log ;, , /f log m, E ; where m is, N, , midpoint and N = Σf, , Example 8.5, Daily income (in Rs) of ten families of a, particular place is given below. Find out GM, 85, 70, 15, 75, 500, 8, 45, 250, 40, 36., Solution:, X, , log X, , 85, , 1.9294, , 70, , 1.8451, , 15, , 1.1761, , 75, , 1.8751, , 500, , 2.6990, , 8, , 0.9031, , 45, , 1.6532, , 250, , 2.3979, , 40, , 1.6021, , 36, , 1.5563, Σ log X =, 17.6373, , Table : 8.3,  Σ log X , G M = Anti log , ; where n = 10,  n ,  17.6373 , G M = Anti log ,  10 , = Anti log(1.7637), , GM = 58.03, Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 183, , 183, , 21-04-2020 12:25:53 PM

Page 190 :
www.tntextbooks.in, , Example 8.6, Calculate the geometric mean of the data, given below giving the number of families, and the income per head of different classes, of people in a village of Kancheepuram, District., No. of, Families, , Income per head in, 1990 (Rs), , Landlords, , 1, , 1000, , Cultivators, , 50, , 80, , Landless labourers, , 25, , 40, , Money- lenders, , 2, , 750, , School teachers, , 3, , 100, , Shop-keepers, , 4, , 150, , Carpenters, , 3, , 120, , Weavers, , 5, , 60, , Class of people, , Income per head No of, Families log X, in 1990 (`), f, X, , 10-20, , 20-30, , 30-40, , 40-50, , No. of, Students, , 8, , 12, , 18, , 8, , 6, , Solution:, Marks, , m, , f, , log m, , f log m, , 0-10, , 5, , 8, , 0.6990, , 5.5920, , 10-20, , 15, , 12, , 1.1761, , 14.1132, , 20-30, , 25, , 18, , 1.3979, , 25.1622, , 30-40, , 35, , 8, , 1.5441, , 12.3528, , 40-50, , 45, , 6, , 1.6532, , 9.9192, , Σf log m = 67.1394, ,  Σ f log m , G M = Anti log , , , N,  67.1394 , = Anti log ,  52 , = Anti log (1.2911), , f log X, , 1000, , 1, , 3.0000, , 3.0000, , Cultivators, , 80, , 50, , 1.9031, , 95.1550, , Landless, labourers, , 40, , 25, , 1.6021, , 40.0525, , Money- lenders, , 750, , 2, , 2.8751, , 5.7502, , School teachers, , 100, , 3, , 2.0000, , 6.0000, , Shopkeepers, , 150, , 4, , 2.1761, , 8.7044, , Carpenters, , 120, , 3, , 2.0792, , 6.2376, , Weavers, , 60, , 5, , 1.7782, , 8.8910, , GM = 19.54, Specific uses of Geometric mean, The most useful application of, geometric mean is to average the rate of, changes. For example, from 2006 to 2008, prices increased by 5%, 10% and 18%, respectively. The average annual increase, ,  Σ f log X , G M = Anti log , , , N,  173.7907 , = Anti log , , , 93 , = Anti log (1.8687 ), , GM = 73.91, ,  5 + 10 + 18, , = 11 as given by, , 3, , is not 11% , , , Σ =173.7907, f log X =, 173.7907, , Table : 8.4, , 184, , 0-10, , Table : 8.5, , Landlords, , N = 93, , Marks, , N = 52, , Solution:, Calculation of Geometric Mean, Class of people, , Example 8.7, Compute the Geometric mean from the, data given below:, , the arithmetic average but 10.9% as, obtained by the geometric mean. This, average is also useful in measuring the, growth of population, because population, increases in geometric progression., , Example 8.8, Compared to the previous year the overhead, expenses went up by 32% in 1995, they, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 184, , 21-04-2020 12:25:59 PM

Page 191 :
www.tntextbooks.in, , increased by 40% in the next year and by 50%, in the following year. Calculate the average, rate of increase in overhead expenses over, the three years., Solution:, In averaging ratios and percentages,, geometric mean is more appropriate. Let us, consider X represents Expenses at the end, of the year., % Rise, , X, , log X, , 32, , 132, , 2.1206, , 40, , 140, , 2.1461, , 50, , 150, , 2.1761, Σ log X =, 6.4428, , Thus, HM =, ,  6.4428 , = Anti log ,  3 , = Anti log ( 2.1476), ,  1, 1, 1 ,  X + X + ..... + X , 1, 2, n, , When the number of items is large the, computation of harmonic mean in the, above manner becomes tedious. To simplify, calculations we obtain reciprocals of the, various items from the tables and apply the, following formulae:, , (i) In individual observations, HM =, , HM =, , Table : 8.6,  Σ log X , GM = Anti log ,  n , , n, , n,  1, 1, 1 , +, +, +, .....,  X, X2, X n , 1, , (or), , n,  1, Σ , X, , where n is number of observations or items, or values., , (ii) In discrete frequency distribution, HM =, , N, where,  f, Σ , X, , GM = 140.5, , N = total frequency= ∑f, , Average rate of increase in overhead, expenses, , 140.5 – 100 = 40.5 %, , (iii) I n continuous frequency, distribution, HM =, , Geometric mean cannot be, calculated if one of the observations is zero., , N,  f, Σ ,  m, , Where m is midpoint and N is total frequency, Example 8.9, , 8.1.5 Harmonic mean, Harmonic mean is defined as the reciprocal, of the arithmetic mean of the reciprocal of the, individual observations. It is denoted by HM., , Calculate the Harmonic Mean of the, following values:, 1, 0.5, 10, 45.0, 175.0, 0.01, 4.0, 11.2., Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 185, , 185, , 21-04-2020 12:26:05 PM

Page 192 :
www.tntextbooks.in, , Solution:, , HM =, , X, 1, , 1.0000, , 0.5, , 2.0000, , 10, , 0.1000, , 45, , 0.0222, , Value, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 175, , 0.0057, , Frequency, , 8, , 12, , 20, , 6, , 4, , 0.01, , 100.0000, , 4.0, , 0.2500, , 11.2, , 0.0893, , Example 8.11, Calculate Harmonic Mean for the following, data given below:, , Solution:, Calculation of Harmonic Mean, Value Mid-Point, , 1, Σ =, 103.4672, X, , Table : 8.7, n=8, n, 8, HM =, =, = 0.077, 103.467,  1, Σ , X, , Example 8.10, From the following data compute the value, of Harmonic Mean, Marks, , 10, , 20, , 25, , 40, , 50, , No. of, students, , 20, , 30, , 50, , 15, , 5, , Solution:, Calculation of Harmonic Mean, Marks, X, , No. of Students, f, , f, x, , 10, , 20, , 2.000, , 20, , 30, , 1.500, , 25, , 50, , 2.000, , 40, , 15, , 0.375, , 50, , 5, , 0.100, , N = 120, ,  f , Σ  =, 5.975, X, , Table : 8.8, 186, , N, 120, =, = 20.08, 5.975,  f, Σ , X, , 1, X, , m, , f, ,  f ,  , m, , 0-10, , 5, , 8, , 1.60, , 10-20, , 15, , 12, , 0.80, , 20-30, , 25, , 20, , 0.80, , 30-40, , 35, , 6, , 0.17, , 40-50, , 45, , 4, , 0.09, , N = 50, ,  f, Σ   =3.46,  m, , Table : 8.9, HM =, , N, 50, =, = 14.45, 3.46,  f, Σ ,  m, , Special applications of Harmonic Mean, The Harmonic Mean is restricted, in its field of usefulness. It is useful for, computing the average rate of increase, of profits of a concern or average speed, at which a journey has been performed, or the average price at which an article, has been sold. The rate usually indicates, the relation between two different types, of measuring units that can be expressed, reciprocally., For example, if a man walked, 20km in 5 hours, the rate of his walking, speed can be expressed, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 186, , 21-04-2020 12:26:07 PM

Page 193 :
www.tntextbooks.in, , of two speeds., , 20 km, = 4 km per hour, 5 hours, , i.e., X =, , where the units of the first term, is a km and the unit of the second term, is an hour or reciprocally,, , 30 km + 20 km, = 25 kmph, 2, , But this is not the correct average., So harmonic mean would be mean suitable, in this situation., Harmonic Mean of 30 and 20 is, , 5 hours 1, = hour per km ., 20 km 4, , HM =, , where the unit of the first term, is an hour and the unit of the second, term is a km., , 2, , 2, 2(60), =, = 24 kmph, 5,  1  1  5,   +    , 20, 30, 60, =, , 8.1.6 Relationship among the averages, , Example 8.12, An automobile driver travels from plain to hill, station 100km distance at an average speed, of 30km per hour. He then makes the return, trip at average speed of 20km per hour what, is his average speed over the entire distance, (200km)?, , In any distribution when the original items, differ in size, the values of AM, GM and HM, would also differ and will be in the following, order, AM ≥ GM ≥ HM, , Solution:, If the problem is given to a layman he is, most likely to compute the arithmetic mean, , If all the numbers X1, X2, …, Xn, are identical then, AM = GM = HM., , Example 8.13, Verify the relationship among AM, GM and HM for the following data, X, , 7, , 10, , 13, , 16, , 19, , 22, , 25, , 28, , f, , 10, , 22, , 24, , 28, , 19, , 9, , 12, , 16, , Solution:, X, , f, , Xf, , logX, , f logX, , f/X, , 7, , 10, 22, 24, 28, 19, 9, 12, 16, , 70, , 220, 312, 448, 361, 198, 300, 448, , 0.8451, 1, 1.1139, 1.2041, 1.2788, 1.3424, 1.3979, 1.4472, , 8.4510, 22.0000, 26.7346, 33.7154, 24.2963, 12.0818, 16.7753, 23.1545, , /f = N =140, , /fX = 2357, , 1.4286, 2.2000, 1.8462, 1.7500, 1.0000, 0.4091, 0.4800, 0.5714, f, / x =9.6852, , 10, 13, 16, 19, 22, 25, 28, , Table : 8.10, , /f log x =167.209, , Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 187, , 187, , 21-04-2020 12:26:10 PM

Page 194 :
www.tntextbooks.in, , ∑ fX, AM =, , 2357, =, = 16.84, 140, , N, , (ii) H,  ere the quantity of petrol consumed, is fixed i.e 2 litres. Here the arithmetic, mean will give the correct answer., ,  ∑ f log X ,  167.209 , 1944) distance, = 15.65 Covered, GM = Anti log , = Anti log , = Anti log(1.Total, , X = Total Petrol Consumed,  140 , N, , , ,  ∑ f log X ,  167.209 , log , = Anti log , = Anti log(1.1944) = 15.65, ,  140 , N, , , HM =, , N, 140, =, = 14.46, 9.6852,  f, Σ , X, , i.e. 16.84 > 15.65 > 14.46, ` AM > GM > HM, , NOTE, The Harmonic Mean is a measure of, central tendency for data expressed as rates, such as kilometres per hour, kilometres, per litre, periods per semester, tonnes per, month etc., , 40 × 2 + 30 × 2, = 35, 4, ∴ Average speed = 35 km per litre., X =, , Example 8.15, A person purchases tomatoes from each of, the 4 places at the rate of 1kg., 2kg., 3kg., and, 4kg. per rupee respectively .On the average,, how many kilograms has he purchased per, rupee?, Solution:, Since we are given rate per rupee, harmonic, mean will give the correct answer., HM=, , Example 8.14, A’s scooter gives an average of 40km a litre, while B’s scooter gives an average of 30km a, litre . Find out the mean, if, , Solution:, (i) Here the distance is constant. Hence, harmonic mean is appropriate., HM =, , =, , =, 188, , n, , 1 1, +, a b, , 2, 1, 1, +, 40 30, 2 × 120, 7, , =, , 2, 7, 120, , = 34.3 km per litre, , 1 1 1 1, + + +, a b c d, 4, =, 1 1 1 1, + + +, 1 2 3 4, , =, , 4 × 12, 25, , = 1.92 kg per rupee., , (i) each one of them travels 120 km., (ii) t he petrol consumed by both of them is, 2 litres per head., , n, , 8.2, , Measures of Dispersion, , Average gives us an idea on the point of the, concentration of the observations about, the central part of the distribution. If we, know the average alone we cannot form a, complete idea about the distribution as will, be clear from the following example., Consider the series:, (i) 7,8,9,10,11 (ii) 3,6,9,12,15 and (iii) 1,5,9,, 13,17. In all these cases we see that n, the, number of observations is 5 and the mean, is 9. If we are given that the mean of 5, observations is 9, we cannot form an idea, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 188, , 21-04-2020 12:26:17 PM

Page 195 :
www.tntextbooks.in, , as to whether it is the average of first series, or second series or third series or of any, other series of 5 observations whose sum, is 45. Thus we see that the measures of, central tendency are inadequate to give us, a complete idea of the distribution. So they, must be supported and supplemented by, some other measures. One such measure, is Dispersion, which provides the nature of, spreadness of the data., , respectively and Q3 – Q1 is called as inter, quartile range., , Literal meaning of dispersion is, “scatteredness” we study dispersion to, have an idea about the homogeneity or, heterogeneity of the distribution. In the, above case we say that series (i) is more, homogeneous (less dispersed) than the, series (ii) or (iii) or we say that series (iii) is, more heterogeneous (more scattered) than, the series (i) or (ii)., , Coefficient of quartile deviation can be, used to compare the degree of variation in, different distributions., , Various measures of dispersion can be, classified into two broad categories., The measures which express the, (a) , spread of observations in terms of, distance between the values of selected, observations. These are also termed as, distance measures., , Example: Range and interquartile range, (or) quartile deviation., (b) The measures which express the spread, of observations in terms of the average, of deviations of observation from some, central value., , Example: Mean deviation and Standard, deviation., , 8.2.1 Quartile Deviation, Quartile Deviation is defined as, 1, QD = 2 (Q3 – Q1). It may also be called as, semi-inter quartile. Where Q1 and Q3 are the, first and third quartiles of the distribution, , (i) Relative measures for QD, Quartile deviation is an absolute measure of, dispersion. The relative measure corresponding, to this measure, called the coefficient of quartile, deviation is calculated as follows:, Coefficient of QD =, , Q3 – Q1, Q3 + Q1, , (ii) C,  omputation of Quartile, Deviation, The process of computing quartile deviation, is very simple since we just have to compute, the values of the upper and lower quartiles, that is Q3 and Q1 respectively., Example 8.16, Calculate the value of quartile deviation and, its coefficient from the following data, Roll No., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , Marks, , 20, , 28, , 40, , 12, , 30, , 15, , 50, , Solution:, Marks are arranged in ascending order, 12, , 15, , 20, , 28, , 30, , 40, , 50, , n = number of observations =7, Q1= Size of b, , ]n + 1g th, 4 l value = Size of, , 7 + 1 th, b 4 l value = Size of 2nd value = 15, , Hence Q1 = 15, , 3 ]n + 1g lth, value = Size of, Q3 = Size of b, 4, # th, b 3 4 8 l value = Size of 6th value = 40, Hence Q3 = 40, Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 189, , 189, , 21-04-2020 12:26:18 PM

Page 196 :
www.tntextbooks.in, , Example 8.18, Compute Quartile deviation from the, following data, , 40 - 15, 1, QD = 2 (Q3 – Q1)=, = 12.5, 2, Coefficient of QD, Q -Q, 40 - 15 25, = Q3 + Q1 = 40 + 15 = 55 = 0.455, 3, 1, , CI 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 – 80, , Hence coefficient of QD = 0.455, , f, , Example 8.17, Compute coefficient of quartile, deviation from the following data, , 12, , 19, , 5, , 10, , 9, , 6, , 6, , Solution:, CI, , f, , cf, , 10 − 20, , 12, , 12, , Marks, , 10, , 20, , 30, , 40, , 50, , 60, , 20 − 30, , 19, , 31, , No. of, Students, , 4, , 7, , 15, , 8, , 7, , 2, , 30 − 40, , 5, , 36, , 40 − 50, , 10, , 46, , 50 − 60, , 9, , 55, , 60 − 70, , 6, , 61, , 70 – 80, , 6, , 67, , Solution:, Marks, , Frequency, , Cumulative Frequency, , X, , f, , cf, , 10, , 4, , 4, , 20, , 7, , 11, , 30, , 15, , 26, , Table : 8.12, , 40, , 8, , 34, , 50, , 7, , 41, , 60, , 2, , 43, , N th, 67 th, Q1 = Size of b 4 l value = b 4 l = 16.75th value, Thus Q1 lies in the class 20 – 30; and the, corresponding values are, N, L = 20, 4 = 16.75, pcf = 12, f = 19, c = 10, N, - pcf #, p c, Q1 = L + f 4, f, , N = ∑ f = 43, Table : 8.11, th, ,  N + 1,  value = Size 11th value = 20, Q1 = Size of ,  4 , th, ,  3 (N + 1), ,  value, Q3 = Size of , , , 4, , = Size of 33rd value = 40., 1, 40 – 20, (Q3 – Q1) =, = 10, 2, 2, Q – Q1, 40 – 20, =, Coefficient of QD = 3, Q3 + Q1, 40 + 20, , QD =, , , 190, , =, , 20, = 0.333, 60, , N = 67, , 16.75 − 12,  ,  × 10 = 20 + 2.5 = 22.5, , , 19, th,  3N , , Q3= Size of   value = 50.25th value,  4 , , Q1 = 20 + , , Thus Q3 lies in the class 50 – 60 and the, corresponding values are, 3N, L= 50, 4 = 50.25, pcf = 46, f = 9, c = 10, 3N, - pcf #, p c, Q3 = L + f 4, f, Q3, , = 50 + :, , 50.25 - 46 D, # 10 = 54.72, 9, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 190, , 21-04-2020 12:26:21 PM

Page 197 :
www.tntextbooks.in, , QD =, =, , 1, (Q3 – Q1), 2, , MD about Median =, , 54.72 – 22.5, = 16.11, 2, , , , \ QD = 16.11, , Mean Deviation (MD) is defined as the, average of the absolute difference between, the items in a distribution and the mean or, median of that series., , (i) C,  omputation of Mean Deviation Individual observations, If X 1 , X 2 , X 3 ,... X n are n given observation, then the mean deviation about mean or, median is as follows, , ∑, , X−X, n, , =, , ∑, , D, , n, , where D = X − X and n is the, number of observations., Σ x − median Σ | D |, =, n, n, D = X − median and n is, , D, , N, , For calculating Mean deviation in continuous, series we have to obtain the midpoints of the, various classes and take the deviations of, these mid points from mean or median., MD about Mean =, MD about Mean =, , ∑f, ∑, , MD about Mean =, , If the Mean deviation is computed from, Median then in that case D shall denote, deviations of the items from Median,, ignoring signs., , (ii) C,  omputation of Mean Deviation Discrete series, In discrete series the formula for calculating, mean deviation is, X−X, N, , =, , ∑f, , D, , N, , where D = X − X by ignoring negative, signs and N is total frequencies., , N, f D, , (or), , where M is a mid value, |D| = M − X and N, is the total frequencies., , where, the number of observations., NOTE, , M−X, , N, , MD about Median =, , ∑f, , ∑f, , N, , (iii) C,  alculation of Mean DeviationContinuous Series, , MD about Median =, , MD about Mean=, , X − Median, , where |D| = |X – Median| by ignoring, negative sign and N is total frequencies., , 8.2.2 Mean deviation, , MD about Mean = �, , =, , ∑f, , ∑f, , ∑f, , M − Median, N, , D, , N, , where M is a mid value, |D| = |M–Median|, (by ignoring negative sign) and N is the total, frequencies., , (iv) R,  elative Measure for Mean, Deviation, The relative measure corresponding to the, mean deviation is called the coefficient of, mean deviation and it is obtained as follows, Coefficient of MD about mean, =�, , Mean � Deviation, � about � Mean, Mean �, , Coefficient of MD about median, =�, , Mean � Deviation, � about � Median, Median, , Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 191, , (or), , 191, , 21-04-2020 12:26:26 PM

Page 198 :
www.tntextbooks.in, , However, in practice the arithmetic mean, is more frequently used in calculating the, mean deviation. If specifically stated to, calculate mean deviation about median,, median can be used., Example 8.19, Calculate the Mean Deviation about mean, and its coefficient of the income groups of, five, given below., Income Rs., , 4000 4200, , 4400, , 4600, , 4800, , Solution:, Calculation of Mean Deviation about Mean, Mean=, , ∑X, n, , =, ,  ( 7 + 1) , = size of , value,  2 , th, , = size of 4th item = 45, X, , |X – Median| = |X – 45|, , 20, , 25, , 30, , 15, , 40, , 5, , 45, , 0, , 55, , 10, , 60, , 15, , 80, , 35, ∑ |X – Median| = 105, , 22000, = 4400, 5, , Table : 8.14, , Income (Rs) X, , |D|=(X–4400), , 4000, , 400, , MD about Median =, , 4200, , 200, , 4400, , 0, , ∑, , , 4600, , 200, , Coefficient of MD about median, , 4800, , 400, , , , ∑ X = 22000, , ∑ D = 1200, , X − Median, n, , Table : 8.13, Median Deviation about Mean MD =, 1200, = 240, MD =, 5, , Coefficient of MD =, , ΣD, n, , 240, = 0.055, 4400, , Example 8.20, Calculate the mean deviation about median, and its relative measure for seven numbers, given below: 55, 45, 40, 20, 60, 80, and 30., Solution:, Arrange the values in ascending order 20,, 30, 40, 45, 55, 60, 80., 192, ,  ( n + 1) , value when n is odd, Median= size of ,  2 , th, , NOTE, , ∑ X − Median, , n, 105, = 15.0, =, 7, , =, , =, , 105, 7, , 15, = 0.33, 45, , Example 8.21, Calculate the Mean deviation about mean, for the following data., Size, , 2, , 4, , 6, , 8, , 10, , 12, , 14, , 16, , frequency, , 2, , 2, , 4, , 5, , 3, , 2, , 1, , 1, , Solution:, Calculation of Mean Deviation about Mean, From the Table 8.15, we get, X=, , ΣfX 160, =, =8, N, 20, , Mean Deviation about Mean, =, , Σf D 56, =, = 2.8, 20, N, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 192, , 21-04-2020 12:26:32 PM

Page 199 :
www.tntextbooks.in, ,  ( n + 1) , Median = size of , value,  2 , th, , X, , f, , fX, , |D| =X– 8, , f|D|, , 2, , 2, , 4, , 6, , 12, , 4, , 2, , 8, , 4, , 8, , , , 6, , 4, , 24, , 2, , 8, , = size of 60.5th item = 55, , 8, , 5, , 40, , 0, , 0, , MD about Median, , 10, , 3, , 30, , 2, , 6, , =, , 12, , 2, , 24, , 4, , 8, , 14, , 1, , 14, , 6, , 6, , 16, , 1, , 16, , 8, , 8, , N = 20, , RfX = 160, ,  (120 + 1) , = size of , 2 , , , ∑f, , X − Median, , th, , =, , N, , value, , ∑f, , D, , N, , Mean deviation about Median, X, , f, , |D| = |X– 55|, , f |D|, , 15, , 12, , 40, , 480, , 25, , 11, , 30, , 330, , Table : 8.15, , 35, , 10, , 20, , 200, , Example 8.22, Calculate the Mean deviation about median, and its relative measure for the following, data., , 45, , 15, , 10, , 150, , 55, , 22, , 0, , 0, , 65, , 13, , 10, , 130, , 75, , 18, , 20, , 360, , 85, , 19, , 30, , 570, , Σf | D | =56, , X, , 15, , 25, , 35, , 45, , 55, , 65, , 75, , 85, , frequency, , 12, , 11, , 10, , 15, , 22, , 13, , 18, , 19, , Solution:, Already the values are arranged in ascending, order then Median is obtained by the, following:, , Σf | D | =2220, , N = 120, , Table : 8.17, MD about Median=, , 2220, =18.5, 120, , X, , f, , cf, , Coefficient of mean deviation about median , MD about Median, , =, Median, , 15, , 12, , 12, , 25, , 11, , 23, , , , 35, , 10, , 33, , 45, 65, , 15, 22, 13, , 48, 70, 83, , 75, , 18, , 101, , 85, , 19, , 120, , 55, , N = 120, Table : 8.16, , =, , 18.5, = 0.34, 55, , Example 8.23, Find out the coefficient of mean deviation, about median in the following series, Age in, years, No. of, persons, , 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80, 20, , 25, , 32, , 40, , 42, , 35, , 8, , Calculations have to be made correct to two, places of decimals., Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 193, , 10, , 193, , 21-04-2020 12:26:36 PM

Page 200 :
www.tntextbooks.in, , Solution:, Calculation for median follows by the, following table, X, , f, , cf, , 0-10, , 20, 25, 32, 40, 42, 35, 10, 8, Table : 8.18, , 20, 45, 77, 117, 159, 194, 204, , 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, , N=212, , Then the mean deviation about median is to, be computed by the following, MD about Median =, , Coefficient of MD about Median, =, , M.D. about median 15.14, =, = 0.4064 = 0.41, Median, 37.25, , (corrected to two decimal places)., , NOTE, The above problem can also be solved for, mean deviation about mean instead of, median., , Exercise 8.1, , N 212, =, =106. Class interval corresponding, 2, 2, , to cumulative frequency 106 is (30 – 40). So,, the corresponding values from the median, class are L = 30, pcf = 77, f = 40 and c =10., N, b 2 l - pcf, p# c, Median= L + f, f, , 1. F,  ind the first quartile and third quartile, for the given observations., 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, ,  106 − 77 , ×10,  40 , , ∴ Median =37.25 (corrected to two places, , of decimals), , 4, , 4.5, , 5, , 5.5, , 6, , 6.5, , 7, , 7.5, , 8, , 10, , 18, , 22, , 25, , 40, , 15, , 10, , 8, , 7, , 3. F,  ind lower quartile, upper quartile, 7th, decile, 5th decile and 60th percentile for, the following frequency distribution., , X, , f, , M, , |D| = |X–37.25|, , f |D|, , 0-10, , 20, , 5, , 32.25, , 645, , Wages, , 10-20, , 25, , 15, , 22.25, , 556.25, , 20-30, , 32, , 25, , 12.25, , 392, , Frequency, , Calculations proceeded for mean deviation, about the median., , Size of, Shares, Frequency, , 2. Find Q1 , Q3, D8 and P67 of the following, data:, , Median = 30 + , , 30-40, , 40, , 35, , 2.25, , 90, , 40-50, , 42, , 45, , 7.75, , 325.5, , 50-60, , 35, , 55, , 17.75, , 621.25, , 60-70, , 10, , 65, , 27.75, , 277.5, , 70-80, , 8, , 75, , 37.75, , 302, , Table : 8.19, 194, , 10-20 20-30 30-40 40-50 50-60 60-70 70-80, 1, , 3, , 11, , 21, , 43, , 32, , 9, , 4. C,  alculate GM for the following table, gives the weight of 31 persons in sample, survey., Weight, (lbs):, Frequency, , Σf | D | =3209.5, , N=212, , ∑ f D 3209.5, =, = 15.14, 212, N, , 130 135 140 145 146 148 149 150 157, 3, , 4, , 6, , 6, , 3, , 5, , 2, , 1, , 1, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 194, , 21-04-2020 12:26:38 PM

Page 201 :
www.tntextbooks.in, , 5. Th,  e price of a commodity increased by, 5% from 2004 to 2005 , 8% from 2005, to 2006 and 77% from 2006 to 2007., Calculate the average increase from, 2004 to 2007?, 6. A,  n aeroplane flies, along the four sides, of a square at speeds of 100,200,300 and, 400 kilometres per hour respectively., What is the average speed of the plane, in its flight around the square., 7. A,  man travelled by car for 3 days. He, covered 480 km each day. On the first, day he drove for 10 hours at 48 km. an, hour. On the second day, he drove for, 12 hours at 40 km an hour and for the, last day he drove for 15 hours at 32 km., What is his average speed?, , 11. C,  alculate the quartile deviation and its, coefficient from the following data:, Age in, Years:, , 20, , 30, , 40, , 50, , 60, , 70, , 80, , No. of, Members:, , 13, , 61, , 47, , 15, , 10, , 18, , 36, , 12. C, alculate quartile deviation and its, relative measure from the following, data:, X, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 50-60, , f, , 5, , 10, , 13, , 18, , 14, , 8, , 13. C,  ompute mean deviation about median, from the following data:, Height in inches, , No. of students, , 58, , 15, , 59, , 20, , 60, , 32, , 61, , 35, , 62, , 35, , 63, , 22, , 64, , 20, , 65, , 10, , 66, , 8, ,  e monthly incomes of 8 families, 8. Th, in rupees in a certain locality are, given below. Calculate the mean, the, geometric mean and the harmonic, mean and confirm that the relations, among them holds true. Verify their, relationships among averages., Family:, , A, , B, , C, , D, , E, , F, , G, , H, , Income, (Rs.):, , 70, , 10, , 50, , 75, , 8, , 25, , 8, , 42, , 9. C,  alculate AM, GM and HM and also, verify their relations among them for, the following data, X, , 5, , 15, , 10, , 30, , 25, , 20, , 35, , 40, , f, , 18, , 16, , 20, , 21, , 22, , 13, , 12, , 16, , 10. C,  alculate AM, GM and HM from, the following data and also find its, relationship:, Marks:, , 0-10, , No. of, students:, , 5, , 10-20 20-30 30-40 40-50 50-60, 10, , 25, , 30, , 20, , 10, , 14. C,  ompute the mean deviation about, mean from the following data:, Class Interval:, , 0-5, , 5-10, , 10-15, , 15-20, , 20-25, , Frequency f, , 3, , 5, , 12, , 6, , 4, , 15. F, ind out the coefficient of mean, deviation about median in the following, series, Age in, years, No. of, persons, , 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80, 8, , 12, , 16, , 20, , 37, , 25, , Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 195, , 19, , 13, , 195, , 21-04-2020 12:26:39 PM

Page 202 :
www.tntextbooks.in, , 8.3, , Probability, , The word ‘probability’ or ‘chance’ is very, commonly used in day-to-day conversation, and generally people have a rough idea, about its meaning. For example, we come, across statements like, “Probably it may rain tomorrow”;, “The chances of teams A and B winning a, certain match are equal”;, All these terms – possible, probable, etc.,, convey the same sense, i.e., the event is not, certain to take place or, in other words, there, is uncertainty about happening of the event, in question. In Layman’s terminology the, , Impossible, , 1-in-6 chance, , Unlikely, , word ‘Probability’ thus can notes that there, is uncertainty about what has happened., However, in mathematics and statistics we, try to present conditions under which we, can make sensible numerical statements, about uncertainty and apply certain, methods of calculating numerical values of, probabilities., Galileo (1564-1642), an Italian mathematician, was the first man to attempt, quantitative measure of probability while, dealing with some problems related to the, theory of dice in gambling. The figure (8.1), given below represents the basic concepts of, probability., , Even chance, , Likely, , Certain, , 4-in-5 chance, , Fig. 8.1, , 8.3.1 Basic concepts of Probability, Recall, (i) Random Experiment, If an experiment or trial can be repeated under, the same conditions, any number of times, and it is possible to count the total number of, outcomes, but individual result ie., individual, outcome is not predictable,, then the, experiment is known as random experiment., Example: Tossing a coin, throwing a die,, selecting a card from a pack of playing, cards, etc., 196, , (ii) Outcome:, The result of a random experiment will be, called an outcome., , (iii) Trial and Event:, Any particular performance of a random, experiment is called a trial and outcome or, combinations of outcomes are termed as, events., , (iv) Exhaustive Events:, The total number of possible outcomes, of a random experiment is known as the, exhaustive events., , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 196, , 21-04-2020 12:26:39 PM

