{"id":13680,"date":"2018-04-10T10:12:09","date_gmt":"2018-04-10T04:42:09","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=13680"},"modified":"2025-10-09T17:07:43","modified_gmt":"2025-10-09T11:37:43","slug":"cbse-sample-papers-class-11-mathematics","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-mathematics\/","title":{"rendered":"CBSE Sample Papers Class 11 Mathematics 2025"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-mathematics\/#CBSE_Sample_Paper_for_Class_11_Mathematics\" >CBSE Sample Paper for Class 11 Mathematics<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-mathematics\/#CBSE_Sample_Papers_Class_11_Mathematics_2024-25%E2%80%93_in_PDF\" >CBSE Sample Papers Class 11 Mathematics 2024-25\u2013 in PDF<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-mathematics\/#Marking_Scheme_for_the_Class_11_exam_2025\" >Marking Scheme for the Class 11 exam 2025<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-mathematics\/#Class_11_%E2%80%93_Mathematics_Sample_Paper_2024-25\" >Class 11 &#8211; Mathematics Sample Paper (2024-25)<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-mathematics\/#Solution\" >Solution<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-mathematics\/#CBSE_Sample_Papers_for_Class_11_2025\" >CBSE Sample Papers for Class 11 2025<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-mathematics\/#Download_Class_11_Sample_Papers_for_All_Subjects_from_myCBSEguide\" >Download Class 11 Sample Papers for All Subjects from myCBSEguide<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Sample_Paper_for_Class_11_Mathematics\"><\/span>CBSE Sample Paper for Class 11 Mathematics<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Download Free Class 11 Mathematics Sample Papers 2024-25 from myCBSEguide <\/strong>As per the newly released <strong>CBSE marking scheme<\/strong> and <strong>blueprint<\/strong> for<a href=\"https:\/\/mycbseguide.com\/cbse-sample-papers-class-11.html\"> <strong>Class 11<\/strong><\/a>, <strong>myCBSEguide<\/strong> provides updated <strong>Class 11 Mathematics sample papers 2024-25<\/strong>. These sample papers are designed to match the latest exam pattern and are available for <strong>free download<\/strong> in <strong>PDF format<\/strong> on the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app<\/strong> and <strong><a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a> website<\/strong>.To excel in the upcoming exams, practicing <strong>CBSE Sample Papers Class 11 Mathematics<\/strong> is essential for students. These sample papers offer a detailed overview of the exam format and the types of questions students can expect. By regularly working through <strong>CBSE Sample Papers Class 11 Mathematics<\/strong>, students can enhance their problem-solving abilities and improve time management skills. Moreover, <strong>CBSE Sample Papers Class 11 Mathematics<\/strong> help students identify key areas of improvement by providing solutions to difficult problems. The latest <strong>CBSE Sample Papers Class 11 Mathematics<\/strong> are designed according to the current syllabus and exam pattern, ensuring that students are fully prepared for the board exams. Make sure to download and practice these sample papers to boost your confidence and achieve high marks in your Class 11 exams. Practicing <strong>CBSE Sample Papers Class 11 Mathematics<\/strong> is one of the most effective ways to prepare for your exams. The <strong>Class 11 Mathematics sample papers<\/strong> with solutions will help students thoroughly understand the exam format, improve problem-solving skills, and enhance exam preparation. By practicing these sample papers, students can gain confidence and boost their chances of scoring high marks in the <strong>Class 11 exams<\/strong>. Download your <strong>CBSE Class 11 Mathematics Sample Papers 2024-25<\/strong> today from <strong>myCBSEguide<\/strong> and start preparing effectively for your exams.<\/p>\n<p style=\"text-align: center;\"><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/1371\/type\/2\">Download Mathematics Sample Papers\u00a0as PDF<\/a><\/strong><\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Sample_Papers_Class_11_Mathematics_2024-25%E2%80%93_in_PDF\"><\/span><strong>CBSE Sample Papers Class 11 Mathematics 2024-25<\/strong>\u2013 in PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>myCBSEguide<\/strong> offers <strong>CBSE Class 11 Mathematics sample papers<\/strong> for the <strong>2024-25 session<\/strong> with solutions in <strong>PDF format<\/strong> for <strong>free download<\/strong>. These <strong>new pattern sample papers<\/strong> are designed to align with the <strong>latest syllabus<\/strong>, <strong>marking scheme<\/strong>, and <strong>blueprint<\/strong> for the current academic year. We recommend that students download these <strong>Class 11 Mathematics model papers<\/strong> and practice them regularly to familiarize themselves with the <strong>updated exam format<\/strong>. The <strong>Class 11 Mathematics sample papers<\/strong> on <strong>myCBSEguide<\/strong> are designed to boost students&#8217; problem-solving skills and exam readiness. These sample papers, available on both the <strong>myCBSEguide app<\/strong> and website, are crafted based on the latest <strong>exam pattern<\/strong> and <strong>blueprint<\/strong> issued by CBSE for the <strong>2024-25 session<\/strong>. Download the <strong>Class 11 Mathematics sample papers<\/strong> today from <strong>myCBSEguide<\/strong> and start practicing to excel in your exams. These papers provide a real-time exam experience, helping students get accustomed to the question format and time constraints.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/11_maths_sample_paper.jpg\" width=\"600\" height=\"300\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"Marking_Scheme_for_the_Class_11_exam_2025\"><\/span><strong>Marking Scheme for the Class 11 exam 2025<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>CBSE has made significant updates to the <strong>marking scheme<\/strong> for the <strong>2024-25 exams<\/strong>, as reflected in the latest <strong>model papers<\/strong>. This year, the exam will primarily feature <strong>subjective questions<\/strong> worth <strong>2<\/strong>, <strong>3<\/strong>, and <strong>5 marks<\/strong>. Additionally, there is a fair weightage of <strong>objective-type questions<\/strong>, including <strong>MCQs<\/strong>, <strong>ARQs<\/strong> (Assertion &amp; Reason), and <strong>CBQs<\/strong> (Case Study Questions). Given this balanced approach, it\u2019s crucial for students not to ignore any section, as each contributes to the overall score. By solving <strong>CBSE Sample Papers Class 11 Mathematics<\/strong>, students can identify their strengths and weaknesses, which is crucial for focused revision.<\/p>\n<p>We strongly advise against the selective study<strong>\u00a0method<\/strong>. Instead, aim to <strong>cover the entire syllabus<\/strong> on time and <strong>revise<\/strong> consistently for better retention. By practicing a variety of question types and focusing on both subjective and objective sections, students can improve their performance in the upcoming <strong>CBSE board exams<\/strong>. The <strong>CBSE Sample Papers Class 11 Mathematics<\/strong> are aligned with the current syllabus, ensuring that the practice is relevant and up-to-date with the latest exam pattern.<\/p>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Class_11_%E2%80%93_Mathematics_Sample_Paper_2024-25\"><\/span><strong>Class 11 &#8211; Mathematics Sample Paper (2024-25)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<hr \/>\n<p style=\"text-align: center;\"><strong>Class 11 &#8211; Mathematics<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<p><b>Maximum Marks: 80<br \/>\nTime Allowed: : 3 hours<\/b><\/p>\n<hr \/>\n<p><b>General Instructions:<\/b><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>This Question paper contains 38 questions. All questions are compulsory.<\/li>\n<li>This Question paper is divided into five Sections &#8211; A, B, C, D and E.<\/li>\n<li>In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs) and Questions no. 19 and 20 are Assertion-Reason based questions of 1 mark each.<\/li>\n<li>In Section B, Questions no. 21 to 25 are Very Short Answer (VSA)-type questions, carrying 2 marks each.<\/li>\n<li>In Section C, Questions no. 26 to 31 are Short Answer (SA)-type questions, carrying 3 marks each.<\/li>\n<li>In Section D, Questions no. 32 to 35 are Long Answer (LA)-type questions, carrying 5 marks each.<\/li>\n<li>In Section E, Questions no. 36 to 38 are Case study-based questions, carrying 4 marks each.<\/li>\n<li>There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 3 questions in Section C, 2 questions in Section D and one subpart each in 2 questions of Section E.<\/li>\n<li>Use of calculators is not allowed.<\/li>\n<\/ol>\n<hr \/>\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section A<\/b><\/li>\n<li>If x is an acute angle and tan x\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{\\sqrt{7}}{\/tex}<\/span>,\u00a0then the value of\u00a0<span class=\"math-tex\">{tex}\\frac{\\operatorname{cosec}^{2} x-\\sec ^{2} x}{\\operatorname{cosec}^{2} x+\\sec ^{2} x}{\/tex}<\/span>\u00a0is\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac {1} {2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac {5} {4}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)2<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac {3} {4}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>Let f (x) = cos<sup>-1<\/sup> 2x\u00a0 then, dom f(x) = ?