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ISC Class XII Board Exam 2017 : Mathematics

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Analysis of Pupil Performance Research Development and Consultancy Division Council for the Indian School Certificate Examinations New Delhi Year 2017 __________________________________________________________________________________ Published by: Research Development and Consultancy Division (RDCD) Council for the Indian School Certificate Examinations Plot No. 35-36, Sector VI Pushp Vihar, Saket New Delhi-110017 Tel: (011) 29564831/33/37 E-mail: council@cisce.org Copyright, Council for the Indian School Certificate Examinations All rights reserved. The copyright to this publication and any part thereof solely vests in the Council for the Indian School Certificate Examinations. This publication and no part thereof may be reproduced, transmitted, distributed or stored in any manner whatsoever, without the prior written approval of the Council for the Indian School Certificate Examinations. FOREWORD This document of the Analysis of Pupils Performance at the ISC Year 12 and ICSE Year 10 Examination is one of its kind. It has grown and evolved over the years to provide feedback to schools in terms of the strengths and weaknesses of the candidates in handling the examinations. We commend the work of Mrs. Shilpi Gupta (Deputy Head) and the Research Development and Consultancy Division (RDCD) of the Council who have painstakingly prepared this analysis. We are grateful to the examiners who have contributed through their comments on the performance of the candidates under examination as well as for their suggestions to teachers and students for the effective transaction of the syllabus. We hope the schools will find this document useful. We invite comments from schools on its utility and quality. Gerry Arathoon Chief Executive & Secretary November 2017 i PREFACE The Council has been involved in the preparation of the ICSE and ISC Analysis of Pupil Performance documents since the year 1994. Over these years, these documents have facilitated the teaching-learning process by providing subject/ paper wise feedback to teachers regarding performance of students at the ICSE and ISC Examinations. With the aim of ensuring wider accessibility to all stakeholders, from the year 2014, the ICSE and the ISC documents have been made available on the Council s website www.cisce.org. The document includes a detailed qualitative analysis of the performance of students in different subjects which comprises of examiners comments on common errors made by candidates, topics found difficult or confusing, marking scheme for each answer and suggestions for teachers/ candidates. In addition to a detailed qualitative analysis, the Analysis of Pupil Performance documents for the Examination Year 2017 have a new component of a detailed quantitative analysis. For each subject dealt with in the document, both at the ICSE and the ISC levels, a detailed statistical analysis has been done, which has been presented in a simple user-friendly manner. It is hoped that this document will not only enable teachers to understand how their students have performed with respect to other students who appeared for the ICSE/ISC Year 2017 Examinations, how they have performed within the Region or State, their performance as compared to other Regions or States, etc., it will also help develop a better understanding of the assessment/ evaluation process. This will help them in guiding their students more effectively and comprehensively so that students prepare for the ICSE/ ISC Examinations, with a better understanding of what is required from them. The Analysis of Pupil Performance document for ICSE for the Examination Year 2017 covers the following subjects: English (English Language, Literature in English), Hindi, History, Civics and Geography (History & Civics, Geography), Mathematics, Science (Physics, Chemistry, Biology), Commercial Studies, Economics, Computer Applications, Economics Applications, Commercial Applications. Subjects covered in the ISC Analysis of Pupil Performance document for the Year 2017 include English (English Language and Literature in English), Hindi, Elective English, Physics (Theory and Practical), Chemistry (Theory and Practical), Biology (Theory and Practical), Mathematics, Computer Science, History, Political Science, Geography, Sociology, Psychology, Economics, Commerce, Accounts and Business Studies. I would like to acknowledge the contribution of all the ICSE and the ISC examiners who have been an integral part of this exercise, whose valuable inputs have helped put this document together. I would also like to thank the RDCD team of Dr. Manika Sharma, Dr. M.K. Gandhi, Ms. Mansi Guleria and Mrs. Roshni George, who have done a commendable job in preparing this document. The statistical data pertaining to the ICSE and the ISC Year 2017 Examinations has been provided by the IT section of the Council for which I would like to thank Col. R. Sreejeth (Deputy Secretary - IT), Mr. M.R. Felix, Education Officer (IT) ICSE and Mr. Samir Kumar, Education Officer (IT) - ISC. Shilpi Gupta Deputy Head - RDCD November 2017 ii CONTENTS Page No. FOREWORD i PREFACE ii INTRODUCTION 1 QUANTITATIVE ANALYSIS 3 QUALITATIVE ANALYSIS 10 This document aims to provide a comprehensive picture of the performance of candidates in the subject. It comprises of two sections, which provide Quantitative and Qualitative analysis results in terms of performance of candidates in the subject for the ISC Year 2017 Examination. The details of the Quantitative and the Qualitative analysis are given below. Quantitative Analysis This section provides a detailed statistical analysis of the following: Overall Performance of candidates in the subject (Statistics at a Glance) State wise Performance of Candidates Gender wise comparison of Overall