Page 203 :
www.tntextbooks.in, , (v) Favourable Events:, The number of cases favourable to an event, in a trial is the number of outcomes which, entail the happening of the event., , (vi) Mutually Exclusive events:, Events are said to be mutually exclusive if, the happening of any one of them precludes, the happening of all the others, ie., if no two, or more of them can happen simultaneously, in the same trial. Symbolically the event A, and B are mutually exclusive if A ∩ B = ∅ ., , (vii) Equally Likely Events:, Events (two or more) of an experiment are, said to be equally likely, if any one of them, cannot be expected to occur in preference to, the others., , (viii) Classical definition of Probability, If a random experiment or trial results in ‘n’, exhaustive, mutually exclusive and equally, likely outcomes (or cases), out of which m, are favourable to the occurrence of an event, E, then the probability ‘p’ of occurrence (or, happening) of E, usually denoted by P(E), is, given by, P = P (E) =, , Number of favourable cases, m, =, Total number of exhastive cases n, , James Bernoulli who was the, first person to obtain a quantitative, measure of uncertainty., , (ix) Properties, , Solution: The total possible outcomes of an, experiment {H,T}, Therefore n = 2, The favourable outcome for getting a head, {H}. Therefore m =1. Thus the required, probability is, P(getting a H}=, , m 1, =, n 2, , (x) Modern Definition of Probability, The modern approach to probability is purely, axiomatic and it is based on the set theory, concepts. In order to study, the theory of, probability with an axiomatic approach it is, necessary to define certain basic concepts., They are, (i), , Sample Space: Each possible outcome, of an experiment that can be repeated, under similar or identical conditions is, called a sample point and the collection, of sample points is called the sample, space, denoted by S., , (ii) Event: Any subset of a sample space is, called an event., (iii) , Mutually Exclusive events:, Two, events A and B are said to be mutually, exclusive events if A + B = z i.e., if A, and B are disjoint sets., Example: Consider S = { 1,2,3,4,5}, Let A = the set of odd numbers = {1,3,5}, and B = the set of even numbers = {2,4}, Then A∩B = z, , (ii) Sum of all the probability equal to 1., , (iv) Therefore the events A and B are, mutually exclusive., , (iii) If P(E) = 0 then E is an impossible event., , (xi) Observation:, , For example : A coin is tossed. Find the, probability of getting a head., , Statement meaning in terms of Set theory, approach., , (i) 0 ≤ P(E) ≤ 1, , Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 197, , 197, , 21-04-2020 12:26:40 PM

Page 204 :
www.tntextbooks.in, , (i) A , B & at least one of the events A or, B occurs, , (ii) The addition theorem may be extended, to any three events A,B,C and we have, , P ( A ∪ B ∪ C ) = P ( A) + P( B ) + P (C ) − P ( A ∩ B ) − P ( A ∩ C ) − P ( B ∩, P ( A ∪ B ∪ C ) = P ( A) + P( B ) + P (C ) − P ( A ∩ B ) − P ( A ∩ C ) − P ( B ∩ C ) + P ( A  B  C ), , (ii) A + B & both events A and B occurs, (iii) A + B & Neither A nor B occur, , (iv) A + B & Event A occurs and B does, not occur, , (xii) D,  efinition of Probability, (Axiomatic approach), , Let E be an experiment. Let S be a sample, space associated with E. With every event, in S we associate a real number denoted by, P(A) called the probability of the event A, satisfying the following axioms., , It is believed that the students might, be familiar with the above concepts, and our present syllabus continues, from the following., , 8.3.2 I ndependent and Dependent, events, (i) Independent Events, , Axiom 3 : If A1 , A2 ,⊃ , An be a sequence of n, mutually exclusive events in S then, , Two or more events are said to be, independent when the outcome of one, does not affect and is not affected by, the, other. For example, if a coin is tossed, twice, the result of the second throw would, in no way be affected by the result of the, first throw., , P ( A1  A2  …  An ) = P ( A1 ) + P ( A2 ) + …+ P ( An ), , (ii) Dependent events, , (xiii) Basic Theorems on probability, , Are those in which the occurrence or nonoccurrence of one event in any one trial, affects the other events in other trials., , Axiom 1 : P(A) ≥ 0, Axiom 2 : P(S) = 1, , Theorem 1:, , P( ∅ ) = 0 i.e., probability of an impossible, event is zero., , For example the probability of drawing a, , Theorem 2:, , But if the card drawn (queen) is not replaced, in the pack, the probability of drawing again, , Let S be the sample space and A be an event, in S, then P( A ) = 1– P(A)., , Theorem 3: Addition Theorem, If A and B are any two events, then, P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B ), , (xiv) Observation:, , (i) If the two events A and B are mutually, exclusive, then A∩B = ∅, ∴ P(A∩B) = 0, ⇒ P(A  B) =P(A) + P(B), 198, , queen from a pack of 52 cards is, , a queen is, , 4, 1, or, ., 52, 13, , 3, ., 51, , 8.3.3 Conditional Probability, If two events A and B are dependent, then, the conditional probability of B given that A, as occurred already is, P ( A  B), ; P(A) ≠ 0, P ( A), P ( A  B), ; P(B) ≠ 0, Similarly P(A/B) =, P ( B), , P(B/A) =, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 198, , 21-04-2020 12:26:46 PM

Page 205 :
www.tntextbooks.in, , (i) Multiplication Theorem:, The probability of the simultaneous happening, of two events A and B is given by, P(A∩B) = P(A).P(B/A) (or), P(A∩B) = P(B).P(A/B), NOTE, If A and B are two independent events, then P(A and B) = P(A∩B) = P(A) P(B),  e theorem can be extended to three or, Th, more independent events. Thus for three, events the theorem states that, , 3, 1, and P(B) = ,, 5, 5, 3 1, 3, then P(A∩B) = × =, 5 5, 25, , Given that P(A) =, , Example 8.26, Three coins are tossed simultaneously., Consider the events A ‘three heads or three tails’,, B ‘atleast two heads’ and C ‘at most two heads’, of the pairs (A,B), (A,C) and (B,C), which are, independent? Which are dependent?, Solution:, Here the sample space of the experiment is, , S = {HHH, HHT, HTH, HTT, THH, TTH,, P(A and B and C) = P ( � A  B  C� ) = � P ( A� ) � P ( B )� P (C ), THT, TTT}, P ( � A  B  C� ) = � P ( A� ) � P ( B )� P (C ), , Example 8.24, An unbiased die is thrown. If A is the event, ‘the number appearing is a multiple of 3’, and B be the event ‘the number appearing is, even’ number then find whether A and B are, independent?, Solution:, We know that the sample space is, S = {1,2,3,4,5,6}, Now, A= {3,6} ; B = { 2,4,6} then (A∩B) ={6}, P(A) =, P(B)=, , 2, 1, =, 6, 3, , 3, 1, 1, = and P(A ∩B) =, 6, 2, 6, , Clearly P(A ∩ B) =P(A) P(B), Hence A and B are independent events., Example 8.25, 3, 1, Let P(A) = and P(B) = . Find P(A ∩B), 5, 5, , if A and B are independent events., , Solution:, Since A and B are independent events then, P(A∩B) = P(A) P(B), , A = {Three heads or Three tails}, = {HHH, TTT}, B = {at least two heads}, = {HHH, HHT, HTH, THH} and, C = {at most two heads} = {HHT, HTH,, HTT, THH, TTH, THT, TTT}, Also (A∩B)= {HHH}; (A∩C) = {TTT} and, (B∩C) ={HHT, HTH, THH}, 2 1, 1, 7, = ; P(B ) = ; P(C) = and, 8 4, 2, 8, 1, 1, 3, P(A∩B)= , P(A∩C) = , P(B∩C) =, 8, 8, 8, 1 1 1, Also P(A). P(B) = ⋅ =, 4 2 8, 1 7 7, P(A). P(C) = ⋅ =, 4 8 32, 1 7 7, and P(B). P(C) = ⋅ =, 2 8 16, , ∴ P(A)=, , Thus, P(A∩B) = P(A). P(B), P(A∩C), , ≠ P(A) . P(C) and, , P(B∩C), , ≠ P(B). P(C), , Hence, the events (A and B) are independent,, and the events (A and C) and (B and C) are, dependent., Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 199, , 199, , 21-04-2020 12:26:54 PM

Page 206 :
www.tntextbooks.in, , Example 8.27, A can solve 90 per cent of the problems given, in a book and B can solve 70 per cent. What, is the probability that at least one of them, will solve a problem selected at random?, Solution:, Given the probability that A will be able to, solve the problem =, , 90, 9, =, and the, 100 10, , probability that B will be able to solve the, 70, 7, =, 100 10, 9, 7, i.e., P(A) =, and P(B) =, 10, 10, 9, 1, =, P ( A ) = 1–P(A) =, 10 10, 7, 3, =, P ( B ) = 1– P(B) =, 10 10, , problem =, , = 1− P ( A  B ) = 1− P ( � A  B � ), , = 1− P ( A � ) ⋅ P ( B ), 3, 97, =, 100 100, , Hence the probability that at least one of, them will solve the problem =, , 2, 2, =, 5+ 2 7, , ∴ The probability that both balls drawn are, , black is given by, P(A∩B) = P(A) P(B/A) =, , 3 2 3, × =, 8 7 28, , Example 8.29, In a shooting test the probability of hitting, the target are, , 3, 1, 2, for A, for B and for C., 4, 2, 3, , If all of them fire at the same target, calculate, the probabilities that, (i) All the three hit the target, (ii) Only one of them hits the target, (iii) At least one of them hits the target, Solution:, , P(at least one solve the problem) = P(A  B), , = 1−, , P(B/A) =, , 97, ., 100, , Example 8.28, A bag contains 5 white and 3 black balls., Two balls are drawn at random one after, the other without replacement. Find the, probability that both balls drawn are black., , 3, 1, 2, , P(B) = , P(C) =, 4, 2, 3, 3, 1, 1, 1, Then P ( A) = 1 − = ; P ( B ) = 1 − = and, 4 4, 2 2, 2, 1, P (C ) = 1 − =, 3 3, , Given P(A) =, , (i) P( A )=(all the three hit the targets), = P(A∩B∩C)=P(A)P(B)P(C), (since A,B,C hits independently), 3 1 2 1, = . . =, 4 2 3 4, (ii) P (only one of them hits the target), , = P {( A ∩ B � ∩ �C )� ∪ ( A ∩ B ∩ �C ) ∪ ( A ∩ B � ∩ C )}, , = P {( A  B �  �C ) + P � ( A  B  �C ) + � P ( A  � B  C )}, , P {( A  B �  �C ) + P � ( A  B  �C ) + � P ( A  � B  C )}, Solution:, Let A, B be the events of getting a black ball,  3 1 1  1 1 1  1 1 2 1, =  . .  + . .  + . .  =, in the first and second draw.,  4 2 3  4 2 3  4 2 3  4, Probability of drawing a black ball in the, (iii) P (at least one of them hit the target), first attempt is, = 1– P(none of them hit the target), 3, 3, =, P(A) =, = 1− P ( A�  � B  C � ), 5+3, , 8, , Probability of drawing the second black ball, given that the first ball drawn is black, 200, , = 1− � P ( A) P ( B ) P (C ), = 1–, , 1 23, =, 24 24, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 200, , 21-04-2020 12:27:08 PM

Page 207 :
www.tntextbooks.in, , Example 8.30, Find the probability of drawing a queen, a, king and a knave (Jack) in that order from, a pack of cards in three consecutive draws,, the card drawn not being replaced., , E2, , En, , ..., , C: the card is drawn is a knave(jack), , Er + A, , 4, 52, , Er, , P (drawing a king card given that a, queen card has been already drawn), , , = P(B/A) =, , 4, 51, , = P(C/AB) =, , n, , 4, 50, , Since they are dependent events, the required, probability of the compound event is, , P(ABC) = P(A) P(B/A) P(C/AB), 4 4 4, 64, × ×, =, 52 51 50 132600, , , , =, , , , = 0.00048, , A, Fig. 8.2, , P (drawing a knave card given that a, queen and a king cards have been drawn), , , E3, , E3 + A, , En + A, , B: the card drawn is a king, , P (drawing a queen card) = P(A) =, , E2 + A, , E1 + A, , ..., , Solution:, Let A : the card drawn is a queen, , E1, , where P ( A) = ∑P( Ei ) P ( A / Ei ), i =1, , Example 8.31, Bag I contains 3 red and 4 blue balls while, another Bag II contains 5 red and 6 blue, balls. One ball is drawn at random from one, of the bags and it is found to be red. Find the, probability that it was drawn from second Bag., , 8.3.4 Baye’s Theorem, , Solution:, Let E1 be the event of choosing the first bag,, E2 the event of choosing the second bag and, A be the events of drawing a red ball. Then P, , If E1 , E2 , E3 ..., En are a set of n mutually, , (E1) = P (E2) =, , exclusive and collectively exhaustive events, , Also P (A/E1) = P (drawing a red ball from, , with P(Ei) ≠ 0 (i = 1,2,3…, n), then for any, , arbitrary event A which is associated with, n, , sample space S =  i =1 Ei such that P(A) > 0,, we have, , P ( Ei / A) =, , P ( Ei ) P ( A / Ei ), , ∑, , P( Ei ) P ( A / Ei ), i =1, n, , ; i = 1,2,3,…, n, , Bag I) =, , 1, ., 2, , 3, 7, , and P (A/E2) = P (drawing a red ball from, Bag II) =, , 5, ., 11, , Now, the probability of drawing a ball from, Bag II, being given that it is red, is P(E2/A)., By using Baye’s theorem, we have, Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 201, , 201, , 21-04-2020 12:27:13 PM

Page 208 :
www.tntextbooks.in, , P(E2/A) =, , P ( E2 ) P ( A / E2 ), , ∑, , 2, , P( Ei ) P ( A / Ei ), , i =1, , P ( E2 ) P ( A / E2 ), =, P ( E1 ) P ( A / E1 ) + P ( E2 ) P ( A / E2 ), 1 5, ., 35, 2, 11, =, =,  1 3   1 5  68,  ×  +  × , 2 7, 2 11, , Example 8.32, X speaks truth 4 out of 5 times. A die is, thrown. He reports that there is a six. What, is the chance that actually there was a six?, Solution:, Let us define the following events., E1 : X speaks truth, E2 : X tells a lie, E : X reports a six, From the data given in the problem, we have, 4, 1, ; P (E2) =, 5, 5, 1, 5, P(E/E1) = ; P(E/E2) =, 6, 6, , P(E1), , =, , The required probability that actually there, was six (by Bayes theorem) is, , chosen at random at the end of a day and, found defective. What is the probability that, it comes from machines A1?, Solution:, P (A1) = P(that the machine A1 produces, screws) =, , 1000 1, =, 6000 6, , P (A2) = P(that the machine A2 produces, screws) =, , 2000 1, =, 6000 3, , P (A3) = P(that the machine A3 produces, screws) =, , 3000 1, =, 6000 2, , Let B be the event that the chosen screw is, defective, ∴ P (B/A1) = P(that defective screw from the, , machine A1) = 0.01, , P (B/A2) = P(that defective screw from the, machine A2) = 0.015 and, P (B/A3) = P(that defective screw from the, machine A3) = 0.02, We have to find P(A1/B), Hence by Baye’s theorem, we get, P (A1/B), =, , P ( A1 ) P ( B / A1 ), P ( A1 ) P ( B / A1 ) + P ( A2 ) P ( B / A 2 ) + P (A 3 ) P (B / A 3 ), , 4 1, ×, P ( E1 ) P ( E / E1 ), 4 1 , 5, 6, =, =   ( 0.01), P(E1/E) =, P ( E1 ) P ( E / E1 ) + P ( E2 ) P ( E / E2 )  4 =1   1 5  9 6,  × +1 ×   1 ,  1, 6 ) +,  5 6   5( 0.01, 0.015) +   ( 0.02), (, , , 4 1,  6,  3,  2, ×, P ( E1 ) P ( E / E1 ), 4, 5, 6, =, =, 0.01, 0.01 1, P ( E / E1 ) + P ( E2 ) P ( E / E2 )  4 1   1 5  9, =, =, =,  × + × , 0.01 + 0.03 + 0.06 0.1 10,  5 6   5 6 , , Example 8.33, A factory has 3 machines A1, A2, A3 producing, 1000, 2000, 3000 screws per day respectively., A1 produces 1% defectives, A2 produces 1.5%, , and A3 produces 2% defectives. A screw is, 202, , Exercise 8.2, 1. A family has two children. What is the, probability that both the children are girls, given that at least one of them is a girl?, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 202, , 21-04-2020 12:27:19 PM

Page 209 :
www.tntextbooks.in, , 2. A die is thrown twice and the sum of the, number appearing is observed to be 6., What is the conditional probability that, the number 4 has appeared at least once?, 3. An unbiased die is thrown twice. Let, the event A be odd number on the first, throw and B the event odd number on, the second throw. Check whether A and, B events are independent., 4. Probability of solving specific problem, 1, , 1, , independently by A and B are and, 2, 3, respectively. If both try to solve the, problem independently, find the, probability that the problem is,  (i) solved, (ii) exactly one of them solves the problem, 5. Suppose one person is selected at random, from a group of 100 persons are given in, the following, Title, , Psychologist, , Socialist, , Democrate, , Total, , Men, , 15, , 25, , 10, , 50, , Women, , 20, , 15, , 15, , 50, , Total, , 35, , 40, , 25, , 100, , , What is the probability that the man, selected is a Psychologist?, 6. Two urns contains the set of balls as given, in the following table, Title, , White, , Red, , Black, , Urn 1, , 10, , 6, , 9, , Urn 2, , 3, , 7, , 15, , One ball is drawn from each urn and find, the probability that,  (i) both balls are red, (ii) both balls are of the same colour., , 7. B,  ag I contains 3 Red and 4 Black balls, while another Bag II contains 5 Red and 6, Black balls. One ball is drawn at random, from one of the bags and it is found to, be red. Find the probability that it was, drawn from Bag I., 8. Th,  e first of three urns contains 7 White, and 10 Black balls, the second contains, 5 White and 12 Black balls and third, contains 17 White balls and no Black, ball. A person chooses an urn at random, and draws a ball from it. And the ball is, found to be White. Find the probabilities, that the ball comes from,    (i) the first urn,   (ii) the second urn,  (iii) the third urn, 9. Th,  ree boxes B1, B2, B3 contain lamp, bulbs some of which are defective. The, defective proportions in box B1, box B2, and box B3 are respectively, , 1 1, 3, , and ., 2 8, 4, , A box is selected at random and a bulb, drawn from it. If the selected bulb, is found to be defective, what is the, probability that box B1 was selected?, , 10. Th,  ree horses A, B, C are in race. A is, twice as likely to win as B and B is twice, as likely to win as C. What are their, respective probabilities of winning?, 11. A,  die is thrown. Find the probability of, getting,  (i) a prime number, (ii) a number greater than or equal to 3, 12. T,  en cards numbered 1 to 10 are placed in, a box, mixed up thoroughly and then one, card is drawn randomly. If it is known, that the number on the drawn card is, Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 203, , 203, , 21-04-2020 12:27:21 PM

Page 210 :
www.tntextbooks.in, , more than 4. What is the probability that, it is an even number?,  ere are 1000 students in a school out of, 13. Th, which 450 are girls. It is known that out, of 450, 20% of the girls studying in class, XI. A student is randomly selected from, 1000 students. What is the probability, that the selected student is from class, XI given that the selected student is a, girl?rom a pack of 52 cards, two cards are, drawn at random. Find the probability, that one is a king and the other is a queen.,  rom a pack of 52 cards, two cards are, 14. F, drawn at random. Find the probability, that one is a king and the other is a queen.,  card is drawn from a pack of playing, 15. A, cards and then another card is drawn, without the first being replaced. What is, the probability of drawing, (i) two aces, (ii)  two spades,  company has three machines A, B, C, 16. A, which produces 20%, 30% and 50% of, the product respectively. Their respective, defective percentages are 7, 3 and 5., From these products one is chosen and, inspected. If it is defective what is the, probability that it has been made by, machine C?10., , Exercise 8.3, Choose the correct answer:, 1. , Which of the following is positional, measure?, (a) Range, (b) Mode, (c) Mean deviation (d) Percentiles, 204, , 2. When calculating the average growth of, economy, the correct mean to use is?, (a) Weighted mean, (b) Arithmetic mean, (c) Geometric mean, (d) Harmonic mean, 3. When an observation in the data is zero,, then its geometric mean is, (a) Negative, (b) Positive, (c) Zero, (d) Cannot be, calculated, 4. The best measure of central tendency is, (a) Arithmetic mean , (b) Harmonic mean, (c) Geometric mean , (d) Median, 5. , The harmonic mean of the numbers, 2,3,4 is, (a), , 12, 13, , (b) 12, , (c), , 36, 13, , (d), , 13, 36, , 6. The geometric mean of two numbers 8, and 18 shall be, (a) 12, (b) 13, (c) 15, (d) 11.08, 7. T, he correct relationship among, A.M.,G.M.and H.M.is, (a) A.M.<G.M.<H.M., (b) G.M.≥A.M.≥H.M., (c) H.M.≥G.M.≥A.M., (d) A.M.≥G.M.≥H.M., 8. Harmonic mean is the reciprocal of, (a) Median of the values., (b) Geometric mean of the values., (c) Arithmetic mean of the reciprocal, of the values., (d) Quartiles of the values., 9. Median is same as, (b) Q2, (a) Q1, , (c) Q3, , (d) D2, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 204, , 21-04-2020 12:27:23 PM

Page 211 :
www.tntextbooks.in, , 10. The median of 10,14,11,9,8,12,6 is, (a) 10, (b) 12, (c) 14, (d) 9, 11. The mean of the values 11,12,13,14 and 15 is, (a) 15, (b) 11, (c) 12.5, (d) 13, 6n, , , then, 12. If the mean of 1,2,3, …, n is, 11, the value of n is, (a) 10, (b) 12, (c) 11, (d) 13, 13. H,  armonic mean is better than other, means if the data are for, (a) Speed or rates., (b) Heights or lengths., (c) Binary values like 0 and 1., (d) Ratios or proportions., 14. The first quartile is also known as, (a) median., (b) lower quartile., (c) mode., (d) third decile, 15. I f Q1 = 30 and Q3 = 50, the coefficient of, quartile deviation is, (a) 20, (b) 40, (c) 10, (d) 0.25, 16. If median = 45 and its coefficient is 0.25,, then the mean deviation about median is, (a) 11.25 (b) 180 (c) 0.0056 (d) 45, 17. Th,  e two events A and B are mutually, exclusive if, (a) P(A∩B) = 0 (b) P(A∩B) =1, (c) P(A  B) = 0 (d) P(A  B) = 1, 18. The events A and B are independent if, (a) P(A∩B) = 0 , (b) P(A∩B) = P(A)×P(B), (c) P(A∩B) = P(A)+P(B), (d) P(A  B) = P(A)×P(B), 19. I f two events A and B are dependent then, the conditional probability of P(B/A) is, (a) P(A) P(B/A), P ( A  B), (c), P ( A), , (b), , P ( A  B), P ( B), , (d) P(A) P(A/B), , 20. Th,  e probability of drawing a spade from, a pack of card is, (a) 1/52 (b) 1/13 (c) 4/13 (d) 1/4, 21. If the outcome of one event does not, influence another event then the two, events are, (a) Mutually exclusive, (b) Dependent, (c) Not disjoint, (d) Independent, 22. Let a sample space of an experiment, n, be S = {E1 , E2 ,…, En } , then ∑P( Ei ) is, i =1, equal to, 1, 1, (d), (a) 0 (b) 1 (c), 2, , 3, , 23. Th,  e probability of obtaining an even, prime number on each die, when a pair, of dice is rolled is, (a) 1/36 (b) 0 (c) 1/3 (d) 1/6, 24. Probability of an impossible event is, (a) 1 (b) 0 (c) 0.2 (d) 0.5,  robability that at least one of the events, 25. P, A, B occur is, (b) P(A∩B), (a) P(A  B), (d) (A  B), (c) P ] A/Bg, , Miscellaneous Problems, 1. Find out the GM for the following, Yield of Rice (tones), , No. of farms, , 7.5-10.5, , 5, , 10.5-13.5, , 9, , 13.5-16.5, , 19, , 16.5-19.5, , 23, , 19.5-22.5, , 7, , 22.5-25.5, , 4, , 25.5-28.5, , 1, , Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 205, , 205, , 21-04-2020 12:27:26 PM

Page 212 :
www.tntextbooks.in, , 2. A,  n investor buys ` 1,500 worth of shares, in a company each month. During the, first four months he bought the shares, at a price of ` 10, ` 15, ` 20 and ` 30 per, share. What is the average price paid, for the shares bought during these four, months? Verify your result., , 7. A committee of two persons is formed, from 3 men and 2 women. What is the, probability that the committee will have, (i) No woman, (ii) One man, (iii) No man, 8., ,  , B and C was 50% , 30% and 20% of the, A, cars in a service station respectively. They, fail to clean the glass in 5% , 7% and 3% of, the cars respectively. The glass of a washed, car is checked. What is the probability that, the glass has been cleaned?, , 9., ,  ata on readership of a magazine indicates, D, that the proportion of male readers over, 30 years old is 0.30 and the proportion, of male reader under 30 is 0.20. If the, proportion of readers under 30 is 0.80., What is the probability that a randomly, selected male subscriber is under 30?, , 3. C,  alculate Mean deviation about median, of the following data., Class interval : 0-10 10-20 20-30 30-40 40-50 50-60, Frequency :, , 6, , 7, , 15, , 16, , 4, , 2, , 4. C,  alculate Mean deviation about Mean, of the following data., X, , 2, , 5, , 6, , 8, , 10, , 12, , f, , 2, , 8, , 10, , 7, , 8, , 5, , 5. C, alculate Quartile deviation and, Coefficient of Quartile deviation of the, following data., Marks:, No. of, students:, , 0, , 10, , 20, , 30, , 40, , 50, , 60, , 70, , 150 142 130 120 72, , 30, , 12, , 4, , 6. In a screw factory machines A, B, C, manufacture respectively 30%, 40% and, 30% of the total output of these 2% , 4%, and 6% percent are defective screws., A screws is drawn at random from the, product and is found to be defective., What is the probability that it was, manufactured by Machine C?, , 206, , 10. G,  un 1 and Gun 2 are shooting at the, same target. Gun 1 shoots on the, average nine shots during the same time, Gun 2 shoots 10 shots. The precision, of these two guns is not the same. On, the average, out of 10 shots from Gun, 2 seven hit the target. In the course, of shooting the target has been hit, by a bullet, but it is not known which, Gunshot this bullet. Find the chance, that the target was hit by Gun 2?, , 11th Std. Business Mathematics and Statistics, , 08_11th_BM-STAT_Ch-8-EM.indd 206, , 21-04-2020 12:27:26 PM

Page 213 :
www.tntextbooks.in, , Summary, , zz, , A measure which divides an array into four equal parts in known as quartiles., , zz, , A measure which divides an array into ten equal parts is known as deciles., , zz, , A measure which divides an array into hundred equal parts is known as percentiles, , zz, , Q2 = D5 = P50 = Median, , zz, , Inter quartile range = Q3 – Q1, , zz, , QD =, , zz, , Harmonic mean =, , zz, , Mean Deviation for Individual series MD =, , zz, , The conditional probability of an event A given the occurrence of the event B is, P^ A + Bh, , P ^B h ! 0, given by P^A/Bh =, P ^B h, , zz, , Baye’s Theorem:, If E1 , E2 , E3 ,..., En are a set of n mutually exclusive and collectively exhaustive events, then for any arbitrary event A which is associated with, with P(Ei) ≠ 0 (i =1,2,3…,n),, n, sample space S =  Ei such that P(A) > 0, we have, , Q3 –Q1, 2, n, 1, 1, 1, 1, +, +, + ⋅⋅⋅, X1 X 2 X 3, Xn, , i =1, , , where, , P( Ei / A) =, , P ( Ei ) P ( A / Ei ), , =, , ∑ i =1P( Ei ) P ( A / Ei ), n, , n, 1, , ∑X, ∑ X−X, n, , =, , ∑D, n, , ; i =1,2,3…, n, , n, , P(A) = P ( A) = ∑P( Ei ) P ( A / Ei ), i =1, , Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 207, , 207, , 21-04-2020 12:27:30 PM

Page 215 :
www.tntextbooks.in, , ICT Corner, Descriptive statistics and probability, Step – 1, Open the Browser type the URL Link, given below (or) Scan the QR Code., GeoGebra Workbook called “11th, BUSINESS MATHEMATICS and, STATISTICS” will appear. In that, there are several worksheets related to, your Text Book., , Expected Outcome ⇒, , Step-2, Select the work sheet “Probability-Bayes theorem” Find each probabilities step by step as, shown and Click on the respective boxes to see the answers., Step 1, , Step 2, , Browse in the link, 11th Business Mathematics and Statistics:, https://ggbm.at/qKj9gSTG (or) scan the QR Code, , Descriptive statistics and probability, , 08_11th_BM-STAT_Ch-8-EM.indd 209, , 209, , 21-04-2020 12:27:30 PM

Page 216 :
www.tntextbooks.in, , Chapter, , 9, , CORRELATION AND REGRESSION ANALYSIS, , Learning Objectives, After studying this chapter students will, be able to understand, • , Concept of Karl Pearson’s correlation, co-efficient and the methods of, computing it., • Spearman’s Rank correlation co-efficient, • Concept of Regression and Regression, co-efficients, • Regression lines both x on y and y on x., , Correlation is the statistical analysis which, measures and analyses the degree or extent, to which two variables fluctuate with, reference to each other., , 9.1.1 Meaning of Correlation, The term correlation refers to the, degree of relationship between two or more, variables. If a change in one variable effects, a change in the other variable, the variables, are said to be correlated., , 9.1.2 Types of correlation, Correlation is classified into many, types, but the important are, (i) Positive, (ii) Negative, , 9.1 Correlation, Introduction, In the previous, Chapter we have studied, , Positive and negative correlation, depends upon the direction of change of the, variables., , the characteristics of only, one variable; example,, marks, weights, heights,, rainfalls,, sales, etc., , prices,, , ages,, , This type, , Karl Pearson, , of analysis is called univariate analysis., Sometimes we may be interested to find if, there is any relationship between the two, variables under study., , For example, the, , price of the commodity and its sale, height, of a father and height of his son, price and, demand, yield and rainfall,, height and weight and so, on. Thus the association, of any two variables is, known as correlation., 210, , Positive Correlation, If two variables tend to move together, in the same direction that is, an increase in, the value of one variable is accompanied by, an increase in the value of the other variable;, or a decrease in the value of one variable is, accompanied by a decrease in the value of, the other variable, then the correlation is, called positive or direct correlation., , Example, (i), (ii), (iii), (iv), ,  e heights and weights of individuals, Th, Price and Supply, Rainfall and Yield of crops, The income and expenditure, , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 210, , 21-04-2020 12:28:06 PM

Page 217 :
www.tntextbooks.in, , Negative Correlation, , Simple correlation, , If two variables tend to move together, in opposite direction so that an increase, or decrease in the values of one variable, is accompanied by a decrease or increase, in the value of the other variable, then the, correlation is called negative or inverse, correlation., , The correlation between two variables, is called simple correlation. The correlation, in the case of more than two variables is, called multiple correlation., , Example, (i), , (i), , Scatter diagram, , (ii) Karl Pearson’s Coefficient of, Correlation, , Price and demand, , (ii) Repayment period and EMI, , 9.1.3 Scatter Diagram, , (iii) Yield of crops and price, , No Correlation, Two variables are said to be, uncorrelated if the change in the value of, one variable has no connection with the, change in the value of the other variable., , For example, We should expect zero correlation, (no correlation) between weight of a person, and the colour of his hair or the height of a, person and the colour of his hair., , Y, , 0, , The following are the mathematical methods, of correlation coefficient, , Let (X1, Y1), (X2, Y2) … (XN , YN ) be, the N pairs of observation of the variables, X and Y. If we plot the values of X along, x - axis and the corresponding values of, Y along y-axis, the diagram so obtained, is called a scatter diagram. It gives, us an idea of relationship between X, and Y. The type of scatter diagram, under a simple linear correlation is given, below., , Y, , Positive Correlation, , X, , 0, , Y, , Negative Correlation X, , 0, , No Correlation, , X, , Fig 9.1, , (i), , If the plotted points show an upward trend, the correlation will be positive., , (ii), , If the plotted points show a downward trend, the correlation will be negative., , (iii) If the plotted points show no trend the variables are said to be uncorrelated., Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 211, , 211, , 21-04-2020 12:28:08 PM