\n<div style=\"margin-left: 20px;\">\n<p>a)[-1,1]\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\left[\\frac{-\\pi}{2}, \\frac{\\pi}{2}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\left[\\frac{-1}{2}, \\frac{1}{2}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\left[\\frac{-\\pi}{4}, \\frac{\\pi}{4}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>A fair coin is tossed repeatedly. The probability of getting a result in the fifth toss different from those obtained in the first four tosses is\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{31}{32}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{1}{32}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{1}{16}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>The derivative of <span class=\"math-tex\">{tex}\\sec ^{-1}\\left(\\frac{1}{2 x^2-1}\\right){\/tex}<\/span> with respect to <span class=\"math-tex\">{tex}\\sqrt {1 &#8211; {x^2}} {\/tex}<\/span>\u00a0at x =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0is\n<div style=\"margin-left: 20px;\">\n<p>a)2<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)4<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)1<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)-2<\/p>\n<\/div>\n<\/li>\n<li>Slope of a line is not defined if the line is\n<div style=\"margin-left: 20px;\">\n<p>a)parallel to the line x \u2013 y<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)parallel to the line x + y<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)parallel to Y axis<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)parallel to X axis<\/p>\n<\/div>\n<\/li>\n<li>If A = {2, 3, 4, 8, 10}, B = {3, 4, 5, 10, 12} and C = {4, 5, 6, 12, 14}, then (A\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0B)\u00a0<span class=\"math-tex\">{tex}\\cap{\/tex}<\/span>\u00a0(A\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0C)\u00a0is equal to\n<div style=\"margin-left: 20px;\">\n<p>a){4, 5, 8, 10, 12}<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b){2, 4, 5, 10, 12}<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c){3, 8, 10, 12}<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d){2, 3, 4, 5, 8, 10, 12}<\/p>\n<\/div>\n<\/li>\n<li>The value of\u00a0<span class=\"math-tex\">{tex}\\frac {\\left(i^{5}+i^{6}+i^{7}+i^{8}+i^{9}\\right)} {(1 + i)}{\/tex}<\/span>\u00a0is\n<div style=\"margin-left: 20px;\">\n<p>a)1<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0(1 + i)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0(1 &#8211; i)<\/p>\n<\/div>\n<\/li>\n<li>Consider the non \u2013 empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is\n<div style=\"margin-left: 20px;\">\n<p>a)both symmetric and transitive<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)transitive but not symmetric<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)neither symmetric nor transitive<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)symmetric but not transitive<\/p>\n<\/div>\n<\/li>\n<li>If x and a are real numbers such that a &gt; 0 and |x| &gt; a, then\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}x \\in(-a, \\infty){\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}x \\in(-\\infty,-a) \\cup(a, \\infty){\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)x <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> (-a, a)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}x \\in[-\\infty, a]{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>The value of cos (36\u00b0-A ) cos(36\u00b0 + A) + cos (54\u00b0 + A) cos(54\u00b0 &#8211;\u00a0A) is\n<div style=\"margin-left: 20px;\">\n<p>a)sin 3A<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)cos 2A<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)sin 2A<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)cos 3A<\/p>\n<\/div>\n<\/li>\n<li>For any two sets A and B,\u00a0<span class=\"math-tex\">{tex}A \\cap (A \\cup B) = &#8230;.{\/tex}<\/span>\n<div style=\"margin-left: 20px;\">\n<p>a)A<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\phi{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\ne{\/tex}<\/span><span class=\"math-tex\">{tex}\\phi{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)B<\/p>\n<\/div>\n<\/li>\n<li>If <span class=\"math-tex\">{tex}\\frac{1}{3}{\/tex}<\/span>,\u00a0x<sub>1<\/sub>, x<sub>2<\/sub>, 9 are in GP then x<sub>2<\/sub> = ?\n<div style=\"margin-left: 20px;\">\n<p>a)3<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Cannot be determined<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)6<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)1<\/p>\n<\/div>\n<\/li>\n<li><span class=\"math-tex\">{tex}\\sqrt 5 \\left\\{ {{{(\\sqrt 5 + 1)}^{4}} &#8211; {{\\left( {\\sqrt 5 &#8211; 1} \\right)}^{4}}} \\right\\}{\/tex}<\/span>is\n<div style=\"margin-left: 20px;\">\n<p>a)0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)212<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)240<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)224<\/p>\n<\/div>\n<\/li>\n<li>If a, b, c are real numbers such that a\u00a0<span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0b, c &lt; 0, then\n<div style=\"margin-left: 20px;\">\n<p>a)ac\u00a0<span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0bc<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)ac &gt; bc<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)ac\u00a0<span class=\"math-tex\">{tex} \\geq{\/tex}<\/span>\u00a0bc<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)ac &lt;\u00a0bc<\/p>\n<\/div>\n<\/li>\n<li>Which of the following is a null set?\n<div style=\"margin-left: 20px;\">\n<p>a)C =\u00a0<span class=\"math-tex\">{tex}\\phi {\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)B = {x : x + 3 = 3}<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)D = {0}<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)A = {x : x &gt; 1 and x &lt; 3}<\/p>\n<\/div>\n<\/li>\n<li>If cos\u00a0<span class=\"math-tex\">{tex}\\theta=\\frac{4}{5}{\/tex}<\/span>\u00a0and cos\u00a0<span class=\"math-tex\">{tex}\\phi=\\frac{12}{13}{\/tex}<\/span>, where\u00a0<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\phi{\/tex}<\/span>\u00a0both lie in quadrant IV, then sin\u00a0<span class=\"math-tex\">{tex}(\\theta+\\phi){\/tex}<\/span>\u00a0= ?\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{16}{65}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{-16}{65}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{-33}{65}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{-56}{65}{\/tex}<\/span><\/p>\n<p><strong>Boost Your Exam Preparation with myCBSEguide App and Examin8 App<\/strong><\/p>\n<div style=\"margin-left: 20px;\">\n<p>For comprehensive exam preparation, download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide<\/strong><\/a>\u00a0app. It offers complete <strong>study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. With <strong>practice questions<\/strong>, <strong>sample papers<\/strong>, and <strong>NCERT solutions<\/strong>, it\u2019s the perfect tool to help you score well and excel in your exams. Regular practice with these sample papers can boost confidence and improve speed in solving problems. <strong>CBSE Sample Papers Class 11 Mathematics<\/strong> also help students understand the mark distribution and the types of questions likely to appear in the final exams.<\/p>\n<p>Teachers can also use the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\"><strong>Examin8<\/strong><\/a>\u00a0app to easily create <strong>customized exam papers<\/strong> featuring their own <strong>name<\/strong> and <strong>logo<\/strong>, offering a personalized experience for their students. Whether you&#8217;re a student looking to prepare effectively or a teacher seeking to design tailored assessments, visit <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a> and <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a> are your go-to resources for academic success.<\/p>\n<p>|z<sub>1\u00a0<\/sub>+ z<sub>2<\/sub>| = |z<sub>1<\/sub>| + |z<sub>2<\/sub>|\u00a0is possible if<\/p>\n<\/div>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>a)z<sub>2<\/sub> =\u00a0<span class=\"math-tex\">{tex}\\overline{z_{1}}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)arg (z<sub>1<\/sub>) = arg (z<sub>2<\/sub>)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)|z<sub>1<\/sub>| = |z<sub>2<\/sub>|<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)z<sub>2<\/sub> =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{z_{1}} {\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>A convex polygon of n sides has n diagonals. Then value of n is\n<div style=\"margin-left: 20px;\">\n<p>a)6<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)8<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)7<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)5<\/p>\n<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> The expansion of (1 + x)<sup>n<\/sup> = <span class=\"math-tex\">{tex}n_{c_0}+n_{c_1}x + n_{c_2} x^2 \\ldots+n_{c_n} x^n{\/tex}<\/span>.<br \/>\n<strong>Reason (R):<\/strong> If x = -1, then the above expansion is zero.<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)Both A and R are true and R is the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Both A and R are true but R is not the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)A is true but R is false.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)A is false but R is true.<\/p>\n<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> The mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17 is 3.<br \/>\n<strong>Reason (R):<\/strong> The mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 is 8.5.<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)Both A and R are true and R is the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Both A and R are true but R is not the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)A is true but R is false.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)A is false but R is true.<\/p>\n<\/div>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section B<\/b><\/li>\n<li>Draw the graphs of the logarithmic functions log<sub>a<\/sub> x, when 0 &lt; a &lt; 1.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Let A = {1, 2}, B = {2, 3, 4}, C = {4, 5}. Find A <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> (B\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0C)<\/li>\n<li>Find the derivative of f (x) = tan x at x = 0<\/li>\n<li>Find the equation of each of the following parabolas\u00a0directrix x = 0, focus at (6, 0)\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>lf y<sub>1<\/sub>, y<sub>2<\/sub>, y<sub>3<\/sub> be the ordinates of a vertices of the triangle inscribed in a parabola y<sup>2<\/sup> = 4ax, then show that the area of the triangle is\u00a0<span class=\"math-tex\">{tex}\\frac{1}{8 a}{\/tex}<\/span>\u00a0| (y<sub>1<\/sub> &#8211; y<sub>2<\/sub>) (y<sub>2<\/sub> &#8211; y<sub>3<\/sub>) (y<sub>3<\/sub> &#8211; y<sub>1<\/sub>) |.<\/li>\n<li>Let A = {x : x <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> N, x is a multiple of 3} and B = {x : x <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> N and x is a multiple of 5}. Write A <span class=\"math-tex\">{tex}\\cap{\/tex}<\/span> B.