Performance Region wise comparison of Performance Comparison of Region wise performance on the basis of Gender Comparison of performance in different Mark Ranges and comparison on the basis of Gender for the top and bottom ranges Comparison of performance in different Grade categories and comparison on the basis of Gender for the top and bottom grades The data has been presented in the form of means, frequencies and bar graphs. Understanding the tables Each of the comparison tables shows N (Number of candidates), Mean Marks obtained, Standard Errors and t-values with the level of significance. For t-test, mean values compared with their standard errors indicate whether an observed difference is likely to be a true difference or whether it has occurred by chance. The t-test has been applied using a confidence level of 95%, which means that if a difference is marked as statistically significant (with * mark, refer to t-value column of the table), the probability of the difference occurring by chance is less than 5%. In other words, we are 95% confident that the difference between the two values is true. t-test has been used to observe significant differences in the performance of boys and girls, gender wise differences within regions (North, East, South and West), gender wise differences within marks ranges (Top and bottom ranges) and gender wise differences within grades awarded (Grade 1 and Grade 9) at the ISC Year 2017 Examination. The analysed data has been depicted in a simple and user-friendly manner. 1 Given below is an example showing the comparison tables used in this section and the manner in which they should be interpreted. Comparison on the basis of Gender Gender Girls Boys *Significant at 0.05 level N 2,538 1,051 Mean 66.1 60.1 SE 0.29 0.42 The t-value 11.91* table shows comparison between the performances of boys and girls in a particular subject. The t-value of 11.91 is significant at 0.05 level (mentioned below the table) with a mean of girls as 66.1 and that of boys as 60.1. It means that there is significant difference between the performance of boys and girls in the subject. The probability of this difference occurring by chance is less than 5%. The mean value of girls is higher The results have also been depicted pictographically. In this case, the girls performed significantly better than the than that of boys. It can be interpreted that girls are performing significantly better than boys. boys. This is depicted by the girl with a medal. Qualitative Analysis The purpose of the qualitative analysis is to provide insights into how candidates have performed in individual questions set in the question paper. This section is based on inputs provided by examiners from examination centres across the country. It comprises of question wise feedback on the performance of candidates in the form of Comments of Examiners on the common errors made by candidates along with Suggestions for Teachers to rectify/ reduce these errors. The Marking Scheme for each question has also been provided to help teachers understand the criteria used for marking. Topics in the question paper that were generally found to be difficult or confusing by candidates, have also been listed down, along with general suggestions for candidates on how to prepare for the examination/ perform better in the examination. 2 STATISTICS AT A GLANCE Total Number of Candidates: 44,084 Mean Marks: Highest Marks: 100 63.4 Lowest Marks: 02 3 PERFORMANCE (STATE-WISE & FOREIGN) West Bengal 62.4 Uttarakhand 65.6 Uttar Pradesh 60.7 Tripura 44.0 Tamil Nadu 74.9 Telangana 67.1 Sikkim 47.4 Rajasthan 67.1 Punjab 63.6 Odisha 51.2 Madhya Pradesh 65.2 Manipur 50.6 Meghalaya 61.8 Maharashtra 76.0 Kerala 65.8 Karnataka 74.0 Jharkhand 59.1 Himachal Pradesh 68.9 Haryana 79.0 Gujarat 72.3 Goa 67.0 Delhi 73.5 Chandigarh 64.6 Chattisgarh 63.0 Bihar 64.1 Assam 84.2 Andhra Pradesh 63.7 Foreign 74.1 The States of Assam and Haryana secured highest mean marks. Mean marks secured by candidates studying in schools abroad were 74.1. 4 GENDER-WISE COMPARISON BOYS GIRLS Mean Marks: 66.2 Mean Marks: 61.7 Candidates: 16,581 Candidates: 27,503 Number of Number of Comparison on the basis of Gender Gender Girls Boys N Mean SE t-value 16,581 27,503 66.2 61.7 0.17 0.15 20.07* *Significant at 0.05 level Girls performed significantly better than boys. 5 REGION-WISE COMPARISON East North Mean Marks: 61.1 Mean Marks: 63.0 Number of Candidates: 18,370 Number of Candidates: 19,535 Highest Marks: 100 Lowest Marks: 03 Highest Marks: 100 Lowest Marks: 03 REGION Mean Marks: 69.1 Mean Marks: 74.3 Number of Candidates: 3,710 Number of Candidates: 2,285 Highest Marks: 100 Lowest Marks: 3 South Mean Marks: 82.8 Number of Candidates: 184 Highest Marks: 100 Lowest Marks: 40 Foreign 6 Highest Marks: 100 Lowest Marks: 2 West Mean Marks obtained by Boys and Girls-Region wise 67.2 61.0 North 62.8 77.0 71.1 67.5 59.9 East South 85.0 81.1 72.4 West Foreign Comparison on the basis of Gender within Region Region North (N) East (E) South (S) West (W) Foreign (F) Gender Girls Boys Girls Boys Girls Boys Girls Boys Girls Boys N Mean SE 6,343 13,192 7,552 10,818 1,659 2,051 947 1,338 80 104 67.2 61.0 62.8 59.9 71.1 67.5 77.0 72.4 85.0 81.1 0.28 0.22 0.26 0.24 0.47 0.47 0.60 0.58 1.43 1.63 *Significant at 0.05 level 7 t-value 17.55* 8.03* 5.45* 5.51* 1.80 MARK RANGES : COMPARISON GENDER-WISE Comparison on the basis of gender in top and bottom mark ranges Marks Range Top Range (81-100) Bottom Range (0-20) *Significant at 0.05 level Gender Girls Boys Girls Boys N Mean SE 5,356 7,796 773 2,531 90.7 91.0 13.7 12.6 0.08 0.06 0.17 0.10 Boys Marks Range (81-100) Girls t-value -2.60* 5.32* All Candidates 91.0 81 - 100 90.7 90.9 Marks Range (81-100) 70.4 61 - 80 70.6 70.5 48.0 41 - 60 Marks Range (0-20) 48.6 48.2 33.5 21 - 40 33.3 33.4 Marks Range (0-20) 12.6 0 - 20 13.7 12.8 8 GRADES AWARDED : COMPARISON GENDER-WISE Comparison on the basis of gender in Grade 1 and Grade 9 Grades Gender Girls Boys Girls Boys Grade 1 Grade 9 *Significant at 0.05 level N Mean SE 3,214 4,830 1,009 3,102 94.5 94.6 17.0 15.4 1.67 1.36 0.58 0.31 Boys Girls t-value -0.08 2.33* All Candidates 94.6 94.5 94.6 1 In Grade 9, 84.5 84.5 84.5 2 74.5 74.6 74.5 3 no significant difference was observed 64.5 64.5 64.5 4 between the average 57.0 57.0 57.0 5 performance of girls and boys in Grade 1. 