Page 218 :
www.tntextbooks.in, , 9.1.4 K,  arl Pearson’s Correlation, Coefficient, Karl Pearson, a great biometrician, and statistician, suggested a mathematical, method for measuring the magnitude of, linear relationship between two variables, say X and Y. Karl Pearson’s method is, the most widely used method in practice, and is known as Pearsonian Coefficient of, Correlation. It is denoted by the symbol ‘r’, and defined as, cov ^ X, Y h, ,, r=, vx v y, 1, where cov(X, Y) = N / ] X - X g_Y - Y i, 2, , N, , 1, ^ Xi - X h, N i/, =1, , vx =, , N, , 2, , N, , i =1, , N, , / ^ Xi - X h2 / ^Yi - Y h2, , i=1, , i=1, , Interpretation of Correlation, coefficient:, Coefficient of correlation lies between, –1 and +1. Symbolically, –1≤ r ≤ + 1, • When r =+1 , then there is perfect positive, correlation between the variables., • When r=–1 , then there is perfect negative, correlation between the variables., 212, , (i) When deviations are taken from Mean, Of all the several mathematical, methods of measuring correlation, the, Karl Pearson’s method, popularly known, as Pearsonian coefficient of correlation, is, most widely used in practice., , ∑ (X, , − X ) (Yi − Y ), =, i =1, , N, , i, , ∑ ( X i − X )2, , N, , ∑ (Yi − Y )2, , ∑ xy, ∑x ∑y, 2, , 2, , =i 1 =i 1, , / ^ Xi - X h^Yi - Y h, , N, , Methods of computing Correlation, Coefficient, , =, r, , Hence the formula to compute Karl, Pearson Correlation coefficient is, N, 1 / ^ X - X h^Y - Y h, i, N i=1 i, r=, N, N, 1 / ^ X - X h2 1 / ^Y - Y h2, N i=1 i, N i=1 i, r=, , Thus, the coefficient of correlation, describes the magnitude and direction of, correlation., , N, , 1, ^Yi - Y h, N i/, =1, , vy =, , • When r=0, then there is no relationship, between the variables, that is the variables, are uncorrelated., , Where x = ^ Xi - X h and y = ^Yi - Y h ;, i = 1, 2 ... N, This method is to be applied only, when the deviations of items are taken from, actual means., , Steps to solve the problems:, (i) Find out the mean of the two series that, is X and Y ., (ii) Take deviations of the two series from, X and Y respectively and denoted, by x and y., (iii) Square the deviations and get the total, of the respective squares of deviation, of x and y respectively and it is denoted, by ∑x2 and ∑y2 ., (iv) Multiply the deviations of x and y and, get the total and it is denoted by ∑xy., (v) Substitute the values of ∑xy, ∑x2 and, ∑y2 in the above formula., , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 212, , 21-04-2020 12:28:11 PM

Page 219 :
www.tntextbooks.in, , Example 9.1, Calculate Karl Pearson’s coefficient of, correlation from the following data:, X:, , 6, , 8, , 12, , 15, , 18, , 20, , 24, , 28, , 31, , Y:, , 10, , 12, , 15, , 15, , 18, , 25, , 22, , 26, , 28, , Solution:, X, , x = (X–18), , x2, , 6, , –12, , 144, , 8, , –10, , 100, , 12, , –6, , 36, , 15, , –3, , 9, , 18, , 0, , 0, , 20, , 2, , 4, , 24, , 6, , 36, , 28, , 10, , 100, , 31, , 13, , 169, , / X = 162, , /x = 0, , /x, , 2, , 431, 431, = 449.582 =+0.959, 598 # 338, , r=, , (ii) W,  hen actual values are taken, (without deviation), , when the values of X and Y are considerably, small in magnitude the following formula, can be used, NRXY - ]RX g]RY g, r=, NRX2 - ]RX g2 # NRY2 - ]RY g2, Example 9.2, Calculate coefficient of correlation from the, following data, , = 598, , X, , 12, , 9, , 8, , 10, , 11, , 13, , 7, , Y, , 14, , 8, , 6, , 9, , 11, , 12, , 3, , Solution:, In both the series items are in small number., Therefore correlation coefficient can also be, calculated without taking deviations from, actual means or assumed mean., , Y, , y = (Y–19), , y2, , xy, , 10, , –9, , 81, , 108, , 12, , –7, , 49, , 70, , X, , Y, , X2, , Y2, , XY, , 15, , –4, , 16, , 24, , 12, , 14, , 144, , 196, , 168, , 15, , –4, , 16, , 12, , 9, , 8, , 81, , 64, , 72, , 18, , –1, , 1, , 0, , 8, , 6, , 64, , 36, , 48, , 25, , 6, , 36, , 12, , 10, , 9, , 100, , 81, , 90, , 22, , 3, , 9, , 18, , 26, , 7, , 49, , 70, , 28, , 9, , 81, , 117, , / Y = 171 / y = 0 / y2 = 338 / xy = 431, Table 9.1, RX 162, RY 171, N=9, X = N = 9 = 18,Y = N = 9 =19, Rxy, r=, Rx 2 Ry 2, where x=(X– X )and y=(Y–Y ), Rxy = 431, Rx2 =598, Ry2 =338, , 11, , 11, , 121, , 121, , 121, , 13, , 12, , 169, , 144, , 156, , 7, , 3, , 49, , 9, , 21, , ∑ X = 70 ∑ Y = 63 ∑ X2 = 728 ∑ Y2 = 651 ∑ XY = 676, , Table 9.2, r=, =, , NRXY - ]RX g]RY g, NRX - ]RX g2 # NRY2 - ]RY g2, 2, , 7 ]676g - ]70g]63g, 7 ]728g - ]70g2 # 7 ]651g - ]63g2, , 322, = 339.48, r = +0.95, , Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 213, , 213, , 21-04-2020 12:28:14 PM

Page 220 :
www.tntextbooks.in, , (iii) W,  hen deviations are taken from, an Assumed mean, When actual means are in fractions, say the, actual means of X and Y series are 20.167 and, 29.23, the calculation of correlation by the, method discussed above would involve too, many calculations and would take a lot of time., In such cases we make use of the assumed, mean method for finding out correlation., When deviations are taken from an assumed, mean the following formula is applicable:, r=, , NR dx dy - ]Rdxg^Rdy h, , Example 9.3, Find out the coefficient of correlation in the, following case and interpret., Height of, father, (in inches), , 65, , 66, , 67, , 67, , 68, , 69, , 71, , 73, , Height of, son, (in inches), , 67, , 68, , 64, , 68, , 72, , 70, , 69, , 70, , Solution:, Let us consider Height of father (in inches), is represented as X and Height of son (in, inches) is represented as Y., X, , dx = (X–67), , dx2, , 65, , -2, , 4, , 66, , -1, , 1, , 67, , 0, , 0, , 67, , 0, , 0, , 68, , 1, , 1, , 69, , 2, , 4, , 71, , 4, , 16, , 73, , 6, , 36, , ∑ X = 546, , ∑ dx = 10, , ∑ dx2 = 62, , NRdx - ]Rdxg # NRdy - ^Rdy h, 2, , 2, , 2, , 2, , Where dx = X–A and dy=Y–B . Here A and, B are assumed mean, NOTE, While applying assumed mean method,, any value can be taken as the assumed, mean and the answer will be the same., However, the nearer the assumed mean, to the actual mean, the lesser will be the, calculations., , Y, , dy = (Y–68), , dy2, , dxdy, , 67, , –1, , 1, , 2, , 68, , 0, , 0, , 0, , 64, , -4, , 16, , 0, , 68, , 0, , 0, , 0, , 72, , 4, , 16, , 4, , 70, , 2, , 4, , 4, , 69, , 1, , 1, , 4, , 70, , 2, , 4, , 12, , ∑ Y = 548, , ∑ dy = 4, , ∑ dy = 42, , ∑ dxdy = 26, , Steps to solve the problems:, , (i) Take the deviations of X series from an, assumed mean, denote these deviations, by dx and obtain the total that is Rdx ., (ii) T,  ake the deviations of Y series from an, assumed mean, denote these deviations, by dy and obtain the total that is Rdy ., , Table 9.3, , (iii) Square dx and obtain the total Rdx2 ., (iv) Square dy and obtain the total Rdy2 .,  ultiply dx and dy and obtain the total, (v) M, Rdx dy ., (vi) S ubstitute the values of R dxdy , R dx,, R dy, R dx2 and R dy2 in the formula, given above., 214, , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 214, , 2, , r=, , N Rdx dy - ]Rdxg^Rdy h, , NRdx2 - ]Rdxg2 # NRdy2 - ^Rdy h, , 2, , Here Rdx = 10, Rdx2 = 62, Rdy = 4,, Rdy2 = 42 and Rdxdy = 26, r=, , ]8 # 26g - ]10 # 4g, ]8 # 62g - ]10g2 # ]8 # 42g - ]4g2, , 21-04-2020 12:28:17 PM

Page 221 :
www.tntextbooks.in, , r=, , 168, 396 # 320, , 168, r = 355.98 =0.472, i.e. r = +0.472, Heights of fathers and their respective sons, are positively correlated., Example 9.4, Calculate the correlation coefficient from, the following data, N=9, / X =45, / Y =108, RX2 =285,, RY2 =1356, RXY =597., Solution:, We know that correlation coefficient, NRXY - ]RX g]RY g, r=, NRX2 - ]RX g2 NRY2 - ]RY g2, =, , 9 ]597g - ]45 # 108g, 9 ]285g - ]45g2 # 9 ]1356g - ^1082 h, , r = +0.95, , Example 9.5, From the following data calculate the, correlation coefficient Rxy =120, Rx2 =90,, Ry2 =640., Solution:, Given Rxy =120, Rx2 =90, Ry2 =640, Rxy, 120, 120, Then r =, 2, 2 = 90 ]640g = 57600, Rx Ry, 120, = 240 =0.5, , 9.2, , Rank Correlation, , 9.2.1 S,  pearman’s Rank Correlation, Coefficient, In 1904, Charles Edward Spearman, a British, psychologist found out the method of, ascertaining the coefficient of correlation by, ranks. This method is based on rank. This, , measure is useful in dealing with qualitative, characteristics, such as intelligence, beauty,, morality, character, etc. It cannot be, measured quantitatively, as in the case of, Pearson’s coefficient of correlation., Rank correlation is applicable only to, individual observations. The result we get, from this method is only an approximate one,, because under ranking method original value, are not taken into account. The formula for, Spearman’s rank correlation which is denoted, by ρ (pronounced as row) is, 6Rd2, (or), t = 1–, N ^N2 - 1h, 6Rd 2, N3 - N, where d = The difference of two ranks, = R X - RY and, t = 1-, , N = Number of paired observations., Rank coefficient of correlation value, lies between –1 and +1., Symbolically, –1≤ρ≤+1., When we come across spearman’s, rank correlation, we may find two types of, problem, (i) When ranks are given, (ii) When ranks are not given, Example 9.6, The following are the ranks obtained by 10, students in Statistics and Mathematics, Statistics, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , Mathematics 1, , 4, , 2, , 5, , 3, , 9, , 7, , 10, , 6, , 8, , Find the rank correlation coefficient., Solution:, Let RX is considered for the ranks of, Statistics and RY is considered for the ranks, of mathematics., Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 215, , 215, , 21-04-2020 12:28:19 PM

Page 222 :
www.tntextbooks.in, , R X - RY, , d2, , RX, , RY, , 1, , 1, , 0, , 0, , 2, , 4, , -2, , 4, , 3, , 2, , 1, , 1, , 4, , 5, , -1, , 1, , 5, , 3, , 2, , 4, , 6, , 9, , -3, , 9, , 7, , 7, , 0, , 0, , 8, , 10, , -2, , 4, , 9, , 6, , 3, , 9, , 10, , 8, , 2, , 4, , d=, , / d2 =36, , RX, , RY, , RZ, , dXY =, R X–RY, , dYZ =, RY–RZ, , dZX =, RZ–R X, , 1, , 2, , 3, , -1, , -1, , 2, , 4, , 6, , 7, , -2, , -1, , 3, , 6, , 5, , 4, , 1, , 1, , -2, , 3, , 4, , 5, , -1, , -1, , 2, , 2, , 7, , 10, , -5, , -3, , 8, , 9, , 10, , 8, , -1, , 2, , -1, , 7, , 9, , 9, , -2, , 0, , 2, , 8, , 3, , 2, , 5, , 1, , -6, , 10, , 8, , 6, , 2, , 2, , -4, , 5, , 1, , 1, , 4, , 0, , -4, , Table 9.4, , d2XY, , d2YZ, , d2ZX, , 1, , 1, , 4, , 4, , 1, , 9, , 1, , 1, , 4, , 1, , 1, , 4, , 25, , 9, , 64, , 1, , 4, , 1, , 4, , 0, , 4, , 25, , 1, , 36, , 4, , 4, , 16, , 16, , 0, , 16, , ∑ d2XY = 82, , ∑ d2YZ = 22, , ∑ d2ZX = 158, , The rank correlation is given by, 6 ]36g, 6Rd2, = 1–, ρ = 1–, 2, N ^N - 1h, 10 ^102 - 1 h, , = 1–0.218, , ` ρ = 0.782, Example 9.7, Ten competitors in a beauty contest are, ranked by three judges in the following, order., First, judge, , 1, , 4, , 6, , 3, , 2, , 9, , 7, , 8, , 10, , 5, , Second, judge, , 2, , 6, , 5, , 4, , 7, , 10, , 9, , 3, , 8, , 1, , Third, judge, , 3, , 7, , 4, , 5, , 10, , 8, , 9, , 2, , 6, , 1, , Use the method of rank correlation, coefficient to determine which pair of judges, has the nearest approach to common taste, in beauty?, Solution:, Let RX, RY , RZ denote the ranks by First judge,, Second judge and third judge respectively, 216, , Table 9.5, 6Rd 2XY, 6 ]82g, =1–, 2, N ^N - 1h, 10 ^102 - 1 h, = 1–0.4969 = 0.5031, , t XY = 1–, , tYZ, , 2, 6RdYZ, 6 ]22g, = 1–, =1–, 2, 10 ^102 - 1 h, N ^N - 1h, , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 216, , 21-04-2020 12:28:21 PM

Page 223 :
www.tntextbooks.in, , 132, = 1– 990 = 1–0.1333=0.8667, t ZX, , 2, 6Rd ZX, 6 ]158g, =1–, = 1–, 2, N ^N - 1h, 10 ^102 - 1 h, , = 1– 0.9576 = 0.0424, Since the rank correlation coefficient, between Second and Third judges i.e., ρYZ, is positive and weight among the three, coefficients. So, Second judge and Third, judge have the nearest approach for common, taste in beauty., Example 9.8, Calculate rank correlation coefficient of the, following data, , 6Rd2, N ^N2 - 1h, 6 ]142g, = 1–, 11 ^112 - 1h, 852, = 1– 1320 = 0.354, , t = 1–, , Exercise 9.1, 1. C,  alculate the correlation co-efficient for, the following data, X, , 5, , 10, , 5, , 11 12, , 4, , 3, , 2, , 7, , 1, , Y, , 1, , 6, , 2, , 8, , 1, , 4, , 6, , 5, , 2, , 5, , 2. F,  ind coefficient of correlation for the, following:, , Subject 1 40 46 54 60 70 80 82 85 87 90 95, , Cost, (`), , 14, , 19, , 24, , 21, , 26, , 22, , 15, , 20, , 19, , Subject2 45 46 50 43 40 75 55 72 65 42 70, , Sales, (`), , 31, , 36, , 48, , 37, , 50, , 45, , 33, , 41, , 39, , 3. C,  alculate coefficient of correlation for, the ages of husbands and their respective, wives:, , Solution:, Let X is considered for Subject 1 and, Y is considered for Subject 2., X, , Y, , RX, , RY, , d = RX–RY, , d2, , 40, , 45, , 1, , 4, , –3, , 9, , 46, , 46, , 2, , 5, , –3, , 9, , 54, , 50, , 3, , 6, , –3, , 9, , 60, , 43, , 4, , 3, , 1, , 1, , 70, , 40, , 5, , 1, , 4, , 16, , 80, , 75, , 6, , 11, , –5, , 25, , 82, , 55, , 7, , 7, , 0, , 0, , 85, , 72, , 8, , 10, , –2, , 4, , 87, , 65, , 9, , 8, , 1, , 1, , 90, , 42, , 10, , 2, , 8, , 64, , 95, , 70, , 11, , 9, , 2, , 4, ∑ d2 = 142, , Table 9.6, , Age of, husbands 23 27 28 29 30 31 33 35 36 39, Age of, wives, , 18 22 23 24 25 26 28 29 30 32, , 4. C,  alculate the coefficient of correlation, between X and Y series from the, following data, Description, , X, , Y, , Number of pairs of observation, , 15, , 15, , Arithmetic mean, , 25, , 18, , Standard deviation, , 3.01, , 3.03, , Sum of squares of deviation, from the arithmetic mean, , 136, , 138, , , Summation of product deviations of, X and Y series from their respective, arithmetic means is 122., Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 217, , 217, , 21-04-2020 12:28:22 PM

Page 224 :
www.tntextbooks.in, , 5. C,  alculate correlation coefficient for, the following data, X, , 25 18 21 24 27 30 36 39 42 48, , Y, , 26 35 48 28 20 36 25 40 43 39, , 6. F,  ind coefficient of correlation for the, following:, X, , 78, , 89, , 96, , 69, , 59, , 79, , 68, , 62, , Y, , 121, , 72, , 88, , 60, , 81, , 87, , 123, , 92, , 7. A,  n examination of 11 applicants for a, accountant post was taken by a finance, company. The marks obtained by the, applicants in the reasoning and aptitude, tests are given below., Applicant, , A, , B, , C, , D, , E, , F, , Reasoning test, , 20, , 50, , 28, , 25, , 70, , 90, , Aptitude test, , 30, , 60, , 50, , 40, , 85, , 90, , Applicant, , G, , H, , I, , J, , K, , Reasoning test, , 76, , 45, , 30, , 19, , 26, , Aptitude test, , 56, , 82, , 42, , 31, , 49, , Calculate Spearman’s rank correlation, coefficient from the data given above., 8. Th,  e following are the ranks obtained, by 10 students in commerce and, accountancy are given below:, Commerce, , 6, , 4, , 3, , 1, , 2, , 7, , 9, , 8 10 5, , Accountancy 4, , 1, , 6, , 7, , 5, , 8 10 9, , 3, , 2, , , To what extent is the knowledge of, students in the two subjects related?, 9. A,  random sample of recent repair jobs, was selected and estimated cost and, actual cost were recorded., 218, , Estimated 300 450 800 250 500 975 475 400, cost, Actual, cost, , 273 486 734 297 631 872 396 457, , , Calculate the value of spearman’s, correlation coefficient., 10. The rank of 10 students of same batch, in two subjects A and B are given, below. Calculate the rank correlation, coefficient., , 9.3, , Rank of, A, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , Rank of, B, , 6, , 7, , 5, , 10, , 3, , 9, , 4, , 1, , 8, , 2, , Regression Analysis, , Introduction:, So far we have studied correlation analysis, which measures the direction and strength, of the relationship between two variables., Here we can estimate or predict the value, of one variable from the given value of, the other variable. For instance, price and, supply are correlated. We can find out the, expected amount of supply for a given price, or the required price level for attaining the, given amount of supply., The term “ regression” literally means, “Stepping back towards the average”. It was, first used by British biometrician Sir Francis, Galton (1822 -1911), in connection with the, inheritance of stature. Galton found that, the offsprings of abnormally tall or short, parents tend to “regress” or “step back”, to the average population height. But the, term “regression” as now used in Statistics is, only a convenient term without having any, reference to biometry., , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 218, , 21-04-2020 12:28:23 PM

Page 225 :
www.tntextbooks.in, , Definition 9.1, Regression analysis is a mathematical, measure of the average relationship between, two or more variables in terms of the original, units of the data., , 9.3.1 D,  ependent and independent, variables, Definition 9.2, In regression analysis there are two types of, variables. The variable whose value is to be, predicted is called dependent variable and, the variable which is used for prediction is, called independent variable., Regression helps us to estimate the value of, one variable, provided the value of the other, variable is given. The statistical method, which helps us to estimate the unknown, value of one variable from the known value, of the related variable is called Regression., , 9.3.2 Regression Equations, Regression equations are algebraic, expressions of the regression lines. Since, there are two regression lines, there are, two regression equations. The regression, equation of X on Y is used to describe the, variation in the values of X for given changes, in Y and the regression equation of Y on X is, used to describe the variation in the values, of Y for given changes in X. Regression, equations of (i) X on Y (ii) Y on X and their, coefficients in different cases are described, as follows., , Case 1: When the actual values are taken, When we deal with actual values of X and Y, variables the two regression equations and, their respective coefficients are written as, follows:, , (i) Regression Equation of X on Y:, X - X = bxy ]Y - Y g, , where X is the mean of X series,, , , , Y is the mean of Y series,, , v, NRXY - ]RX g]RY g, b xy=r v x =, is known, y, NRY2 - ]RY g2, as the regression coefficient of X on Y, and, r is the correlation coefficient between X and Y,, σx and σy are standard deviations of X and Y, respectively., (ii) R,  egression Equation of Y on X;, Y - Y = b yx ] X - X g, , where X is the mean of X series,, , , , Y is the mean of Y series,, , vy, NRXY - (RX) (RY), is known, b yx = r vx =, NRY2 - (RY) 2, as the regression coefficient of Y on X, and, r is the correlation coefficient between, X and Y, σx and σy are standard deviations of, X and Y respectively., , Case 2: D,  eviations taken from, Arithmetic means of X and Y, The calculation can very much be simplified, instead of dealing with the actual values of, X and Y, we take the deviations of X and Y, series from their respective means. In such, a case the two regression equations and, their respective coefficients are written as, follows:, (i) R,  egression Equation of X on Y:, X - X = bxy (Y - Y), where X is the mean of X series,, Y is the mean of Y series,, Rxy, vx, is known as the regression, bxy=r vy =, Ry 2, coefficient of X on Y, x = (X– X ) and y = (Y–Y ), , , Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 219, , 219, , 21-04-2020 12:28:25 PM

Page 226 :
www.tntextbooks.in, , (ii) Regression Equation of Y on X;, Y–Y = byx ] X - X g, , where X is the mean of X series,, Y is the mean of Y series,, vy Rxy, is known as the regression, byx= r vx =, Rx 2, coefficient of Y on X, x = ] X - X g and y = (Y–Y ), Note: Instead of finding out the values of, correlation coefficient σx, σy, etc, we can, find the value of regression coefficient by, calculating ∑xy and ∑y2 ., , Case 3: D,  eviations taken from, Assumed Mean, , When actual means of X and Y variables, are in fractions the calculations can be, simplified by taking the deviations from the, assumed means. The regression equations, and their coefficients are written as follows, (i) Regression Equation of Y on X:, Y - Y = b yx ] X - X g, NRdxdy - ]Rdxg^Rdy h, b yx =, NRdx2 - ]Rdxg2, , (ii) Regression Equation of X on Y:, X - X =bxy ]Y - Y g, bxy =, , NRdxdy - ]Rdxg^Rdy h, 2, NRdy2 - ^Rdy h, , (iii) B,  oth the regression coefficients are of, same sign., Example 9.9, Calculate the regression coefficient and, obtain the lines of regression for the, following data, X, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , Y, , 9, , 8, , 10, , 12, , 11, , 13, , 14, , Solution:, X, , Y, , X2, , Y2, , XY, , 1, , 9, , 1, , 81, , 9, , 2, , 8, , 4, , 64, , 16, , 3, , 10, , 9, , 100, , 30, , 4, , 12, , 16, , 144, , 48, , 5, , 11, , 25, , 121, , 55, , 6, , 13, , 36, , 169, , 78, , 7, , 14, , 49, , 196, , 98, , ∑ X = 20 ∑ Y = 77 ∑ X2 = 140 ∑ Y2 = 875 ∑ XY = 334, , Table 9.7, RX, 28, RY, 77, X = N = 7 =4 , Y = N = 7 =11, , Regression coefficient of X on Y:, NRXY - ]RX g]RY g, N RY2 - ]RY g2, 7 ]334g - ]28g]77g, =, 7 ]875g - ]77g2, 2338 - 2156, 182, = 6125 - 5929 = 196, , bxy =, , where dx=X–A, dy=Y–B, A and B are, assumed means or arbitrary values are taken, from X and Y respectively., , ` bxy = 0.929, , Properties of Regression Coefficients, , (i) Regression equation of X on Y:, , (i) , Correlation, Coefficient, is, the, geometric mean between the regression, coefficients r =! bxy # b yx ., , X– X = bxy(Y-Y ), , (ii) If one of the regression coefficients is, greater than unity, the other must be, less than unity., 220, , X–4 = 0.929(Y–11 ), X–4 = 0.929Y–10.219,  e regression equation X on Y is, ` Th, X= 0.929Y–6.219., , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 220, , 21-04-2020 12:28:27 PM

Page 227 :
www.tntextbooks.in, , (ii) Regression coefficient of Y on X:, byx =, =, =, , v, Rxy, -6, bxy = r vx = 2 = 50 =–0.12  , Ry, y, , NRXY - ]RX g]RY g, N RX2 - ]RX g2, , X–13 = –0.12 (Y–41 ), , 7 ]334g - ]28g]77g, 7 ]140g - ]28g2, 2338 - 2156, 980 - 784, 182, 196, 0.929, , X–13 = –0.12Y+4.92, X = –0.12Y+17.92, X, , x = (X –13), , x2, , 10, , –3, , 9, , 12, , –1, , 1, , 13, , 0, , 0, , Y–Y = byx(X– X ), , 12, , –1, , 1, , Y–11 = 0.929 (X–4), , 16, , 3, , 9, , 15, , 2, , 4, , ∑ X = 78, , ∑ x=0, , ∑ x2 = 24, , =, ` byx =, , (iii) Regression equation of Y on X:, , Y = 0.929X–3.716+11, Y = 0.929X+7.284, ` The regression equation of Y on X is, Y= 0.929X + 7.284, Example 9.10, Calculate the two regression equations of X, on Y and Y on X from the data given below,, taking deviations from a actual means of X, and Y., Price (`), , 10, , 12, , 13, , 12, , 16, , 15, , Amount, demanded, , 40, , 38, , 43, , 45, , 37, , 43, , Estimate the likely demand when the price, is ` 20., Solution:, Calculation of Regression equation, (i) Regression equation of X on Y:, v, X– X = r v x (Y–Y ), y, From the Table 9.8, we get, 246, 78, X = 6 =13 , Y = 6 =41, , Y, , y = (Y –41), , y2, , xy, , 40, , –1, , 1, , 3, , 38, , –3, , 9, , 3, , 43, , 2, , 4, , 0, , 45, , 4, , 16, , –4, , 37, , –4, , 16, , –12, , 43, , 2, , 4, , 4, , ∑ Y = 246, , ∑ y=0, , ∑ y2 = 50 ∑ xy = –6, , Table 9.8, (ii) Regression Equation of Y on X:, vy, Y–Y = r v (X– X ), x, v y Rxy, 6, byx = r v =, 2 = – 24 =–0.25, Rx, x, Y–41 = –0.25 (X–13 ), Y–41 = –0.25 X+3.25, Y = –0.25 X+44.25, Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 221, , 221, , 21-04-2020 12:28:30 PM

Page 228 :
www.tntextbooks.in, , When X is 20, Y will be, , = 5102, 5414, , Y = –0.25 (20)+44.25, , = 0.942, , = –5+44.25, = 39.25 (when the price is ` 20, the, likely demand is 39.25), Example 9.11, Obtain regression equation of Y on X and, estimate Y when X = 55 from the following:, X, , 40, , 50, , 38, , 60, , 65, , 50, , 35, , Y, , 38, , 60, , 55, , 70, , 60, , 48, , 30, , Solution:, , byx = 0.942, (ii) Regression equation of Y on X:, Y - Y = b yx ] X - X g, , Y–51.57 = 0.942(X–48.29 ), Y = 0.942X–45.49+51.57=0.942 #, –45.49+51.57, Y, , = 0.942X+6.08, ,  e regression equation of Y on X is, ` Th, Y= 0.942X+6.08, , X, , Y, , dx =, (X–48), , dx2, , dy =, (Y–50), , dy2, , dxdy, , 40, , 38, , –8, , 64, , –12, , 144, , 96, , Estimation of Y when X= 55, , 50, , 60, , 2, , 4, , 10, , 100, , 20, , Y= 0.942(55)+6.08=57.89, , 38, , 55, , –10, , 100, , 5, , 25, , –50, , 60, , 70, , 12, , 144, , 20, , 400, , 240, , 65, , 60, , 17, , 289, , 10, , 100, , 170, , 50, , 48, , 2, , 4, , -2, , 4, , -4, , Example 9.12, Find the means of X and Y variables and the, coefficient of correlation between them from, the following two regression equations:, , 35, , 30, , -13, , 169, , -20, , 400, , 260, , 2Y–X–50 = 0, , ∑ X=, 338, , ∑ Y=, 361, , ∑ dx = ∑ dx2 = ∑ dy = ∑ dy2 = ∑ dxxy =, 2, 774, 11, 1173, 732, , Table 9.9, RX 338, X = N = 7 = 48.29, RY, 361, Y = N = 7 = 51.57, (i) Regression coefficients of Y on X:, b yx =, =, , NRdx dy - ]Rdxg^Rdy h, N Rdx2 - ]Rdxg2, , 7 ]732g - ]2 g]11g, 7 ]774g - ]2 g2, , 5124 - 22, = 5418 - 4, 222, , 3Y–2X–10 = 0., Solution:, We are given, 2Y–X–50 = 0, , ... (1), , 3Y–2X–10 = 0, , ... (2), , Solving equation (1) and (2), We get Y= 90, Putting the value of Y in equation (1), We get X = 130, Hence X = 130 andY =90, , Calculating correlation coefficient, Let us assume equation (1) be the regression, equation of Y on X., , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 222, , 21-04-2020 12:28:31 PM

Page 229 :
www.tntextbooks.in, , 2Y = X+50, 1, 1, Y = 2 X+25 therefore byx= 2, Clearly equation (2) would be treated as, regression equation of X on Y, 3Y–2X–10 = 0, 2X = 3Y–10, 3, 3, X = 2 Y–5 therefore bxy= 2, The Correlation coefficient r = ± bxy # b yx, r=, , 1#3, 2 2 = 0.866, , It may be noted that in the above problem, one of the regression coefficient is greater, than 1 and the other is less than 1. Therefore, our assumption on given equations are, correct., Example 9.13, Find the means of X and Y variables and the, coefficient of correlation between them from, the following two regression equations:, 4X–5Y+33 = 0, 20X–9Y–107 = 0, , ... (1), , 20X–9Y–107 = 0, , ... (2), , We get X = 13, Hence X = 13 and Y = 17, , 1, X = 4 (5Y–33), 5, 33, X = 4 Y– 4, 5, bxy = 4 = 1.25, Let us assume equation (2) be the regression, equation of Y on X, 1, Y = 9 (20X–107), 20, 107, Y = 9 X– 9, 20, byx = 9 =2.22, But this is not possible because both the, regression coefficient are greater than 1. So, our above assumption is wrong. Therefore, treating equation (1) has regression equation, of Y on X and equation (2) has regression, equation of X on Y. So we get, 4, byx = 5 =0.8 and, 9, bxy = 20 =0.45, r=±, , 4X–5Y+33 = 0, , Putting the value of Y in equation (1), , 4X = 5Y–33, , The Correlation coefficient, , Solution:, We are given, , We get Y = 17, , Let us assume equation (1) be the regression, equation of X on Y, , 9Y = 20X–107, , NOTE, , Solving equation (1) and (2), , Calculating correlation coefficient, , r=, , bxy # bxy, 0.45 # 0.8 =0.6, , Example 9.14, The following table shows the sales and, advertisement expenditure of a form, Title, , Sales, , Advertisement expenditure, ( ` in Crores), , Mean, , 40, , 6, , SD, , 10, , 1.5, , Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 223, , 223, , 21-04-2020 12:28:33 PM