<\/li>\n<li>If a, b, c are in G.P. write the area of the triangle formed by the line ax + by + c = 0 with the coordinates axes.<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section C<\/b><\/li>\n<li>Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, add a minimum number of ordered pairs so that the enlarged relation is symmetric, transitive and reflexive.<\/li>\n<li>Solve\u00a0the inequality\u00a0<span class=\"math-tex\">{tex}\\frac{(2 x-1)}{3} \\geq \\frac{(3 x-2)}{4}-\\frac{(2-x)}{5}{\/tex}<\/span> for real x.<\/li>\n<li>If the origin is the centroid of the triangle PQR with vertices P(2a, 2, 6), Q(-4, 3b, -10) and R(8, 14, 2c), then find the values of a, b and c.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Find the equation of the curve formed by the set of all points which are equidistant from the points A (-1, 2, 3) and B(3,\u00a02, 1).<\/li>\n<li>Using g binomial theorem, expand\u00a0<span class=\"math-tex\">{tex}\\left\\{\\left(x+y\\right)^5\\;+\\left(x-y\\right)^5\\right\\}{\/tex}<\/span>\u00a0and hence find the value of\u00a0<span class=\"math-tex\">{tex}\\left\\{(\\sqrt{2}+1)^{5}+(\\sqrt{2}-1)^{5}\\right\\}{\/tex}<\/span>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Simplify (x + 2y)<sup>8<\/sup> + (x &#8211; 2y)<sup>8<\/sup><\/li>\n<li>Find the real values of x and y for which <span class=\"math-tex\">{tex}\\left(\\frac{x-1}{3+i}+\\frac{y-1}{3-i}\\right)=i{\/tex}<\/span>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Solve the equation Izl + z = (2 + i) for complex value of z.<\/li>\n<li>If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12,16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}, find: A &#8211; B?<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section D<\/b><\/li>\n<li>A sample space consists of 9 elementary outcomes e<sub>1<\/sub>, e<sub>2<\/sub>, &#8230;, e<sub>9<\/sub> whose probabilities are<br \/>\nP(e<sub>1<\/sub>) = P(e<sub>2<\/sub>) = .08, P(e<sub>3<\/sub>) = P(e<sub>4<\/sub>) = P(e<sub>5<\/sub>) = 0.1<br \/>\nP(e<sub>6<\/sub>) = P(e<sub>7<\/sub>) = .2, P(e<sub>8<\/sub>) = P(e<sub>9<\/sub>) = .07<br \/>\nSuppose A = {e<sub>1<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>}, B = {e<sub>2<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>, e<sub>9<\/sub>}<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Calculate P (A), P (B), and P (A <span class=\"math-tex\">{tex}\\cap{\/tex}<\/span>\u00a0B)<\/li>\n<li>Using the addition law of probability, calculate P (A\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span> B)<\/li>\n<li>List the composition of the event A \u222a B, and calculate P (A <span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0B) by adding the probabilities of the elementary outcomes.<\/li>\n<li>Calculate P (<span class=\"math-tex\">{tex}\\bar B{\/tex}<\/span>) from P (<span class=\"math-tex\">{tex}\\bar B{\/tex}<\/span>), also calculate P (<span class=\"math-tex\">{tex}\\bar B{\/tex}<\/span>) directly from the elementary outcomes of\u00a0<span class=\"math-tex\">{tex}\\bar B{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n<li>The\u00a0differentiation of sec x with respect to x is sec x tan x.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Evaluate :\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{x \\to \\sqrt {10} } \\frac{{\\sqrt {7 &#8211; 2x} &#8211; (\\sqrt 5 &#8211; \\sqrt 2 )}}{{{x^2} &#8211; 10}}{\/tex}<\/span><\/li>\n<li>The sum of first three terms of a G.P. is <span class=\"math-tex\">{tex}\\frac { 13 } { 12 }{\/tex}<\/span> and their product is -1. Find the common ratio and the terms.<\/li>\n<li>If 2 tan<span class=\"math-tex\">{tex}\\alpha{\/tex}<\/span> = 3 tan <span class=\"math-tex\">{tex}\\beta{\/tex}<\/span>, prove that tan (<span class=\"math-tex\">{tex}\\alpha{\/tex}<\/span> &#8211; <span class=\"math-tex\">{tex}\\beta{\/tex}<\/span>) =\u00a0<span class=\"math-tex\">{tex}\\frac{\\sin 2 \\beta}{5-\\cos 2 \\beta}{\/tex}<\/span>.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Prove that\u00a0<span class=\"math-tex\">{tex}\\cos 2 x \\cdot \\cos \\frac { x } { 2 } &#8211; \\cos 3 x \\cdot \\cos \\frac { 9 x } { 2 } = \\sin 5 x \\cdot \\sin \\frac { 5 x } { 2 }{\/tex}<\/span><\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section E<\/b><\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nIndian track and field athlete Neeraj Chopra, who competes in the Javelin throw, won a gold medal at Tokyo Olympics. He is the first track and field athlete to win a gold medal for India at the Olympics.<br \/>\n<img decoding=\"async\" style=\"height: 82px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1661839613-g5dbdq.jpg\" \/><br \/>\n<img decoding=\"async\" style=\"height: 190px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1661839684-8unf4t.jpg\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Name the shape of path followed by a javelin. If equation of such a curve is given by x<sup>2<\/sup> = -16y, then find the coordinates of foci. (1)<\/li>\n<li>Find the equation of directrix and length of latus rectum of parabola x<sup>2<\/sup> = -16y. (1)<\/li>\n<li>Find the equation of parabola with Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis and also find equation of directrix. (2)<br \/>\n<strong>OR<\/strong><br \/>\nFind the equation of the parabola with focus (2, 0) and directrix x = -2 and also length of latus rectum. (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nConsider the data<\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>x<sub>i<\/sub><\/td>\n<td>4<\/td>\n<td>8<\/td>\n<td>11<\/td>\n<td>17<\/td>\n<td>20<\/td>\n<td>24<\/td>\n<td>32<\/td>\n<\/tr>\n<tr>\n<td>f<sub>i<\/sub><\/td>\n<td>3<\/td>\n<td>5<\/td>\n<td>9<\/td>\n<td>5<\/td>\n<td>4<\/td>\n<td>3<\/td>\n<td>1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Find the standard deviation. (1)<\/li>\n<li>Find the variance. (1)<\/li>\n<li>Find the mean. (2)<br \/>\n<strong>OR<\/strong><br \/>\nWrite the formula of variance? (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nAshish is writing examination. He is reading question paper during reading time. He reads instructions carefully. While reading instructions, he observed that the question paper consists of 15 questions divided in to two parts &#8211; part I containing 8 questions and part II containing 7 questions.<br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1666152323-nsfw67.jpg\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>If Ashish is required to attempt 8 questions in all selecting at least 3 from each part, then in how many ways can he select these questions (1)<\/li>\n<li>If Ashish is required to attempt 8 questions in all selecting 3 from I part, then in how many ways can he select these questions (1)<\/li>\n<li>If Ashish is required to attempt 8 questions in all selecting 4 from part I and 4 from part II, then in how many ways can he select these questions (2)<br \/>\n<strong>OR<\/strong><br \/>\nIf Ashish is required to attempt 8 questions in all selecting 6 from one section and remaining from another section, then in how many ways can he select these questions (2)<br \/>\n<strong>Download myCBSEguide App for Comprehensive Exam Preparation &amp; Practice<\/strong>To boost your exam preparation and practice more questions, download the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app<\/strong>. It offers <strong>complete study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>, including <strong>sample papers<\/strong>, <strong>NCERT solutions<\/strong>, and <strong>chapter-wise tests<\/strong>. With these resources, students can easily prepare and excel in their exams, gaining confidence and mastering key concepts.In addition, teachers can leverage the <strong><a href=\"https:\/\/examin8.com\/\">Examin8<\/a> app<\/strong> to create <strong>customized exam papers<\/strong> with their <strong>own branding<\/strong>, including <strong>name<\/strong> and <strong>logo<\/strong>. This makes it easier to assess student performance and offer personalized learning experiences. Whether you are a student aiming for top scores or a teacher creating unique assessments, visit <strong><a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a> Website<\/strong>\u00a0and <strong><a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong>\u00a0are the perfect tools for success.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center; page-break-before: always;\"><strong>Class 11 &#8211; Mathematics<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Solution\"><\/span><strong>Solution <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ol style=\"padding-left: 20px;\">\n<li style=\"text-align: center; display: block;\"><b>Section A <\/b><\/li>\n<li>(d)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac {3} {4}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>We have:<br \/>\ntan x\u00a0<span class=\"math-tex\">{tex}=\\frac{1}{\\sqrt{7}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\tan ^{2} x=\\frac{1}{7}{\/tex}<\/span><br \/>\nNow, dividing the numerator and the denominator of\u00a0<span class=\"math-tex\">{tex}\\frac{\\operatorname{cosec}^{2} x-\\sec ^{2} x}{\\operatorname{cosec}^{2} x+\\sec ^{2} x}{\/tex}<\/span>\u00a0by cosec<sup>2<\/sup> x:<br \/>\n<span class=\"math-tex\">{tex}\\frac{1-\\tan ^{2} x}{1+\\tan ^{2} x}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{1-\\frac{1}{7}}{1+\\frac{1}{7}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{6}{8}=\\frac{3}{4}{\/tex}<\/span><\/li>\n<li>(c)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\left[\\frac{-1}{2}, \\frac{1}{2}\\right]{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>f(x) = cos<sup>-1<\/sup> 2x<br \/>\nThe domain for function cos<sup>-1<\/sup>\u00a0x is [-1,1] and range is [0, <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>]\nWhen a function is multiplied by an integer, the domain of the function is decreased by the same number.