52.1 52.0 52.1 6 47.0 47.0 47.0 7 Grade 9 42.4 42.5 42.5 8 9 9 15.4 17.0 15.8 QUALITATIVE ANALYSIS SECTION A (80 Marks) Question 1 [10x3] 6 2 is symmetrix, find the value of x. 2 15 10 If y 2x k = 0 touches the conic 3x2 5y2 = 15, find the value of k. (ii) If the matrix (iii) Prove that (iv) Using L Hospital s Rule, evaluate: (i) (v) (vi) (vii) 1 2 1 1 1+ = 1 ( 2 1 1 Evaluate: 2 2 4 . sec ) Evaluate: 4 0 log(1 + ) By using the data =25, =30, b yx = 1 6 and b xy = 0 4, find: (a) The regression equation y on x. (b) What is the most likely value of y when x = 60? (c) What is the coefficient of correlation between x and y? (viii) A problem is given to three students whose chances of solving it are respectively. Find the probability that the problem is solved. (ix) If a + ib = prove that 2 + 2 = 1 (x) + Solve: = 1 + 10 2 = 2 2 1 1 , 4 5 1 3 Comments of Examiners (i) Most candidates used the concept of singular matrix instead of symmetric matrix. Also, errors were made by some candidates in solving quadratic equation. (ii) Many candidates while expressing the given equation into standard form of Hyperbola made mistakes. In addition, candidates used incorrect formula of condition of tangency. Some candidates used elimination method but made errors in solving quadratic equation to find the value of k by equating discriminant. (iii)Some candidates used = cos 2 but left it half way through. Some candidates attempted it by writing in terms of tan-1 function but failed in the conversion. (iv) This part of the question was well attempted by most candidates. (v) Candidates made errors by making inappropriate substitution. Some candidates applied correct substitution (1/ = ) but made errors in further simplification. Many attempted it by incorrect method of integration by parts so could not get the correct solution. (vi) Many candidates made errors in applying the properties of definite integrals correctly and some candidates attempted it by integration by parts. (vii) Some candidates considered incorrect regression equations due to lack of understanding of the regression concept. (viii) Most candidates calculated complement of given probabilities but could not complete the solution due to lack of understanding of the concept of Probability. (ix) Many candidates took rationalization factor correctly but errors were committed in the process of simplifying the expression. Some candidates not identify real and imaginary parts in the complex number. (x) Many candidates could not separate the variables to solve the differential equation. Some candidates attempted it by linear equation method but made errors in calculating integrating factor and further. In most of the cases, solution had been written without a constant c . 11 Suggestions for teachers - Explain basic concepts of Singular and Symmetric matrices to students. Also provide practice on simple applications based on the above. - Clarify tangency condition to the students giving sufficient number of examples. Give ample practice for solving different types of problems by applying the tangency condition. - Explain clearly the conversion of inverse circular functions (one to another form) to students. Give adequate practice for conversion through diagram and by using formulae. - Thorough practice for calculating Limits of Indeterminate Forms 0 by application of 0 L Hospital s Rule should be given to the students. - Practice should also be given to the students for Graded questions of integration by substitution. - Teach properties of definite integrals well to the students and give them adequate practice to learn to apply them appropriately. - Explain clearly the Regression equations Y on X and X on Y and the basic difference between them. Give different types of problems for sufficient practice. - More practice is required in simplifying the expressions by applying the concept of rationalization. - Emphasize on mutually exclusive events and independent events. Give plenty of practice to understand the application of ( ) = 1 ( ). ( ). ( ). - Adequate practice should be given on various types of differential equations. MARKING SCHEME Question 1 (i) 6 2 15 2 15 2 = 6 2 10 10 2 = 2 15 x2 + 2x 15 = 0 x2 + 5x 3 + 15 = 0 (x + 5) (x 3) = 0 (ii) x = 5 3 3x2 5y2 = 15 and Tangent: y 2 = 0 2 5 2 =1 3 y = 2x + k a2 = 5, b2 = 3, m = 2 and constant c = k. Conditions for tangency: c2 = a2 m2 2 k2 = 5 4 3 = 17 (iii) k = 17 1 1 LHS = 2 1 1+ 1 2 1 LHS = 2 1 1+ 2 1 Let x = tan2 1 = 2 1 ( 2 ) = 2 2 = tan2 = x = (iv) = 1 (Alternate correct method is also . ) ( 2 . sec ) ( ) 2 1 2 2 2 0 2 + sin 2 1 (v) = 1 = 1 1 1 Evaluate: 2 2 Let t = 1 12 1 dt = 2 dx 2 = = 1 2 2 = + 4 sin 2 1 (vi) 2 4 Evaluate: 0 /4 I = 0 2 + log(1 + ) 0 ( ) = 0 ( ) log[1 + tan )] /4 = 0 1 2 1 2 log[1 + tan( 4 )] = /4 0 log 2 log(1 + )] /4 I = + I =2I = 0 log 2 = log 2[ ] 0 = log 2 . 4 (vii) I = 8.log2 4 = 25, = 30, b yx = 1 6, b xy = 0 4 Regression equation y on x: y = b yx (x ) y 30 = 16 10 ( 25) y - 30 = 1 6x 40 y = 1 6x 10 when x = 60 y = (1 6)(60)-10 = 96 10 = 86 Coefficient of correlation: r = . = (1 6)(0 4) = 0 64 = + 0 8 both regression coefficients +ve. 1 1 1 (viii) P(A) = 4 , ( ) = 5 ( ) = 3 1 1 1 P( ) = 1 , ( ) = 1 , ( ) = 1 = 3 4 4 = 4 5 5 = 2 3 3 3 4 2 2 Probability of none of them solve the problem: 4 . 5 . 