Page 230 :
www.tntextbooks.in, , Coefficient of correlation r= 0.9. Estimate, the likely sales for a proposed advertisement, expenditure of Rs. 10 crores., , X–0.8Y–17.6 = 0 (or) X=0.8Y+17.6, , Solution:, Let the sales be X and advertisement, expenditure be Y, , The regression line Y on X be, vy, Y–Y = r v (X– X ), x, 4, Y–103 = (0.4) 8 (X–100), Y–103 = 0.2 (X–100), , Given X =40,Y =6, σx=10, σy=1.5 and r=0.9, , Y–103 = 0.2 X–20, , Equation of line of regression x on y is, v, X– X = r v x (Y–Y ), y, 10, X–40 = (0.9) 1.5 (Y–6), X–40 = 6Y–36, X = 6Y+4, When advertisement expenditure is, 10 crores i.e., Y=10 then sales X=6(10)+4=64, which implies sales is 64., Example 9.15, There are two series of index numbers P for, price index and S for stock of the commodity., The mean and standard deviation of P are, 100 and 8 and of S are 103 and 4 respectively., The correlation coefficient between the two, series is 0.4. With these data obtain the, regression lines of P on S and S on P., Solution:, Let us consider X for price P and Y for stock, S. Then the mean and SD for P is considered, as X =100 and σx=8. respectively and, the mean and SD of S is considered as, Y =103 and σy= 4. The correlation coefficient, between the series is r(X,Y) = 0.4, Let the regression line X on Y be, v, X - X = r v x (Y–Y ), y, 8, X–100 = (0.4) 4 (Y–103), X–100 = 0.8(Y–103), 224, , Y = 0.2 X+83 (or) 0.2 X–Y+83=0, Example 9.16, For 5 pairs of observations the following, results are obtained ∑X=15, ∑Y=25, ∑X2 =55,, ∑Y2 =135, ∑XY=83 Find the equation of the, lines of regression and estimate the value of, X on the first line when Y=12 and value of Y, on the second line if X=8., Solution:, RX 15, RY 25, Here N=5, X = N = 5 =3 , Y = N = 5 = 5, and the regression coefficient, bxy=, , NRXY - RXRY 5 ]83g - ]15g]25g, =, =0.8, 5 ]135g - ]25g2, NRY2 - ]RY g2, , The regression line of X on Y is, X– X, , = bxy (Y–Y ), , X–3 = 0.8(Y–5 ), X = 0.8Y–1, When Y = 12, the value of X is estimated as, X = 0.8 (12)–1=8.6, The regression coefficient, NRXY - RXRY, NRX2 - ]RX g2, 5 ]83g - ]15g]25g, =, =0.8, 5 ]55g - ]15g2, byx =, , Thus byx = 0.8 then the regression line Y on X is, Y–Y, , = byx (X– X ), , Y–5 = 0.8(X–3), Y = 0.8X+2.6, , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 224, , 21-04-2020 12:28:36 PM

Page 231 :
www.tntextbooks.in, , When X = 8 the value of Y is estimated as, Y = 0.8(8)+2.6, Y = 9, Example 9.17, The two regression lines are 3X+2Y=26 and, 6X+3Y=31. Find the correlation coefficient., Solution:, Let the regression equation of Y on X be, 3X+2Y = 26, 2Y = –3X+26, 1, Y = 2 (–3X+26), Y = –1.5X+13, vy, = –1.5, r v, x, vy, Implies by x = r v =–1.5, x, Let the regression equation of X on Y be, 6X+3Y = 31, 1, X = 6 (–3Y+31)=–0.5Y+5.17, v, r v x = –0.5, y, v, Implies bxy = r v x =–0.5, y, r = ± bxy .b yx, = – ]- 1.5g . ]- 0.5g (Since both, the regression coefficient are, negative r is negative), ` r = –0.866, Example 9.18, In a laboratory experiment on correlation, research study the equation of the two, regression lines were found to be 2X–Y+1=0, and 3X–2Y+7=0 . Find the means of X and, Y. Also work out the values of the regression, coefficient and correlation between the two, variables X and Y., , Solution:, Solving the two regression equations we get, mean values of X and Y., 2X–Y = –1, , ... (1), , 3X–2Y = –7, , ... (2), , Solving equation (1) and equation (2) We, get X=5 and Y=11., Therefore the regression line passing, through the means X =5 and Y =11., The regression equation of Y on X is 3X–2Y=–7, 2Y = 3X+7, 1, Y = 2 (3X+7), 3, 7, Y = 2 X+ 2, 3, ` byx = 2 (>1), The regression equation of X on Y is, 2X–Y = –1, 2X = Y–1, 1, X = 2 ]Y - 1g, 1, 1, X = 2 Y- 2, 1, ` bxy = 2, The regression coefficients are positive, r = ±, , bxy .b yx = ±, , =, , 3#1, 2 2, , =, , 3, 4, , 3#1, 2 2, , = 0.866, ` r = 0.866, Example 9.19, For the given lines of regression 3X–2Y=5and, X–4Y=7. Find, (i) Regression coefficients, (ii) Coefficient of correlation, Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 225, , 225, , 21-04-2020 12:28:38 PM

Page 232 :
www.tntextbooks.in, , Solution:, (i) First convert the given equations Y on, X and X on Y in standard form and find, their regression coefficients respectively., Given regression lines are , 3X–2Y = 5, , ... (1), , X–4Y = 7, , ... (2), , Let the line of regression of X on Y is, 3X–2Y = 5, 3X = 2Y+5, 1, X = 3 (2Y+5), 1, X = 3 (2Y+5), 2, 5, X = 3 Y+ 3, ` Regression coefficient of X on Y is, 2, bxy = 3 (<1), Let the line of regression of Y on X is, X–4Y = 7, –4Y = –X+7, 4Y = X–7, 1, Y = 4 (X–7), 1, 7, Y = 4 X– 4, ` Regression coefficient of Y on X is, 1, byx = 4 (<1), (ii) Coefficient of correlation, Since the two regression coefficients are, positive then the correlation coefficient is, also positive and it is given by, r =, =, =, , b yx . bxy, 2 1, 3. 4, 1, 6, , = 0.4082, ` r = 0.4082, 226, , Exercise 9.2, 1. From the data given below:, Marks in, Economics:, , 25, , 28, , 35, , 32, , 31, , Marks in, Statistics:, , 43, , 46, , 49, , 41, , 36, , Marks in, Economics:, , 36, , 29, , 38, , 34, , 32, , Marks in, Statistics:, , 32, , 31, , 30, , 33, , 39, , Find (a) The two regression equations,, (b) The coefficient of correlation between, marks in Economics and statistics,, (c) The mostly likely marks in Statistics, when the marks in Economics is 30., 2. Th,  e heights (in cm.) of a group of fathers, and sons are given below:, Heights of 158 166 163 165 167 170 167 172 177 181, fathers:, Heights of 163 158 167 170 160 180 170 175 172 175, Sons :, , Find the lines of regression and estimate, the height of son when the height of the, father is 164 cm., 3. Th,  e following data give the height in, inches (X) and the weight in lb. (Y) of, a random sample of 10 students from a, large group of students of age 17 years:, X, , 61 68 68 64 65 70 63 62 64 67, , Y 112 123 130 115 110 125 100 113 116 125, , Estimate weight of the student of a, height 69 inches., 4. Obtain the two regression lines from the, following data N=20, ∑X=80, ∑Y=40,, ∑X2=1680, ∑Y2=320 and ∑XY=480., , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 226, , 21-04-2020 12:28:40 PM

Page 233 :
www.tntextbooks.in, , 5. G,  iven the following data, what will be the, possible yield when the rainfall is 29., Details, , Rainfall, , Production, , Mean, , 25``, , 40 units per acre, , Standard Deviation, , 3``, , 6 units per acre, , , Coefficient of correlation between, rainfall and production is 0.8., 6. Th, e following data relate to, advertisement expenditure (in lakh of, rupees) and their corresponding sales, (in crores of rupees), Advertisement 40, expenditure, Sales, , 38, , 50, , 38, , 60, , 65, , 50, , 35, , 60, , 55, , 70, , 60, , 48, , 30, , Estimate the sales corresponding to, advertising expenditure of ` 30 lakh., 7. You are given the following data:, Details, Arithmetic Mean, Standard Deviation, , X, 36, 11, , Y, 85, 8, , If the Correlation coefficient between, X and Y is 0.66, then find (i) the two, regression coefficients, (ii) the most, likely value of Y when X = 10., 8. F,  ind the equation of the regression line, of Y on X, if the observations (Xi, Yi), are the following (1,4) (2,8) (3,2) (4,12), (5, 10) (6, 14) (7, 16) (8, 6) (9, 18)., 9. A,  survey was conducted to study the, relationship between expenditure on, accommodation (X) and expenditure, on Food and Entertainment (Y) and the, following results were obtained:, Details, , Mean, , SD, , Expenditure on Accommodation (`), , 178, , 63.15, , Expenditure on Food and, Entertainment (`), , 47.8, , 22.98, , Coefficient of Correlation, , 0.43, , Write down the regression equation and, estimate the expenditure on Food and, Entertainment, if the expenditure on, accommodation is ` 200.,  or 5 observations of pairs of (X, Y) of, 10. F, variables X and Y the following results, are obtained. ∑X = 15, ∑Y = 25, ∑X2 = 55,, ∑Y2 = 135, ∑XY = 83. Find the equation, of the lines of regression and estimate, the values of X and Y if Y = 8; X = 12., 11. The two regression lines were found to, be 4X–5Y+33 = 0 and 20X–9Y–107 = 0., Find the mean values and coefficient of, correlation between X and Y., 12. The equations of two lines of regression, obtained in a correlation analysis are the, following 2X = 8–3Y and 2Y = 5–X. Obtain, the value of the regression coefficients, and correlation coefficient., , Exercise 9.3, Choose the correct answer, 1. Example for positive correlation is, (a) Income and expenditure, (b) Price and demand, (c) Repayment period and EMI, (d) Weight and Income, 2. If the values of two variables move in, same direction then the correlation is, said to be, (a) Negative , (b) Positive, (c) Perfect positive , (d) No correlation, 3. If the values of two variables move in, opposite direction then the correlation, is said to be, Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 227, , 227, , 21-04-2020 12:28:40 PM

Page 235 :
www.tntextbooks.in, , 15. I f X and Y are two variates, there can be, atmost, (a) One regression line, (b) two regression lines, (c) three regression lines, (d) more regression lines, , 22. The term regression was introduced by, (a) R.A. Fisher, (b) Sir Francis Galton, (c) Karl Pearson, (d) Croxton and Cowden, , 16. The lines of regression of X on Y estimates, (a) X for a given value of Y, (b) Y for a given value of X, (c) X from Y and Y from X, (d) none of these, , 23. If r=–1 , then correlation between the, variables, (a) perfect positive , (b) perfect negative, (c) negative, (d) no correlation, , 17. Scatter diagram of the variate values, (X,Y) give the idea about, (a) functional relationship, (b) regression model, (c) distribution of errors, (d) no relation, , 24. The coefficient of correlation describes, (a) the magnitude and direction, (b) only magnitude, (c) only direction, (d) no magnitude and no direction, , 18. If regression co-efficient of Y on X is 2,, then the regression co-efficient of X on Y is, 1, (b) 2, (a) # 2, 1, (d) 1, (c) > 2, 19. If two variables moves in decreasing, direction then the correlation is, (a) positive, (b) negative, (c) perfect negative (d) no correlation, 20. Th,  e person suggested a mathematical, method for measuring the magnitude, of linear relationship between two, variables say X and Y is, (a) Karl Pearson, (b) Spearman, (c) Croxton and Cowden, (d) Ya Lun Chou,  e lines of regression intersect at the, 21. Th, point, (a) (X,Y), (b) ^ X , Y h, (c) (0,0), (d) (σx,σy), , 25. If Cov(x,y)=–16.5, σ 2x =2.89, σ 2y =100., Find correlation coefficient., (a) –0.12, (b) 0.001, (c) –1, (d –0.97, , Miscellaneous Problems, 1. F,  ind the coefficient of correlation for, the following data:, X, , 35, , 40, , 60, , 79, , 83, , 95, , Y, , 17, , 28, , 30, , 32, , 38, , 49, , 2. C,  alculate the coefficient of correlation, from the following data:, ∑X=50, ∑Y=–30, ∑X2 =290, ∑Y2 =300,, ∑XY=–115, N=10, 3. C, alculate the correlation coefficient, from the data given below:, X, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , Y, , 9, , 8, , 10, , 12, , 11, , 13, , 14, , 16, , 15, , Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 229, , 229, , 21-04-2020 12:28:43 PM

Page 236 :
www.tntextbooks.in, , 4. C, alculate the correlation coefficient, from the following data:, ∑X=125, ∑Y=100, ∑X2 =650, ∑Y2 =436, dx2, ∑XY=520, N=25, 5. A,  random sample of recent repair jobs, was selected and estimated cost , actual, cost were recorded., , 10. The following information is given., Details, , X (in `), , Y (in `), , Estimated 30, cost, , 45, , 80, , 25, , 50, , 97, , 47, , 40, , Arithmetic Mean, , 6, , 8, , Actual cost 27, , 48, , 73, , 29, , 63, , 87, , 39, , 45, , Standard Deviation, , 5, , 40, 3, , , Calculate the value of spearman’s, correlation., 6. Th,  e following data pertains to the, marks in subjects A and B in a certain, examination. Mean marks in A = 39.5,, Mean marks in B= 47.5 standard, deviation of marks in A =10.8 and, Standard deviation of marks in B = 16.8., coefficient of correlation between marks, in A and marks in B is 0.42. Give the, estimate of marks in B for candidate, who secured 52 marks in A., 7. X and Y are a pair of correlated variables., Ten observations of their values (X,Y), have the following results., , ∑X=55,, , ∑XY=350, ∑X =385, ∑Y=55, Predict the, 2, , value of y when the value of X is 6., 8. Find the line regression of Y on X, , 9., , inspection expenditure amounts to, ` 82 ∑X=424, ∑Y=363, ∑X2 =21926,, ∑Y2 =15123, ∑XY=12815, N=10. Here, X is the expenditure on inspection, Y is, the defective parts delivered., , X, , 1, , 2, , 3, , 4, , 5, , 8, , 10, , Y, , 9, , 8, , 10, , 12, , 14, , 16, , 15, ,  sing the following information you, U, are requested to (i) obtain the linear, regression of Y on X (ii) Estimate the, level of defective parts delivered when, , 230, , , Coefficient of correlation between X, 8, and Y is 15 . Find (i) The regression, Coefficient of Y on X (ii) The most likely, value of Y when X = ` 100., , Case Study-1, Mr. Bean visited a departmental store in, Triplicane at Chennai on 1st March 2018, and chooses 15 different types of food, items that include nutrition information, on its packaging. For each food, Mr. Bean observed and identified the, amount of fat (gms) and the sodium content, ( mgs/100gms) per serving and recorded in, the following table., Sl., No., , Product Items, , Fat, Sodium, (gm/100gms) (mg / 100gms), , 1., , Dates, , 0.4, , 74.4, , 2., , Appalam, , 0.26, , 1440, , 3., , Energy drink, , 1.8, , 136, , 4., , Gulabjamun, Powder, , 10.4, , 710, , 5., , Atta, , 2.2, , 4.97, , 6., , Athi fruits, , 0.14, , 2, , 7., , Alu Muttar mix, , 5, , 440, , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 230, , 21-04-2020 12:28:43 PM

Page 237 :
www.tntextbooks.in, , 8., , Popcorn, , 2.32, , 51.38, , 9., , Perungayum, , 0.37, , 40, , 10., , Mushroom, , 31, , 11.73, , 11., , Friut juice, , 0.1, , 74, , 12., , Choclate, , 0.8, , 0.09, , 13., , Rava, , 9, , 575, , 14., , Biscuit, , 19.7, , 498, , 15., , Snacks, , 33.5, , 821, , Mr. Bean wants to establish some statistical, relationship between the above mentioned, food contents. Here the variable under, study are X and Y which represents the, amount of Fat content and the amount, of Sodium content in each food items, respectively. Thus Mr. Bean gets a pair of, values (X,Y) for each food item. Mr. Bean, further found the average fat content in all, the 15 food items is X =7.8 (gms/100g), and the average sodium content in all the, items is Y =325.23(mg/100g). Further, it, was identified that the minimum amount, of fat contained in Tropicana fruit juice is, 0.1(g/100gms) and the maximum amount, of fat contained in Lays is 33.5(g/100gms)., Thus the fat contained in all the 15 food, items is ranging from 0.1(g/100gms) to, 33.5(g/100gms). Similarly, the minimum, amount of sodium content contained in, Kelloggys Choco is 0.09(mg/100gm) and, the maximum amount of sodium content, in Bhindhu appalam is 1440(mg/100gm)., So, the measure of range of fat content and, sodium content in all the 15 items are 33.4gm, and 1439.91(mg/100gm) respectively., Besides, Mr. Bean is interested in knowing, , the variation of each individual item from, the mean of all observations. He attempted, another measure of dispersion known, as standard deviation. The measure of, standard deviation indicates that there is an, average deviation of 11.3 (g/100gms) in fat, content and 420.14(mg/100gms) in sodium, content of all the 15 food items. Further,, Mr. Bean is keen on finding the association, between the variables X and Y. So, the, correlation analysis has been carried out., The correlation coefficient r(X,Y)=0.2285, indicates that there is 22.86% positive, association between sodium content and, the amount of fat content. Thus from this, study Mr. Bean inferred and convinced that, the nutrition information on the packaging, of each food item is sufficient., , Case Study-2, We collected data relating to the gold price, (per gram) in two places namely Chennai, Market and Mumbai Market for ten days, from 20th Feb 2018 to 1st March 2018 and the, same is recorded below., Date, , 20th, Feb, , 21st, Feb, , 22nd, Feb, , 23rd, Feb, , 24th, Feb, , Chennai X, , 2927, , 2912, , 2919, , 2912, , 2921, , Mumbai Y, , 2923, , 2910, , 2907, , 2920, , 2919, , Date, , 25th, Feb, , 26th, Feb, , 27th, Feb, , 28th, Feb, , 1st, Mar, , Chennai X, , 2921, , 2927, , 2924, , 2908, , 2893, , Mumbai Y, , 2919, , 2925, , 2918, , 2902, , 2895, , Do we agree that the price of gold in Chennai, market has its impact on Mumbai market?, Let X denotes the gold price per gm in, Chennai market and Y denotes the gold, Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 231, , 231, , 21-04-2020 12:28:44 PM

Page 238 :
www.tntextbooks.in, , price per gm in Mumbai market. The actual, observations given in the table indicates, that the gold price ranges from ` 2893 to `, 2927 in Chennai market and the gold price, range from ` 2895 (per gram) to ` 2925, (per gram) in Mumbai market. Further it is, to be noted that there is some oscillations, in the gold price rate dated between 20th, Feb to 24th Feb and 25th Feb price is remain, same as the previous day that is Feb 24th., It may be due to holiday of Gold markets., It is clear from the above table the price of, gold rate go on rapidly decreasing from, 27th Feb to 1st March. The same fluctuations, is observed in Mumbai market from, Feb 20th to 24th and remain same on 24th, and 25th and rapidly decreasing from, Feb 27th to 1st March . So, the market trend in, respect of gold rate is same in two markets., We found that the average gold price in, Chennai market during these 10 days is, ` 2916.4 (per gm) and the Mumbai market, is ` 2913.8 (per gm). The variation among, the prices of gold in 10 days. To identify the, variation of each individual observation, from the mean of all observations. We make, use of the best measure known as standard, , 232, , deviation. In this study it is found that the, price of gold has a average deviation of, ` 10 (approximately) in Chennai market and, ` 9 (approximately) in Mumbai market., To verify the consistency of the prices, between the two cities, we attempt, coefficient of variation which expresses, the percentage of variation. Coefficient of, variation of gold price in Chennai market is, v, 10, CVX= X ×100= 2916.4 ×100=0.343%, X, Similarly, Coefficient of variation of gold, price in Mumbai market is, v, 9, CVY= YY ×100= 2913.8 ×100=0.31%, Comparison, , of, , these, , coefficients, , of, , variations we inferred that the Mumbai, market has consistent or more stable in, price of gold., Further to examine the linear relationship, between the two variables X and Y. We carry, out correlation analysis results r(X,Y) = 0.8682., It indicates that there is a positive correlation, in price of gold between Mumbai market and, Chennai market. Do you think finding a, regression line makes sense here?, , 11th Std. Business Mathematics and Statistics, , 09_11th_BM-STAT_Ch-9-EM.indd 232, , 21-04-2020 12:28:44 PM

Page 239 :
www.tntextbooks.in, , Summary, zz The term correlation refers to the degree of relationship between two or more, variables., , zz Scatter diagram is a graphic device for finding correlation between two variables., zz Karl Pearson correlation coefficient r(x,y)=, , cov ^ X, Y h, vx v y, , zz Correlation coefficient r lies between –1 and 1. (i.e) –1 # r # 1, zz When r=+1 , then the correlation is perfect positive, zz When r=–1 , then the correlation is perfect negative, zz When r=0, then there is no relationship between the variables, (i.e) the variables, are uncorrelated., , zz Rank correlation deals with qualitative characteristics., zz Spearman’s rank correlation coefficient formula ρ is given by, t = 1–, , 6Rd2, where d = The difference between two ranks = R X–RY and, N ^N2 - 1h, , N = Number of paired observations., , zz C orrelation represents linear relationship between the variables but the regression, helps to estimate (or predict) one variable by using the other variable., , zz Regression lines of, (i) X on Y is X– X =bxy(Y– Y ), , (ii)Y on X is Y– Y =byx (X– X ), , zz The two regression lines passing through their respective means of X and Y, zz Calculation of the regression coefficients., vy, v, bxy=r v x and byx=r v, y, x, , zz The Properties of regression coefficients., (i) r =, , b yx # bxy, , (ii) both the regression coefficients cannot be greater than one., (iii) Both the regression coefficients have same sign., , Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 233, , 233, , 21-04-2020 12:28:45 PM

Page 241 :
www.tntextbooks.in, , ICT Corner, Correlation and regression analysis, Step – 1, Open the Browser type the URL Link given, below (or) Scan the QR Code., GeoGebra Workbook called “11th BUSINESS, MATHEMATICS and STATISTICS” will, appear. In that there are several worksheets, related to your Text Book., Expected Outcome ⇒, Step-2, Select the work sheet “Regression lines” work out, the problem for the data given and workout as given and verify the steps. See the graph of, regression line x on y and regression line y on x and check the intersection of these two lines., (Mean of x, mean of y) you can change the data x and Y in the spread sheet for new problem., Step 1, , Step 2, , Browse in the link, 11th Business Mathematics and Statistics, https://ggbm.at/qKj9gSTG (or) scan the QR Code, Correlation and Regression analysis, , 09_11th_BM-STAT_Ch-9-EM.indd 235, , 235, , 21-04-2020 12:28:45 PM

Page 242 :
Chapter, , www.tntextbooks.in, , 10, , OPERATIONS RESEARCH, , Learning Objectives, After studying this chapter, the students, will be able to understand, , •, , Formulation of Linear Programming, Problem, , •, , Solution of LPP by graphical, method., , •, , Construction of network of a, project, , •, , Project completion time by Critical, Path Method(CPM), , Introduction, , During the world war II,, the military management, in England recruited a team, of scientists, engineers, and mathematicians to, study the strategic and, L.V. Kantorovich, tactical problems of air, and land defence. Their objective was to, determine the best utilization of limited, military resources like ammunition, food, and other things needed for war. This group, of scientists formed the first operations, research team. The name operations research, was apparently coined because the team, was dealing with the research on (military), operations. Operations research team, helped in developing strategies for mining, operations, inventing new flight pattern and, planning of sea mines. Following the end, of war the success of military team attracted, 236, , the attention of industrial managers who, were seeking solutions to their complex type, of problems., It is not possible to give, un i for m ly a cce pt abl e, definition of operations, research. The following is, the definition of operations, research published on, behalf of UK operational research society., “operations research is the application of, scientific methods to complex problems, arising from operations involving large, system of men, machines, materials and, money in industry, business, government, and defence.”, An operations research model is defined as, any abstract or idealized representation of, real life system or situation. The objective, of the model is to identify significant factors, and inter-relationship. Here we study only, two models namely linear programming, problem and network analysis., , 10.1 Linear Programming Problem, , The Russian Mathematician L.V. Kantorovich, applied mathematical model to solve linear, programming problems. He pointed out in, 1939 that many classes of problems which arise, in production can be defined mathematically, and therefore can be solved numerically., This decision making technique was further, developed by George B. Dantziz. He formulated, the general linear programming problem, , 11th Std. Business Mathematics and Statistics, , 10_11th_BM-STAT_Ch-10-EM.indd 236, , 21-04-2020 12:32:46 PM

Page 243 :
www.tntextbooks.in, , and developed simplex method (1947) to, solve complex real time applications. Linear, programming is one of the best optimization, technique from theory, application and, computation point of view., , Definition:, Linear Programming Problem(LPP) is, a mathematical technique which is used, to, optimize (maximize or minimize), the objective function with the limited, resources., Mathematically,, the, general, linear, programming problem (LPP) may be stated, as follows., Maximize or Minimize, Z = c1 x1 + c2 x2 + … + cn xn, , Decision variable:, The decision variables are the variables,, which has to be determined xj , j = 1,2,3,…,n,, to optimize the objective function., , Constraints:, There are certain limitations on the use of, limited resources called constraints., n, , / aij x j # ^qr = qr $h bi ,, , j=1, , i = 1,2,3,…, m, , are the constraints., , Solution:, , A set of values of decision variables xj,, j=1,2,3,…, n satisfying all the constraints, of the problem is called a solution to that, problem., , Feasible solution:, , Subject to the conditions (constraints), , a21 x1 + a22 x2 + f + a2n xn # ^qr = qr $h b2, , …, , …., , …., , …., , …, , A set of values of the decision variables that, satisfies all the constraints of the problem, and non-negativity restrictions is called a, feasible solution of the problem., , …, , …., , …., , …., , …, , Optimal solution:, , a11 x1 + a12 x2 + f + a1n xn # ^qr = qr $h b1, , am1 x1 + am2 x2 + ... + amn xn ^# qr = or $h bm, x1, x2 …….xn $ 0, , Short form of LPP, , Maximize or Minimize Z =, Subject to, , n, , Feasible region:, , n, , / cjxj, , j=1, , / aij x j # ^qr = qr $h bi,, , j=1, , i = 1,2,3,…,m... (1), and xj≥0 , , Any feasible solution which maximizes or, minimizes the objective function is called, an optimal solution., , ... (2), , The common region determined by all, the constraints including non-negative, constraints xj ≥0 of a linear programming, problem is called the feasible region (or, solution region) for the problem., , Some useful definitions:, Objective function:, , A function Z=c1 x1 + c2x2 + …+ cnxn which is, to be optimized (maximized or minimized), is called objective function., , Linear Programming Problem helps, a farmer to produce the best crop by, minimising risk and maximizing profit., Operations Research, , 10_11th_BM-STAT_Ch-10-EM.indd 237, , 237, , 21-04-2020 12:32:49 PM

Page 244 :
www.tntextbooks.in, , 10.1.1 M,  athematical formulation of a, linear programming problem:, The procedure for mathematical formulation, of a linear programming problem consists, of the following steps., (i) Identify the decision variables., (ii) Identify the objective function to be, maximized or minimized and express it, as a linear function of decision variables., (iii) I dentify the set of constraint conditions, and express them as linear inequalities, or equations in terms of the decision, variables., , Linear Programming Problem is used, in determining shortest routes for, travelling salesman., Example 10.1, A furniture dealer deals only two items viz.,, tables and chairs. He has to invest `10,000/and a space to store atmost 60 pieces. A table, cost him ` 500/– and a chair ` 200/–. He can, sell all the items that he buys. He is getting, a profit of ` 50 per table and ` 15 per chair., Formulate this problem as an LPP, so as to, maximize the profit., Solution:, (i) Variables:, Let x1 and x2 denote the number of, tables and chairs respectively., (ii) Objective function:, Profit on x1 tables = 50 x1, Profit on x2 chairs = 15 x2, Total profit = 50 x1+15x2, Let Z = 50 x1+15 x2 , which is the, objective function., 238, , Since the total profit is to be maximized,, we have to maximize Z=50x1+15x2, (iii) Constraints:, The dealer has a space to store atmost, 60 pieces, x1+x2 ≤ 60, The cost of x1 tables = ` 500 x1, The cost of x2 tables = ` 200 x2, Total cost = 500 x1 + 200 x2 ,which, cannot be more than 10000, 500 x1 + 200 x2 ≤ 10000, 5 x1+ 2x2 ≤ 100, (iv) Non-negative restrictions:, , Since the number of tables and chairs, cannot be negative, we have x1 ≥0, x2 ≥0, , Thus, the mathematical formulation of, the LPP is, Maximize Z=50 x1+15x2, , Subject to the constrains, x1+x2 ≤60, , 5x1+2x2≤100, , x1, x2≥0, , Example 10.2, A company is producing three products P1, P2, and P3, with profit contribution of ` 20, ` 25, and ` 15 per unit respectively. The resource, requirements per unit of each of the products, and total availability are given below., P1, , P2, , P3, , Total availability, , 6, , 3, , 12, , 200, , Machine hours/unit 2, , 5, , 4, , 350, , 2kg, , 1kg, , 100kg, , Product, Man hours/unit, , Material/unit, , 1kg, , Formulate the above as a linear programming, model., , 11th Std. Business Mathematics and Statistics, , 10_11th_BM-STAT_Ch-10-EM.indd 238, , 21-04-2020 12:32:49 PM

Page 246 :
www.tntextbooks.in, , Example 10.4, A soft drink company has two bottling plants, C1 and C2. Each plant produces three different, soft drinks S1,S2 and S3. The production of the, two plants in number of bottles per day are:, Product, , Plant, C1, , C2, , S1, , 3000, , 1000, , S2, , 1000, , 1000, , S3, , 2000, , 6000, , A market survey indicates that during the, month of April there will be a demand for, 24000 bottles of S1, 16000 bottles of S2 and, 48000 bottles of S3. The operating costs,, per day, of running plants C1 and C2 are, respectively ` 600 and ` 400. How many days, should the firm run each plant in April so, that the production cost is minimized while, still meeting the market demand? Formulate, the above as a linear programming model., Solution:, (i) Variables:, , Let x1 be the number of days required, to run plant C1 and x2 be the number of, days required to run plant C2., , Objective function: Minimize, Z = 600 x1 + 400 x2, (ii) Constraints: 3000 x1 + 1000 x2 ≥ 24000, (since there is a demand of 24000, bottles of drink A, production should, not be less than 24000), 1000 x1 + 1000 x2 ≥ 16000, 2000 x1+ 6000 x2 ≥ 48000, (iii) Non-negative restrictions: Since be, the number of days required of a firm, are non-negative, we have x1, x2 ≥0., 240, , Thus we have the following LP model., Minimize Z = 600 x1 + 400 x2 subject, to 3000 x1 + 1000 x2 ≥ 24000, 1000 x1 + 1000 x2 ≥ 16000, 2000 x1 + 6000 x2 ≥ 48000 and x1, x2 ≥0, , 10.1.2 S,  olution of LPP by graphical, method, After formulating the linear programming, problem, our aim is to determine the values, of decision variables to find the optimum, (maximum or minimum) value of the, objective function. Linear programming, problems which involve only two variables, can be solved by graphical method. If the, problem has three or more variables, the, graphical method is impractical., The major steps involved in this method are, as follows:, (i) State the problem mathematically, (ii) W,  rite all the constraints in the form of, equations and draw the graph, (iii) Find the feasible region, (iv) , Find the coordinates of each vertex, (corner points) of the feasible region., The coordinates of the vertex can be, obtained either by inspection or by, solving the two equations of the lines, intersecting at the point, (v) B,  y substituting these corner points in, the objective function we can get the, values of the objective function, (vi) If the problem is maximization then, the maximum of the above values is, the optimum value. If the problem is, minimization then the minimum of the, above values is the optimum value., , 11th Std. Business Mathematics and Statistics, , 10_11th_BM-STAT_Ch-10-EM.indd 240, , 21-04-2020 12:32:49 PM