<br \/>\nSo, domain of cos<sup>-1<\/sup> x is [-1, 1]\nMultiply function by 2<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0domain of cos<sup>-1 <\/sup>2x is\u00a0<span class=\"math-tex\">{tex}\\left[\\frac{-1}{2}, \\frac{1}{2}\\right]{\/tex}<\/span><\/li>\n<li>(d)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{1}{16}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>Probability of getting head in a single toss, P(H) =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><br \/>\nProbability of getting tail in a single toss, P(T) =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Required probability<br \/>\n= P(HHHHT or TTTTH)<br \/>\n= P(HHHHT) + P(TTTTH)<br \/>\n= P(H)\u00a0<span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>\u00a0P(H)\u00a0<span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>\u00a0P(H)\u00a0<span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>\u00a0P(H)\u00a0<span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>\u00a0P(T) + P(T)\u00a0<span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>\u00a0P(T)\u00a0<span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>\u00a0P(T)\u00a0<span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>\u00a0P(T)\u00a0<span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>\u00a0P(H)<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)+\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right){\/tex}<\/span><br \/>\n= 2\u00a0<span class=\"math-tex\">{tex}\\times \\frac{1}{32}=\\frac{1}{16}{\/tex}<\/span><\/li>\n<li>(b)\n<p style=\"display: inline;\">4<\/p>\n<p><b>Explanation: <\/b>Substitute x = cos\u00a0<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{d(2 \\theta)}{d(\\sin \\theta)}=\\frac{\\frac{d(2 \\theta)}{d t}}{\\frac{d(\\sin \\theta)}{d t}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02 sec\u00a0<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02(2) = 4<\/li>\n<li>(c)\n<p style=\"display: inline;\">parallel to Y axis<\/p>\n<p><b>Explanation: <\/b>Vertical lines have undefined\u00a0slopes. Hence a line which is parallel to Y-axis has undefined slopes.<\/li>\n<li>(d)\n<p style=\"display: inline;\">{2, 3, 4, 5, 8, 10, 12}<\/p>\n<p><b>Explanation: <\/b>Given A = {2, 3, 4, 8, 10}, B = {3, 4, 5, 10, 12} and C = {4, 5, 6, 12, 14}<br \/>\nHere A\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0B = {2, 3, 4, 5, 8, 10, 12} and\u00a0A\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0C = {2, 3, 4, 5, 6, 8, 10, 12, 14}<br \/>\nNow (A\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0B)\u00a0<span class=\"math-tex\">{tex}\\cap{\/tex}<\/span>\u00a0(A\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0C) = {2, 3, 4, 5, 8, 10, 12}<\/li>\n<li>(b)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0(1 + i)<\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0(1 + i)<br \/>\n<span class=\"math-tex\">{tex}\\frac {i^{5}+i^{6}+i^{2}+i^{8}+i^{9}} {1 + i}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{i-1-i+1+i}{1+i}{\/tex}<\/span>\u00a0[As, i<sup>5<\/sup> = i, i<sup>6<\/sup> = -1, i<sup>7<\/sup> = -i, i<sup>8<\/sup> = 1, i<sup>9<\/sup> = i<br \/>\n<span class=\"math-tex\">{tex}=\\frac {i} {i + 1}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{i}{i+1} \\times \\frac{i-1}{i-1}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{i(i-1)}{i^{2}-1}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{i^{2}-i}{-2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{1}{2}{\/tex}<\/span>\u00a0(1 + i)<\/li>\n<li>(b)\n<p style=\"display: inline;\">transitive but not symmetric<\/p>\n<p><b>Explanation: <\/b>Consider the non \u2013 empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is not symmetric, because aRb means a is brother of b, then, it is not necessary that b is also brother of a , it can be the sister of a. Therefore, bRa is not true. Therefore, the relation is not symmetric . Again, if aRb and bRc\u00a0is true, then aRc is also true. Therefore, R is transitive only.<\/li>\n<li>(b)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}x \\in(-\\infty,-a) \\cup(a, \\infty){\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>|x| &gt; a<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x &lt; -a or\u00a0x &gt; a<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>x\u00a0<span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0(<span class=\"math-tex\">{tex}-\\infty{\/tex}<\/span>, -a)\u00a0\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0(a, <span class=\"math-tex\">{tex}\\infty{\/tex}<\/span>)<\/li>\n<li>(b)\n<p style=\"display: inline;\">cos 2A<\/p>\n<p><b>Explanation: <\/b>cos(36<sup>o<\/sup> &#8211; A) cos(36<sup>o<\/sup> + A) + cos(54<sup>o<\/sup> + A) cos(54<sup>o<\/sup> &#8211; A)<br \/>\n= cos(36<sup>o<\/sup> &#8211; A) cos(36<sup>o<\/sup> + A) + sin[90<sup>o<\/sup> &#8211; (54<sup>o<\/sup> + A)] sin [90<sup>o<\/sup> &#8211; (54<sup>o<\/sup> &#8211; A)] [Since\u00a0<span class=\"math-tex\">{tex}\\sin \\left(90^{\\circ}-\\theta\\right)=\\cos \\theta{\/tex}<\/span>]\n= cos(36<sup>o<\/sup> &#8211; A) cos(36<sup>o<\/sup> + A) + sin(36<sup>o<\/sup> &#8211; A) sin(36<sup>o<\/sup> + A)<br \/>\n= cos(36<sup>o<\/sup> + A &#8211; 36<sup>o<\/sup> + A) [Using cos(A &#8211; B) formula]\n= cos 2A<\/li>\n<li>(a)\n<p style=\"display: inline;\">A<\/p>\n<p><b>Explanation: <\/b>Since,\u00a0<span class=\"math-tex\">{tex}A \\subseteq A \\cup B{\/tex}<\/span>, therefore,\u00a0<span class=\"math-tex\">{tex}A \\cap(A \\cup B){\/tex}<\/span>\u00a0= A<\/li>\n<li>(a)\n<p style=\"display: inline;\">3<\/p>\n<p><b>Explanation: <\/b>Here, we have a = <span class=\"math-tex\">{tex}1 \\over 3{\/tex}<\/span> and let r be the common ratio.<br \/>\nThen,\u00a0<span class=\"math-tex\">{tex}x_{1}=a r=\\frac{1}{3} r{\/tex}<\/span>,\u00a0<span class=\"math-tex\">{tex}, x_{2}=a r^{2}=\\frac{1}{3} r^{2} {\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}9=a r^{3}=\\frac{1}{3} r^{3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\quad r^{3}=27=3^{3} \\Rightarrow r=3{\/tex}<\/span><br \/>\nTherefore,\u00a0<span class=\"math-tex\">{tex}x_{2}=\\frac{1}{3} \\times 3^{2}=\\frac{1}{3} \\times 9=3{\/tex}<\/span>.<\/li>\n<li>(c)\n<p style=\"display: inline;\">240<\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\sqrt{5}\\left\\{(\\sqrt{5}+1)^{4}-(\\sqrt{5}-1)^{4}\\right\\}{\/tex}<\/span><br \/>\nLet a =\u00a0<span class=\"math-tex\">{tex}\\sqrt{5}{\/tex}<\/span>\u00a0and b = 1<br \/>\nThen\u00a0<span class=\"math-tex\">{tex}(\\sqrt{5}+1)^{4}-(\\sqrt{5}-1)^{4}{\/tex}<\/span>\u00a0= (a + 4)<sup>4<\/sup> &#8211; (a &#8211; b)<sup>4\u00a0<\/sup><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{llll} ^{4} C_{0} a^{4}+^{4} C_{1} a^{3} b+^{4} C_{2} a^{2} b^{2} \\end{array}^{2}\\right.{\/tex}<\/span>\u00a0+\u00a0<span class=\"math-tex\">{tex}\\left.^{4} C_{3} \\quad a^{1} b^{3}+^{4} C_{4} \\quad b^{4}\\right]{\/tex}<\/span>\u00a0&#8211;\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{llll} ^{4} C_{0} a^{4}-^{4} C_{1} a^{3} b+^{4} C_{2} a^{2} b^{2} \\end{array}^{2}\\right.{\/tex}<\/span>\u00a0&#8211;\u00a0<span class=\"math-tex\">{tex}\\left.^{-4} C_{3} a^{1} b^{3}+^{4} C_{4} b^{4}\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}2\\left[^{4} C_{1}a^{3} b+^{4} C_{3} a^{1} b^{3}\\right]{\/tex}<\/span><br \/>\n= 2[4a<sup>3<\/sup>b + 4a<sup>1<\/sup>b<sup>3<\/sup>] = 8ab[a<sup>2<\/sup> + b<sup>2<\/sup>]\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}(\\sqrt{5}+1)^{4}-(\\sqrt{5}-1)^{4}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}8 \\cdot \\sqrt{5} \\cdot 1\\left[(\\sqrt{5})^{2}+1^{2}\\right]{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}48 \\sqrt{5}{\/tex}<\/span><br \/>\nHence\u00a0<span class=\"math-tex\">{tex}\\sqrt{5}\\left\\{(\\sqrt{5}+1)^{4}-(\\sqrt{5}-1)^{4}\\right\\}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\sqrt{5} .48 \\sqrt{5}{\/tex}<\/span>\u00a0= 240<\/li>\n<li>(c)\n<p style=\"display: inline;\">ac\u00a0<span class=\"math-tex\">{tex} \\geq{\/tex}<\/span>\u00a0bc<\/p>\n<p><b>Explanation: <\/b>The sign of the inequality is to be reversed\u00a0(<span class=\"math-tex\">{tex}\\le{\/tex}<\/span>\u00a0to\u00a0<span class=\"math-tex\">{tex}\\ge{\/tex}<\/span>\u00a0or\u00a0<span class=\"math-tex\">{tex}\\ge{\/tex}<\/span>\u00a0to\u00a0<span class=\"math-tex\">{tex}\\le{\/tex}<\/span>)\u00a0if both sides of an inequality are multiplied by the same negative real number.<\/li>\n<li>(a)\n<p style=\"display: inline;\">C =\u00a0<span class=\"math-tex\">{tex}\\phi {\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\phi {\/tex}<\/span> is denoted as null set.<\/li>\n<li>(d)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{-56}{65}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>In quadrant IV,\u00a0<span class=\"math-tex\">{tex}\\cos \\theta&gt;0, \\sin \\theta&lt;0{\/tex}<\/span>,\u00a0<span class=\"math-tex\">{tex}\\cos \\phi&gt;0{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\sin \\phi&lt;0{\/tex}<\/span><br \/>\nNow,\u00a0<span class=\"math-tex\">{tex}\\cos \\theta=\\frac{4}{5} \\Rightarrow \\sin ^{2} \\theta{\/tex}<\/span><span class=\"math-tex\">{tex}=\\left(1-\\cos ^{2} \\theta\\right){\/tex}<\/span><span class=\"math-tex\">{tex}=\\left(1-\\frac{16}{25}\\right)=\\frac{9}{25}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\sin \\theta=-\\sqrt{\\frac{9}{25}}=\\frac{-3}{5}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\cos \\phi=\\frac{12}{13} \\Rightarrow \\sin ^{2} \\theta{\/tex}<\/span><span class=\"math-tex\">{tex}=\\left(1-\\cos ^{2} \\phi\\right){\/tex}<\/span><span class=\"math-tex\">{tex}=\\left(1-\\frac{144}{169}\\right)=\\frac{25}{169}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\sin \\phi=-\\sqrt{\\frac{25}{169}}=\\frac{-5}{13}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\sin (\\theta+\\phi)=\\sin \\theta{\/tex}<\/span><span class=\"math-tex\">{tex} \\cos \\phi+\\cos\\theta \\sin \\phi{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\left(\\frac{-3}{5} \\times \\frac{12}{13}\\right)+\\left(\\frac{-5}{13} \\times \\frac{4}{5}\\right){\/tex}<\/span><span class=\"math-tex\">{tex}=\\left(\\frac{-36}{65}-\\frac{20}{65}\\right)=\\frac{-56}{65}{\/tex}<\/span><\/p>\n<p><strong>Enhance Your Exam Preparation with myCBSEguide and Examin8 Apps<\/strong><\/p>\n<p>To boost your exam preparation, download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide.com<\/strong><\/a> app, your one-stop solution for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. The app offers <strong>complete study material<\/strong>, including <strong>practice questions<\/strong>, <strong>sample papers<\/strong>, <strong>NCERT solutions<\/strong>, and <strong>chapter-wise test papers<\/strong>, all designed to help you achieve top scores. By regularly practicing <strong>CBSE Sample Papers Class 11 Mathematics<\/strong>, students can build confidence and improve their time-management skills.