3 = 5 2 Probability of the problem is being solved = 1 - 5 = 3 5 (Alternate correct method is also . ) 13 (ix) + a + ib = + | + | = 2 + 2 = a2 + b2 = 1 2 + 2 + + a + ib = + a + ib = ( + )2 2 + 2 2 2 a = 2 + 2 , (x) =1 2 + 2 = = 2 = 2 2 2 2 2 + 2 2 + 2 + 2 2 2 + 2 = (1 + y) x(y + 1) = (1+y) (1-x) 1+ = (1 ) 1+ = (1 ) + log(1 + ) = 2 2 + (Alternate correct method is also . ) Question 2 (a) Using properties of determinants, prove that: [5] + + = (a + b + c)(a c)2 + (b) Given that: 1 1 0 2 2 4 A = 2 3 4 = 4 2 4 , find AB. 0 1 2 2 1 5 Using this result, solve the following system of equation: x y = 3, 2x + 3y + 4z = 17 and y + 2z = 7 14 [5] Comments of Examiners (a) Some candidates did not apply the properties of determinants in the right order. Many candidates used 1 1 + 2 + 3 but failed to take common (a+b+c). In addition, candidates used row/column operation second time but failed to simplify the determinant. Suggestions for teachers - (b) Many candidates did not know how to use the product A.B to solve the given equations. Made errors in substituting 1 Some . A few 1 = 6 . - candidates solved the equations by finding 1 by conventional method i.e det candidates solved it using Cramer s Rule. Explain every property of the determinants with proper examples. Teach this type of questions in the class by discussion method. Stress upon developing logical and reasoning skills to apply the correct property. Basic concepts of inverse of a matrix and its properties need to be explained with examples. Adequate practice should be given to students for solving equations by inverse matrix method. MARKING SCHEME Question 2 (a) + + + + + R1 : R1+ R2 + R3 = (a + b+ c) 1 1 2 + + C 1 : C 1 C 2 (a + b+ c) C 3 : C 3 2C 2 0 + + 1 2( + + ) + + 0 + 2 (a + b+ c) [ 1{( )( + 2 ) ( )( )}] (a + b+ c) (a ) [ + 2 + ] (a + b+ c) (a )2 (b) 1 A.B = 2 0 1 3 1 6 = 0 0 0 2 2 4 4 4 2 4 2 2 1 5 0 0 6 0 = 6I 0 6 15 x y = 3, 2x + 3y + 4z = 17, y + 2z = 7 1 = 2 0 AB = 6I 1 0 3 3 4 = 17 1 2 7 1 A. 6 = 2 2 4 A = 6 = 6 4 2 4 2 1 5 3 2 2 4 3 1 x = A-1 17 = 6 4 2 4 17 7 2 1 5 7 12 2 1 = 6 2 = 1 24 4 -1 1 1 x = 2, y = 1, = 4 Question 3 (a) Solve the equation for x: (b) [5] sin x + sin (1 x) = cos x, x 0 -1 -1 -1 If A, B and C are the elements of Boolean algebra, simplify the expression (A +B ) (A + C ) + B (B + C). Draw the simplified circuit. [5] Comments of Examiners (a) Many candidates made errors in applying the formula of sin 1 + sin 1 . In addition, errors took place in simplifying and solving higher degree algebraic equations. Some candidates converted all terms into Inverse tangent function form and could not handle the resulting equations. (b) Many candidates made errors after applying the distributive property and some candidates were not clear about addition and multiplication sign. Many candidates made errors in drawing the circuit diagram. 16 Suggestions for teachers each Inverse Circular Functions laws thoroughly to the students. Also, give them adequate practice for inter conversion of the functions. More practice should be given in solving different types of higher order equations. - Explain laws of Boolean algebra comprehensively to the students and ample practice should be given on simplification of different types of Boolean expressions and then drawing their simplified circuits. MARKING SCHEME Question 3 (a) 1 + 1 (1 ) = 1 1 + 1 (1 ) = 1 2 1 (1 ) = 2 1 2 1 = sin[ 2 2 1 ] 1 = cos(2 1 ) 1 = [ 1 (1 2 2 )] 1 = 1 2 2 2 2 = 0 (2 1) = 0 = P (b) (A + B ) (A+ C ) + B (B + C) A A + A C + B A + B C + B B + B C (A A = 0, B B= 0) A C + B A + B (C + C) ( C + C = 1) A C + B A + B A C + B C A B Question 4 (a) Verify Lagrange s mean value theorem for the function: [5] f(x) = x (1 log x) and find the value of c in the interval [1, 2] (b) Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 3y2 4x = 8. 17 [5] Comments of Examiners (a) Many candidates got confused about closed interval and open interval while proving Lagrange s Mean Value theorem. Some candidates committed errors in solving logarithmic equations as they could not make a distinction between logarithmic function with base 10 and base e . (b) Many candidates failed to convert the given equation into standard form of conic section. The candidates also made errors while applying the concepts such as shifting the origin and translation of axes. Suggestions for teachers - Teach Lagrange s Mean Value theorem to the students in detail. Also discuss difference between closed and open intervals with their significance by sketching the curve etc. - Explain basic concepts and properties of logarithmic functions with examples in the class. - Explain to the students, conics, their equations, basic terms related to them thoroughly giving appropriate examples. Also, give adequate practice for simplification of general conic equation into standard conic equation, shifting of origin and translation of axes to compute foci, directrix and eccentricity, etc. MARKING SCHEME Question 4 (a) (i) ( ) = (1 ) is continuous function in the closed interval [1,2] 1 (ii) 1 ( ) = +(1 log x) = log x 1 ( )existing in open (1, 2) (iii) (1) = 1(1 log 1) = 1 (2) = 2(1 2) = (1) (2) All three conditions of Lagrange s mean values are satisfied, there exists at least one value of c such that: ( ) ( ) 2(1 2) 1 log = 2 1 log = 2 2 2 1 1 ( ) = log = 2 2 1 = 2 2 log log = 4 = 4 (1,2) 18 (b) 2 3 2 4 = 8 2 4 + 4 3 2 = 8 + 4 ( 2)2 3 2 = 12 ( 2)2 2 =1 12 4 2 = 12, 2 = 4 2 = 2 ( 2 1) 4 = 12( 2 1) 1 + 1 = 2 3 2 = 4 3 2 = 3 Centre (2, 0) Focus ( , ) = ( , 0) = , = 0 2 , = 0 2 = 12 . 3 = 4 + 2, = 0 = 6 2, = 0 (6, 0) and ( 2, 0) Equation of directrix: = 3 2 = 12 . 