Page 247 :
www.tntextbooks.in, , Example 10.5, Solve the following LPP, , x2, , 24, 22, (0, 21), , 20, , Maximize Z = 2 x1 +5x2 subject to the, conditions x1+ 4x2 ≤ 24 3x1+x2 ≤ 21, x1+x2 ≤ 9and x1, x2 ≥ 0, Solution:, First we have to find the feasible region, using the given conditions., Since both the decision variables x1 and x2, are non-negative, the solution lies in the, first quadrant., Write all the inequalities of the constraints, in the form of equations., Therefore we have the lines x1+ 4x2=24;, 3x1 + x2 = 21; x1 + x2= 9, x1+ 4x2= 24 is a line passing through the, points (0 , 6) and (24 , 0)., [(0,6) is obtained by taking x1=0 in, x1 + 4x2 = 24 , (24 , 0) is obtained by taking, x2 = 0 in x1+ 4x2 = 24]., Any point lying on or below the line, x1 + 4x2 = 24 satisfies the constraint, x1 + 4x2≤ 24 ., 3 x1 +x2= 21 is a line passing through the, points (0, 21) and (7, 0). Any point lying on, or below the line 3 x1 + x2 = 21 satisfies the, constraint 3 x1 + x2 ≤ 21., x1+ x2 = 9 is a line passing through the, points (0 , 9) and ( 9 , 0) .Any point lying, on or below the line x1 + x2 = 9 satisfies the, constraint x1+ x2 ≤ 9., Now we draw the graph., , 18, 16, 14, 12, 10, 8, 6, , (0, 9), D(0, 6) C(4, 5), , 4, B(6, 3), , 2, , F.R., (7, 0), , 0, , 2, , 4, , 6, , (24, 0), , (9, 0), , 8, , 10, , 12, , 14, , 16, , 18, , 20, , 22, , x1, , 26, , Fig 10.1, , The feasible region satisfying all the, conditions is OABCD. The co-ordinates, of the points are O(0,0) A(7,0);B(6,3) [the, point B is the intersection of two lines, x1+ x2= 9 and 3 x1+ x2= 21];C(4,5) [the point, C is the intersection of two lines x1+ x2 = 9, and x1+ 4x2 = 24] and D(0,6)., Corner points, , Z = 2x1 + 5x2, , O(0,0), , 0, , A(7,0), , 14, , B(6,3), , 27, , C(4,5), , 33, , D(0,6), , 30, , Table 10.2, , Maximum value of Z occurs at C. Therefore, the solution is x1 =4, x2 = 5, Z max = 33., Example 10.6, Solve the following LPP by graphical method, Minimize z = 5x1+4x2 Subject to constraints, 4x1+ x2 ≥ 40 ; 2x1+3x2 ≥ 90 and x1, x2 > 0, Solution:, Since both the decision variables x1 and x2, are non-negative, the solution lies in the, first quadrant of the plane., Operations Research, , 10_11th_BM-STAT_Ch-10-EM.indd 241, , 24, , 241, , 21-04-2020 12:32:49 PM

Page 248 :
www.tntextbooks.in, , Consider the equations 4x1+x2 = 40 and, 2 x1+3 x2 = 90., , Example 10.7, , 4x1+x2 = 40 is a line passing through the, points (0,40) and (10,0). Any point lying, on or above the line 4x1+x2= 40 satisfies the, constraint 4x1+ x2 ≥ 40., , Maximize Z= 2 x1 +3x2 subject to constraints, x1 + x2 ≤ 30 ; x2 ≤ 12; x1 ≤ 20 and, x1, x2≥ 0., , Solve the following LPP., , 2x1+3x2 = 90 is a line passing through the, points (0,30) and (45,0). Any point lying on, or above the line 2 x1+3x2= 90 satisfies the, constraint 2x1+3x2 ≥ 90., Draw the graph using the given constraints., x2, , 40 C(0, 40), 36, , Therefore we have the lines, , 32, (0, 30), B(3, 28), , x1+x2=30; x2 =12; x1= 20, , 24, , x1+x2 =30 is a line passing through the points, (0,30) and (30,0), , 20, 16, 12, , x2 = 12 is a line parallel to x1–axis, , 8, , x1 = 20 is a line parallel to x2–axis., , 4, A(45, 0), , (10, 0), , 0, , Since both the decision variables x1 and x2, are non-negative, the solution lies in the first, quadrant of the plane., Write all the inequalities of the constraints, in the form of equations., , 44, , 28, , Solution:, We find the feasible region using the given, conditions., , 4, , 8, , 12, , 16, , 20, , 24, , 28, , 32, , 36, , 40, , 44, , 48, , x1, , Fig 10.2, , The feasible region is ABC (since the, problem is of minimization type we are, moving towards the origin., Corner points, , z = 5x1+4x2, , A(45,0), , 225, , The feasible region satisfying all the, conditions x1+ x2≤ 30; x2≤ 12 ; x1≤ 20 and, x1, x2 ≥ 0 is shown in the following graph., x2, , 30 (0, 30), 28, 26, 24, 22, 20, 18, 16, , B(3,28), , 127, , 14, 12, , C(0,40), , 160, , Table 10.3, , The minimum value of Z occurs at B(3,28)., Hence the optimal solution is x1 = 3, x2 = 28, and Zmin=127., 242, , D(0, 12), , C(18, 12), , 10, , x1, , B(20, 10), , 8, 6, 4, 2, (30, 0), , A(20, 0), , 0, , 2, , 4, , 6, , 8, , 10, , 12, , 14, , 16, , 18, , 20, , 22, , 24, , 26, , 28, , 30, , x1, , Fig 10.3, , 11th Std. Business Mathematics and Statistics, , 10_11th_BM-STAT_Ch-10-EM.indd 242, , 21-04-2020 12:32:49 PM

Page 249 :
www.tntextbooks.in, , The feasible region satisfying all the, conditions is OABCD., , Now we draw the graph satisfying the, conditions x1 – x2 < –1; –x1+x2 < 0 and x1, x2 ≥ 0., , The co-ordinates of the points are O(0,0);, A(20,0); B(20,10); C(18,12) and D(0,12)., , There is no common region(feasible region), satisfying all the given conditions. Hence, the given LPP has no solution., , Corner points, , Z = 2x1 + 3x2, , O(0,0), , 0, , A(20,0), , 40, , B(20,10), , 70, , C(18,12), , 72, , D(0,12), , 36, , Exercise 10.1, , Table 10.4, , Maximum value of Z occurs at C. Therefore, the solution is x1 = 18 , x2= 12, Z max = 72., Example 10.8, Maximize Z = 3x1 + 4x2 subject to x1 – x2 < –1;, –x1+x2 < 0 and x1, x2 ≥ 0., Solution:, x2, , 9, 8, 7, , (0, 30), B(3, 28), , 6, 5, , –, x1, , 4, , =, x2, , –1, , +, , –x 1, , 3, , =, x2, , 0, , 2, 1, , (0, 1), , x1, , (0, –1), , –2, , 0 (0, 0), –1 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 1. A company produces two types of pens A, and B. Pen A is of superior quality and pen, B is of lower quality . Profits on pens A, and B are ` 5 and ` 3 per pen respectively., Raw materials required for each pen A is, twice as that of pen B. The supply of raw, material is sufficient only for 1000 pens, per day . Pen A requires a special clip and, only 400 such clips are available per day., For pen B, only 700 clips are available per, day . Formulate this problem as a linear, programming problem., 2. , A company produces two types of, products say type A and B. Profits on, the two types of product are ` 30/- and, ` 40/- per kg respectively. The data on, resources required and availability of, resources are given below., Requirements, , Capacity, available per, Product A Product B, month, , 12, , Fig 10.4, , Since both the decision variables x1, x2 are, non-negative ,the solution lies in the first, quadrant of the plane., Consider the equations x1– x2 = –1 and, – x1 + x2 = 0., x1– x2 =–1 is a line passing through the, points (0,1) and (–1,0)., –x1 + x2 = 0 is a line passing through the, point (0,0)., , Raw material, (kgs), , 60, , 120, , 12000, , Machining, hours / piece, , 8, , 5, , 600, , Assembling, (man hours), , 3, , 4, , 500, , Formulate this problem as a linear, programming problem to maximize the, profit., 3. A,  company manufactures two models of, voltage stabilizers viz., ordinary and a­ utoOperations Research, , 10_11th_BM-STAT_Ch-10-EM.indd 243, , 243, , 21-04-2020 12:32:50 PM

Page 250 :
www.tntextbooks.in, , cut. All components of the stabilizers are, purchased from outside sources , assembly, and testing is carried out at company’s, own works. The assembly and testing, time required for the two models are 0.8, hour each for ordinary and 1.20 hours, each for auto-cut. Manufacturing capacity, 720 hours at present is available per week., The market for the two models has been, surveyed which suggests maximum weekly, sale of 600 units of ordinary and 400 units, of auto-cut. Profit per unit for ordinary, and auto-cut models has been estimated at, ` 100 and ` 150 respectively. Formulate the, linear programming problem., 4. S olve the following linear programming, problems by graphical method., (i) Maximize Z = 6x1 + 8x2 subject, to constraints 30x1+20x2 # 300;, 5x1+10x2 # 110; and x1, x2 > 0 ., (ii) Maximize Z = 22x1 + 18x2 subject to, constraints 960x1 + 640x2 # 15360 ;, x1 + x2 # 20 and x1 , x2 $ 0 ., (iii) Minimize Z = 3x1 + 2x2 subject, to the constraints 5x1+ x2≥10;, x1+ x2≥6; x1+ 4 x2 ≥12 and x1, x2≥0., (iv) Maximize Z = 40x1 + 50x2 subject to, constraints 3x1 + x2 # 9; x1 + 2x2 # 8, and x1 , x2 $ 0, (v) Maximize Z = 20x1 + 30x2 subject, to, constraints, 3x1 + 3x2 # 36 ;, 5x1 + 2x2 # 50 ; 2x1 + 6x2 # 60 and, x1 , x2 $ 0, (vi) Minimize Z = 20x1 + 40x2 subject, to the constraints 36x1 + 6x2 $ 108,, 3x1 + 12x2 $ 36, 20x1 + 10x2 $ 100, and x1 , x2 $ 0, 244, , 10.2, , Network Analysis, , Networks are diagrams easily visualized in, transportation system like roads, railway, lines, pipelines, blood vessels, etc., A project will consist of a number of jobs, and particular jobs can be started only after, finishing some other jobs. There may be, jobs which may not depend on some other, jobs. Network scheduling is a technique, which helps to determine the various, sequences of jobs concerning a project and, the project completion time. There are two, basic planning and control techniques that, use a network to complete a pre-determined, schedule. They are Program Evaluation and, Review Technique (PERT) and Critical Path, Method (CPM). The critical path method, (CPM) was developed in 1957 by JE Kelly, of Ramington R and M.R. Walker of Dupon, to help schedule maintenance of chemical, plants. CPM technique is generally applied, to well known projects where the time, schedule to perform the activities can, exactly be determined., , Some important definitions in network, Activity:, An activity is a task or item of work to be, done, that consumes time, effort, money or, other resources. It lies between two events,, called the starting event and ending event., An activity is represented by an arrow, indicating the direction in which the events, are to occur., , Event:, The beginning and end points of an activity, are called events (or nodes). Event is a, point in time and does not consume any, resources. The beginning and completion, , 11th Std. Business Mathematics and Statistics, , 10_11th_BM-STAT_Ch-10-EM.indd 244, , 21-04-2020 12:32:53 PM

Page 251 :
www.tntextbooks.in, , of an activity are known as tail event and, head event respectively. Event is generally, represented by a numbered circle. The, head event, called the jth event, has always, a number higher than the tail event, called, the ith event, ie., j > i., , Predecessor Activity:, Activities which must be completed, before a particular activity starts are called, predecessor activities. If an activity A is, predecessor of an activity B, it is denoted by, A<B. (i.e.,) activity B can start only if activity, A is completed., , Successor Activity:, An activity that cannot be started until one, or more of the other activities are completed,, but immediately succeed them is referred to, as successor activity., , (iii) , Nodes are numbered to identify an, activity uniquely. Tail node (starting, point) should be lower than the head, node (end point) of an activity., (iv) Arrows should not cross each other., (v) Arrows should be kept straight and not, curved or bent., (vi) Every node must have atleast one activity, preceding it and atleast one activity, following it except for the node at the, beginning and at the end of the network., , Numbering the Events, After the network is drawn in a logical, sequence, every event is assigned a number., The number sequence must be such as to, reflect the flow of the network. In event, numbering, the following rules should be, observed:, , 10.2.1 Construction of network:, , (i) Event numbers should be unique., (ii) Event numbering should be carried out, on a sequential basis from left to right., (iii) The initial event is numbered 0 or 1., (iv) The head of an arrow should always bear, a number higher than the one assigned, at the tail of the arrow., (v) Gap should be left in the sequence of, event numbering to accommodate, subsequent inclusion of activities, if, necessary., , Rules for constructing network, , Remark: The above procedure of assigning, , Network:, Network is a diagrammatic representation, of various activities concerning a project, arranged in a logical manner., , Path:, A path is defined as a set of nodes connected, by lines which begin at the initial node and, end at the terminal node of the network., , For the construction of a network, generally,, the following rules are followed:, , numbers to various events of a network is, known as Fulkerson’s Rule., , (i) E,  ach activity is represented by one and, only one arrow. (i.e) only one activity, can connect any two nodes., , Example 10.9, Draw the logic network for the following:, ,  o two activities can be identified by, (ii) N, the same head and tail events., , Activities C and D both follow A, activity, E follows C, activity F follows D, activity, E and F precedes B., Operations Research, , 10_11th_BM-STAT_Ch-10-EM.indd 245, , 245, , 21-04-2020 12:32:53 PM

Page 252 :
www.tntextbooks.in, , Solution:, The required network for the above, information., E, , 3, , C, A, , 1, , 2, , B, , 5, , D, , 6, , Solution:, Using the precedence relationships and, following the rules of network construction,, the required network diagram is shown in, following figure, , F, , 4, , D, , 2, , Fig 10.5, , 1, , Example 10.10, Develop a network based on the following, information:, A, , B, , C, , D, , Immediate, predecessor:, , -, , -, , A, , B, , E, , F, , C,D C,D, , G, , H, , E, , F, , 3, , C, , 2, , E, , 1, , F, , 7, , J, , 9, K, , B, D < F; C < G and B < H., , Solution:, Using the precedence relationships and, following the rules of network construction,, the required network is shown in following, figure., H, , 2, , B, , H, , 6, , Fig 10.6, , DUM, , MY, , A, , D, , 3, , F, , 5, , 6, , E, , C, , G, 4, , Dummy activity:, , Fig 10.8, , An activity which does not consume any, resource or time, but merely depict the, technological dependence is called a dummy, activity. It is represented by dotted lines., Example 10.11, Draw a network diagram for the project, whose activities and their predecessor, relationships are given below:, Activity : A B, , C D E, , F, , G H, , I, , Predecessor activity:, , -, , B, , C D, , F H,I F,G, , 246, , G, , DUMMY, , I, , Example 10.12, Construct a network diagram for the, following situation:, , 1, , -, , 6, , Fig 10.7, , 7, , D, , 3, , E, F, , G, , 5, , 4, , B, , 4, , A < D, E;, , Solution:, Using, the, immediate, precedence, relationships and following the rules of, network construction, the required network, is shown in following figure., A, , H, , 8, , B, C, , Activity :, , 5, , A, , A B, , J, , K, , 10.2.2 Critical path analysis, For each activity an estimate must be made, of time that will be spent in the actual, accomplishment of that activity. Estimates, may be expressed in hours, days, weeks, or any other convenient unit of time. The, time estimate is usually written in the, network immediately above the arrow. For, the purpose of calculating various times of, events and activities, the following terms, shall be used in critical path calculations:, Ei = Earliest start time of event i, , 11th Std. Business Mathematics and Statistics, , 10_11th_BM-STAT_Ch-10-EM.indd 246, , 21-04-2020 12:32:53 PM

Page 253 :
www.tntextbooks.in, , Lj = Latest start time of event j, tij = Duration of activity (i,j), The next step after making the time estimates, is the calculations which are done in the, following ways:, (i) Forward Pass Calculations, (ii) Backward Pass Calculations., , Forward pass calculations:, We start from the initial node 1 with starting, time of the project as zero. Proceed through, the network visiting nodes in an increasing, order of node number and end at the final, node of the network. At each node, we, calculate earliest start times for each activity, by considering Ei as the earliest occurrence, of node i., , The method may be summarized as, below:, Step 1: Set E1 = 0 ; i = 1 (initial node)., Step 2: Set the earliest start time(EST), for each activity that begins at node i as, ESij = Ei; for all activities (i, j) that start at, node i., Step 3: Compute the earliest finish time, (EFT) of each activity that begins at node i, by adding the earliest start time of the, activity to the duration of the activity. Thus, EFij = ESij + tij = Ei + tij., Step 4: Move on to next node, say node j(j > i), and compute the earliest start time at node j,, using E j = max i " EFij , = maxi " Ei + tij , for, all immediate predecessor activities., Step 5: If j = n (final node), then the, earliest finish time for the project is given by, En = max " EFij , = max " En - 1 + tij , ., , Backward pass calculations:, We start from the final (last) node n of the, network, proceed through the network visiting, nodes in the decreasing order of node numbers, and end at the initial node 1. At each node, we, calculate the latest finish time and latest start, time for each activity by considering Lj as the, latest occurrence of node j. The method may, be summarized below:, Step 1: Ln = En; for j = n, Step 2: Set the latest finish time (LFT)of, each activity that ends at node j as LFij = L j, Step 3: Compute the latest start time (LST), of all activities ending at node j, subtracting, the duration of each activity from the, latest finish time of the activity. Thus,, LSij = LFij - tij = L j - tij, Step 4: Proceed backward to the next, node i( i<j) in the sequence and compute, the latest occurrence time at node i using, Li = min j " LSij , = min j " L j - tij , ., , Step 5: If j = 1 (initial node), then, L1 = min " LSij , = min " L2 - tij ,, , Critical path:, The longest path connected by the activities in, the network is called the critical path. A path, along which it takes the longest duration., For the activity (i,j), to lie on the critical, path, following conditions must be satisfied:, (i) Ei = Li and E j = L j, (ii) E j - Ei = L j - Li = tij, Example 10.13, Compute the earliest start time, earliest, finish time ,latest start time and latest finish, time of each activity of the project given, below:, Operations Research, , 10_11th_BM-STAT_Ch-10-EM.indd 247, , 247, , 21-04-2020 12:32:55 PM

Page 254 :
www.tntextbooks.in, , Activity, Duration, ( in days), , 1-2, , 1-3, , 2-4, , 2-5, , 3-4, , 4-5, , 8, , 4, , 10, , 2, , 5, , 3, , Solution:, Earliest start time (EST) and latest finish, time (LFT) of each activity are given ithe, following network., E=8, L=8, , 8, , 5, , 1, , E = 21, L = 21, , Critical path is 1-2-4-5 and project, completion time is 21 days., Example 10.14, Calculate the earliest start time, earliest finish, time, latest start time and latest finish time of, each activity of the project given below and, determine the Critical path of the project and, duration to complete the project., , 3, , 10, , E=0, L=0, , 2, , 2, , The path connected by the critical activities, is the critical path (the longest path)., , 4, , 5, , Activity, , 4, , 3, E=4, L = 13, , 1-2 1-3 1-5 2-3 2-4 3-4 3-5 3-6 4-6 5-6, , Duration, ( in week), , E = 18, L = 18, , 8, , 7, , 10, , 3, , 5, , 10, , E=8, L=8, , E1 = 0, , L5 = 21, , E2 = E1 + t12 = 0 + 8 = 8, , L4 = L5 – t45 = 21–3 = 18, , E3 = E1 + t13 = 0 + 4 = 4, , L3 = L4 – t34 = 18–5 =13, , E4 = E2 +t24 or E3 + t34, = 8 + 10 = 18, (take E2 + t24 or E3 + t34, whichever is maximum), , L2 = L5 –t25 or L4–t24, = 18–10= 8, (take L5 –t25 or L4– t24, whichever is minimum), , 7, , 4, , (take E2 + t25 or E4+t45 (take L2 – t12 or L3 – t13, whichever is maximum), whichever is minimum), Here the critical path is 1-2-4-5, which is, denoted by double lines., EFT=EST+tij LST=LFT–, LFT, tij, , E = 18, L = 18, , 10, , 2, , 4, , 8, 4, 3, E=0, L=0, , 1, , 7, , E = 12, L = 15, , 7, 3, , 10, , 12, , 6, , 5, , E = 25, L = 25, , 4, , 5, , E5 = (E2+t25 or E4+t45) L1= L2 –t12 or L3- t13, = 18 + 3 = 21 = 8 – 8 = 0, , EST, , 4, , Solution:, , Fig 10.9, , Activity Duration, (tij), , 12, , E = 17, L = 21, , Fig 10.10, Activity Duration, (in week), , EST, , EFT, , LST, , LFT, , 1-2, , 8, , 0, , 8, , 0, , 8, , 1-3, , 7, , 0, , 7, , 8, , 15, , 1-5, , 12, , 0, , 12, , 9, , 21, , 2-3, , 4, , 8, , 12, , 11, , 15, , 2-4, , 10, , 8, , 18, , 8, , 18, , 3-4, , 3, , 12, , 15, , 15, , 18, , 1–2, , 8, , 0, , 8, , 0, , 8, , 1–3, , 4, , 0, , 4, , 9, , 13, , 2–4, , 10, , 8, , 18, , 8, , 18, , 2–5, , 2, , 8, , 10, , 19, , 21, , 3–4, , 5, , 4, , 9, , 13, , 18, , 3-5, , 5, , 12, , 17, , 16, , 21, , 4–5, , 3, , 18, , 21, , 18, , 21, , 3-6, , 10, , 12, , 22, , 15, , 25, , Table 10.5, , 4-6, , 7, , 18, , 25, , 18, , 25, , The longest duration to complete this project, is 21 days., , 5-6, , 4, , 17, , 21, , 21, , 25, , 248, , Table 10.6, , 11th Std. Business Mathematics and Statistics, , 10_11th_BM-STAT_Ch-10-EM.indd 248, , 21-04-2020 12:32:56 PM

Page 255 :
www.tntextbooks.in, , Here the critical path is 1–2–4–6, , the critical path and the project, completion time., , The project completion time is 25 weeks, 6., Exercise 10.2, 1., ,  raw the network for the project whose, D, activities with their relationships are, given below:, , , Activities, A,D,E, can, start, simultaneously; B,C>A;, G,F>D,C;, H>E,F., 2., , 3., ,  raw the event oriented network for, D, the following data:, Events, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , Immediate, Predecessors, , –, , 1, , 1, , 2,3, , 3, , 4,5, , 5,6, , 4., ,  onstruct the network for each the, C, projects consisting of various activities, and their precedence relationships are, as given below:, Activity, , A B C D E F G H I, , J K, , Immediate, Predecessors – – – A B B C D E H,I F,G, , 5., , Activity 1-2 1-3 2-4 3-4 3-5 4-9 5-6 5-7 6-8 7-8 8-10 9-10, Time, ,  onstruct the network for the project, C, whose activities are given below., Activity, Duration, (in week), , 0-1 1-2 1-3 2-4 2-5 3-4 3-6 4-7 5-7 6-7, 3, , 8 12 6, , 3, , 3, , 8, , 5, , 3, , 8, , Calculate the earliest start time, earliest, finish time, latest start time and latest, finish time of each activity. Determine, , 4, , 1, , 1, , 1, , 6, , 5, , 4, , 8, , 1, , 2, , 7., , 7, ,  raw the network and calculate the, D, earliest start time, earliest finish time,, latest start time and latest finish time of, each activity and determine the Critical, path of the project and duration to, complete the project., Jobs, Duration, , 8., , 1-2 1-3 2-4 3-4 3-5 4-5 4-6 5-6, 6, , 5, , 10, , 3, , 4, , 6, , 2, , 9, , The following table gives the activities, of a project and their duration in days, Activity, , 1-2, , 1-3, , 2-3, , 2-4, , 3-4, , 3-5, , 4-5, , Duration, , 5, , 8, , 6, , 7, , 5, , 4, , 8, , Construct the network and calculate, the earliest start time, earliest finish, time, latest start time and latest finish, time of each activity and determine the, Critical path of the project and duration, to complete the project., 9., , , A Project has the, schedule, Activity, Duration, (in days), , following time, , 1-2 1-6 2-3 2-4 3-5 4-5 6-7 5-8 7-8, 7, , 6, , 14, , 5, , 11, , 7, , 11, , Operations Research, , 10_11th_BM-STAT_Ch-10-EM.indd 249, , 5, , Construct the network and calculate, the earliest start time, earliest finish, time, latest start time and latest finish, time of each activity and determine the, Critical path of the project and duration, to complete the project., ,  onstruct the network for the projects, C, consisting of various activities and their, precedence relationships are as given, below:, , A,B,C can start simultaneously A<F, E;, B<D, C; E, D<G, , A project schedule has the following, characteristics, , 4, , 18, , 249, , 21-04-2020 12:32:56 PM

Page 256 :
www.tntextbooks.in, , Construct the network and calculate, the earliest start time, earliest finish, time, latest start time and latest finish, time of each activity and determine the, Critical path of the project and duration, to complete the project., 10. Th,  e following table use the activities, in a construction projects and relevant, information, Activity, , 1-2, , 1-3, , 2-3, , 2-4, , 3-4, , 4-5, , Duration, (in days), , 22, , 27, , 12, , 14, , 6, , 12, , , Draw the network for the project, ,calculate the earliest start time, earliest, finish time, latest start time and latest, finish time of each activity and find, the critical path. Compute the project, duration., Exercise 10.3, , Choose the correct answer, , 2, 10, , 1, , 25, 3, , , , 12, 4, , (a) 1 – 2 – 4 – 5, (c) 1 – 2 – 3 – 5, , 5, , 10, 5, , 8, , (b) 1– 3– 5, (d) 1 – 2 – 3 – 4 – 5, , 2. M,  aximize: z = 3x1 + 4x2 subject to, 2x1 + x2 # 40, 2x1 + 5x2 # 180, x1 , x2 $ 0 ., In the LPP, which one of the following, is feasible corner point?, (a) x1 = 18, x2 = 24, (b) x1 = 15, x2 = 30, 250, , 3. One of the conditions for the activity, (i, j) to lie on the critical path is, (a) E j - Ei = L j - Li = tij, (b) Ei - E j = L j - Li = tij, (c) E j - Ei = Li - L j = tij, (d) E j - Ei = L j - Li ! tij, 4. In constructing the network which one, of the following statement is false?, (a) Each activity is represented by one, and only one arrow. (i.e.) only one, activity can connect any two nodes., (b) Two activities can be identified by, the same head and tail events., (c) Nodes are numbered to identify an, activity uniquely. Tail node (starting, point) should be lower than the, head node (end point) of an activity., (d) Arrows should not cross each other., , 1. Th,  e critical path of the following, network is, 20, , (c) x1 = 2 $ 5, x2 = 35, (d) x1 = 20 $ 5, x2 = 19, , 5. , In a network while numbering the, events which one of the following, statement is false?, (a) Event numbers should be unique., (b) Event numbering should be carried, out on a sequential basis from left to, right., (c) The initial event is numbered 0 or 1., (d) The head of an arrow should always, bear a number lesser than the one, assigned at the tail of the arrow., 6. A solution which maximizes or minimizes, the given LPP is called, , 11th Std. Business Mathematics and Statistics, , 10_11th_BM-STAT_Ch-10-EM.indd 250, , 21-04-2020 12:32:57 PM

Page 257 :
www.tntextbooks.in, , (a) a solution, (b) a feasible solution, (c) an optimal solution, (d) none of these, , 11. I n the context of network, which of the, following is not correct, (a) A n e t w o r k i s a g r a p h i c a l, representation ., , 7. In the given graph the coordinates of, M1 are, X2, , 2x, 4x 1+ 2, B, , 12. The objective of network analysis is to, , 2x, , 1 +5x, , 2, , (a) Minimize total project cost, , =1, , C, , 20, , (b) Minimize total project duration, A, X1, , O, , (a) x1 = 5, x2 = 30, , (b) x1 = 20, x2 = 16, , (c) x1 = 10, x2 = 20 (d) x1 = 20, x2 = 30, 8. Th,  e maximum value of the objective, subject to, function Z = 3x + 5y, x ≥ 0,y ≥ 0 and, the constraints, 2x + 5y # 10 is, (a) 6, , (b) 15, , (c) An arrow diagram is essentially a, closed network, (d) An arrow representing an activity, may not have a length and shape, , = 80, , M1, , (b) , A project network cannot have, multiple initial and final nodes, , (c) 25, , (d) 31, , (c) , Minimize production delays,, interruption and conflicts, (d) All the above,  etwork problems have advantage in, 13. N, terms of project, (a) Scheduling, (c) Controlling, , (b) Planning, (d) All the above, , 14. In critical path analysis, the word CPM, mean, , (a) Critical path method, 9. T, , he minimum value of the, objective function Z = x + 3y subject, (b) Crash project management, to, the, constraints, 2x + y # 20, x + 2y # 20, , x Critical, 20, (c), project management, 2x + y # 20, x + 2y # 20, x 2 0 and y > 0 is, (d) Critical path management, (a) 10, (b) 20, (c) 0, (d) 5, 15. Given an L.P.P. maximize Z = 2x1 + 3x2, 10. Which of the following is not correct?, subject to the constrains x1 + x2 # 1, (a) Objective that we aim to maximize, , 5x1 + 5x2 $ 0 and x1 ≥ 0, x2 ≥0 using, or minimize, graphical method, we observe, (b) Constraints that we need to specify, (a) No feasible solution, (c) Decision variables that we need to, (b) unique optimum solution, determine, (c) multiple optimum solution, (d) , Decision variables are to be, (d) none of these, unrestricted., Operations Research, , 10_11th_BM-STAT_Ch-10-EM.indd 251, , 251, , 21-04-2020 12:33:00 PM

Page 258 :
www.tntextbooks.in, , Miscellaneous Problems, 1. A,  firm manufactures two products, A and B on which the profits earned per, unit are ` 3 and ` 4 respectively. Each, product is processed on two machines, M1 and M2. Product A requires one, minute of processing time on M1 and, two minutes on M2, While B requires, one minute on M1 and one minute on, M2. Machine M1 is available for not, more than 7 hrs 30 minutes while M2 is, available for 10 hrs during any working, day. Formulate this problem as a linear, programming problem to maximize, the profit.,  firm manufactures pills in two sizes, 2. A, A and B. Size A contains 2 mgs of, aspirin, 5 mgs of bicarbonate and 1 mg, of codeine. Size B contains 1 mg. of, aspirin, 8 mgs. of bicarbonate and 6, mgs. of codeine. It is found by users that, it requires atleast 12 mgs. of aspirin,, 74 mgs.of bicarbonate and 24 mgs., of codeine for providing immediate, relief. It is required to determine the, least number of pills a patient should, take to get immediate relief. Formulate, the problem as a standard LLP., 3. S olve the following linear programming, problem graphically., Maximise Z = 4x1 + x2 subject to the, constraints x1 + x2 # 50; 3x1 + x2 # 90, and x1 $ 0, x2 $ 0 ., 4. S olve the following linear programming, problem graphically., Minimize Z = 200x1 + 500x2 subject, to the constraints: x1 + 2x2 $ 10;, 3x1 + 4x2 # 24 and x1 $ 0, x2 $ 0 ., 5., , S olve the following linear programming, problem graphically., , 252, , Maximize Z = 3x1 + 5x2 subject to the, constraints: x1 + x2 # 6, x1 # 4; x2 # 5,, and x1, x2 $ 0 ., 6., , S olve the following linear programming, problem graphically. Maximize, Z = 60x1 + 15x2 subject to the, constraints: x1 + x2 # 50; 3x1 + x2 # 90, and x1, x2 $ 0 ., , 7., , raw a network diagram for the, D, following activities., Activity code A B C D E F G H I, , J, , K, , Predecessor, – A A A B C C C,D E,F G,H I,J, activity, , 8., , 9., ,  raw the network diagram for the, D, following activities, Activity code, , A, , B, , C, , D, , E, , F, , G, , Predecessor, activity, , –, , –, , A, , A, , B, , C, , D,E, ,  Project has the, A, schedule, , following time, , Activity, , 1-2, , 2-3, , 2-4, , 3-5, , 4-6, , 5-6, , Duration, (in days), , 6, , 8, , 4, , 9, , 2, , 7, , Draw the network for the project, calculate, the earliest start time, earliest finish time,, latest start time and latest finish time of, each activity and find the critical path., Compute the project duration., 10. Th, e following table, characteristics of project, Activity, Duration, (in days), , gives, , the, , 1-2 1-3 2-3 3-4 3-5 4-6 5-6 6-7, 5, , 10, , 3, , 4, , 6, , 6, , 5, , 5, , Draw the network for the project ,calculate, the earliest start time, earliest finish time,, latest start time and latest finish time of, each activity and find the critical path., Compute the project duration., , 11th Std. Business Mathematics and Statistics, , 10_11th_BM-STAT_Ch-10-EM.indd 252, , 21-04-2020 12:33:02 PM