<\/p>\n<p>For <strong>teachers<\/strong>, the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\"><strong>Examin8.com<\/strong><\/a> app provides a powerful tool to create <strong>customized exam papers<\/strong> with their own <strong>name<\/strong> and <strong>logo<\/strong>, making it easy to assess and track students&#8217; progress. Whether you&#8217;re a student aiming for high marks or a teacher designing unique assessments, both <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a> Website and <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a> Website offer the resources you need for success. The <strong>CBSE Sample Papers Class 11 Mathematics<\/strong> align with the latest syllabus and marking scheme, making them the perfect tool for exam preparation.<\/p>\n<p>(b)<\/li>\n<li>\n<p style=\"display: inline;\">arg (z<sub>1<\/sub>) = arg (z<sub>2<\/sub>)<\/p>\n<p><b>Explanation: <\/b>Let z<sub>1<\/sub>\u00a0= r<sub>1<\/sub>\u00a0(cos <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>1<\/sub>\u00a0+ i sin <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>1<\/sub>) and z<sub>2<\/sub>\u00a0= r<sub>2<\/sub>\u00a0(cos <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>2<\/sub>\u00a0+ i sin <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>2<\/sub>)<br \/>\nSince |z<sub>1<\/sub>\u00a0+ z<sub>2<\/sub>| = |z<sub>1<\/sub>| + |z<sub>2<\/sub>|<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0z<sub>1<\/sub>\u00a0+ z<sub>2<\/sub>\u00a0= r<sub>1<\/sub>\u00a0cos <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>1<\/sub>\u00a0+ ir<sub>1<\/sub>\u00a0sin <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>1<\/sub>+ r<sub>2<\/sub>\u00a0cos <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>2<\/sub>\u00a0+ ir<sub>2<\/sub>\u00a0sin <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>2<\/sub><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0|z<sub>1<\/sub> + z<sub>2<\/sub>|<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\sqrt{\\mathrm{r}_{1}^{2} \\cos ^{2} \\theta_{1}+\\mathrm{r}_{2}^{2} \\cos ^{2} \\theta_{2}+2 \\mathrm{r}_{1} \\mathrm{r}_{2} \\cos \\theta_{1} \\cos \\theta_{2}+\\mathrm{r}_{1}^{2} \\sin ^{2} \\theta_{1}+\\mathrm{r}_{2}^{2} \\sin ^{2} \\theta_{2}+2 \\mathrm{r}_{1} \\mathrm{r}_{2} \\sin \\theta_{1} \\sin \\theta_{2}}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\sqrt{\\mathrm{r}_{1}^{2}+\\mathrm{r}_{2}^{2}+2 \\mathrm{r}_{1} \\mathrm{r}_{2} \\cos \\left(\\theta_{1}-\\theta_{2}\\right)}{\/tex}<\/span><br \/>\nBut |z<sub>1<\/sub>\u00a0+ z<sub>2<\/sub>| = |z<sub>1<\/sub>| + |z<sub>2<\/sub>|<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\sqrt{\\mathrm{r}_{1}^{2}+\\mathrm{r}_{2}^{2}+2 \\mathrm{r}_{1} \\mathrm{r}_{2} \\cos \\left(\\theta_{1}-\\theta_{2}\\right)}=\\mathrm{r}_{1}+\\mathrm{r}_{2}{\/tex}<\/span><br \/>\nSquaring both sides,<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \\cos \\left(\\theta_{1}-\\theta_{2}\\right)=r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02r<sub>1<\/sub>r<sub>2<\/sub>\u00a0\u2013 2r<sub>1<\/sub>r<sub>2<\/sub>\u00a0cos (<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>1<\/sub>\u00a0\u2013 <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>2<\/sub>) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a01 \u2013 cos (<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>1<\/sub>\u00a0\u2013 <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>2<\/sub>) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0cos\u00a0(<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>1<\/sub>\u00a0\u2013 <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>2<\/sub>) = 1<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>1<\/sub>\u00a0\u2013 <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>2<\/sub>) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>1<\/sub>\u00a0= <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><sub>2<\/sub><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0arg (z<sub>1<\/sub>) = arg (z<sub>2<\/sub>)<\/li>\n<li>(d)\n<p style=\"display: inline;\">5<\/p>\n<p><b>Explanation: <\/b>If you join every distinct pair of vertices you will get <span class=\"math-tex\">{tex}^{ n }{ C_{ 2\\quad } }{\/tex}<\/span><span style=\"font-size: 0.9em;\">\u00a0lines.<\/span><br \/>\nThese <span class=\"math-tex\">{tex}^{ n }{ C_{ 2\\quad } }{\/tex}<\/span><span style=\"font-size: 0.9em;\">\u00a0lines account for the n sides of the polygon as well as for the diagonals.<\/span><\/p>\n<p>So the number of diagonals is given by\u00a0<span class=\"math-tex\">{tex}^{n} C_{2}-n=\\frac{n(n-1)}{2}-n=\\frac{n(n-3)}{2}{\/tex}<\/span><\/p>\n<p>But given number of \u00a0sides = number of diagonals =n<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow n=\\frac{n(n-3)}{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow 2 n=n(n-3){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow n-3=2{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow n=5{\/tex}<\/span><\/li>\n<li>(b)\n<p style=\"display: inline;\">Both A and R are true but R is not the correct explanation of A.<\/p>\n<p><b>Explanation: <\/b><strong>Assertion:<\/strong><br \/>\n(1 + x)<sup>n <\/sup>= <span class=\"math-tex\">{tex}n_{c_0}+n_{c_1}x + n_{c_2} x^2 \\ldots+n_{c_n} x^n{\/tex}<\/span><br \/>\n<strong>Reason:<\/strong><br \/>\n(1 + (-1))<sup>n <\/sup>= <span class=\"math-tex\">{tex}n_{c_0}{ 1}^n+n_{c_1}(1)^{n-1}(-1)^1+n_{c_2}(1)^{n-2}(-1)^2{\/tex}<\/span> +&#8230; + <span class=\"math-tex\">{tex}{ }^n c_n(1)^{n-n}(-1)^n{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}n_{c_8}-n_{c_1}+n_{c_2}-n_{c_3}{\/tex}<\/span> + &#8230; (-1)<sup>n<\/sup><span class=\"math-tex\">{tex}n_{c_n}{\/tex}<\/span><br \/>\nEach term will cancel each other<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> (1 + (-1))<sup>n<\/sup> = 0<br \/>\nReason is also the but not the correct explanation of Assertion.<\/li>\n<li>(c)\n<p style=\"display: inline;\">A is true but R is false.<\/p>\n<p><b>Explanation: <\/b><strong>Assertion <\/strong>Mean of the given series<br \/>\n<span class=\"math-tex\">{tex}\\bar{x} =\\frac{\\text { Sum of terms }}{\\text { Number of terms }}=\\frac{\\sum x_i}{n}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{4+7+8+9+10+12+13+17}{8}{\/tex}<\/span> = 10<\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\">xi<\/th>\n<th scope=\"col\">|xi &#8211; <span class=\"math-tex\">{tex}\\bar x{\/tex}<\/span>|<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">|4 &#8211; 10| = 6<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">7<\/td>\n<td style=\"text-align: center;\">|7 &#8211; 10| = 3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">8<\/td>\n<td style=\"text-align: center;\">|8 &#8211; 10| = 2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">9<\/td>\n<td style=\"text-align: center;\">|9 &#8211; 10| = 1<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">10<\/td>\n<td style=\"text-align: center;\">|10 &#8211; 10| = 0<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">12<\/td>\n<td style=\"text-align: center;\">|12 &#8211; 10| = 2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">13<\/td>\n<td style=\"text-align: center;\">|13 &#8211; 10| = 3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">17<\/td>\n<td style=\"text-align: center;\">|17 &#8211; 10| = 7<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\sum x_i{\/tex}<\/span> = 80<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\sum \\left|x_i-\\bar{x}\\right|{\/tex}<\/span> = 24<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> Mean deviation about mean<br \/>\n<span class=\"math-tex\">{tex}=\\frac{\\Sigma\\left|x_i-\\bar{x}\\right|}{n}=\\frac{24}{8}{\/tex}<\/span> = 3<br \/>\n<strong>Reason<\/strong> Mean of the given series<br \/>\n<span class=\"math-tex\">{tex}\\bar{x}= \\frac{\\text { Sum of terms }}{\\text { Number of terms }}=\\frac{\\sum x_i}{n}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\u00a0\\frac{38+70+48+40+42+55}{+63+46+54+44}{\/tex}<\/span> = 50<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> Mean deviation about mean<br \/>\n<span class=\"math-tex\">{tex}= \\frac{\\Sigma\\left|x_i-\\bar{x}\\right|}{n}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{84}{10}{\/tex}<\/span> = 8.4<br \/>\nHence, Assertion is true and Reason is false.<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section B <\/b><\/li>\n<li style=\"clear: both;\"><img decoding=\"async\" style=\"width: 150px; height: 111px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/4QKPGAU.png\" alt=\"\" data-imgur-src=\"4QKPGAU.png\" \/><br \/>\nWe know that when 0 &lt; a &lt; 1, the function a<sup>x<\/sup> is strictly decreasing, i.e., different values of x give different values of a<sup>x<\/sup>. So, the function is one-one. Also, its range is R. So, it is onto.<br \/>\nThus, the function a<sup>x<\/sup> is invertible. Its inverse is the log function reflected by the line y = x, as shown. The graph clearly passes through (1, 0).<br \/>\nRequired graph of the function shown in the above fig.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Given, A = {1, 2}, B = {2, 3, 4} and C = {4, 5}<br \/>\nB\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0C =\u00a0{2, 3, 4} <span class=\"math-tex\">{tex}\\cup{\/tex}<\/span> {4, 5}<br \/>\n= {2, 3, 4, 5}<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0A\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> (B <span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0C) =\u00a0{1, 2} <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0{2, 3, 4, 5}<br \/>\n=\u00a0{(1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5)}<\/li>\n<li>Derivative of a function f(x) at any real number a is given by f'(a)=\u00a0<span class=\"math-tex\">{tex}\\lim \\limits_{h \\rightarrow 0}{\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{f(a+h)-f(a)}{h}{\/tex}<\/span>\u00a0\u00a0{where h is a very small positive number}<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>derivative of cos x at x = 0 is given as<br \/>\nf'(0) =\u00a0<span class=\"math-tex\">{tex}\\lim \\limits_{h \\rightarrow 0}{\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{\\mathrm{f}(0+\\mathrm{h})-\\mathrm{f}(0)}{\\mathrm{h}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0f&#8217; (0) =\u00a0<span class=\"math-tex\">{tex}\\lim \\limits_{h \\rightarrow 0}{\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{\\tan (\\mathrm{h})-\\tan 0}{\\mathrm{h}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0f&#8217; (0) =\u00a0<span class=\"math-tex\">{tex}\\lim \\limits_{h \\rightarrow 0}{\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{\\tanh }{\\mathrm{h}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0we can\u2019t find the limit by direct substitution as it gives <span class=\"math-tex\">{tex}\\frac 00{\/tex}<\/span> (indeterminate form)<br \/>\nUse the formula:\u00a0<span class=\"math-tex\">{tex}\\lim \\limits_{x \\rightarrow 0}{\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{\\tan x }{\\mathrm{x}}{\/tex}<\/span>\u00a0= 1\u00a0{sandwich theorem}<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0f\u2019(0) = 1<\/li>\n<li style=\"clear: both;\">Given that directrix = 0 and focus (6, 0)<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0by definition,<br \/>\nThe equation of the parabola is<br \/>\nMP<sup>2<\/sup> = PF<sup>2<\/sup><br \/>\n(x \u2013 6)<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0= x<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x<sup>2<\/sup>\u00a0+ 36 \u2013 12x + y<sup>2<\/sup>\u00a0= x<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>y<sup>2<\/sup>\u00a0\u2013 12x + 36 = 0<br \/>\nHence, the required equations is y<sup>2<\/sup>\u00a0\u2013 12x + 36 = 0<br \/>\n<img decoding=\"async\" id=\"Picture 13\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncertexem\/11\/math\/ch-11\/image173.png\" \/><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Let A (x<sub>1<\/sub>, y<sub>1<\/sub>), B (x<sub>2<\/sub>, y<sub>2<\/sub>) and C (x<sub>3<\/sub>,\u00a0y<sub>3<\/sub>) vertices of triangle ABC.