2 2 = 3 = 3 + 2 = 5 1 Question 5 (a) If y = cos (sin x), show that: 2 (b) 2 [5] 2 + tan x + y cos x = 0 Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube. 19 [5] Comments of Examiners (a) First order differentiation was attempted correctly by many candidates but some made errors in second order derivative and framing the required equation. Several candidates failed to differentiate using chain rule. (b) Many candidates did not express volume and surface area of cube with square base by correct formula. Most of the candidates did not show the working of second order derivative and failed to 2 prove 2 > 0. Suggestions for teachers Ample practice of derivatives of all forms of functions should be given to the students. Ensure that the students are familiar with the terms area, perimeter, surface and volume of 2-dimensional and 3-dimensional figures referred to in the syllabus. - Train students to express the function to be optimized in terms of a single variable by using the given data. For minimum value f'(x)=0 and f''(x) > 0. MARKING SCHEME Question 5 (a) = cos( ) : 2 + . + . 2 = 0 2 = sin( ) . 2 = [ 2 . cos( ) sin( ) . ] 2 2 = 2 . + . sin( ) 2 2 1 2 = . sin . . 2 (b) 2 + + 2 = 0 2 Let the length and breadth be x units (square base), height of cuboid be = h unit. = 2 = 2 S = 2( 2 + + ) = (2 2 ) + 4 S = 2 2 + 4 . 2 1 = 2 2 + 4 . 1 = 4 4 . 2 4 = 0 4 = 2 x3 = v 20 x3 = x2h x=h 2 8 =4+ 3 2 2 8 at x = h 2 = 4 + 3 > 0 Surace area is minimum when x = h i.e. When it is a cube, surface area will be minimum. Question 6 (a) (b) Evaluate: [5] sin 2 (1+sin ) (2+sin ) Draw a rough sketch of the curve y2 = 4x and find the area of the region enclosed by the curve and the line y = x. [5] Comments of Examiners (a) Some candidates made errors in substitution. In most of the cases, candidates who used appropriate substitution did not split the function into partial fractions to find the integral of each term. (b) Many candidates got the correct solution for this question. A few candidates solved the question with incorrect limits. Candidates made mistakes in sketching the curve, finding limits and required area bounded by the curves. Suggestions for teachers - Give students a thorough understanding and good practice of Integration using substitution, partial fractions and special integrals. Instruct students not to leave their answer in terms of a new variable. - Train students to sketch the curve to understand the requirement like the area required to be found, the points of intersection and the limits of the definite integral. Adequate practice of solving (sketching) different types of curves should be given. MARKING SCHEME Question 6 (a) 2 (1 + ) (2 + ) 2 = (1+ ) (2+ ) 2 = (1+ )2+ ) Let sin x = t Cos x = dt = + (1 + )(2 + ) 1 + 2 + t = (2 + ) + (1 + ) 1 = +A A= 1 B=2 21 1 = 1+ + 2 2+ = 2[ log(1 + ) + 2 log(2 + )] (b) = 2 log(1 + ) + 4 log(2 + ) + (Alternate correct method is also ) y2 = 4x X 0 1 1 y 0 2 -2 4 4 (4,4) +4 -4 y=x (0,0) X 0 1 2 4 y 0 1 2 4 Area of the shaded region 4 4 = 2 0 0 2 4 = 2. 3 . 3 2 0 4 = 3 43 2 = 32 3 8= 1 2 8 3 1 2 [42 ] [ 2 ]40 sq units Question 7 (a) Calculate the Spearman s rank correlation coefficient for the following data and interpret the result: X Y (b) 35 40 54 60 80 75 95 90 73 70 73 75 35 38 91 95 83 75 81 70 Find the line of best fit for the following data, treating x as dependent variable (Regression equation x on y): X Y 14 14 12 23 13 17 14 24 16 18 Hence, estimate the value of x when y = 16. 22 10 25 13 23 12 24 [5] [5] Comments of Examiners Suggestions for teachers (a) Many candidates made errors in computing ranks for the scores. Some candidates did not have an idea of repetition of rank and correction factor. Many used wrong formula for Spearman s correlation coefficient. Most of the candidates did not interpret the result. (b) Many candidates made errors in finding the regression equation X on Y. Candidates were not clear about the formulae used for b yx and b xy. Many candidates failed to get the correct value of for = 16. Some candidates started solving with the help of normal equations but did not get the result. - Provide plenty of practice for solving problems on correlation. Train students to use the appropriate formula as per the requirement of the question. Accuracy levels are to be maintained at highest level as such simplifications involve simple calculations only. MARKING SCHEME Question 7 (a) x 35 54 80 95 73 73 35 91 83 81 y 40 60 75 90 70 75 38 95 75 70 R1 1 5 3 6 10 4 5 4 5 1 5 9 8 7 R2 2 3 7 9 4 5 7 1 10 7 4 5 D2 0 25 0 1 1 0 6 25 0 25 1 1 6 25 2 =17 D=R 1 -R 2 -0 5 0 -1 1 0 -2 5 0 5 -1 1 2 5 1 C.F = 12 ( 3 ) 1 = 12 [(23 2) + (23 2) + (23 2) + (33 3)] 1 = 12 [18 + 24] = 3 5 R=1 6 2 + . ( 2 1) = 1 6[17+3 5] 10 99 Very high positive correlation (b) x 14 12 13 14 y 14 23 17 24 dx 1 -1 0 1 dy -7 2 -4 3 = 867 990 = 0 875 dx2 1 1 0 1 23 dy2 49 4 16 9 dx.dy -7 -2 0 3 16 10 13 12 18 25 23 24 3 -3 0 -1 -3 4 2 3 9 9 0 1 9 16 4 9 -9 -12 0 -3 = 104 = 168 = 0 = 0 2 = 22 2 = 116 = 30 = 13 = 21 b xy = 2 = 30 116 Required regression line: x - = b xy (y - ) 30 x 13 = 116 ( 21) 116x + 30y 2138=0 When y = 16 x = 14 29 Question 8 (a) (b) In a class of 60 students, 30 opted for Mathematics, 32 opted for Biology and 24 opted for both Mathematics and Biology. If one of these students is selected at random, find the probability that: (i) The student opted for Mathematics or Biology. (ii) The student has opted neither Mathematics nor Biology. (iii) The student has opted Mathematics but not Biology. Bag A contains 1 white, 2 blue and 3 red balls. Bag B contains 3 white, 3 blue and 2 red balls. Bag C contains 2 white, 3 blue and 4 red balls. One bag is selected at random and then two balls are drawn from the selected bag. Find the probability that the balls drawn are white and red. [5] [5] Comments of Examiners (a) Many candidates used Venn diagrams but incorrect values of ( ). Some candidates 24 24 noted as ( ) = 60 instead of ( ) = 60 Most of the candidates could not write the required combinations correctly. Many candidates wrote incorrectly ( ) = ( ) + ( ). Some candidates did not understand third part of the question. (b) Many candidates did not write this part correctly. They mixed the probabilities of one white and one red. Many candidates wrote the result without multiplying the result with 1/3. 