Page 259 :
www.tntextbooks.in, , Summary, , zz, , Linear programming problem(LPP) is a mathematical modeling technique which, is used to allocate limited available resources in order to optimize (maximize or, minimize) the objective function., , zz, , Short form of LPP, Maximize or Minimize Z =, Subject to, , n, , ∑a, j =1, , ij, , n, , / cjxj, , j=1, , x j ≤ (or = or ³ )bi, i = 1, 2, 3, …, m , , and xj ≥ 0, , zz, , Objective function: A function Z = c1x 1 +c 2 x 2  +…,  .+c n x n which is to be, optimized (maximized or minimized) is called objective function., , zz, , Decision variable: The decision variables are the variables we seek to determine, to optimize the objective function. xj , j = 1,2,3,…,n , are the decision variables., , zz, , Solution: A set of values of decision variables xj,j =1,2,3,…,n satisfying all the, constraints of the problem is called a solution to that problem., , zz, , Feasible solution: A set of values of the decision variables that satisfies all the, constraints of the problem and non-negativity restrictions is called a feasible, solution of the problem., , zz, , Optimal solution: Any feasible solution which maximizes or minimizes the, objective function is called an optimal solution., , zz, , Feasible region: The common region determined by all the constraints including, non-negative constraints xj≥0 of a linear programming problem is called the, feasible region ( or solution region) for the problem., , zz, , Linear programming problems which involve only two variables can be solved by, graphical method., , zz, , It should be noted that the optimal value of LPP occurs at the corner points of the, feasible region, , zz, , Network is a diagrammatic representation of various activities concerning a, project arranged in a logical manner., , Operations Research, , 10_11th_BM-STAT_Ch-10-EM.indd 253, , 253, , 21-04-2020 12:33:02 PM

Page 261 :
www.tntextbooks.in, , ICT Corner, Operational Research, Step – 1, Open the Browser type the URL Link given, below (or) Scan the QR Code., GeoGebra Workbook called “11th BUSINESS, MATHEMATICS and STATISTICS”, will appear. In that there are several worksheets, related to your Text Book., , Expected Outcome ⇒, Step-2, Select the work sheet “Linear Programming Problem” Move the slider on Right side to see the steps, for working Linear Programming Problem. Work out the problem as given. Graphical representation, is given on left side. Also refer the worksheet “Inequality video” in the work book., Step 1, , Step 2, , Browse in the link, 11th Business Mathematics and Statistics, https://ggbm.at/qKj9gSTG (or) scan the QR Code, , Operations Research, , 10_11th_BM-STAT_Ch-10-EM.indd 255, , 255, , 21-04-2020 12:33:03 PM

Page 262 :
www.tntextbooks.in, , ANSWERS, 1. MATRICES AND DETERMINANTS, Exercise 1.1, 1.(i) M11 =- 1 M12 = 0 M21 = 20 M22 = 5, , A11 =- 1 A12 = 0 A21 =- 20 A22 = 5, (ii) M11 =- 12 M12 = 2 M13 = 23 M21 =- 16 M22 =- 4 M23 = 14, , M31 =- 4 M32 =- 6 M33 = 11, , A11 =- 12 A12 =- 2 A13 = 23 A21 = 16 A22 =- 4 A23 =- 14, , A31 =- 4 A32 = 6 A33 = 11, 2. 20, , 1, 3. 2, , 4. –30, , 13, 5. 2 , 2 6. 0, , Exercise 1.2, V, RS, 7 - 3 - 3WW, S, 3 1, 4 -3, S, W, 1 >3 - 1H, 0 WW 3.(i) 1 =, H, G, 2. SS- 1 1, (ii), 1. >, 5 -2 1, 10 1 3, SS, W, -1 2, 1 WW, S- 1 0, X, T, RS, VW, RS, V, SS10 - 10 2 WW, SS- 22 - 46 - 7 WWW, 1, 1, 5 - 4WW , (iii) 10 SS 0, (iv) 151 SS- 13 14 - 11WW, SS, W, SS, W, 0, 2 WW, S0, S 5 - 17 - 19WW, T, X, RST, V X, 3 WW, SS0 3, W, 1, 7, 7. 9 SS3 2 - 7WW 10. m = 4, 11. p = 2, q =- 3, SS, WW, S3 - 1 - 1W, T, X, Exercise 1.3, 1. x = 1, y = 1 , , 2.(i) x = 3, y =- 2, z = 1, , (ii) x =- 1, y = 2, z = 3 (iii) x = 1, y =- 1, z = 2, 4. 3000, 1000, 2000 , 256, , 5. 13,2,5, , 3. 17.95, 43.08, 103.85, 6. 700, 600, 300, , 11th Std. Business Mathematics and Statistics, , 11_11th_BM-STAT_Answer-EM.indd 256, , 21-04-2020 12:08:38 PM

Page 263 :
www.tntextbooks.in, , Exercise 1.4, 1. It is viable, , 2. It is not viable, , 3. It is viable, , 4. A = 27.82 tonnes B = 98.91 tonnes, 6. 34.16, 17.31,, , 5. 181.62, 84.32, , 7. 42 and 78, Exercise 1.5, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , (b), , (d), , (b), , (b), , (c), , (c), , (c), , (c), , (c), , (d), , (c), , (b), , (a), , 14, , 15, , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, , (c), , (c), , (c), , (b), , (a), , (d), , (b), , (b), , (d), , (d), , (b), , (a), , Miscellaneous Problems, RS, V, - 1WW, 0, SS2, W, 1, 0 WW, 1. x = 3. x =- 1 2. 0, 6. SS5, SS, W, 1, 3 WW, S0, T, X, 1 3 1G, 7. 5 =, 8. x = 1, y = 2, z = 3 9. x = 20, y = 30, z = 50, -2 1, 10. `1200 Crores, `1600 Crores, 2. ALGEBRA, Exercise 2.1, 13, 10, 1. x - 2 - x - 1 , , 3, 1, 2. x - 2 + x + 1, , 1, 1, 1, 1, 1, 4., 2, +, x, 1, x, 2, x, 1, x, g 9], g 3 ] x + 2g, g 2 ] + 1g, 9], 2], -4, -44, 414, 3, 9, 41, 1 6. 2 +, −, +, 5., +, 2, +, +, 9 ]x 2g 99]]xx 21gg 93]]]xx +-121ggg 39]]x - 1gg2 3 ]x - 1g2x - 2 ]x - 2g2 ]x - 2g3, 4 (x - 2), -7, 1, 1, 9, 8. 5 (x + 2) +, 7. 2x + 2 + 2 (x + 2), 5 (x2 + 1), x, 1-x +, 1, 3, 3, 1, 9. 16 (x - 1) - 16 (x + 3) +, 10., 2, 2, 5 (x + 4) 5 (x + 1), 4 (x + 3), 3., , Exercise 2.2, 1. 64, , 2. 10, , 3. 3    4. 336   5. 85, Exercise 2.3, , 1. 6, 7!, 5. (a) 2, , 2. 14400, (b) 7!, , 3. 1680, , 13!, 4. 4! 3! 2! 2!, , 6. Rank of the word CHAT = 9., Answers, , 11_11th_BM-STAT_Answer-EM.indd 257, , 257, , 21-04-2020 12:08:41 PM

Page 264 :
www.tntextbooks.in, , Exercise 2.4, 1. 4, , 2. 8C4 + 8C3 = 9C4 = 126, , 3. 210 cards, , 4. 20, , 5. 25200, , 13!, 7. 7! 6! # 9! # 9!, , 8. 11, , 6. 671, , 9.(i) 1365, , 10.(i) 186, , (ii) 1001, , (iii) 364, , (ii) 186, Exercise 2.6, , 1.(i) 16a 4 - 96a3 b + 216a2 b2 - 216ab3 + 81b 4, 7x6 21x5 35x 4 35x3 21x2 7x, 1, (ii) x7 + y + 2 + 3 + 4 + 5 + 6 + 7, y, y, y, y, y, y, 20 15, 6, 1, (iii) x6 + 6x3 + 15 + 3 + 6 + 9 + 12, x, x, x, x, 2.(i) 10, 40,60,401 (ii)995009990004999    3. 11440 x9 y4, 11C5, x, , 4.(i) 11C5 x ,, 5.(i) 9C6, , 26, 36, , (ii), , 8C4 (81) 12, 16 x, , (iii), , (ii) - 32 (15C5), , - 10C5 ^6 h5, x5, , (iii) 7920., , Exercise 2.7, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , (d), , (c), , (d), , (c), , (a), , (c), , (d), , (b), , (b), , (c), , (b), , (d), , (b), , 14, , 15, , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, , (a), , (c), , (a), , (c), , (a), , (c), , (a), , (b), , (c), , (d), , (b), , (a), , Miscellaneous Problems, 3, 2, 1. x - 1 + x + 3 , 1, 5, 3, 3. x + 2 + x - 3 ]x - 3g2, 5.(i) 7, (ii) 7, 2, 24, 6.(a) 720, (b) 7776, 7. 74 ways, , 258, , 8. 85 ways, , 3, 2, 2. x - 1 - x - 2, 3, 2x - 1, 4. x - 2 + 2, x -x+1, (iii) 12, (c) 120, 9. 210, , (d) 6, 10. 544, , 11th Std. Business Mathematics and Statistics, , 11_11th_BM-STAT_Answer-EM.indd 258, , 21-04-2020 12:08:44 PM

Page 266 :
www.tntextbooks.in, , 3., Axis, , Vertex, , Focus, , y=4, , V(1, 4), , F(3, 4), , 4., , Problem, (a), , y2 = 20x, (b), , x2 = 8y, x2, , (c), , =- 16y, , Equation of, directrix, x+1=0, , Length of, latus rectum, 8 units, , Axis, , Vertex, , Focus, , Equation of, directrix, , Length of, latusrectum, , y=0, , V(0, 0), , F(5, 0), , x=–5, , 20 units, , x=0, , V(0, 0), , F(0, 2), , y = –2, , 8 units, , x=0, , V(0, 0), , F(0, –4), , y=4, , 16 units, , 5. (x - 15) 2 = 5 (y - 55). The output and the average cost at the vertex are 15 kg. and ` 55., 6. when x = 5 months, , Exercise 3.7, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , (b), , (c), , (c), , (c), , (b), , (c), , (a), , (c), , (a), , (b), , (c), , (a), , (d), , 14, , 15, , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, , (a), , (a), , (d), , (a), , (c), , (b), , (b), , (a), , (d), , (b), , (b), , (b), , Miscellaneous problems, 1. 3x + y – 8 =0 2. y = 6x + 3000, , 4. p = 1, p = 2, , 6. a = 9, b = 8, , 7. (–1, –2) is on the line, (1, 0) is above the line and (–3, –4) is below the line, 9, 8. (–2, –7), 9. y2 =- 2 x, 10. Axis : y = 2, Vertex : (1, 2), Focus : (2, 2),, Equation of directrix : x = 0 and Length of latus rectum : 4 units, 4. TRIGONOMETRY, Exercise 4.1, 5r, 6, , r, 1.(i) 3, , (ii), , 2.(i) 22c30l, , (ii) 324⁰, , 260, , 4r, (iii) 3, , (iv), , (iii) - 171c48l, , (iv) 110c, , - 16r, 9, , 11th Std. Business Mathematics and Statistics, , 11_11th_BM-STAT_Answer-EM.indd 260, , 21-04-2020 12:08:53 PM

Page 267 :
www.tntextbooks.in, , 3.(i) 1st quadrant (ii) 3rd quadrant, 4.(i), , - 3, 2, , (v), , 2, , 1.(i), , - 3, 2, -7, 10. 2, , (ii), , 2, 3, , (iii), , (iv) 1 , , Exercise 4.2, 3 +1, 3 -1, o, (ii) -e, (iii), 2 2, 3 +1, 3, (ii) 2, (iii) cos 80c, - 16, 16, (ii) 65, (iii) 33, , 2 2, 3 -1, , 2.(i) sin 92c, 3.(i), , (iii) 2nd quadrant., , - 33, 65, , 2, 6. 11 , a + b lies in 1st quadrant., 9, 9.(i) 13, , (iv), , 3, 2, , 14., , 2 -1, , 7. ! 2 - 1, , 828, (ii) - 2197, , 44, 117, 10. - 125 , - 44, Exercise 4.3, , 3, 1, (ii) 2 b- sin 2A + 2 l, , A, A, 1, 1.(i) 2 b cos 4 - cos 2 l , , 2A, 1, (iii) 2 b sin 4A - sin 3 l, 3A, A, 2.(i) 2 sin 2 cos 2 , , (iv) 1 (sin10i-sin4i), 2, (ii) 2 cos 3A cos A, i, 1 b sin 4A - sin 2A 3, 2 sin3 l2, 2 (iv), , (iii) 2 sin 2i cos 4i , , i, sin 2, , Exercise 4.4, 1.(i), , -r, 6, , (ii), , -r, 4, , r, (iii) 6, , 1, 4. x = 6 , 4, 1, (ii), 6.(i) 5, 10, , (iv), , 1, 5. x = 2, , 3r, 4, , r x, 10. 4 + 2, , 7. -33, 65, Exercise 4.5, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , (b), , (a), , (c), , (b), , (b), , (d), , (a), , (d), , (a), , (c), , (c), , (d), , (b), , 14, , 15, , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, , (c), , (b), , (a), , (c), , (b), , (c), , (c), , (b), , (c), , (a), , (b), , (d), , Answers, , 11_11th_BM-STAT_Answer-EM.indd 261, , 261, , 21-04-2020 12:08:57 PM

Page 268 :
www.tntextbooks.in, , Miscellaneous Problems, 4., , -1, 3, and, , 10, 10, , 7., , 15 + 2 2, 12, , 6+ 2, 4, , 6.(i), , 56, 9. 33, , (ii) 2 – 3, , r, 10. 4 - x, , 5. DIFFERENTIAL CALCULUS, Exercise 5.1, 1.(i) Odd function (ii), , Even function, , (iii) Neither even nor odd function, (v) Neither even nor odd function, 6.(i) e, , (ii) 0, , 7.(i), , y, , 2–, x,, x<, 2, , (–2, 4), , (2, 12), , y=16–x 2, , y, , y=, , (–3, 7), , (iv) 0, , (ii), , (1, 15), , (–2, 12), , 1 - x 3x + 1, 5. 3 + x , x - 1, , 2. k = 0, (iii) 3e, , (0, 16), , (–1, 15), , (iv) Even function, , (–1, 3), (0, 2), , (3, 7), , (1, 1), , x′, (–4, 0), , (4, 0), , x′, , y, , =x, , ,, -2, , x>, , 2, , (5, 3), , (4, 2), , (3, 1), , (2, 0), , x, , y′, x, , y′, y, , (iii), , (3, 9), , y, , (iv), , 2x, , y=e, , y=x2, x>0, , (2, 4), , x′, (–1, –1), , (1, 1), , (0, 1), , x, , (0, 0), , x′, , O, , x, , y′, , (–2, –4), y=–x2, x<0, , (–3, –9), , 262, , y′, , 11th Std. Business Mathematics and Statistics, , 11_11th_BM-STAT_Answer-EM.indd 262, , 21-04-2020 12:08:58 PM

Page 269 :
www.tntextbooks.in, , (v), , (vi), , y, , y=, , e –2x, , y=1, x>0, , (0, 1), , O, , x′, , y, , x, , y′, , x, , y=0 x=0, , x′, , O, y=–1, x<0, , y′, , Exercise 5.2, 10, 3, , (ii) 0, , 1, 15, (v) 16 a- 24, , (vi) 9, , 1.(i), , 2. a = ± 1, , 1, (iii) 2, 28, 4. 5, , n=7, , 3., , (iv) 1, , 5. f(–2) = 0, , Exercise 5.3, 1.(a) Not a continuous function at x = 2, , (b) continuous function at x = 3, , Exercise 5.4, 1, (iii) x + 1, , (ii) - e-x, , (i) 2x, , Exercise 5.5, - 20, + 64 - 52 (iii) 1 - 31 4 + e x, x5, x, x, 2 x 3, x, , 1.(i) 12x3 - 6x2 + 1 (ii), 3, (iv) -, , + x2, x2, , (v) x2 e x ^ x + 3h, , (vii) 4x3 - 3 cos x - sin x , 2.(i), , xe x, (1 + x) 2, , (ii), , 2 ^1 - x2h, ^ x 2 - x + 1h2, , 3.(i) x cos x + sin x (ii) e x ^sin x + cos x h, 1, (iii) e x `1 + x + x + log x j, , 4.(i) sin 2x, , (ii) - sin 2x, , (v) n ^ax2 + bx + chn - 1 ^2ax + bh, , (vi) 3x2 - 4x - 1, 1, x2, ex, (iii), ^1 + e x h2, , (viii) 1 -, , (iv) cos 2x, , (v) e x x2 ^ x + 3h, , x, 1 + x2, -x, (vii), 2, ^1 + x h 1 + x 2, , -3, (iii) 2 cos x sin 2x (iv), (vi) 2x ^cos ^ x2hh, , Answers, , 11_11th_BM-STAT_Answer-EM.indd 263, , 263, , 21-04-2020 12:09:03 PM

Page 270 :
www.tntextbooks.in, , Exercise 5.6, y - 2x, (ii) 2y - x, , y, 1.(i) - x, , (iii) –, , (x2 + ay), (y2 + ax), , 4 1 - 4x +3y, 3. 3 c 1 + 4x -3y m, , Exercise 5.7, sin x, 1.(i) xsin x 8 x + cos x log xB, , (ii) ^sin x hx 6x cot x + log ^sin x h@, , (iii) ^sin x htanx 61 + sec2 x log ^sin x h@, , ^ x - 1h^ x - 2h, 1 + 1 - 1 - 2x + 1, $, ., ^ x - 3h^ x 2 + x + 1h x - 1 x - 2 x - 3 x 2 + x + 1, , 1, (iv) 2, , Exercise 5.8, 1, 1.(i) - 2, t, , (ii) t cost, , 2. – tan x, , i, (iv) cot 2, , (iii) - tan θ, , sin 2x, , 3., , 2x, , Exercise 5.9, 1, (ii) - 2 + a x ^log ah2, x, , 1.(i) 9y, , - 1a cosec3 i, , (iii), , Exercise: 5.10, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , (d), , (c), , (a), , (b), , (a), , (a), , (c), , (a), , (d), , (a), , (a), , (d), , (b), , 14, , 15, , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, , (a), , (a), , (a), , (c), , (b), , (d), , (a), , (a), , (c), , (a), , (b), , (d), , Miscellaneous Problems, 2., , y, , (0, 9), , (–1, 8), , 6. (i) continuous at, , (1, 8), , (–2, 5), , = 1; differentiable at x = 1, , (ii) not continuous at x = 2; not differentiable at x = 2, , (2, 5), , y=9–x 2, , (–3, 0), , 1, 4. - 10, , (3, 0), , x′, , x, y′, , 264, , 11th Std. Business Mathematics and Statistics, , 11_11th_BM-STAT_Answer-EM.indd 264, , 21-04-2020 12:09:06 PM

Page 271 :
www.tntextbooks.in, , 6. APPLICATIONS OF DIFFERENTIATION, Exercise 6.1, 1, , 7, , 2, 1, 7, 1. AC = 10 x − 4x − 20 + x , AVC = x 2 − 4x − 20 , AFC = ,, 10, x, 3, 1, 7, MC = x 2 − 8x − 20 , MAC = x − 4 −, , 10, 5, x2, , 2., , C =`, , 29 MC = ` 2, 121, , AC = `, ,, 3, 6, 12, , 3. AC = x 2 − 2 , MC = 3x 2 − 2 , AR = 14 − x , MR = 14 − 2x, , 2, x, , 4. ηd =, =, 5(i) ηndd =, , 6. ηs =, , 2a − bx, a, , x=, 3b, x 2bx, , (ii) ηndd =, , a2 − bx 2, ,x =, x2bx 2, , 4p 2, 36, ,, , 2, 2p + 5 23, , 7. MR =, , p, ,1 , 2 (p − b ), , 2, 11. ηd =, =4, x, , 8. ηs = , , 12. P = −, , a, 3b, , 50 − 2x, , 10,, 10, 00, 5, , x2, −x, 120, Rs 147.9, + 160x − 120 , AP =, + 160 −, ,, ,` 147.90, 100, 100, x, , 120, 1, x, + 2 ,, , ` 11.19, .19, , MP = − + 160 , `159.80, MAP = −, 50, , 13. x = – 8, 2, , 100, , 15. nd =, , x, , p, 3, , nd = > 1 ⇒ elastic., 10 − p, 2, , 16. PE=` 30, xE = 40 units, , 17. x = 2100 units, p = ` 130, , 18. x = 6 units, Exercise 6.2, 1. , AC is increasing when x > 5, 3. , P is maximum when x ≈ 46, maximum profit ` 1107.68, 4. Revenue is maximum when x = 220, 5. Minimum value is –71, maximum value is 54, Answers, , 11_11th_BM-STAT_Answer-EM.indd 265, , 265, , 21-04-2020 12:09:07 PM

Page 272 :
www.tntextbooks.in, , Exercise 6.3, 1., Items, , EOQ in, units, , A, , 2000, , B, , 200, , C, , 2627, , Minimum, Inventory, Cost, , EOQ in `, , EOQ in, years of, supply, , Number of, order, per year, , 40, , 2.5, , 0.4, , 200, , 0.5, , 2, , 525.40, , 0.19, , 5.26, , `4, `20, , `52.54, , (ii) ≈ `20, 066 per week., , 2 (i) ≈ 913 units per order, , Exercise 6.4, 1., , ∂z, ∂z, = a (cy + d ),, = c (ax + b ), ∂x, ∂y, , Exercise 6.5, 1. 23, 25, , 3. –2, 8, , 4, 5. - 3 , - 8, , 4. 0.8832, 2.2081, , 6., , 10, 3, ,79, 79, , Exercise 6.6, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , (d), , (a), , (b), , (a), , (b), , (a), , (b), , (c), , (b), , (a), , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , (a), , (c), , (b), , (b), , (a), , (b), , (b), , (b), , (b), , (c), , Miscellaneous Problems, 10, −10, − 4x 2 + 3x 3 , MC = −12x 2 + 12x 3 , MAC = 2 − 8x + 9x 2, x, x, 23, 1, 2.(i) ηs =, (ii) ηnd d==, xx, x +1, , 1., , AC =, , 3., , ηs =, n, s, , 4, 4p 2, ,, 2p 2 + 5 7, , 6. TC is increasing in [0.20] ∪ [30, ∞) and decreasing in, [20, 30], x = 3, TC is minimum, , 7., , 7. FINANCIAL MATHEMATICS, Exercise 7.1, 1. ` 68,429, , 2. ` 1,20,800, , 3. ` 18,930, , 4. ` 500, , 5. ` 23,79,660, , 6. ` 14,736, , 7. ` 8,433, , 8. ` 1,17,612, , 9. ` 14,339, 266, , 10. ` 1,000, , 11th Std. Business Mathematics and Statistics, , 11_11th_BM-STAT_Answer-EM.indd 266, , 21-04-2020 12:09:08 PM

Page 273 :
www.tntextbooks.in, , Exercise 7.2, , ` 8,184, , 1., , 2., , ` 2,250, , 3., , 900 shares, 1, 2, , 4.(i) 242, , (ii) ` 3630, , (iii) 12 %, , 5.(i) ` 5,000, , (ii) ` 480, , (iii) 9.6%, , ` 7000, ` 6500, , 7., , 8., , 9.(i) ` 945, , 6. ` 897.50, , 99 shares, , (ii) ` 960 2nd investment is better., , 10.(i) 1400, (ii) 1400. For the same investment both stocks fetch the, same income. Therefore they are equivalent shares., Exercise 7.3, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , (a), , (b), , (c), , (c), , (d), , (c), , (a), , (b), , (a), , (b), , (b), , (b), , (a), , (a), , (b), , Miscellaneous Problems, 1. ` 9,282, , 2. ` 15,638, , 3. ` 4,327; ` 1,25,780; ` 29,334; ` 1,39,560., 5. ` 12,500, , 4. Required months ≈ 24, , 6. ` 13,250; ` 36,443.75; Machine B may be purchased, 7. 270, 216, 300, 450, 8. ` 700, ` 900, ` 200, 2.5%, , 9. 1,500 shares, ` 625 10. 20%, , 8. DESCRIPTIVE STATISTICS AND PROBABILITY, Exercise 8.1, 1. Q1 = 6 , Q3 = 18 , , 2. Q1 = 5, Q3 = 6.5, D8 = 6.5 and P67 = 6, , 3. Q1= 47.14, Q3= 63.44, D5 = 55.58, D7 = 61.56 and P60 = 58.37, 4. GM= 142.5 lbs , , 5. GM= 26.1%, , 6. 192 km/hr , , 7. 38.92 km/hr, , 8. AM=36 GM=25.46 HM=17.33, 9. AM=21.96 GM=18.31 HM=14.32, Answers, , 11_11th_BM-STAT_Answer-EM.indd 267, , 267, , 21-04-2020 12:09:08 PM

Page 274 :
www.tntextbooks.in, , 10. AM=33, GM=29.51, HM=24.10, 11. Q1 = 30, Q3 = 70, QD = 20 , Coefficient of QD = 0.4, 12. QD= 11.02 , Coefficient of QD = 0.3384, 13. Median = 61, MD =1.71, 14. Mean = 13, MD = 4.33, 15. Median = 45.14 , MD = 14.83, Exercise 8.2, 1., , 1, 3, , 2., , 2, 5, , 3. A and B are independent events, , 4.(i), , 2, 3, , (ii), , 5. 3/10, , 6.(i) 42/625, , (ii) 207/625, , 7. 33/68, , 8.(i) 7/29, , (ii) 5/29, , 9. 4/11, 11.(i), , 1, 2, , (iii) 17/29, , 10. P(A)=4/7 P(B) = 2/7 P(C) = 1/7, , 1, 2, , (ii), , 14. 0.012, , 2, 3, , 12., , 15.(i) 1/221, , 1, 2, , 13. 0.2, , (ii) 1/17, , 16. P(C/D)= 0.5208, , Exercise 8.3, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , (d), , (c), , (d), , (a), , (c), , (a), , (d), , (c), , (b), , (a), , (d), , (c), , (a), , 14, , 15, , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, , (b), , (d), , (a), , (a), , (b), , (c), , (d), , (d), , (b), , (a), , (b), , (a), , Miscellaneous Problems, 1. 16.02 tons, , 2. 16, , 4. Mean=7.5, MD =2.3, 6. 0.45, 8. 0.948, 268, , 7.(i) 3/10, 9. 0.727, , 3. Median=28, MD = 10.16, 5. QD=10, Coefficient of QD=0.5, (ii) 3/5, , (iii) 1/10, , 10. 0.393, , 11th Std. Business Mathematics and Statistics, , 11_11th_BM-STAT_Answer-EM.indd 268, , 21-04-2020 12:09:08 PM

Page 275 :
www.tntextbooks.in, , 9. CORRELATION AND REGRESSION ANALYSIS, Exercise 9.1, 1. 0.575, , 2. 0.947, , 3. 0.996, , 4. 0.891, , 5. 0.225, , 6. –0.0735, , 7. 0.9, , 8. 0.224, , 9. 0.905, , 10. –0.37, Exercise 9.2, , 1.(a) Y=–0.66X+59.12; X=–0.234Y+40.892, (b) r=-0.394, , (c) Y= 39.32, , 2. , Y=0.6102X+66.12; X=0.556Y+74.62 Height of son is 166.19, 3. , Y=2.3X–35.67; weight of the student is 125.79 lb, 4. , Y=0.24X+1.04; X=1.33Y+1.34, 5. Y=1.6X; estimated yield = 46.4 unit per acre, 6. Y=0.942X+6.08; Estimated sales =34.34 ( in crores of rupees), 7. Y=0.48X+67.72; X=0.91Y–41.35;Y=72.52, 8. byx=1.33; Y=1.33X+3.35, 9. Y=0.1565X+19.94; estimated expenditure on food and entertainment ( Y) is 51.24, 10. , X=0.8Y–1 and estimated value of X is 5.4 when Y=8, , Y=0.8X+2.6 and estimated value of Y is 12.2 when X=12, 1, 2, , 3, 2, , 11. X = 13; Y =17 and r = 0.6, , 12. b xy = − ;byx = − ; r = −0.866, Exercise 9.3, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , (a), , (b), , (a), , (b), , (c), , (a), , (a), , (b), , (a), , (b), , (a), , (a), , (c), , 14, , 15, , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, , (a), , (b), , (a), , (a), , (a), , (a), , (a), , (b), , (b), , (b), , (a), , (d), Answers, , 11_11th_BM-STAT_Answer-EM.indd 269, , 269, , 21-04-2020 12:09:08 PM

Page 276 :
www.tntextbooks.in, , Miscellaneous Problems, 1. 0.906, , 2. 0.382, , 3. 0.95, , 5. 0.905, , 6. , Y=0.653X+21.71; Marks in B = 55.67, , 7. Y=0.576X+2.332; Y =5.788, , 4. 0.667, , 8. Y=0.867X+7.913, , 9. Y=–0.652X+63.945, Y=10.481≈10, , 10. byx=1.422, Y = 141.67, , 10. OPERATIONS RESEARCH, Exercise: 10.1, 1. Maximize Z = 5x 1 +3x 2 subject to constraints 2x 1 +x 2 ≤ 1000 ; x 1 £ 400; x 2 £ 700, and x 1, x 2 ³ 0 ., Z = 30x 1 +40x 2 subject to constraints, 8x 1 + 5x 2 ≤ 600; 3x 1 + 4x 2 ≤ 500 and x1, x2 ≥ 0., , 2. Maximize, , 60x 1 + 120x 2 ≤ 12000, , subject to constraints 0.8 x 1 +1 .2x 2 ≤ 720; x 1 ≤ 600;, , 3. Maximize, x2 ≤ 400 and x1, x2 ≥0., , 4.(i) , x1 = 4 ; x2 = 9 and Zmax = 96, , (ii) x1 = 8; x2 = 12 and Zmax = 392, , (iii) , x1 = 1; x2 = 5 and Zmin = 13, , (iv) x1 = 2; x2 = 3 and Zmax = 230, , (v) , x1 = 3; x2 = 9 and Zmax = 330, , (vi) x1 = 4; x2 = 2 and Zmin = 160, , Exercise: 10.2, 1. , 2, , C, , A, , D, , 1, , B, , E, , 3, , 5, , G, F, , H, , 4, , , , 270, , 11th Std. Business Mathematics and Statistics, , 11_11th_BM-STAT_Answer-EM.indd 270, , 21-04-2020 12:09:09 PM

Page 277 :
www.tntextbooks.in, , 4, , 2, , 2., , 6, , 1, , 7, 5, , 3, , F, , 5, , 2, , A, 3., , E, , B, , 1, , 3, , 4, , D, , C, D, , 2, , G, , H, , 5, , 7, , A, 4., , B, , 1, , E, , 3, , G, E=14, L=28, , 5, , H, , 3, , J, , 6, , 1, , 3, , 2, , 8, , E=0, L=0, , K, 8, , E=11, L=20, , 0, , 9, , F, 4, , 5., , I, , 6, , C, , 3, , J, , 4, , E=3, L=3, , E=18, L=26, , 7, , E=31, L=31, , 5, , 8, 3, , 12, 3, , E=15, L=15, , 6, , 8, , E=23, L=23, , Critical path 0-1-3-6-7 and the duration is 31 weeks, Answers, , 11_11th_BM-STAT_Answer-EM.indd 271, , 271, , 21-04-2020 12:09:09 PM