<br \/>\nSince (x<sub>1<\/sub>, y<sub>1<\/sub>), (x<sub>2<\/sub>, y<sub>2<\/sub>) and (x<sub>3<\/sub>, y<sub>3<\/sub>) lie on the parabola.<br \/>\nHence,\u00a0<span class=\"math-tex\">{tex}y_{1}^{2}=4 a x_{1}, y_{2}^{2}=4 a x_{2}{\/tex}<\/span>\u00a0\u00a0and\u00a0<span class=\"math-tex\">{tex}y_{3}^{2}=4 a x_{3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}x_{1}=\\frac{y_{1}^{2}}{4 a}{\/tex}<\/span>,\u00a0<span class=\"math-tex\">{tex}x_{2}=\\frac{y_{2}^{2}}{4 a}{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}x_{3}=\\frac{y_{3}^{2}}{4 a}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Area of <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABC =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0[x<sub>1<\/sub> (y<sub>2<\/sub> &#8211; y<sub>3<\/sub>) + x<sub>2<\/sub> (y<sub>3<\/sub> &#8211; y<sub>1<\/sub>) + x<sub>3<\/sub> (y<sub>1<\/sub> &#8211; y<sub>2<\/sub>)]\n=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left[\\frac{y_{1}^{2}}{4 a}\\left(y_{2}-y_{3}\\right)\\right.{\/tex}<\/span>\u00a0+\u00a0<span class=\"math-tex\">{tex}\\frac{y_{2}^{2}}{4 a}\\left(y_{3}-y_{1}\\right){\/tex}<\/span>\u00a0+\u00a0<span class=\"math-tex\">{tex}\\left.\\frac{y_{3}^{2}}{4 a}\\left(y_{1}-y_{2}\\right)\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{8 a}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left[y_{1}^{2}\\left(y_{2}-y_{3}\\right)\\right.{\/tex}<\/span>\u00a0+\u00a0<span class=\"math-tex\">{tex}\\left(y_{2}^{2} y_{3}-y_{2} y_{3}^{2}\\right){\/tex}<\/span>\u00a0&#8211;\u00a0<span class=\"math-tex\">{tex}\\left.y_{1}\\left(y_{2}^{2}-y_{3}^{2}\\right)\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{8 a}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left[y_{1}^{2}\\left(y_{2}-y_{3}\\right)+y_{2} y_{3}\\right.{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left.\\left(y_{2}-y_{3}\\right)-y_{1}\\left(y_{2}^{2}-y_{3}^{2}\\right)\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{8 a}{\/tex}<\/span>\u00a0(y<sub>2<\/sub>\u00a0&#8211; y<sub>3<\/sub>)<span class=\"math-tex\">{tex}\\left[y_{1}^{2}+y_{2} y_{3}-y_{1}\\left(y_{2}+y_{3}\\right)\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{8 a}{\/tex}<\/span>\u00a0(y<sub>2<\/sub>\u00a0&#8211; y<sub>3<\/sub>)\u00a0<span class=\"math-tex\">{tex}\\left[\\left(y_{1}^{2}-y_{1} y_{2}\\right)+\\left(y_{2} y_{3}-y_{1} y_{3}\\right)\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{8 a}{\/tex}<\/span>\u00a0(y<sub>2<\/sub>\u00a0&#8211; y<sub>3<\/sub>)\u00a0[y<sub>1<\/sub>\u00a0(y<sub>1<\/sub>\u00a0&#8211; y<sub>2<\/sub>) &#8211; y<sub>3<\/sub> (y<sub>1<\/sub>\u00a0&#8211; y<sub>2<\/sub>)]\n=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{8 a}{\/tex}<\/span>\u00a0(y<sub>2<\/sub> &#8211; y<sub>3<\/sub>) (y<sub>1<\/sub>\u00a0&#8211; y<sub>2<\/sub>) (y<sub>1<\/sub>\u00a0&#8211; y<sub>3<\/sub>)<br \/>\n=\u00a0&#8211;\u00a0<span class=\"math-tex\">{tex}\\frac{1}{8 a}{\/tex}<\/span>\u00a0(y<sub>1<\/sub>\u00a0&#8211; y<sub>2<\/sub>) (y<sub>2<\/sub> &#8211; y<sub>3<\/sub>) (y<sub>3<\/sub> &#8211; y<sub>1<\/sub>)<br \/>\nTherefore, Area of <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABC\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{8 a}{\/tex}<\/span>\u00a0| (y<sub>1<\/sub>\u00a0&#8211; y<sub>2<\/sub>) (y<sub>2<\/sub> &#8211; y<sub>3<\/sub>) (y<sub>3<\/sub> &#8211; y<sub>1<\/sub>) |<\/li>\n<li>According to the question, we can write,<br \/>\nA = {x : x <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> N, x is a multiple of 3}<br \/>\n= {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, &#8230;}<br \/>\nB = {x : x <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> N and x is a multiple of 5}<br \/>\n= {5, 10, 15, 20, 25, 30, 35, 40, 45,&#8230;}<br \/>\nThus, we have<br \/>\nA\u00a0<span class=\"math-tex\">{tex}\\cap{\/tex}<\/span>\u00a0B = {15, 30, 45, &#8230;}<br \/>\n= {x : x\u00a0<span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0N, where x is a multiple of 15}<\/li>\n<li>The point of intersection of the line ax + by + c = 0\u00a0with the coordinate axis are\u00a0<span class=\"math-tex\">{tex}\\left(-\\frac{c}{a}, 0\\right){\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\left(0,-\\frac{c}{b}\\right){\/tex}<\/span>.<br \/>\nTherefore, the vertices of the triangle are\u00a0\u00a0<span class=\"math-tex\">{tex}\\left(-\\frac{c}{a}, 0\\right){\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\left(0,-\\frac{c}{b}\\right){\/tex}<\/span>.<br \/>\nSuppose A\u00a0be the area of the required triangle.<br \/>\n<span class=\"math-tex\">{tex}A=\\frac{1}{2}\\left|\\begin{array}{ccc} 0 &amp; 0 &amp; 1 \\\\ \\frac{-c}{a} &amp; 0 &amp; 1 \\\\ 0 &amp; \\frac{-c}{b} &amp; 1 \\end{array}\\right|{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}A=\\frac{1}{2}\\left|-\\frac{c}{a} \\times \\frac{-c}{b}\\right|{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}=\\frac{1}{2}\\left|\\frac{c^{2}}{a b}\\right|{\/tex}<\/span><br \/>\nIt is given that a, b and c are in GP.<br \/>\n<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span>\u00a0b<sup>2<\/sup> = ac<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow A=\\frac{1}{2}\\left|\\frac{b^{4}}{a^{2} \\times a b}\\right|{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}=\\frac{1}{2}\\left|\\frac{b}{a}\\right|^{3}{\/tex}<\/span>\u00a0square units.<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section C <\/b><\/li>\n<li>We have,<br \/>\nA = {1, 2, 3} and R {(1, 2) (2, 3)}<br \/>\nNow,<br \/>\nTo make R reflexive, we will add (1, 1) (2, 2) and (3, 3) to get<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0R1 = {(1, 2), (2, 3), (1, 1), (2, 2), (3, 3)} is reflexive<br \/>\nAgain to make R&#8217; symmetric we shall add (3, 2) and (2, 1)<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0R&#8221; = {(1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (3, 2), (2, 1)} is reflexive and symmetric<br \/>\nNow,<br \/>\nTo R&#8221; transitive we shall add (1, 3) and (3, 1)<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0R&#8221;&#8216; = {(1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (3, 2), (2, 1)} is reflexive and symmetric<br \/>\nNow,<br \/>\nTo make R&#8221; transitive we shall add (1, 3) and (3, 1)<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0R&#8221;&#8216; {(1, 2), (2, 3), (1, 1) ,(2, 2), (3, 3), (3, 2), (2, 1), (1, 3) ,(3, 1)}<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0R&#8221;&#8216; is reflexive, symetric and transitive.<\/li>\n<li>Here <span class=\"math-tex\">{tex}\\frac{{(2x &#8211; 1)}}{3} \\geqslant \\frac{{(3x &#8211; 2)}}{4} &#8211; \\frac{{(2 &#8211; x)}}{5}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\frac{{2x}}{3} &#8211; \\frac{1}{3} \\geqslant \\frac{{3x}}{4} &#8211; \\frac{2}{4} &#8211; \\frac{2}{5} + \\frac{x}{5}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\frac{{2x}}{3} &#8211; \\frac{3x}{4}-\\frac{x}{5}\\geqslant \\ \\frac{-2}{4} &#8211; \\frac{2}{5} + \\frac{1}{3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\frac{{40x &#8211; 45x &#8211; 12x}}{{60}} \\geqslant \\frac{{ &#8211; 30 &#8211; 24 + 20}}{{60}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\frac{{ &#8211; 17x}}{{60}} \\geqslant \\frac{{ &#8211; 34}}{{60}}{\/tex}<\/span><br \/>\nMultiplying both sides by 60, we have<br \/>\n<span class=\"math-tex\">{tex} &#8211; 17x \\geqslant &#8211; 34{\/tex}<\/span><br \/>\nDividing both sides by -17,\u00a0we have<br \/>\n<span class=\"math-tex\">{tex}\\frac{{ &#8211; 17x}}{{ &#8211; 17}} \\leqslant \\frac{{ &#8211; 34}}{{ &#8211; 17}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow x \\leqslant 2{\/tex}<\/span><br \/>\nThus the solution set is <span class=\"math-tex\">{tex}\\left( { &#8211; \\infty ,2} \\right]{\/tex}<\/span><\/li>\n<li style=\"clear: both;\">Here P(2a, 2, 6), <span class=\"math-tex\">{tex}Q( &#8211; 4,3b, &#8211; 10){\/tex}<\/span> and R(8, 14, 2c) are vertices of triangle PQR.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Coordinates of centriod of <span class=\"math-tex\">{tex}\\Delta PQR{\/tex}<\/span> is <span class=\"math-tex\">{tex}\\left( {\\frac{{2a &#8211; 4 + 8}}{3},\\frac{{2 + 3b + 14}}{3},\\frac{{6 &#8211; 10 + 2c}}{3}} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\left( {\\frac{{2a +4}}{3},\\frac{{3b + 16}}{3},\\frac{{2c &#8211; 4}}{3}} \\right){\/tex}<\/span><br \/>\nBut is it given that coordinates of centroid is (0, 0, 0)<br \/>\n<span class=\"math-tex\">{tex}\\frac{{2a + 4}}{3} = 0 \\Rightarrow{\/tex}<\/span> 2a + 4 = 0 <span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> a = -2<br \/>\n<span class=\"math-tex\">{tex}\\frac{{3b + 16}}{3} = 0 \\Rightarrow{\/tex}<\/span> 3b + 16 = 0 <span class=\"math-tex\">{tex}\\Rightarrow b = \\frac{{ &#8211; 16}}{3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{{2c &#8211; 4}}{3} = 0 \\Rightarrow{\/tex}<\/span> 2c &#8211; 4 = 0 <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> c = 2<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Consider, C(x, y, z) point equidistant from points A(-1, 2, 3) and B(3, 2, 1).