24 Suggestions for teachers - Explain the probability properties/laws and their applications thoroughly. Lay stress on making the students understand the problem and identify the cases that satisfy the situation, conditions, conditional probability and condition-based problems. MARKING SCHEME Question 8 (a) Let A: event that candidates opted Mathematics. Let B: event that candidates opted Biology. 30 (b) 32 24 P(A) = 60 ( ) = 60 ( ) = 60 (i) P(A ) = P(A)+ P(B) - ( ) 30 32 24 19 = 60 + 60 60 = 30 (ii) ( ) = P(A ) = 1 - P(A ) 19 11 = 1- 30 = 30 (iii) Probability of student has opted Mathematics but not Biology = ( ) (A ) 30 24 6 1 = 60 60 = 60 = 10 Probability of selecting each bag = 1/3. A B C White 1 3 2 Blue 2 3 3 Red 3 2 4 Prob of (1W&1R 1 3 1 6 2 3 1 2 1 8 2 2 1 4 1 9 2 1 1 3 1 The probability: 3 1 1 3 = 3 15 + 1 1 = 3 5 + 3 2 3 28 + 2 6 2 2 4 36 + 9 = 14 + 401 1890 3 1 2 1 8 2 + 2 1 4 1 9 2 = 0.212 Question 9 (a) Prove that locus of z is circle and find its centre and radius if imaginary. (b) Solve: ( 2 2 ) + ( 2 + 2 ) = 0 25 1 is purely [5] [5] Comments of Examiners (a) Many candidates took 1 = but could not simplify it further. Some candidates took instead of 1. 1 +1 Suggestions for teachers - Many candidates made errors in - simplification and identifying the real part, and equating it to zero. Some candidates made mistakes while noting down the centre and with correct radius. (b) Many candidates attempted it by the method of Homogeneous equations by substituting y=VX but could not proceed further. Many candidates made errors in separating the variables x and y and a few candidates did not write the constant C. - The concept of Rationalization done in previous classes must be revised. Familiarise students with locus of the standard equation of Circle, Ellipse, Hyperbola etc. Give more practice for sketching of straight lines and curves (circle, conics, etc.). Clarify all the forms of differential equations and provide sufficient practice in solving problems based on these equations. MARKING SCHEME Question 9 (a) Let z = x + iy + + ( 1) = = 1 + 1 ( 1) + Rationalise denominator = = + ( 1) ( 1)+ ( 1) ( 1) ( 1)+ ( 1) + ( 1)( 1) ( 1)2 + 2 1 is purely imaginary Real part = 0 ( 1) + ( 1) = 0 1 1 (b) 2 + 2 = 0 which is a circle having its centre (2, , 2 ) and radius = 1 2 ( 2 2 ) + ( 2 + 2 ) = 0 2 (1 ) + 2 (1 + ) = 0 1 1 + + = 0 2 2 1 1 + 2 + 2 = 1 1 log + log = 1 1 log( ) + + 26 SECTION B (20 Marks) Question 10 (a) (b) If , , are three mutually perpendicular vectors of equal magnitude, prove that ( + + ) is equally inclined with vectors , . Find the value of for which the four points with position vectors 6 7 , 16 19 4k , 6k and 2 5 +10k are coplanar. [5] [5] Comments of Examiners (a) Many candidates used x + + instead of . + + and could not simplify further. Many candidates were unable to express the + . + formula for cos = | |. + + . (b) Some candidates made errors in finding the vectors . Some candidates expanded the , and scalar triple product determinant incorrectly. A few candidates did not know the basic difference between position vector and vector. Suggestions for teachers xplain in detail about vectors [Scalar (dot) product of vectors, Cross product of vectors, their properties, Scalar triple product, Proofs of Formulae (Using Vectors)Sine rule, Cosine rule, Projection formula, Area of a triangle etc.] - Teach Scalar triple product and its applications with the help of practical examples. - Advise students to read the instructions given in the question carefully. - MARKING SCHEME Question 10 (a) , are mutually perpendicular. . = . = . = 0 | | = = | | = ( ) 2 + + = + + . + + = . + . + . + 2( . + . + . ) 2 = | |2 + + | |2 = k2+ k2 + k2 = 3k2 + + = 3 k Let + + make angles , and with vectors , respectively, then + + . = + + . | | + + . = + + . + + . = + + . | | Cos = Cos = Cos = + . . + . 3 . . + . + . 3 . + . . + . 3 . = = = 2 3 2 2 3 2 2 3 2 = = = = = = 1 1 3 1 3 1 3 1 3 27 (b) = 6 7 =16 19 4k = 6k = 2 5 +10k = 10 12 4k = = = 6 + ( +7) 6k = 4 + 2 + 10k = Points A, B, C and D are coplanar = 0 10 12 4 6 + 7 6 = 0 4 2 10 10[ 10 ( + 7) + 12] + 12 [ 60 24] 4[ 12 + 4( + 7)] = 0 10[ 10 + 82] + 12[ 84] 4[4 + 16] = 0 84 = 252 =3 Question 11 (a) (b) 4 +3 +1 Show that the lines 1 = 4 = 7 coordinates of their point of intersection. 1 2 = +1 3 = +10 8 intersect. Find the Find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points A (3, 2, 1) and B (1, 4, 2). Comments of Examiners [5] [5] Suggestions for teachers (a) Many candidates formed correct equations from the general points and some of them found the values of constants and correctly. However, they did not do the verification to prove that the lines are intersecting. Many candidates used incorrect formula to find the shortest distance, to prove the lines are intersecting. Some candidates applied vector method and proved lines are intersecting but failed to calculate the point of intersection. (b) Many candidates made errors in applying the basic concept that the direction ratios of normal to the plane and direction ratios of line perpendicular to the plane are proportional. 28 - Explain three-dimensional geometry thoroughly especially: the different methods for finding the shortest distance between two lines. if two lines intersect, the method to find the coordinates of their point of intersection. the concept of expressing the equation of line and plane in cartesian and vector form and vice versa. different types and methods of finding the equation of a plane and straight line in various forms. condition for a line and plane to be perpendicular, condition for two planes to be perpendicular. MARKING SCHEME Question 11 (a) x 4 1 = y+3 4 = Z+1 7 = (1) and x 1 2 = y+1 3 General point on (1) P, ( +4, 4 3 7 1) = z+10 8 = (2) General point on (2) Q, (2 + 1, 3 1, 8 10) If the lines intersect for some values of and the points P and Q coincide. + 4 = 2 + 1 2 = 3 (i) 4 3 = 3 1 4 + 3 = 2 (ii) 7 1 = 8 10 7 8 = 9 (iii) Solving (i) and (ii) = 1 and = 2 and Substituting and in equation (iii) 7 1 8 2 = 9 