Page 278 :
www.tntextbooks.in, , E=4, L=9, , 2, 4, , E=5, L=10, , E=10, L=15, , 5, , 7, , 9, , 4, , E=22, L=22, , 10, , 5, 1, , 6., , 1, , E=0, L=0, , E=15, L=15, , 1, , 6, , 5, , 6, , 4, , E=11, L=16, , Critical path 1-3-5-7-8-10 and duration 22 time units, E=6, L=6, , 10, , 2, , 6, , E=0, L=0, , 7., , 1, , E=7, L=7, , , , E=17, L=17, , 7, , 8, , 3, , E=1, L=1, , 8, , 2, , E = 16, L = 16, 4, , E = 31, L = 31, , 2, , 6, , 1, , 3, , 6, , 9, , 1, 4, , 3, E=5, L = 18, , 5, E = 22, L = 22, , Critical path 1-2-4-5-6 and duration time taken is 31 days, E=0, L=0, , E=5, L=5, , 5, , 2, , E = 16, L = 16, , 7, , 1, , 8, , 4, 6, , 8, , E = 11, L = 11, , 3, , 5, , 8, , 4, , 5, , E = 24, L = 24, , Critical path 1-2-3-4-5 and duration time taken is 24 days, 272, , 11th Std. Business Mathematics and Statistics, , 11_11th_BM-STAT_Answer-EM.indd 272, , 21-04-2020 12:09:09 PM

Page 279 :
www.tntextbooks.in, , E=7, L=7, , 9., , 14, , 5, , 1, , E = 32, L = 32, , 11, , 3, , 2, , 7, , E =0, L=0, , E = 21, L = 21, , 5, , 7, , 4, , 4, , E = 36, L = 36, 8, , 6, 18, 6, , 7, , 11, , E=6, L=7, , E = 17, L = 18, , Critical path 1-2-3-5-8 and duration time taken is 36 days, , E=0, L=0, , 10., , E=22, L=22, , 22, , 1, , E=40, L=40, , 14, , 2, , 12, , 4, , 12, , 5, , E=52, L=52, , 6, , 27, E=34, L=34, , 3, , Critical path 1-2-3-4-5 and duration time taken is 52 days, Exercise-10.3, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , (d), , (c), , (a), , (b), , (d), , (c), , (c), , (b), , (c), , (d), , (d), , (b), , (d), , (a), , (a), , Miscellaneous problems, 1. Maximize Z = 3x 1 +4x 2 subject to constraints x 1 +x 2 ≤ 450 ; 2x 1 +x 2 ≤ 600 and, x 1, x 2 ³ 0, , 2. Minimize, , Z = x 1 +x 2 subject, x 1 + 6x 2 ≥ 24 and x 1, x 2 ³ 0, , to, , constraints, , 2x 1 +x 2 ≥ 12; 5x 1 + 8x 2 ≥ 74;, , 3. , x1=30; x2 = 0 and Z max = 120, , 4. x1 = 4; x2 = 3 and Z min = 2300, , 5. x1 = 1 ; x2 = 5 and Z_max= 28, , 6. x1 = 20; x2=30 and Z max = 1650, Answers, , 11_11th_BM-STAT_Answer-EM.indd 273, , 273, , 21-04-2020 12:09:09 PM

Page 280 :
www.tntextbooks.in, , E, , 3, , 6, , B, 7., , A, , 1, , I, , F, , C, , 2, , 4, , G, , D, 5, , , , 8., , 7, , C, , 2, , 4, , F, D, , 1, , B, , 3, , E=14, L=14, , 6, , G, , 5, , E, , 9, , 3, , 5, , E=23, L=23, , 7, , 8, 9., , 6, , 1, , E=6, L=6, , 2, , E=0, L=0, , 9, , J, , H, , A, , K, , 8, , 6, , 4, 4, , E=30, L=30, , 2, , E=10, L=28, , Critical path 1-2-3-5-6 and duration time taken is 30 days., E=5, L=7, , 2, , 5, , 3, , 4, , E=14, L=15, , E=21, L=21, , 6, 6, , 4, , , , E=0, L=0, , 10, , 3, E=10, L=10, , 6, , 7, , 5, , 10., 1, , 5, , E=26, L=26, , 5, , E=16, L=16, , Critical path 1-3-5-6-7 and duration time taken is 26 days., 274, , 11th Std. Business Mathematics and Statistics, , 11_11th_BM-STAT_Answer-EM.indd 274, , 21-04-2020 12:09:10 PM

Page 281 :
www.tntextbooks.in, , LOGARITHM TABLE, , Mean Difference, 1.0, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 0.0000, , 0.0043, , 0.0086, , 0.0128, , 0.0170, , 0.0212, , 0.0253, , 0.0294, , 0.0334, , 0.0374, , 4, , 8, , 12, , 17, , 21, , 25, , 29, , 33, , 37, , 1.1, , 0.0414, , 0.0453, , 0.0492, , 0.0531, , 0.0569, , 0.0607, , 0.0645, , 0.0682, , 0.0719, , 0.0755, , 4, , 8, , 11, , 15, , 19, , 23, , 26, , 30, , 34, , 1.2, , 0.0792, , 0.0828, , 0.0864, , 0.0899, , 0.0934, , 0.0969, , 0.1004, , 0.1038, , 0.1072, , 0.1106, , 3, , 7, , 10, , 14, , 17, , 21, , 24, , 28, , 31, , 1.3, , 0.1139, , 0.1173, , 0.1206, , 0.1239, , 0.1271, , 0.1303, , 0.1335, , 0.1367, , 0.1399, , 0.1430, , 3, , 6, , 10, , 13, , 16, , 19, , 23, , 26, , 29, , 1.4, , 0.1461, , 0.1492, , 0.1523, , 0.1553, , 0.1584, , 0.1614, , 0.1644, , 0.1673, , 0.1703, , 0.1732, , 3, , 6, , 9, , 12, , 15, , 18, , 21, , 24, , 27, , 1.5, , 0.1761, , 0.1790, , 0.1818, , 0.1847, , 0.1875, , 0.1903, , 0.1931, , 0.1959, , 0.1987, , 0.2014, , 3, , 6, , 8, , 11, , 14, , 17, , 20, , 22, , 25, , 1.6, , 0.2041, , 0.2068, , 0.2095, , 0.2122, , 0.2148, , 0.2175, , 0.2201, , 0.2227, , 0.2253, , 0.2279, , 3, , 5, , 8, , 11, , 13, , 16, , 18, , 21, , 24, , 1.7, , 0.2304, , 0.2330, , 0.2355, , 0.2380, , 0.2405, , 0.2430, , 0.2455, , 0.2480, , 0.2504, , 0.2529, , 2, , 5, , 7, , 10, , 12, , 15, , 17, , 20, , 22, , 1.8, , 0.2553, , 0.2577, , 0.2601, , 0.2625, , 0.2648, , 0.2672, , 0.2695, , 0.2718, , 0.2742, , 0.2765, , 2, , 5, , 7, , 9, , 12, , 14, , 16, , 19, , 21, , 1.9, , 0.2788, , 0.2810, , 0.2833, , 0.2856, , 0.2878, , 0.2900, , 0.2923, , 0.2945, , 0.2967, , 0.2989, , 2, , 4, , 7, , 9, , 11, , 13, , 16, , 18, , 20, , 2.0, , 0.3010, , 0.3032, , 0.3054, , 0.3075, , 0.3096, , 0.3118, , 0.3139, , 0.3160, , 0.3181, , 0.3201, , 2, , 4, , 6, , 8, , 11, , 13, , 15, , 17, , 19, , 2.1, , 0.3222, , 0.3243, , 0.3263, , 0.3284, , 0.3304, , 0.3324, , 0.3345, , 0.3365, , 0.3385, , 0.3404, , 2, , 4, , 6, , 8, , 10, , 12, , 14, , 16, , 18, , 2.2, , 0.3424, , 0.3444, , 0.3464, , 0.3483, , 0.3502, , 0.3522, , 0.3541, , 0.3560, , 0.3579, , 0.3598, , 2, , 4, , 6, , 8, , 10, , 12, , 14, , 15, , 17, , 2.3, , 0.3617, , 0.3636, , 0.3655, , 0.3674, , 0.3692, , 0.3711, , 0.3729, , 0.3747, , 0.3766, , 0.3784, , 2, , 4, , 6, , 7, , 9, , 11, , 13, , 15, , 17, , 2.4, , 0.3802, , 0.3820, , 0.3838, , 0.3856, , 0.3874, , 0.3892, , 0.3909, , 0.3927, , 0.3945, , 0.3962, , 2, , 4, , 5, , 7, , 9, , 11, , 12, , 14, , 16, , 2.5, , 0.3979, , 0.3997, , 0.4014, , 0.4031, , 0.4048, , 0.4065, , 0.4082, , 0.4099, , 0.4116, , 0.4133, , 2, , 3, , 5, , 7, , 9, , 10, , 12, , 14, , 15, , 2.6, , 0.4150, , 0.4166, , 0.4183, , 0.4200, , 0.4216, , 0.4232, , 0.4249, , 0.4265, , 0.4281, , 0.4298, , 2, , 3, , 5, , 7, , 8, , 10, , 11, , 13, , 15, , 2.7, , 0.4314, , 0.4330, , 0.4346, , 0.4362, , 0.4378, , 0.4393, , 0.4409, , 0.4425, , 0.4440, , 0.4456, , 2, , 3, , 5, , 6, , 8, , 9, , 11, , 13, , 14, , 2.8, , 0.4472, , 0.4487, , 0.4502, , 0.4518, , 0.4533, , 0.4548, , 0.4564, , 0.4579, , 0.4594, , 0.4609, , 2, , 3, , 5, , 6, , 8, , 9, , 11, , 12, , 14, , 2.9, , 0.4624, , 0.4639, , 0.4654, , 0.4669, , 0.4683, , 0.4698, , 0.4713, , 0.4728, , 0.4742, , 0.4757, , 1, , 3, , 4, , 6, , 7, , 9, , 10, , 12, , 13, , 3.0, , 0.4771, , 0.4786, , 0.4800, , 0.4814, , 0.4829, , 0.4843, , 0.4857, , 0.4871, , 0.4886, , 0.4900, , 1, , 3, , 4, , 6, , 7, , 9, , 10, , 11, , 13, , 3.1, , 0.4914, , 0.4928, , 0.4942, , 0.4955, , 0.4969, , 0.4983, , 0.4997, , 0.5011, , 0.5024, , 0.5038, , 1, , 3, , 4, , 6, , 7, , 8, , 10, , 11, , 12, , 3.2, , 0.5051, , 0.5065, , 0.5079, , 0.5092, , 0.5105, , 0.5119, , 0.5132, , 0.5145, , 0.5159, , 0.5172, , 1, , 3, , 4, , 5, , 7, , 8, , 9, , 11, , 12, 12, , 3.3, , 0.5185, , 0.5198, , 0.5211, , 0.5224, , 0.5237, , 0.5250, , 0.5263, , 0.5276, , 0.5289, , 0.5302, , 1, , 3, , 4, , 5, , 6, , 8, , 9, , 10, , 3.4, , 0.5315, , 0.5328, , 0.5340, , 0.5353, , 0.5366, , 0.5378, , 0.5391, , 0.5403, , 0.5416, , 0.5428, , 1, , 3, , 4, , 5, , 6, , 8, , 9, , 10, , 11, , 3.5, , 0.5441, , 0.5453, , 0.5465, , 0.5478, , 0.5490, , 0.5502, , 0.5514, , 0.5527, , 0.5539, , 0.5551, , 1, , 2, , 4, , 5, , 6, , 7, , 9, , 10, , 11, , 3.6, , 0.5563, , 0.5575, , 0.5587, , 0.5599, , 0.5611, , 0.5623, , 0.5635, , 0.5647, , 0.5658, , 0.5670, , 1, , 2, , 4, , 5, , 6, , 7, , 8, , 10, , 11, , 3.7, , 0.5682, , 0.5694, , 0.5705, , 0.5717, , 0.5729, , 0.5740, , 0.5752, , 0.5763, , 0.5775, , 0.5786, , 1, , 2, , 3, , 5, , 6, , 7, , 8, , 9, , 10, , 3.8, , 0.5798, , 0.5809, , 0.5821, , 0.5832, , 0.5843, , 0.5855, , 0.5866, , 0.5877, , 0.5888, , 0.5899, , 1, , 2, , 3, , 5, , 6, , 7, , 8, , 9, , 10, , 3.9, , 0.5911, , 0.5922, , 0.5933, , 0.5944, , 0.5955, , 0.5966, , 0.5977, , 0.5988, , 0.5999, , 0.6010, , 1, , 2, , 3, , 4, , 5, , 7, , 8, , 9, , 10, , 4.0, , 0.6021, , 0.6031, , 0.6042, , 0.6053, , 0.6064, , 0.6075, , 0.6085, , 0.6096, , 0.6107, , 0.6117, , 1, , 2, , 3, , 4, , 5, , 6, , 8, , 9, , 10, , 4.1, , 0.6128, , 0.6138, , 0.6149, , 0.6160, , 0.6170, , 0.6180, , 0.6191, , 0.6201, , 0.6212, , 0.6222, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 4.2, , 0.6232, , 0.6243, , 0.6253, , 0.6263, , 0.6274, , 0.6284, , 0.6294, , 0.6304, , 0.6314, , 0.6325, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 4.3, , 0.6335, , 0.6345, , 0.6355, , 0.6365, , 0.6375, , 0.6385, , 0.6395, , 0.6405, , 0.6415, , 0.6425, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 4.4, , 0.6435, , 0.6444, , 0.6454, , 0.6464, , 0.6474, , 0.6484, , 0.6493, , 0.6503, , 0.6513, , 0.6522, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 4.5, , 0.6532, , 0.6542, , 0.6551, , 0.6561, , 0.6571, , 0.6580, , 0.6590, , 0.6599, , 0.6609, , 0.6618, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 4.6, , 0.6628, , 0.6637, , 0.6646, , 0.6656, , 0.6665, , 0.6675, , 0.6684, , 0.6693, , 0.6702, , 0.6712, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 7, , 8, , 4.7, , 0.6721, , 0.6730, , 0.6739, , 0.6749, , 0.6758, , 0.6767, , 0.6776, , 0.6785, , 0.6794, , 0.6803, , 1, , 2, , 3, , 4, , 5, , 5, , 6, , 7, , 8, , 4.8, , 0.6812, , 0.6821, , 0.6830, , 0.6839, , 0.6848, , 0.6857, , 0.6866, , 0.6875, , 0.6884, , 0.6893, , 1, , 2, , 3, , 4, , 4, , 5, , 6, , 7, , 8, , 4.9, , 0.6902, , 0.6911, , 0.6920, , 0.6928, , 0.6937, , 0.6946, , 0.6955, , 0.6964, , 0.6972, , 0.6981, , 1, , 2, , 3, , 4, , 4, , 5, , 6, , 7, , 8, , 5.0, , 0.6990, , 0.6998, , 0.7007, , 0.7016, , 0.7024, , 0.7033, , 0.7042, , 0.7050, , 0.7059, , 0.7067, , 1, , 2, , 3, , 3, , 4, , 5, , 6, , 7, , 8, , 5.1, , 0.7076, , 0.7084, , 0.7093, , 0.7101, , 0.7110, , 0.7118, , 0.7126, , 0.7135, , 0.7143, , 0.7152, , 1, , 2, , 3, , 3, , 4, , 5, , 6, , 7, , 8, , 5.2, , 0.7160, , 0.7168, , 0.7177, , 0.7185, , 0.7193, , 0.7202, , 0.7210, , 0.7218, , 0.7226, , 0.7235, , 1, , 2, , 2, , 3, , 4, , 5, , 6, , 7, , 7, , 5.3, , 0.7243, , 0.7251, , 0.7259, , 0.7267, , 0.7275, , 0.7284, , 0.7292, , 0.7300, , 0.7308, , 0.7316, , 1, , 2, , 2, , 3, , 4, , 5, , 6, , 6, , 7, , 5.4, , 0.7324, , 0.7332, , 0.7340, , 0.7348, , 0.7356, , 0.7364, , 0.7372, , 0.7380, , 0.7388, , 0.7396, , 1, , 2, , 2, , 3, , 4, , 5, , 6, , 6, , 7, , Tables, , 12_11th_BM-STAT_LogTable_EM.indd 275, , 275, , 21-04-2020 12:07:18 PM

Page 282 :
www.tntextbooks.in, , LOGARITHM TABLE, , Mean Difference, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 5.5, , 0.7404 0.7412 0.7419 0.7427 0.7435 0.7443 0.7451 0.7459 0.7466 0.7474, , 0, , 1, , 2, , 2, , 3, , 4, , 5, , 5, , 6, , 7, , 5.6, 5.7, 5.8, 5.9, 6.0, , 0.7482, 0.7559, 0.7634, 0.7709, 0.7782, , 0.7490, 0.7566, 0.7642, 0.7716, 0.7789, , 0.7497, 0.7574, 0.7649, 0.7723, 0.7796, , 0.7505, 0.7582, 0.7657, 0.7731, 0.7803, , 0.7513, 0.7589, 0.7664, 0.7738, 0.7810, , 0.7520, 0.7597, 0.7672, 0.7745, 0.7818, , 0.7528, 0.7604, 0.7679, 0.7752, 0.7825, , 0.7536, 0.7612, 0.7686, 0.7760, 0.7832, , 0.7543, 0.7619, 0.7694, 0.7767, 0.7839, , 0.7551, 0.7627, 0.7701, 0.7774, 0.7846, , 1, 1, 1, 1, 1, , 2, 2, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 5, 4, 4, 4, , 5, 5, 5, 5, 5, , 6, 6, 6, 6, 6, , 7, 7, 7, 7, 6, , 6.1, 6.2, 6.3, 6.4, 6.5, , 0.7853, 0.7924, 0.7993, 0.8062, 0.8129, , 0.7860, 0.7931, 0.8000, 0.8069, 0.8136, , 0.7868, 0.7938, 0.8007, 0.8075, 0.8142, , 0.7875, 0.7945, 0.8014, 0.8082, 0.8149, , 0.7882, 0.7952, 0.8021, 0.8089, 0.8156, , 0.7889, 0.7959, 0.8028, 0.8096, 0.8162, , 0.7896, 0.7966, 0.8035, 0.8102, 0.8169, , 0.7903, 0.7973, 0.8041, 0.8109, 0.8176, , 0.7910, 0.7980, 0.8048, 0.8116, 0.8182, , 0.7917, 0.7987, 0.8055, 0.8122, 0.8189, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 4, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 5, 5, 5, 5, , 6, 6, 5, 5, 5, , 6, 6, 6, 6, 6, , 6.6, 6.7, 6.8, 6.9, 7.0, , 0.8195, 0.8261, 0.8325, 0.8388, 0.8451, , 0.8202, 0.8267, 0.8331, 0.8395, 0.8457, , 0.8209, 0.8274, 0.8338, 0.8401, 0.8463, , 0.8215, 0.8280, 0.8344, 0.8407, 0.8470, , 0.8222, 0.8287, 0.8351, 0.8414, 0.8476, , 0.8228, 0.8293, 0.8357, 0.8420, 0.8482, , 0.8235, 0.8299, 0.8363, 0.8426, 0.8488, , 0.8241, 0.8306, 0.8370, 0.8432, 0.8494, , 0.8248, 0.8312, 0.8376, 0.8439, 0.8500, , 0.8254, 0.8319, 0.8382, 0.8445, 0.8506, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 3, 2, 2, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 5, 4, 4, 4, , 5, 5, 5, 5, 5, , 6, 6, 6, 6, 6, , 7.1, 7.2, 7.3, 7.4, 7.5, , 0.8513, 0.8573, 0.8633, 0.8692, 0.8751, , 0.8519, 0.8579, 0.8639, 0.8698, 0.8756, , 0.8525, 0.8585, 0.8645, 0.8704, 0.8762, , 0.8531, 0.8591, 0.8651, 0.8710, 0.8768, , 0.8537, 0.8597, 0.8657, 0.8716, 0.8774, , 0.8543, 0.8603, 0.8663, 0.8722, 0.8779, , 0.8549, 0.8609, 0.8669, 0.8727, 0.8785, , 0.8555, 0.8615, 0.8675, 0.8733, 0.8791, , 0.8561, 0.8621, 0.8681, 0.8739, 0.8797, , 0.8567, 0.8627, 0.8686, 0.8745, 0.8802, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 3, , 4, 4, 4, 4, 4, , 5, 5, 5, 5, 5, , 5, 5, 5, 5, 5, , 7.6, 7.7, 7.8, 7.9, 8.0, , 0.8808, 0.8865, 0.8921, 0.8976, 0.9031, , 0.8814, 0.8871, 0.8927, 0.8982, 0.9036, , 0.8820, 0.8876, 0.8932, 0.8987, 0.9042, , 0.8825, 0.8882, 0.8938, 0.8993, 0.9047, , 0.8831, 0.8887, 0.8943, 0.8998, 0.9053, , 0.8837, 0.8893, 0.8949, 0.9004, 0.9058, , 0.8842, 0.8899, 0.8954, 0.9009, 0.9063, , 0.8848, 0.8904, 0.8960, 0.9015, 0.9069, , 0.8854, 0.8910, 0.8965, 0.9020, 0.9074, , 0.8859, 0.8915, 0.8971, 0.9025, 0.9079, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 4, 4, 4, 4, , 5, 5, 5, 5, 5, , 8.1, 8.2, 8.3, 8.4, 8.5, , 0.9085, 0.9138, 0.9191, 0.9243, 0.9294, , 0.9090, 0.9143, 0.9196, 0.9248, 0.9299, , 0.9096, 0.9149, 0.9201, 0.9253, 0.9304, , 0.9101, 0.9154, 0.9206, 0.9258, 0.9309, , 0.9106, 0.9159, 0.9212, 0.9263, 0.9315, , 0.9112, 0.9165, 0.9217, 0.9269, 0.9320, , 0.9117, 0.9170, 0.9222, 0.9274, 0.9325, , 0.9122, 0.9175, 0.9227, 0.9279, 0.9330, , 0.9128, 0.9180, 0.9232, 0.9284, 0.9335, , 0.9133, 0.9186, 0.9238, 0.9289, 0.9340, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 4, 4, 4, 4, 4, , 5, 5, 5, 5, 5, , 8.6, 8.7, 8.8, 8.9, 9.0, , 0.9345, 0.9395, 0.9445, 0.9494, 0.9542, , 0.9350, 0.9400, 0.9450, 0.9499, 0.9547, , 0.9355, 0.9405, 0.9455, 0.9504, 0.9552, , 0.9360, 0.9410, 0.9460, 0.9509, 0.9557, , 0.9365, 0.9415, 0.9465, 0.9513, 0.9562, , 0.9370, 0.9420, 0.9469, 0.9518, 0.9566, , 0.9375, 0.9425, 0.9474, 0.9523, 0.9571, , 0.9380, 0.9430, 0.9479, 0.9528, 0.9576, , 0.9385, 0.9435, 0.9484, 0.9533, 0.9581, , 0.9390, 0.9440, 0.9489, 0.9538, 0.9586, , 1, 0, 0, 0, 0, , 1, 1, 1, 1, 1, , 2, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 4, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 4, 4, 4, 4, , 9.1, 9.2, 9.3, 9.4, 9.5, , 0.9590, 0.9638, 0.9685, 0.9731, 0.9777, , 0.9595, 0.9643, 0.9689, 0.9736, 0.9782, , 0.9600, 0.9647, 0.9694, 0.9741, 0.9786, , 0.9605, 0.9652, 0.9699, 0.9745, 0.9791, , 0.9609, 0.9657, 0.9703, 0.9750, 0.9795, , 0.9614, 0.9661, 0.9708, 0.9754, 0.9800, , 0.9619, 0.9666, 0.9713, 0.9759, 0.9805, , 0.9624, 0.9671, 0.9717, 0.9763, 0.9809, , 0.9628, 0.9675, 0.9722, 0.9768, 0.9814, , 0.9633, 0.9680, 0.9727, 0.9773, 0.9818, , 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 4, 4, 4, 4, 4, , 9.6, 9.7, 9.8, 9.9, , 0.9823, 0.9868, 0.9912, 0.9956, , 0.9827, 0.9872, 0.9917, 0.9961, , 0.9832, 0.9877, 0.9921, 0.9965, , 0.9836, 0.9881, 0.9926, 0.9969, , 0.9841, 0.9886, 0.9930, 0.9974, , 0.9845, 0.9890, 0.9934, 0.9978, , 0.9850, 0.9894, 0.9939, 0.9983, , 0.9854, 0.9899, 0.9943, 0.9987, , 0.9859, 0.9903, 0.9948, 0.9991, , 0.9863, 0.9908, 0.9952, 0.9996, , 0, 0, 0, 0, , 1, 1, 1, 1, , 1, 1, 1, 1, , 2, 2, 2, 2, , 2, 2, 2, 2, , 3, 3, 3, 3, , 3, 3, 3, 3, , 4, 4, 4, 3, , 4, 4, 4, 4, , 276, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 11th Std. Business Mathematics and Statistics, , 12_11th_BM-STAT_LogTable_EM.indd 276, , 21-04-2020 12:07:19 PM

Page 283 :
www.tntextbooks.in, , ANTI LOGARITHM TABLE, , Mean Difference, 0.00, , 0, 1.000, , 1, 1.002, , 2, 1.005, , 3, 1.007, , 4, 1.009, , 5, 1.012, , 6, 1.014, , 7, 1.016, , 8, 1.019, , 9, 1.021, , 1, 0, , 2, 0, , 3, 1, , 4, 1, , 5, 1, , 6, 1, , 7, 2, , 8, 2, , 9, 2, , 0.01, 0.02, 0.03, 0.04, 0.05, , 1.023, 1.047, 1.072, 1.096, 1.122, , 1.026, 1.050, 1.074, 1.099, 1.125, , 1.028, 1.052, 1.076, 1.102, 1.127, , 1.030, 1.054, 1.079, 1.104, 1.130, , 1.033, 1.057, 1.081, 1.107, 1.132, , 1.035, 1.059, 1.084, 1.109, 1.135, , 1.038, 1.062, 1.086, 1.112, 1.138, , 1.040, 1.064, 1.089, 1.114, 1.140, , 1.042, 1.067, 1.091, 1.117, 1.143, , 1.045, 1.069, 1.094, 1.119, 1.146, , 0, 0, 0, 0, 0, , 0, 0, 0, 1, 1, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 1, 1, 1, 2, 2, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 0.06, 0.07, 0.08, 0.09, 0.10, , 1.148, 1.175, 1.202, 1.230, 1.259, , 1.151, 1.178, 1.205, 1.233, 1.262, , 1.153, 1.180, 1.208, 1.236, 1.265, , 1.156, 1.183, 1.211, 1.239, 1.268, , 1.159, 1.186, 1.213, 1.242, 1.271, , 1.161, 1.189, 1.216, 1.245, 1.274, , 1.164, 1.191, 1.219, 1.247, 1.276, , 1.167, 1.194, 1.222, 1.250, 1.279, , 1.169, 1.197, 1.225, 1.253, 1.282, , 1.172, 1.199, 1.227, 1.256, 1.285, , 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 2, 2, 3, 3, 3, , 0.11, 0.12, 0.13, 0.14, 0.15, , 1.288, 1.318, 1.349, 1.380, 1.413, , 1.291, 1.321, 1.352, 1.384, 1.416, , 1.294, 1.324, 1.355, 1.387, 1.419, , 1.297, 1.327, 1.358, 1.390, 1.422, , 1.300, 1.330, 1.361, 1.393, 1.426, , 1.303, 1.334, 1.365, 1.396, 1.429, , 1.306, 1.337, 1.368, 1.400, 1.432, , 1.309, 1.340, 1.371, 1.403, 1.435, , 1.312, 1.343, 1.374, 1.406, 1.439, , 1.315, 1.346, 1.377, 1.409, 1.442, , 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 2, 2, 3, 3, 3, , 3, 3, 3, 3, 3, , 0.16, 0.17, 0.18, 0.19, 0.20, , 1.445, 1.479, 1.514, 1.549, 1.585, , 1.449, 1.483, 1.517, 1.552, 1.589, , 1.452, 1.486, 1.521, 1.556, 1.592, , 1.455, 1.489, 1.524, 1.560, 1.596, , 1.459, 1.493, 1.528, 1.563, 1.600, , 1.462, 1.496, 1.531, 1.567, 1.603, , 1.466, 1.500, 1.535, 1.570, 1.607, , 1.469, 1.503, 1.538, 1.574, 1.611, , 1.472, 1.507, 1.542, 1.578, 1.614, , 1.476, 1.510, 1.545, 1.581, 1.618, , 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 2, 2, 2, 3, 3, , 3, 3, 3, 3, 3, , 3, 3, 3, 3, 3, , 0.21, 0.22, 0.23, 0.24, 0.25, , 1.622, 1.660, 1.698, 1.738, 1.778, , 1.626, 1.663, 1.702, 1.742, 1.782, , 1.629, 1.667, 1.706, 1.746, 1.786, , 1.633, 1.671, 1.710, 1.750, 1.791, , 1.637, 1.675, 1.714, 1.754, 1.795, , 1.641, 1.679, 1.718, 1.758, 1.799, , 1.644, 1.683, 1.722, 1.762, 1.803, , 1.648, 1.687, 1.726, 1.766, 1.807, , 1.652, 1.690, 1.730, 1.770, 1.811, , 1.656, 1.694, 1.734, 1.774, 1.816, , 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 3, 3, 3, 3, 3, , 3, 3, 4, 4, 4, , 0.26, 0.27, 0.28, 0.29, 0.30, , 1.820, 1.862, 1.905, 1.950, 1.995, , 1.824, 1.866, 1.910, 1.954, 2.000, , 1.828, 1.871, 1.914, 1.959, 2.004, , 1.832, 1.875, 1.919, 1.963, 2.009, , 1.837, 1.879, 1.923, 1.968, 2.014, , 1.841, 1.884, 1.928, 1.972, 2.018, , 1.845, 1.888, 1.932, 1.977, 2.023, , 1.849, 1.892, 1.936, 1.982, 2.028, , 1.854, 1.897, 1.941, 1.986, 2.032, , 1.858, 1.901, 1.945, 1.991, 2.037, , 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 3, 3, 3, 3, 3, , 3, 3, 4, 4, 4, , 4, 4, 4, 4, 4, , 0.31, 0.32, 0.33, 0.34, 0.35, , 2.042, 2.089, 2.138, 2.188, 2.239, , 2.046, 2.094, 2.143, 2.193, 2.244, , 2.051, 2.099, 2.148, 2.198, 2.249, , 2.056, 2.104, 2.153, 2.203, 2.254, , 2.061, 2.109, 2.158, 2.208, 2.259, , 2.065, 2.113, 2.163, 2.213, 2.265, , 2.070, 2.118, 2.168, 2.218, 2.270, , 2.075, 2.123, 2.173, 2.223, 2.275, , 2.080, 2.128, 2.178, 2.228, 2.280, , 2.084, 2.133, 2.183, 2.234, 2.286, , 0, 0, 0, 1, 1, , 1, 1, 1, 1, 1, , 1, 1, 1, 2, 2, , 2, 2, 2, 2, 2, , 2, 2, 2, 3, 3, , 3, 3, 3, 3, 3, , 3, 3, 3, 4, 4, , 4, 4, 4, 4, 4, , 4, 4, 4, 5, 5, , 0.36, 0.37, 0.38, 0.39, 0.40, , 2.291, 2.344, 2.399, 2.455, 2.512, , 2.296, 2.350, 2.404, 2.460, 2.518, , 2.301, 2.355, 2.410, 2.466, 2.523, , 2.307, 2.360, 2.415, 2.472, 2.529, , 2.312, 2.366, 2.421, 2.477, 2.535, , 2.317, 2.371, 2.427, 2.483, 2.541, , 2.323, 2.377, 2.432, 2.489, 2.547, , 2.328, 2.382, 2.438, 2.495, 2.553, , 2.333, 2.388, 2.443, 2.500, 2.559, , 2.339, 2.393, 2.449, 2.506, 2.564, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 3, 3, 3, 3, 4, , 4, 4, 4, 4, 4, , 4, 4, 4, 5, 5, , 5, 5, 5, 5, 5, , 0.41, 0.42, 0.43, 0.44, 0.45, , 2.570, 2.630, 2.692, 2.754, 2.818, , 2.576, 2.636, 2.698, 2.761, 2.825, , 2.582, 2.642, 2.704, 2.767, 2.831, , 2.588, 2.649, 2.710, 2.773, 2.838, , 2.594, 2.655, 2.716, 2.780, 2.844, , 2.600, 2.661, 2.723, 2.786, 2.851, , 2.606, 2.667, 2.729, 2.793, 2.858, , 2.612, 2.673, 2.735, 2.799, 2.864, , 2.618, 2.679, 2.742, 2.805, 2.871, , 2.624, 2.685, 2.748, 2.812, 2.877, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 3, 3, 3, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 4, 4, 4, 4, 5, , 5, 5, 5, 5, 5, , 5, 6, 6, 6, 6, , 0.46, 0.47, 0.48, 0.49, , 2.884, 2.951, 3.020, 3.090, , 2.891, 2.958, 3.027, 3.097, , 2.897, 2.965, 3.034, 3.105, , 2.904, 2.972, 3.041, 3.112, , 2.911, 2.979, 3.048, 3.119, , 2.917, 2.985, 3.055, 3.126, , 2.924, 2.992, 3.062, 3.133, , 2.931, 2.999, 3.069, 3.141, , 2.938, 3.006, 3.076, 3.148, , 2.944, 3.013, 3.083, 3.155, , 1, 1, 1, 1, , 1, 1, 1, 1, , 2, 2, 2, 2, , 3, 3, 3, 3, , 3, 3, 4, 4, , 4, 4, 4, 4, , 5, 5, 5, 5, , 5, 5, 6, 6, , 6, 6, 6, 6, , Tables, , 12_11th_BM-STAT_LogTable_EM.indd 277, , 277, , 21-04-2020 12:07:19 PM