<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0AC = BC<br \/>\n<span class=\"math-tex\">{tex}\\sqrt{(x+1)^{2}+(y-2)^{2}+(z-3)^{2}}=\\sqrt{(x-3)^{2}+(y-2)^{2}+(z-1)^{2}}{\/tex}<\/span><br \/>\nSquaring both sides,<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> (x + 1)<sup>2<\/sup>\u00a0+ (y &#8211; 2)<sup>2<\/sup>\u00a0+ (z &#8211; 3)<sup>2<\/sup>\u00a0 = (x &#8211; 3)<sup>2<\/sup>\u00a0+ (y &#8211; 2)<sup>2<\/sup>\u00a0+ (z &#8211; 1)<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x<sup>2<\/sup>\u00a0+2x + 1 + y<sup>2<\/sup>\u00a0&#8211; 4y + 4 + z<sup>2<\/sup>\u00a0&#8211; 6z + 9 = x<sup>2<\/sup>\u00a0&#8211; 6x + 9 + y<sup>2<\/sup>\u00a0&#8211; 4y + 4\u00a0+ z<sup>2<\/sup>\u00a0-2z +1<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 8x &#8211; 4z = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 2x &#8211; z = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> z = 2x<br \/>\n<img decoding=\"async\" style=\"width: 200px; height: 184px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/KaP5bbO.png\" alt=\"\" data-imgur-src=\"KaP5bbO.png\" \/><br \/>\nEquation of curve is z = 2x<\/li>\n<li style=\"clear: both;\"><span class=\"math-tex\">{tex}\\style{font-family:Tahoma}{\\style{font-size:8px}{\\begin{array}{l}We\\;have\\\\\\left(x+y\\right)^5\\;+\\left(x-y\\right)^5\\;=\\;2\\;\\left[{}^5C_0\\;x^5\\;+^5C_2\\;x^3y^2\\;+^5C_4\\;x^1y^4\\right]\\;\\\\\\;\\;\\;\\;\\;\\;\\;\\;=\\;2\\;(x^5+10x^3y^2\\;+5xy^4\\\\Putting\\;\\;x\\;=\\;\\sqrt2\\;and\\;y\\;=\\;1,\\;we\\;get\\\\(\\sqrt2+1)^5+{(\\sqrt2-1{)^5\\;=\\;2\\left[\\left(\\sqrt2\\right)^5\\;+10\\left(\\sqrt2\\right)^3\\;+5\\sqrt2\\right]}}\\\\\\;\\;\\;\\;\\;=\\;2\\left[4\\;\\sqrt2\\;+20\\;\\sqrt2+\\;5\\sqrt2\\right]\\\\\\;\\;\\;=\\;58\\sqrt2\\\\\\\\\\\\\\\\\\end{array}}}{\/tex}<\/span>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>(x + 2y)<sup>8<\/sup> + (x &#8211; 2y)<sup>8<\/sup>\u00a0<span class=\"math-tex\">{tex} = 2{[^8}{C_0}{x^8}{ + ^8}{C_2}{x^5}{(2y)^2}{ + ^8}{C_4}{x^4}{(2y)^4}{\/tex}<\/span><span class=\"math-tex\">{tex}{ + ^8}{C_6}{x^2}{(2y)^6}{ + ^8}{C_8}{(2y)^8}]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\style{font-family:Tahoma}{\\style{font-size:8px}{\\begin{array}{l}\\because{(x+a)^n}+{(x-a)^n}=2\\lbrack^nC_0x^n+^nC_2x^{n-2}a^2+^nC_4x^{n-4}a^4+&#8230;&#8230;&#8230;.\\rbrack\\\\\\\\\\\\\\\\\\end{array}}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow {(x + 2y)^8} + {(x &#8211; 2y)^8}{\/tex}<\/span><span class=\"math-tex\">{tex} = 2[{x^8} + 28{x^6} \\times 4{y^2} + 70{x^4} \\times 16{y^4}{\/tex}<\/span><span class=\"math-tex\">{tex} + 28{x^2} \\times 64{y^6} + 256{y^8}]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 2[{x^8} + 112{x^6}{y^2} + 1120{x^4}{y^4}{\/tex}<\/span><span class=\"math-tex\">{tex} + 1792{x^2}{y^6} + 256{y^8}]{\/tex}<\/span><\/li>\n<li style=\"clear: both;\">We have<br \/>\n<span class=\"math-tex\">{tex}\\frac{x-1}{3+i}+\\frac{y-1}{3-i}=i{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{(x-1)}{(3+i)} \\times \\frac{(3-i)}{(3-i)}+\\frac{(y-1)}{(3-i)} \\times \\frac{(3+i)}{(3+i)}=i{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{(x-1)(3-i)}{\\left(9-i^{2}\\right)}+\\frac{(y-1)(3+i)}{\\left(9-i^{2}\\right)}=i{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{3(x-1)-i(x-1)}{10}+\\frac{3(y-1)+i(y-1)}{10}=i{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 3(x &#8211; 1) &#8211; i(x &#8211; 1) + 3(y &#8211; 1) + i(y &#8211; 1) = 10i<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> (3x &#8211; 3 + 3y &#8211; 3) + i(y &#8211; 1 &#8211; x + 1) = 10i<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> (3x + 3y &#8211; 6) + i(y &#8211; x) = 10i<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 3(x + y &#8211; 2 ) = 0 and y &#8211; x = 10 [equating real parts and the imaginary parts separately]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x + y &#8211; 2 = 0 and y &#8211; x = 10<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x + y = 2 an d -x + y = 10<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x = &#8211; 4 and y = 6 [by\u00a0solving x + y = 2 and\u00a0&#8211; x + y = 10].<br \/>\nHence, the Real value of\u00a0x and y\u00a0 is -4 and 6<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Let z = (x + iy). Then,<br \/>\nlzl + z = 2 + i<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> lx + iy| + (x + iy) = 2 + i<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\{\\sqrt{x^{2}+y^{2}}{\/tex}<\/span>\u00a0+ x} + iy = (2 + i)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\sqrt{x^{2}+y^{2}}+x = 2{\/tex}<\/span> and y = 1 [equating real parts and imaginary parts separately]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> y = 1 and <span class=\"math-tex\">{tex}\\sqrt{x^{2}+1^{2}}+x=2 \\Rightarrow y=1{\/tex}<\/span> and <span class=\"math-tex\">{tex}\\sqrt{x^{2}+1}=(2-x){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> y = and x<sup>2<\/sup> + 1 = (2 &#8211; x)<sup>2<\/sup> <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> y = 1 and x<sup>2<\/sup> + 1 = x<sup>2<\/sup> &#8211; 4x + 4<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 4x = 3 and y = 1 <span class=\"math-tex\">{tex}\\Rightarrow x=\\frac{3}{4}{\/tex}<\/span> and y = 1.<br \/>\nHence, <span class=\"math-tex\">{tex}z=\\left(\\frac{3}{4}+i\\right){\/tex}<\/span> is the desired solution.<\/li>\n<li>Here A = {3, 6,9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}A &#8211; B = {3,6,9,12,15,18,21} &#8211; {4,8,12,16,20}<br \/>\n= { 3, 6, 9, 15, 18, 21}<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section D <\/b><\/li>\n<li>Given that:Sample Space S = {e<sub>1<\/sub>, e<sub>2<\/sub>, e<sub>3<\/sub>, e<sub>4<\/sub>, e<sub>5,<\/sub>\u00a0e<sub>6<\/sub>, e<sub>7<\/sub>, e<sub>8<\/sub>, e<sub>9<\/sub>}<br \/>\nA = {e<sub>1<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>} and B = {e<sub>2<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>, e<sub>9<\/sub>}<br \/>\nP(e<sub>1<\/sub>) = P(e<sub>2<\/sub>) = .08, P(e<sub>3<\/sub>) = P(e<sub>4<\/sub>) = P(e<sub>5<\/sub>) = .1<br \/>\nP(e<sub>6<\/sub>) = P(e<sub>7<\/sub>) = .2, P(e<sub>8<\/sub>) = P(e<sub>9<\/sub>) = .07<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>To find: P(A), P(B) and P(A\u00a0<span class=\"math-tex\">{tex}\\cap{\/tex}<\/span>\u00a0B)\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>A = {e<sub>1<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>}<br \/>\nOn adding the probabilities of elements of A, we get<br \/>\nP(A) = P(e<sub>1<\/sub>) + P(e<sub>5<\/sub>) + P(e<sub>8<\/sub>)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0P(A) = 0.08 + 0.1 + 0.07 [given]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0P(A) = 0.25<\/li>\n<li>B = {e<sub>2<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>, e<sub>9<\/sub>}<br \/>\nSimilarly, on adding the probabilities of elements of B, we get<br \/>\nP(B) = P(e<sub>2<\/sub>) + P(e<sub>5<\/sub>) + P(e<sub>8<\/sub>) + P(e<sub>9<\/sub>)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0P(B) = 0.08 + 0.1 + 0.07 + 0.07 [given]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0P(B) = 0.32<\/li>\n<li>Now, we have to find P(A\u00a0<span class=\"math-tex\">{tex}\\cap{\/tex}<\/span>\u00a0B)<br \/>\nA = {e<sub>1<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>} and B = {e<sub>2<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>, e<sub>9<\/sub>}<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0A\u00a0\u22c2\u00a0B = {e<sub>5<\/sub>, e<sub>8<\/sub>}<br \/>\nOn adding the probabilities of elements of A\u00a0\u22c2\u00a0B, we get<br \/>\nP(A\u00a0\u22c2\u00a0B) = P(e<sub>5<\/sub>) + P(e<sub>8<\/sub>)<br \/>\n= 0.1 + 0.07 = 0.17<\/li>\n<\/ol>\n<\/li>\n<li>To find: P(A\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0B)\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>By General Addition Rule: P(A\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0B) = P(A) + P(B) \u2013 P(A\u00a0<span class=\"math-tex\">{tex}\\cap{\/tex}<\/span>\u00a0B)<br \/>\nfrom part (a), we have<br \/>\nP(A) = 0.25, P(B) = 0.32 and P(A\u00a0<span class=\"math-tex\">{tex}\\cap{\/tex}<\/span>\u00a0B) = 0.17<br \/>\nPutting the values, we get<br \/>\nP(A\u00a0<span class=\"math-tex\">{tex}\\cup{\/tex}<\/span>\u00a0B) = 0.25 + 0.32 \u2013 0.17 = 0.40<\/li>\n<\/ol>\n<\/li>\n<li>A = {e<sub>1<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>} and B = {e<sub>2<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>, e<sub>9<\/sub>}<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0A\u00a0\u22c3\u00a0B = {e<sub>1<\/sub>, e<sub>2<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>, e<sub>9<\/sub>}<br \/>\nAgain, on adding the probabilities of elements of A\u00a0\u22c3\u00a0B, we get<br \/>\nP(A\u00a0\u22c3\u00a0B) = P(e<sub>1<\/sub>) + P(e<sub>2<\/sub>) + P(e<sub>5<\/sub>) + P(e<sub>8<\/sub>) + P(e<sub>9<\/sub>)<br \/>\n= 0.08 +0.08 + 0.1 + 0.07 + 0.07 = 0.40<\/li>\n<li>To find: P(<span class=\"math-tex\">{tex}\\bar B{\/tex}<\/span>)\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>By Complement Rule, we have<br \/>\nP (<span class=\"math-tex\">{tex}\\bar B{\/tex}<\/span>) = 1 &#8211; P(<strong><span class=\"math-tex\">{tex}\\ B{\/tex}<\/span>)<\/strong><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0P (<span class=\"math-tex\">{tex}\\bar B{\/tex}<\/span>) = 1 &#8211; 0.32 = 0.68<\/li>\n<li>Given:\u00a0B = {e<sub>2<\/sub>, e<sub>5<\/sub>, e<sub>8<\/sub>, e<sub>9<\/sub>}<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\bar B{\/tex}<\/span>\u00a0=\u00a0{e<sub>1<\/sub>, e<sub>3<\/sub>, e<sub>4<\/sub>, e<sub>6<\/sub>, e<sub>7<\/sub>}<br \/>\nP(<span class=\"math-tex\">{tex}\\bar B{\/tex}<\/span>)\u00a0=\u00a0P(e<sub>1<\/sub>) + P(e<sub>3<\/sub>) +\u00a0P(e<sub>4<\/sub>)\u00a0+\u00a0P(e<sub>4<\/sub>)\u00a0+\u00a0P(e<sub>6<\/sub>) +\u00a0P(e<sub>17<\/sub>)<br \/>\n= 0.08 + 0.1 + 0.1 + 0.2 + 0.2 [given] = 0.68<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p><span id=\"cke_bm_2613E\" style=\"display: none;\">\u00a0<\/span><\/li>\n<li style=\"clear: both;\">Let f(x) = sec x. Then, f(x + h) = sec (x + h)<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{d}{dx}{\/tex}<\/span>\u00a0(f (x)) =\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{f(x + h) &#8211; f(x)}}{h}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{d}{dx}{\/tex}<\/span>\u00a0(f (x)) =\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{\\sec (x + h) &#8211; \\sec x}}{h}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{d}{dx}{\/tex}<\/span>\u00a0(f (x)) =\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{\\frac{1}{{\\cos (x + h)}} &#8211; \\frac{1}{{\\cos x}}}}{h}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{d}{dx}{\/tex}<\/span>\u00a0(f (x)) =\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{\\cos x &#8211; \\cos (x + h)}}{{h\\cos x\\cos (x + h)}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{d}{dx}{\/tex}<\/span>\u00a0(f (x)) =\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{2\\sin \\left( {\\frac{{x + x + h}}{2}} \\right)\\sin \\left( {\\frac{{x + h &#8211; x}}{2}} \\right)}}{{h\\cos x\\cos (x + h)}}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left[ {\\because \\cos C &#8211; \\cos D = 2\\sin \\left( {\\frac{{C + D}}{2}} \\right)\\sin \\left( {\\frac{{D &#8211; C}}{2}} \\right)} \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{d}{dx}{\/tex}<\/span>\u00a0(f (x)) =\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{2\\sin \\left( {\\frac{{2x + h}}{2}} \\right)\\sin \\left( {\\frac{h}{2}} \\right)}}{{h\\cos x\\cos (x + h)}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{d}{dx}{\/tex}<\/span>\u00a0(f (x)) =\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{\\sin \\left( {\\frac{{2x + h}}{2}} \\right)}}{{\\cos x\\cos (x + h)}} \\times \\mathop {\\lim }\\limits_{h \\to 0} \\frac{{\\sin (h\/2)}}{{(h\/2)}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{d}{dx}{\/tex}<\/span>\u00a0(f (x)) =\u00a0<span class=\"math-tex\">{tex}\\frac{\\sin x}{\\cos x \\cos x} \\times{\/tex}<\/span>\u00a01 = tan x sec x.