Satisfied the (iii) equation. i.e. the lines intersect (b) point of intersection is P (5, 7, 6) Any plane passing through the point (1, 2, 1) a(x 1) + ( + 2) + ( 1) = 0 D.R. of normal to the plane: a, b, c This plane is perpendicular to the line joining points A (3, 2, 1), B (1, 4, 2) . : 3 1, 2 4, 1 2 . . , 2 2, 1 D. R of normal to the plane and D.R line perpendicular to the plane are proportional. = = = 2 2 1 a = 2k, b = 2k, c = k Required equation of the plane is: 2k (x 12 ( + 2) ( 1) = 0 2x 2 2 4 + 1 = 0 2x 2 5 = 0 Question 12 (a) A fair die is rolled. If face 1 turns up, a ball is drawn from Bag A. If face 2 or 3 turns up, a ball is drawn from Bag B. If face 4 or 5 or 6 turns up, a ball is drawn from Bag C. Bag A contains 3 red and 2 white balls, Bag B contains 3 red and 4 white balls and Bag C contains 4 red and 5 white balls. The die is rolled, a Bag is picked up and a ball is drawn. If the drawn ball is red, what is the probability that it is drawn from Bag B? [5] (b) An urn contains 25 balls of which 10 balls are red and the remaining green. A ball is drawn at random from the urn, the colour is noted and the ball is replaced. If 6 balls are drawn in this way, find the probability that: (i) All the balls are red. [5] (ii) Not more than 2 balls are green. (iii) Number of red balls and green balls are equal. 29 Comments of Examiners (a) Some candidates made errors in computing the probabilities of selecting Bag A, Bag B and Bag C as 1/3 instead of 1/6, 2/6 and 3/6 respectively and were unable to get the result. Some candidates did not apply the theorem correctly. (b) Many candidates made errors in computing the values of and and applying the binomial probability formula ( = ) = . Most of the candidates were not clear in writing the correct Binomial Distribution. Suggestions for teachers - Emphasize that probabilities are - ratios, not numbers. Clarify the concept of conditional probability which helps the students to understand the advanced concept of Bayes theorem. Ensure extensive drill on concepts mean and variance of binomial probability distribution for thorough understanding. MARKING SCHEME Question 12 (a) E 1 : Event that bag A is picked E 2 : Event that bag B is picked E 3 : Event that bag C is picked P(E 2 ) = 2 6 P(E 3 ) = 3 6 P(E 1 ) = 1 6 3 3 4 P(R/E 1 ) = 5 P(R/E 2 ) = 7 P(R/E 3 ) = 9 By applying Bay s theorem: P(E 2 /R) = ( 2 ). ( / 2 ) ( 1 ) (( / 1 ) + ( 2 ). ( / 2 )+ ( 3 ). ( / 3 ) = (b) 23 . 67 13 23 34 . + . + . 65 67 69 90 = 293 Number of red balls: 10, number of green balls: 15 10 P(R) = 25 15 P(G) = 25 = 3 5 = 5 P = 2 5 q = 3 5 Number of ways of drawing a red ball: X = 0, 1, 2, 3, 4, 5, 6 P(x=r) = .Pr.qn-r i) All are red 2 ii) 2 6 3 0 2 6 P(X=6) = 6C 6 5 5 = 5 Not more than 2 are green: = P(x=4) + P(x=5)+P(x=6) 2 4 3 2 2 5 3 1 2 6 3 0 P(x 4) = 6 4 5 5 + 6 5 5 5 + 6 6 5 5 30 iii) Equal number of balls 2 3 3 3 P(x=3) = 6C 3 5 . 5 SECTION C (20 Marks) Question 13 (a) A machine costs 60,000 and its effective life is estimated to be 25 years. A sinking fund is to be created for replacing the machine at the end of its life time when its scrap value is estimated as 5,000. The price of the new machine is estimated to be 100% more than the price of the present one. Find the amount that should be set aside at the end of each year, out of the profits, for the sinking fund if it accumulates at an interest of 6% per annum compounded annually. [5] (b) A farmer has a supply of chemical fertilizer of type A which contains 10% nitrogen and 6% phosphoric acid and of type B which contains 5% nitrogen and 10% phosphoric acid. After soil test, it is found that at least 7 kg of nitrogen and same quantity of phosphoric acid is required for a good crop. The fertilizer of type A costs 5.00 per kg and the type B costs 8.00 per kg. Using Linear programming, find how many kilograms of each type of the fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph. [5] Comments of Examiners (a) Some candidates made errors in calculating the price of a new machine after 25 years. Some candidates applied incorrect formula. Many candidates made simplification errors in computing the value of a . (b) Many candidates made errors in calculating the conditions using constraints. Some candidates were unable to represent feasible reason on the graph as such they may not have clarity of the concept of solving linear equations graphically. Some candidates solved it in terms of decimals and mostly went incorrect in simplifications. 31 Suggestions for teachers - - - - Explain the difference between the present value of an annuity and the amount of an annuity and the use of the appropriate formula. Emphasize on basic concept of sinking fund, present value, maturity value. The optimum function and all possible constraints in the form of in-equations must be put down from what is stated in the problem. Train students to solve the constraints equations in pairs to obtain all feasible points leading to optimal value (maximum or minimum) of desired function. MARKING SCHEME Question 13 (a) Amount required at the end of 25 years = 60,000 + 60,000 5,000 = 1,15,000 6 n = 25, i= 100 = 0 06 A= [(1 + ) 1] 1,15,000 = 0 06 [(1 06)25 1] (b) a [(1 06)25 1] = 115000 0 06 =6900 a(3.291871)= 6900, and a = 2096.073 Let x kg of type A fertilizer and y kg of type B fertilizer required. Type A cost per kg = Rs 5, type B cost per kg = Rs.8 C = 5x + 8y subject to the following constraints: 10% of x + 5% of y 7 and 6% of x + 10% of y 7, x 0 0 10x + 5y 700 2x + y 140 1 6x + 10y 700 3x + 5y 350 2 2x + y 140 x = 70, y = 0 (70, 0) x = 0, y = 140 (0, 140) 350 350 3x + 5y 350 x = 3 , y = 0 ( 3 , 0) x = 0, y = 70 (0, 70) Solving 2x + y = 140 and 3x + 5y = 350 we get x = 50 and y = 40 (50, 40) (0,140) 0,70 50,40 70,0 350 , 0 3 The shaded region is the feasible region. C = 5x + 8y 350 350 1 At ( 3 , 0) C = 5 3 + 0 =583 3 (50, 40) C = 5 50+8 40 = 570 (0, 140) C = 5 0+8 140 = 1120 So, it is minimum at (50, 40) Minimum cost is Rs 570 for 50 kg of type A and 40 kg of type B fertilisers procured. 