Page 284 :
www.tntextbooks.in, , ANTI LOGARITHM TABLE, Mean Difference, 0.50, , 0, 3.162, , 1, 3.170, , 2, 3.177, , 3, 3.184, , 4, 3.192, , 5, 3.199, , 6, 3.206, , 7, 3.214, , 8, 3.221, , 9, 3.228, , 1, 1, , 2, 1, , 3, 2, , 4, 3, , 5, 4, , 6, 4, , 7, 5, , 8, 6, , 9, 7, , 0.51, 0.52, 0.53, 0.54, 0.55, , 3.236, 3.311, 3.388, 3.467, 3.548, , 3.243, 3.319, 3.396, 3.475, 3.556, , 3.251, 3.327, 3.404, 3.483, 3.565, , 3.258, 3.334, 3.412, 3.491, 3.573, , 3.266, 3.342, 3.420, 3.499, 3.581, , 3.273, 3.350, 3.428, 3.508, 3.589, , 3.281, 3.357, 3.436, 3.516, 3.597, , 3.289, 3.365, 3.443, 3.524, 3.606, , 3.296, 3.373, 3.451, 3.532, 3.614, , 3.304, 3.381, 3.459, 3.540, 3.622, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 5, 5, 5, 5, , 5, 5, 6, 6, 6, , 6, 6, 6, 6, 7, , 7, 7, 7, 7, 7, , 0.56, 0.57, 0.58, 0.59, 0.60, , 3.631, 3.715, 3.802, 3.890, 3.981, , 3.639, 3.724, 3.811, 3.899, 3.990, , 3.648, 3.733, 3.819, 3.908, 3.999, , 3.656, 3.741, 3.828, 3.917, 4.009, , 3.664, 3.750, 3.837, 3.926, 4.018, , 3.673, 3.758, 3.846, 3.936, 4.027, , 3.681, 3.767, 3.855, 3.945, 4.036, , 3.690, 3.776, 3.864, 3.954, 4.046, , 3.698, 3.784, 3.873, 3.963, 4.055, , 3.707, 3.793, 3.882, 3.972, 4.064, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 3, 3, 4, 4, 4, , 4, 4, 4, 5, 5, , 5, 5, 5, 5, 6, , 6, 6, 6, 6, 6, , 7, 7, 7, 7, 7, , 8, 8, 8, 8, 8, , 0.61, 0.62, 0.63, 0.64, 0.65, , 4.074, 4.169, 4.266, 4.365, 4.467, , 4.083, 4.178, 4.276, 4.375, 4.477, , 4.093, 4.188, 4.285, 4.385, 4.487, , 4.102, 4.198, 4.295, 4.395, 4.498, , 4.111, 4.207, 4.305, 4.406, 4.508, , 4.121, 4.217, 4.315, 4.416, 4.519, , 4.130, 4.227, 4.325, 4.426, 4.529, , 4.140, 4.236, 4.335, 4.436, 4.539, , 4.150, 4.246, 4.345, 4.446, 4.550, , 4.159, 4.256, 4.355, 4.457, 4.560, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 5, 5, 5, 5, , 6, 6, 6, 6, 6, , 7, 7, 7, 7, 7, , 8, 8, 8, 8, 8, , 9, 9, 9, 9, 9, , 0.66, 0.67, 0.68, 0.69, 0.70, , 4.571, 4.677, 4.786, 4.898, 5.012, , 4.581, 4.688, 4.797, 4.909, 5.023, , 4.592, 4.699, 4.808, 4.920, 5.035, , 4.603, 4.710, 4.819, 4.932, 5.047, , 4.613, 4.721, 4.831, 4.943, 5.058, , 4.624, 4.732, 4.842, 4.955, 5.070, , 4.634, 4.742, 4.853, 4.966, 5.082, , 4.645, 4.753, 4.864, 4.977, 5.093, , 4.656, 4.764, 4.875, 4.989, 5.105, , 4.667, 4.775, 4.887, 5.000, 5.117, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 4, , 4, 4, 4, 5, 5, , 5, 5, 6, 6, 6, , 6, 7, 7, 7, 7, , 7, 8, 8, 8, 8, , 9, 9, 9, 9, 9, , 10, 10, 10, 10, 11, , 0.71, 0.72, 0.73, 0.74, 0.75, , 5.129, 5.248, 5.370, 5.495, 5.623, , 5.140, 5.260, 5.383, 5.508, 5.636, , 5.152, 5.272, 5.395, 5.521, 5.649, , 5.164, 5.284, 5.408, 5.534, 5.662, , 5.176, 5.297, 5.420, 5.546, 5.675, , 5.188, 5.309, 5.433, 5.559, 5.689, , 5.200, 5.321, 5.445, 5.572, 5.702, , 5.212, 5.333, 5.458, 5.585, 5.715, , 5.224, 5.346, 5.470, 5.598, 5.728, , 5.236, 5.358, 5.483, 5.610, 5.741, , 1, 1, 1, 1, 1, , 2, 2, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 5, 5, 5, 5, , 6, 6, 6, 6, 7, , 7, 7, 8, 8, 8, , 8, 9, 9, 9, 9, , 10, 10, 10, 10, 10, , 11, 11, 11, 12, 12, , 0.76, 0.77, 0.78, 0.79, 0.80, , 5.754, 5.888, 6.026, 6.166, 6.310, , 5.768, 5.902, 6.039, 6.180, 6.324, , 5.781, 5.916, 6.053, 6.194, 6.339, , 5.794, 5.929, 6.067, 6.209, 6.353, , 5.808, 5.943, 6.081, 6.223, 6.368, , 5.821, 5.957, 6.095, 6.237, 6.383, , 5.834, 5.970, 6.109, 6.252, 6.397, , 5.848, 5.984, 6.124, 6.266, 6.412, , 5.861, 5.998, 6.138, 6.281, 6.427, , 5.875, 6.012, 6.152, 6.295, 6.442, , 1, 1, 1, 1, 1, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 5, 6, 6, 6, , 7, 7, 7, 7, 7, , 8, 8, 8, 9, 9, , 9, 10, 10, 10, 10, , 11, 11, 11, 11, 12, , 12, 12, 13, 13, 13, , 0.81, 0.82, 0.83, 0.84, 0.85, , 6.457, 6.607, 6.761, 6.918, 7.079, , 6.471, 6.622, 6.776, 6.934, 7.096, , 6.486, 6.637, 6.792, 6.950, 7.112, , 6.501, 6.653, 6.808, 6.966, 7.129, , 6.516, 6.668, 6.823, 6.982, 7.145, , 6.531, 6.683, 6.839, 6.998, 7.161, , 6.546, 6.699, 6.855, 7.015, 7.178, , 6.561, 6.714, 6.871, 7.031, 7.194, , 6.577, 6.730, 6.887, 7.047, 7.211, , 6.592, 6.745, 6.902, 7.063, 7.228, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 5, , 6, 6, 6, 6, 7, , 8, 8, 8, 8, 8, , 9, 9, 9, 10, 10, , 11, 11, 11, 11, 12, , 12, 12, 13, 13, 13, , 14, 14, 14, 15, 15, , 0.86, 0.87, 0.88, 0.89, 0.90, , 7.244, 7.413, 7.586, 7.762, 7.943, , 7.261, 7.430, 7.603, 7.780, 7.962, , 7.278, 7.447, 7.621, 7.798, 7.980, , 7.295, 7.464, 7.638, 7.816, 7.998, , 7.311, 7.482, 7.656, 7.834, 8.017, , 7.328, 7.499, 7.674, 7.852, 8.035, , 7.345, 7.516, 7.691, 7.870, 8.054, , 7.362, 7.534, 7.709, 7.889, 8.072, , 7.379, 7.551, 7.727, 7.907, 8.091, , 7.396, 7.568, 7.745, 7.925, 8.110, , 2, 2, 2, 2, 2, , 3, 3, 4, 4, 4, , 5, 5, 5, 5, 6, , 7, 7, 7, 7, 7, , 8, 9, 9, 9, 9, , 10, 10, 11, 11, 11, , 12, 12, 12, 13, 13, , 13, 14, 14, 14, 15, , 15, 16, 16, 16, 17, , 0.91, 0.92, 0.93, 0.94, 0.95, , 8.128, 8.318, 8.511, 8.710, 8.913, , 8.147, 8.337, 8.531, 8.730, 8.933, , 8.166, 8.356, 8.551, 8.750, 8.954, , 8.185, 8.375, 8.570, 8.770, 8.974, , 8.204, 8.395, 8.590, 8.790, 8.995, , 8.222, 8.414, 8.610, 8.810, 9.016, , 8.241, 8.433, 8.630, 8.831, 9.036, , 8.260, 8.453, 8.650, 8.851, 9.057, , 8.279, 8.472, 8.670, 8.872, 9.078, , 8.299, 8.492, 8.690, 8.892, 9.099, , 2, 2, 2, 2, 2, , 4, 4, 4, 4, 4, , 6, 6, 6, 6, 6, , 8, 8, 8, 8, 8, , 9, 10, 10, 10, 10, , 11, 12, 12, 12, 12, , 13, 14, 14, 14, 15, , 15, 15, 16, 16, 17, , 17, 17, 18, 18, 19, , 0.96, 0.97, 0.98, 0.99, , 9.120, 9.333, 9.550, 9.772, , 9.141, 9.354, 9.572, 9.795, , 9.162, 9.376, 9.594, 9.817, , 9.183, 9.397, 9.616, 9.840, , 9.204, 9.419, 9.638, 9.863, , 9.226, 9.441, 9.661, 9.886, , 9.247, 9.462, 9.683, 9.908, , 9.268, 9.484, 9.705, 9.931, , 9.290, 9.506, 9.727, 9.954, , 9.311, 9.528, 9.750, 9.977, , 2, 2, 2, 2, , 4, 4, 4, 5, , 6, 7, 7, 7, , 8, 9, 9, 9, , 11, 11, 11, 11, , 13, 13, 13, 14, , 15, 15, 16, 16, , 17, 17, 18, 18, , 19, 20, 20, 20, , 278, , 11th Std. Business Mathematics and Statistics, , 12_11th_BM-STAT_LogTable_EM.indd 278, , 21-04-2020 12:07:19 PM

Page 285 :
Tables, , 12_11th_BM-STAT_LogTable_EM.indd 279, , Free World U, , 0.00, 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09, 0.10, 0.11, 0.12, 0.13, 0.14, 0.15, 0.16, 0.17, 0.18, 0.19, 0.20, 0.21, 0.22, 0.23, 0.24, 0.25, 0.26, 0.27, 0.28, 0.29, 0.30, 0.31, 0.32, 0.33, 0.34, 0.35, 0.36, 0.37, 0.38, 0.39, 0.40, 0.41, 0.42, 0.43, 0.44, 0.45, 0.46, 0.47, 0.48, 0.49, 0.50, , 0, 1.00000000, 1.01005017, 1.02020134, 1.03045453, 1.04081077, 1.05127110, 1.06183655, 1.07250818, 1.08328707, 1.09417428, 1.10517092, 1.11627807, 1.12749685, 1.13882838, 1.15027380, 1.16183424, 1.17351087, 1.18530485, 1.19721736, 1.20924960, 1.22140276, 1.23367806, 1.24607673, 1.25860001, 1.27124915, 1.28402542, 1.29693009, 1.30996445, 1.32312981, 1.33642749, 1.34985881, 1.36342511, 1.37712776, 1.39096813, 1.40494759, 1.41906755, 1.43332941, 1.44773461, 1.46228459, 1.47698079, 1.49182470, 1.50681779, 1.52196156, 1.53725752, 1.55270722, 1.56831219, 1.58407398, 1.59999419, 1.61607440, 1.63231622, 1.64872127, , 1, 2.71828183, 2.74560102, 2.77319476, 2.80106583, 2.82921701, 2.85765112, 2.88637099, 2.91537950, 2.94467955, 2.97427407, 3.00416602, 3.03435839, 3.06485420, 3.09565650, 3.12676837, 3.15819291, 3.18993328, 3.22199264, 3.25437420, 3.28708121, 3.32011692, 3.35348465, 3.38718773, 3.42122954, 3.45561346, 3.49034296, 3.52542149, 3.56085256, 3.59663973, 3.63278656, 3.66929667, 3.70617371, 3.74342138, 3.78104339, 3.81904351, 3.85742553, 3.89619330, 3.93535070, 3.97490163, 4.01485005, 4.05519997, 4.09595540, 4.13712044, 4.17869919, 4.22069582, 4.26311452, 4.30595953, 4.34923514, 4.39294568, 4.43709552, 4.48168907, , 2, 7.38905610, 7.46331735, 7.53832493, 7.61408636, 7.69060920, 7.76790111, 7.84596981, 7.92482312, 8.00446891, 8.08491516, 8.16616991, 8.24824128, 8.33113749, 8.41486681, 8.49943763, 8.58485840, 8.67113766, 8.75828404, 8.84630626, 8.93521311, 9.02501350, 9.11571639, 9.20733087, 9.29986608, 9.39333129, 9.48773584, 9.58308917, 9.67940081, 9.77668041, 9.87493768, 9.97418245, 10.07442466, 10.17567431, 10.27794153, 10.38123656, 10.48556972, 10.59095145, 10.69739228, 10.80490286, 10.91349394, 11.02317638, 11.13396115, 11.24585931, 11.35888208, 11.47304074, 11.58834672, 11.70481154, 11.82244685, 11.94126442, 12.06127612, 12.18249396, , 4, 54.59815003, 55.14687056, 55.70110583, 56.26091125, 56.82634281, 57.39745705, 57.97431108, 58.55696259, 59.14546985, 59.73989170, 60.34028760, 60.94671757, 61.55924226, 62.17792293, 62.80282145, 63.43400030, 64.07152260, 64.71545211, 65.36585321, 66.02279096, 66.68633104, 67.35653981, 68.03348429, 68.71723217, 69.40785184, 70.10541235, 70.80998345, 71.52163562, 72.24044001, 72.96646850, 73.69979370, 74.44048894, 75.18862829, 75.94428657, 76.70753934, 77.47846293, 78.25713442, 79.04363170, 79.83803341, 80.64041898, 81.45086866, 82.26946350, 83.09628536, 83.93141691, 84.77494167, 85.62694400, 86.48750910, 87.35672301, 88.23467268, 89.12144588, 90.01713130, , 1, , 5, 148.41315910, 149.90473615, 151.41130379, 152.93301270, 154.47001503, 156.02246449, 157.59051632, 159.17432734, 160.77405593, 162.38986205, 164.02190730, 165.67035487, 167.33536962, 169.01711804, 170.71576832, 172.43149032, 174.16445561, 175.91483748, 177.68281099, 179.46855293, 181.27224188, 183.09405819, 184.93418407, 186.79280352, 188.67010241, 190.56626846, 192.48149130, 194.41596245, 196.36987535, 198.34342541, 200.33680997, 202.35022839, 204.38388199, 206.43797416, 208.51271029, 210.60829787, 212.72494645, 214.86286770, 217.02227542, 219.20338555, 221.40641620, 223.63158768, 225.87912250, 228.14924542, 230.44218346, 232.75816591, 235.09742437, 237.46019276, 239.84670737, 242.25720686, 244.69193226, , 7, 1096.63315843, 1107.65450490, 1118.78661775, 1130.03061019, 1141.38760663, 1152.85874278, 1164.44516577, 1176.14803425, 1187.96851851, 1199.90780061, 1211.96707449, 1224.14754609, 1236.45043347, 1248.87696691, 1261.42838910, 1274.10595517, 1286.91093291, 1299.84460280, 1312.90825825, 1326.10320561, 1339.43076439, 1352.89226737, 1366.48906071, 1380.22250409, 1394.09397087, 1408.10484820, 1422.25653720, 1436.55045304, 1450.98802511, 1465.57069720, 1480.29992758, 1495.17718919, 1510.20396976, 1525.38177199, 1540.71211367, 1556.19652784, 1571.83656296, 1587.63378304, 1603.58976783, 1619.70611293, 1635.98443000, 1652.42634686, 1669.03350774, 1685.80757337, 1702.75022115, 1719.86314538, 1737.14805735, 1754.60668558, 1772.24077593, 1790.05209184, 1808.04241446, , 8, 2980.95798704, 3010.91711288, 3041.17733294, 3071.74167327, 3102.61319033, 3133.79497129, 3165.29013436, 3197.10182908, 3229.23323664, 3261.68757023, 3294.46807528, 3327.57802989, 3361.02074508, 3394.79956514, 3428.91786799, 3463.37906548, 3498.18660376, 3533.34396362, 3568.85466082, 3604.72224646, 3640.95030733, 3677.54246627, 3714.50238251, 3751.83375209, 3789.54030817, 3827.62582144, 3866.09410048, 3904.94899215, 3944.19438198, 3983.83419453, 4023.87239382, 4064.31298371, 4105.16000827, 4146.41755226, 4188.08974147, 4230.18074313, 4272.69476640, 4315.63606270, 4359.00892620, 4402.81769423, 4447.06674770, 4491.76051155, 4536.90345519, 4582.50009296, 4628.55498456, 4675.07273551, 4722.05799763, 4769.51546949, 4817.44989687, 4865.86607325, 4914.76884030, , 9, 8103.08392758, 8184.52127494, 8266.77708126, 8349.85957218, 8433.77705601, 8518.53792457, 8604.15065402, 8690.62380571, 8777.96602703, 8866.18605226, 8955.29270348, 9045.29489144, 9136.20161642, 9228.02196918, 9320.76513183, 9414.44037876, 9509.05707757, 9604.62469001, 9701.15277293, 9798.65097920, 9897.12905874, 9996.59685944, 10097.06432815, 10198.54151171, 10301.03855791, 10404.56571656, 10509.13334045, 10614.75188643, 10721.43191645, 10829.18409859, 10938.01920817, 11047.94812878, 11158.98185341, 11271.13148552, 11384.40824018, 11498.82344515, 11614.38854204, 11731.11508747, 11849.01475419, 11968.09933225, 12088.38073022, 12209.87097633, 12332.58221972, 12456.52673161, 12581.71690655, 12708.16526367, 12835.88444790, 12964.88723127, 13095.18651418, 13226.79532664, 13359.72682966, Grade 11, Algebra II, Evaluating Exponential Functions with Tables, , 6, 403.42879349, 407.48332027, 411.57859573, 415.71502938, 419.89303489, 424.11303004, 428.37543686, 432.68068157, 437.02919472, 441.42141115, 445.85777008, 450.33871517, 454.86469450, 459.43616068, 464.05357086, 468.71738678, 473.42807483, 478.18610609, 482.99195635, 487.84610621, 492.74904109, 497.70125129, 502.70323202, 507.75548350, 512.85851094, 518.01282467, 523.21894011, 528.47737788, 533.78866383, 539.15332908, 544.57191013, 550.04494881, 555.57299245, 561.15659385, 566.79631138, 572.49270901, 578.24635639, 584.05782889, 589.92770766, 595.85657969, 601.84503787, 607.89368106, 614.00311413, 620.17394801, 626.40679981, 632.70229281, 639.06105657, 645.48372697, 651.97094627, 658.52336322, 665.14163304, , EXPONENTIAL FUNCTION TABLE, , 3, 20.08553692, 20.28739993, 20.49129168, 20.69723259, 20.90524324, 21.11534442, 21.32755716, 21.54190268, 21.75840240, 21.97707798, 22.19795128, 22.42104440, 22.64637964, 22.87397954, 23.10386686, 23.33606458, 23.57059593, 23.80748436, 24.04675355, 24.28842744, 24.53253020, 24.77908622, 25.02812018, 25.27965697, 25.53372175, 25.79033992, 26.04953714, 26.31133934, 26.57577270, 26.84286366, 27.11263892, 27.38512547, 27.66035056, 27.93834170, 28.21912671, 28.50273364, 28.78919088, 29.07852706, 29.37077111, 29.66595227, 29.96410005, 30.26524426, 30.56941502, 30.87664275, 31.18695817, 31.50039231, 31.81697651, 32.13674244, 32.45972208, 32.78594771, 33.11545196, , EXPONENTIAL FUNCTION TABLE, , www.tntextbooks.in, , 279, , 21-04-2020 12:07:20 PM

Page 286 :
280, , 12_11th_BM-STAT_LogTable_EM.indd 280, , e, , 1, 4.52673079, 4.57222520, 4.61817682, 4.66459027, 4.71147018, 4.75882125, 4.80664819, 4.85495581, 4.90374893, 4.95303242, 5.00281123, 5.05309032, 5.10387472, 5.15516951, 5.20697983, 5.25931084, 5.31216780, 5.36555597, 5.41948071, 5.47394739, 5.52896148, 5.58452846, 5.64065391, 5.69734342, 5.75460268, 5.81243739, 5.87085336, 5.92985642, 5.98945247, 6.04964746, 6.11044743, 6.17185845, 6.23388666, 6.29653826, 6.35981952, 6.42373677, 6.48829640, 6.55350486, 6.61936868, 6.68589444, 6.75308880, 6.82095847, 6.88951024, 6.95875097, 7.02868758, 7.09932707, 7.17067649, 7.24274299, 7.31553376, , 2, 12.30493006, 12.42859666, 12.55350614, 12.67967097, 12.80710378, 12.93581732, 13.06582444, 13.19713816, 13.32977160, 13.46373804, 13.59905085, 13.73572359, 13.87376990, 14.01320361, 14.15403865, 14.29628910, 14.43996919, 14.58509330, 14.73167592, 14.87973172, 15.02927551, 15.18032224, 15.33288702, 15.48698510, 15.64263188, 15.79984295, 15.95863401, 16.11902095, 16.28101980, 16.44464677, 16.60991822, 16.77685067, 16.94546082, 17.11576554, 17.28778184, 17.46152694, 17.63701820, 17.81427318, 17.99330960, 18.17414537, 18.35679857, 18.54128746, 18.72763050, 18.91584631, 19.10595373, 19.29797176, 19.49191960, 19.68781664, 19.88568249, , 4, 90.92181851, 91.83559798, 92.75856108, 93.69080012, 94.63240831, 95.58347983, 96.54410977, 97.51439421, 98.49443016, 99.48431564, 100.48414964, 101.49403213, 102.51406411, 103.54434758, 104.58498558, 105.63608216, 106.69774243, 107.77007257, 108.85317981, 109.94717245, 111.05215991, 112.16825267, 113.29556235, 114.43420168, 115.58428453, 116.74592590, 117.91924196, 119.10435004, 120.30136866, 121.51041752, 122.73161752, 123.96509078, 125.21096065, 126.46935173, 127.74038985, 129.02420211, 130.32091690, 131.63066389, 132.95357405, 134.28977968, 135.63941441, 137.00261319, 138.37951234, 139.77024956, 141.17496392, 142.59379590, 144.02688737, 145.47438165, 146.93642350, , 5, 247.15112707, 249.63503719, 252.14391102, 254.67799946, 257.23755591, 259.82283632, 262.43409924, 265.07160579, 267.73561971, 270.42640743, 273.14423800, 275.88938323, 278.66211763, 281.46271848, 284.29146582, 287.14864256, 290.03453439, 292.94942992, 295.89362064, 298.86740097, 301.87106828, 304.90492296, 307.96926838, 311.06441098, 314.19066029, 317.34832892, 320.53773265, 323.75919042, 327.01302438, 330.29955991, 333.61912567, 336.97205363, 340.35867907, 343.77934066, 347.23438048, 350.72414402, 354.24898027, 357.80924171, 361.40528437, 365.03746787, 368.70615541, 372.41171388, 376.15451382, 379.93492954, 383.75333906, 387.61012424, 391.50567075, 395.44036816, 399.41460993, , 2, , 7, 1826.21354282, 1844.56729405, 1863.10550356, 1881.83002516, 1900.74273134, 1919.84551337, 1939.14028156, 1958.62896539, 1978.31351375, 1998.19589510, 2018.27809772, 2038.56212982, 2059.05001984, 2079.74381657, 2100.64558942, 2121.75742858, 2143.08144525, 2164.61977185, 2186.37456223, 2208.34799189, 2230.54225819, 2252.95958057, 2275.60220079, 2298.47238312, 2321.57241461, 2344.90460528, 2368.47128836, 2392.27482054, 2416.31758219, 2440.60197762, 2465.13043529, 2489.90540804, 2514.92937342, 2540.20483383, 2565.73431683, 2591.52037541, 2617.56558819, 2643.87255970, 2670.44392068, 2697.28232827, 2724.39046634, 2751.77104573, 2779.42680452, 2807.36050830, 2835.57495047, 2864.07295251, 2892.85736422, 2921.93106408, 2951.29695948, , for a given value x ., , 9, 13493.99431650, 13629.61121401, 13766.59108401, 13904.94762458, 14044.69467150, 14185.84619960, 14328.41632413, 14472.41930224, 14617.86953434, 14764.78156558, 14913.17008727, 15063.04993840, 15214.43610708, 15367.34373205, 15521.78810420, 15677.78466809, 15835.34902351, 15994.49692704, 16155.24429358, 16317.60719802, 16481.60187677, 16647.24472945, 16814.55232047, 16983.54138073, 17154.22880929, 17326.63167502, 17500.76721836, 17676.65285301, 17854.30616767, 18033.74492783, 18214.98707751, 18398.05074107, 18582.95422504, 18769.71601992, 18958.35480204, 19148.88943544, 19341.33897375, 19535.72266207, 19732.05993893, 19930.37043823, 20130.67399118, 20332.99062831, 20537.34058145, 20743.74428576, 20952.22238178, 21162.79571750, 21375.48535043, 21590.31254971, 21807.29879823, , Grade 11, Algebra II, Evaluating Exponential Functions with Tables, , eˣ, , 8, 4964.16308832, 5014.05375679, 5064.44583482, 5115.34436165, 5166.75442718, 5218.68117245, 5271.12979019, 5324.10552531, 5377.61367541, 5431.65959136, 5486.24867780, 5541.38639368, 5597.07825281, 5653.32982444, 5710.14673375, 5767.53466250, 5825.49934952, 5884.04659134, 5943.18224271, 6002.91221726, 6063.24248804, 6124.17908811, 6185.72811120, 6247.89571226, 6310.68810809, 6374.11157799, 6438.17246436, 6502.87717335, 6568.23217547, 6634.24400628, 6700.91926702, 6768.26462527, 6836.28681562, 6904.99264036, 6974.38897011, 7044.48274457, 7115.28097317, 7186.79073580, 7259.01918349, 7331.97353916, 7405.66109828, 7480.08922969, 7555.26537625, 7631.19705565, 7707.89186111, 7785.35746218, 7863.60160548, 7942.63211550, 8022.45689535, , This table is for the exponential function, , 6, 671.82641759, 678.57838534, 685.39821149, 692.28657804, 699.24417382, 706.27169460, 713.36984313, 720.53932925, 727.78086990, 735.09518924, 742.48301872, 749.94509711, 757.48217064, 765.09499302, 772.78432554, 780.55093713, 788.39560446, 796.31911202, 804.32225214, 812.40582517, 820.57063945, 828.81751148, 837.14726595, 845.56073585, 854.05876253, 862.64219579, 871.31189399, 880.06872411, 888.91356183, 897.84729165, 906.87080695, 915.98501008, 925.19081248, 934.48913473, 943.88090667, 953.36706749, 962.94856581, 972.62635979, 982.40141722, 992.27471561, 1002.24724229, 1012.31999453, 1022.49397962, 1032.77021496, 1043.14972818, 1053.63355724, 1064.22275054, 1074.91836700, 1085.72147619, , EXPONENTIAL FUNCTION TABLE, , 3, 33.44826778, 33.78442846, 34.12396761, 34.46691919, 34.81331749, 35.16319715, 35.51659315, 35.87354085, 36.23407593, 36.59823444, 36.96605281, 37.33756782, 37.71281662, 38.09183673, 38.47466605, 38.86134287, 39.25190586, 39.64639407, 40.04484696, 40.44730436, 40.85380653, 41.26439411, 41.67910816, 42.09799016, 42.52108200, 42.94842598, 43.38006484, 43.81604174, 44.25640028, 44.70118449, 45.15043887, 45.60420832, 46.06253823, 46.52547444, 46.99306323, 47.46535137, 47.94238608, 48.42421507, 48.91088652, 49.40244911, 49.89895197, 50.40044478, 50.90697767, 51.41860130, 51.93536683, 52.45732595, 52.98453084, 53.51703423, 54.05488936, , (~ 2.71828183) is the base of the natural logarithm, loge (x ) = ln(x )., , Free World U, , The number, , 0.51, 0.52, 0.53, 0.54, 0.55, 0.56, 0.57, 0.58, 0.59, 0.60, 0.61, 0.62, 0.63, 0.64, 0.65, 0.66, 0.67, 0.68, 0.69, 0.70, 0.71, 0.72, 0.73, 0.74, 0.75, 0.76, 0.77, 0.78, 0.79, 0.80, 0.81, 0.82, 0.83, 0.84, 0.85, 0.86, 0.87, 0.88, 0.89, 0.90, 0.91, 0.92, 0.93, 0.94, 0.95, 0.96, 0.97, 0.98, 0.99, , 0, 1.66529119, 1.68202765, 1.69893231, 1.71600686, 1.73325302, 1.75067250, 1.76826705, 1.78603843, 1.80398842, 1.82211880, 1.84043140, 1.85892804, 1.87761058, 1.89648088, 1.91554083, 1.93479233, 1.95423732, 1.97387773, 1.99371553, 2.01375271, 2.03399126, 2.05443321, 2.07508061, 2.09593551, 2.11700002, 2.13827622, 2.15976625, 2.18147227, 2.20339643, 2.22554093, 2.24790799, 2.27049984, 2.29331874, 2.31636698, 2.33964685, 2.36316069, 2.38691085, 2.41089971, 2.43512965, 2.45960311, 2.48432253, 2.50929039, 2.53450918, 2.55998142, 2.58570966, 2.61169647, 2.63794446, 2.66445624, 2.69123447, , EXPONENTIAL FUNCTION TABLE, , www.tntextbooks.in, , 11th Std. Business Mathematics and Statistics, , 21-04-2020 12:07:20 PM