32 Question 14 (a) (b) The demand for a certain product is represented by the equation p = 500 + 25x unit. Find: 2 3 [5] in rupees where x is the number of units and p is the price per (i) Marginal revenue function. (ii) The marginal revenue when 10 units are sold. A bill of 60,000 payable 10 months after date was discounted for 57,300 on 30th June,2007. If the rate of interest was 11 % per annum, on what date was the bill drawn? [5] Comments of Examiners (a) Many candidates made errors in applying the concept of Revenue Function. Some candidates made errors in differentiating the revenue function. In addition, calculation errors made in finding the Marginal Revenue function at x=10. (b) Most candidates could not solve this question completely. Many candidates erred in calculating the value n . Some candidates attempted to solve by taking true discount instead of banker s discount. Some candidates had no idea of calculating the date, when the bill was discounted. Suggestions for teachers - Explain the application of 1st order & order differentiation in 2nd Commerce and Economics thoroughly to develop a better understanding. - Clarify each term like Cost function, Breakeven point, Profit function, demand function Revenue function and Marginal revenue function, etc. - All relevant terms and formulae need to be taught well for complete understanding. In addition, good practice of different types of questions needs to be given. MARKING SCHEME Question 14 (a) Demand function P = 500 + 25x Revenue for function = R(x) = P.x 2 3 = (500 + 25x 2 3 ) 3 = 500x + 25x2 3 Marginal revenue function: = ( ) = 500 + 50x 3 3 = 500 + 50x x2 Marginal revenue when x = 10 = 500+500 100 = 900 33 (b) Bill amount = Rs 60,000 Discounted on 30th June 2007 Period of the bill = 10 months Discounted value of the bill = 57300 Bankers discounts = 2700 BD = A n i; 4 45 1 2700 = 60,000 n 4 100 1 2 n = 2700 45 600 = 5 of year = 146 days Due date of Bill: 30th June 2007 + 146 days July 31 August 31 September 30 October 31 November 23 146 rd Due date of the Bill = 23 November Nominal due date = 20th November Date of drawing the Bill is 20th January. Question 15 (a) The price relatives and weights of a set of commodities are given below: Commodity Price relatives Weights A 125 x B 120 2x C 127 y [5] D 119 y+3 If the sum of the weights is 40 and weighted average of price relatives index number is 122, find the numerical values of x and y. (b) Construct 3 yearly moving averages from the following data and show on a graph against the original data: Year 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 Annual sales in lakhs 18 22 20 26 30 22 24 28 32 35 34 [5] Comments of Examiners (a) Some candidates made errors in obtaining the equations in terms of x and y. Many candidates used incorrect formula for weighted average of price relatives index number. In addition, some candidates could not solve the two equations to find the values of x and y. (b) Most candidates calculated moving averages correctly but plotting of moving averages was not correct. Some candidates did not use the averages. In some cases, the graphs drawn were not neat. Suggestions for teachers - Train students to identify whether the question is based on aggregate or average method. Thorough knowledge of formula should be given to students. - For plotting, moving averages correct to one decimal place is sufficient. The axes should be labelled. Plotting and sketching should be as neat as possible and the graph should be given a caption. - Extensive practice is necessary in plotting points, choosing coordinate axes. - Advise students to read the question carefully and choose the correct scale. MARKING SCHEME Question 15 (a) Commodity PR=R Weights A 125 x B 120 2x C 127 y D 119 y+3 3x + 2y + 3 = 40 3x + 2y = 37 ------------------ (1) 365 +246 +357 Index number: 122 = 40 365x + 246y = 4880 357 365x +246y = 4523 ----- (2) Solving (1) & (2) x = 7, y = 8 (b) RW 125x 240x 127y 119y+357 Year Annual sales 3 yearly moving total 3 yearly moving average 2000 2001 18 -- -- 22 60 20 2002 20 68 22 7 2003 26 76 25 3 2004 30 78 26 2005 22 76 25 3 2006 24 74 24 7 2007 28 84 28 2008 32 95 31 7 2009 35 -- -- 35 1 unit=1 year 1 unit= 4 lakhs sales Note: For questions having more than one correct answer/solution, alternate correct answers/ solutions, apart from those given in the marking scheme, have also been accepted. 36 GENERAL COMMENTS Topics found difficult by candidates Concepts in which candidates got confused Suggestions for candidates Conic Section in general Integrals, definite Integrals and curve sketching Vectors and Interchange of vector equation to Cartesian form (vice versa) of plane and Straight-line equations of 3D Geometry and their applications Conditional Probability and their applications, Binomial Probability distribution Complex numbers and Inverse circular functions Maxima and Minima. Hyperbola and Ellipse: Their standard form and other relations Lagrange s Mean Value Theorem Product and sum rule of probability and concepts of dependent and independent events Definite Integrals and their properties Dot and Cross product of vectors Present value of annuity and Amount of annuity at the end of the period Avoid selective study. Study the entire syllabus thoroughly and revise from time to time. Concepts of Class XI must be revised and integrated with the Class XII syllabus. Concepts of each chapter/topic must be clear. Formulae of related topics must be learnt. Revise all topics and formulae involved and make a chapter wise or topicwise list of these. Time management is important while attempting the paper. Practice solving papers within a stipulated time. Make wise choices from the options available in the question paper and manage time wisely. 37

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