\u00a0<span class=\"math-tex\">{tex}\\left[ {\\because \\mathop {\\lim }\\limits_{h \\to 0} \\frac{{\\sin (h\/2)}}{{(h\/2)}} = 1} \\right]{\/tex}<\/span><br \/>\nHence,\u00a0<span class=\"math-tex\">{tex}\\frac{d}{dx}{\/tex}<\/span>\u00a0(sec x) = sec x tan x.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>We have,<br \/>\n<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{x \\to \\sqrt {10} } \\frac{{\\sqrt {7 &#8211; 2x} &#8211; (\\sqrt 5 &#8211; \\sqrt 2 )}}{{{x^2} &#8211; 10}}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{x \\to \\sqrt {10} } \\frac{{\\sqrt {7 &#8211; 2x} &#8211; \\sqrt {{{(\\sqrt 5 &#8211; \\sqrt 2 )}^2}} }}{{{x^2} &#8211; 10}}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left(\\text { form } \\frac{0}{0}\\right){\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{x \\to \\sqrt {10} } \\frac{{\\sqrt {7 &#8211; 2x} &#8211; \\sqrt {7 &#8211; 2\\sqrt {10} } }}{{{x^2} &#8211; 10}}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left(\\text { form } \\frac{0}{0}\\right){\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{x \\to \\sqrt {10} } \\frac{{\\sqrt {7 &#8211; 2x} &#8211; \\sqrt {7 &#8211; 2\\sqrt {10} } }}{{{x^2} &#8211; 10}}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{{\\sqrt {7 &#8211; 2x} + \\sqrt {7 &#8211; 2\\sqrt {10} } }}{{\\sqrt {7 &#8211; 2x} + \\sqrt {7 &#8211; 2\\sqrt {10} } }}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{x \\to \\sqrt {10} }{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{{(7 &#8211; 2x) &#8211; (7 &#8211; 2\\sqrt {10} )}}{{(x &#8211; \\sqrt {10} )(x + \\sqrt {10} )\\left\\{ {\\sqrt {7 &#8211; 2x} + \\sqrt {7 &#8211; 2} \\sqrt {10} } \\right\\}}}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{x \\to \\sqrt {10} }{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{{ &#8211; 2x + 2\\sqrt {10} }}{{(x &#8211; \\sqrt {10} )(x + \\sqrt {10} )\\left\\{ {\\sqrt {7 &#8211; 2x} + \\sqrt {7 &#8211; 2} \\sqrt {10} } \\right\\}}}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{x \\to \\sqrt {10} }{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{{ &#8211; 2(x &#8211; \\sqrt {10} )}}{{(x &#8211; \\sqrt {10} )(x + \\sqrt {10} )\\{ \\sqrt {7 &#8211; 2x} + \\sqrt {7 &#8211; 2\\sqrt {10} } \\} }}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{x \\to \\sqrt {10} }{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{{ &#8211; 2}}{{(x + \\sqrt {10} )\\left\\{ {\\sqrt {7 &#8211; 2x} + \\sqrt {7 &#8211; 2\\sqrt {10} } } \\right\\}}}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\mathop {\\lim }\\limits_{x \\to \\sqrt {10} }{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{{ &#8211; 2}}{{2\\sqrt {10} \\left\\{ {\\sqrt {7 &#8211; 2\\sqrt {10} } + \\sqrt {7 &#8211; 2\\sqrt {10} } } \\right\\}}}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{{ &#8211; 1}}{{\\sqrt {10} \\times 2 \\times \\sqrt {7 &#8211; 2\\sqrt {10} } }}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{{ &#8211; 1}}{{2\\sqrt {10} (\\sqrt 5 &#8211; \\sqrt 2 )}}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left[\\because(\\sqrt{5}-\\sqrt{2})^{2}=7-2 \\sqrt{10}\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{{ &#8211; 1}}{{2\\sqrt {10} }} \\times \\frac{{(\\sqrt 5 + \\sqrt 2 )}}{3}{\/tex}<\/span>\u00a0= &#8211;<span class=\"math-tex\">{tex}\\frac{(\\sqrt{5}+\\sqrt{2})}{6 \\sqrt{10}}{\/tex}<\/span><\/li>\n<li>Let the three numbers be\u00a0<span class=\"math-tex\">{tex}\\frac { a } { r }{\/tex}<\/span>, a and ar.<br \/>\nTheir sum =\u00a0<span class=\"math-tex\">{tex}\\frac { a } { r }{\/tex}<\/span>\u00a0+ a + ar =\u00a0<span class=\"math-tex\">{tex}\\frac { 13 } { 12 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}a \\left[ \\frac { 1 } { r } + 1 + r \\right] = \\frac { 13 } { 12 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}a \\left[ \\frac { 1 + r + r ^ { 2 } } { r } \\right] = \\frac { 13 } { 12 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a (1 + r + r<sup>2<\/sup>) =\u00a0<span class=\"math-tex\">{tex}\\frac { 13 } { 12 } r{\/tex}<\/span>\u00a0&#8230;(i)<br \/>\nTheir product =\u00a0<span class=\"math-tex\">{tex}\\frac { a } { r }{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0a<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> ar = -1\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a<sup>3<\/sup>\u00a0= -1<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a = -1 [taking cube root on both sides] &#8230;(ii)<br \/>\nOn putting the value of a in Eq. (i), we get<br \/>\n(- 1) [1 + r + r<sup>2<\/sup>] =\u00a0<span class=\"math-tex\">{tex}\\frac { 13 } { 12 } r{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0&#8211; 1 &#8211; r &#8211; r<sup>2<\/sup>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac { 13 } { 12 } r{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0&#8211; 12 &#8211; 12r &#8211; 12r<sup>2<\/sup>\u00a0= 13r<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a012r<sup>2<\/sup>\u00a0+ 25r + 12 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a012r<sup>2\u00a0<\/sup>+ 16r + 9r + 12 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a04r(3r + 4) + 3(3r + 4) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(4r + 3) (3r + 4) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0Either 3r + 4 = 0 or 4r + 3 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0r =\u00a0<span class=\"math-tex\">{tex}-\\frac { 4 } { 3 }{\/tex}<\/span>\u00a0or r =\u00a0<span class=\"math-tex\">{tex}\\frac { &#8211; 3 } { 4 }{\/tex}<\/span><br \/>\nWhen a\u00a0 = &#8211; 1 and r = &#8211;\u00a0<span class=\"math-tex\">{tex}\\frac { 4 } { 3 }{\/tex}<\/span>, then the numbers are<br \/>\n<span class=\"math-tex\">{tex}\\frac { &#8211; 1 } { &#8211; 4 \/ 3 } , &#8211; 1 , &#8211; 1 \\times \\frac { &#8211; 4 } { 3 } i . e . , \\frac { 3 } { 4 } , &#8211; 1 , \\frac { 4 } { 3 }{\/tex}<\/span><br \/>\nAnd when a = &#8211; 1 and r = &#8211;\u00a0<span class=\"math-tex\">{tex}\\frac { 3 } { 4 }{\/tex}<\/span>, then the numbers are<br \/>\n<span class=\"math-tex\">{tex}\\frac { &#8211; 1 } { &#8211; 3 \/ 4 } , &#8211; 1 , &#8211; 1 \\times \\frac { &#8211; 3 } { 4 } \\text { i.e., } \\frac { 4 } { 3 } , &#8211; 1 , \\frac { 3 } { 4 }{\/tex}<\/span><\/li>\n<li style=\"clear: both;\">LHS = tan (<span class=\"math-tex\">{tex}\\alpha{\/tex}<\/span> &#8211; <span class=\"math-tex\">{tex}\\beta{\/tex}<\/span>) =\u00a0<span class=\"math-tex\">{tex}\\frac{\\sin 2 \\beta}{5-\\cos 2 \\beta}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\tan \\alpha-\\tan \\beta}{1+\\tan \\alpha \\tan \\beta}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\frac{3}{2} \\tan \\beta-\\tan \\beta}{1+\\frac{3}{2} \\tan \\beta \\tan \\beta}{\/tex}<\/span>\u00a0&#8230; [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a02 tan<span class=\"math-tex\">{tex}\\alpha{\/tex}<\/span>\u00a0= 3 tan<span class=\"math-tex\">{tex}\\beta{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0tan<span class=\"math-tex\">{tex}\\alpha{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac{3}{2}{\/tex}<\/span>tan<span class=\"math-tex\">{tex}\\alpha{\/tex}<\/span>]\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\tan \\beta\\left(\\frac{3}{2}-1\\right)}{1+\\frac{3}{2} \\tan ^{2} \\beta}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\frac{1}{2} \\tan \\beta}{1+\\frac{3}{2} \\tan ^{2} \\beta}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\frac{1}{2} \\frac{\\sin \\beta}{\\cos \\beta}}{1+\\frac{3}{2} \\cdot\\left(\\frac{\\sin \\beta}{\\cos \\beta}\\right)^{2}}{\/tex}<\/span>\u00a0&#8230; [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0tan<span class=\"math-tex\">{tex}\\beta{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac{\\sin \\beta}{\\cos \\beta}{\/tex}<\/span>]\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\frac{\\sin \\beta}{2 \\cos \\beta}}{1+\\frac{3 \\sin ^{2} \\beta}{2 \\cos ^{2} \\beta}}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\frac{\\sin \\beta}{2 \\cos \\beta}}{\\frac{2 \\cos ^{2} \\beta+3 \\sin ^{2} \\beta}{2 \\cos ^{2} \\beta}}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{2 \\cos ^{2} \\beta \\sin \\beta}{2 \\cos \\beta\\left(2 \\cos ^{2} \\beta+3 \\sin ^{2} \\beta\\right)}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{2 \\cos \\beta \\sin \\beta}{2\\left(2 \\cos ^{2} \\beta+3 \\sin ^{2} \\beta\\right)}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\sin 2 \\beta}{2\\left(2 \\cos ^{2} \\beta\\right)+3\\left(2 \\sin ^{2} \\beta\\right)}{\/tex}<\/span>\u00a0.. {<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0sin 2x = 2(sin x)(cos x)}<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\sin 2 \\beta}{2(1+\\cos 2 \\beta)+3(1-\\cos 2 \\beta)}{\/tex}<\/span>\u00a0&#8230; {<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a02 cos<sup>2<\/sup>\u00a0x = 1 + cos 2x and\u00a02 sin<sup>2<\/sup>\u00a0x = 1 &#8211;\u00a0cos 2x}<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\sin 2 \\beta}{2+2 \\cos 2 \\beta+3-3 \\cos 2 \\beta}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\sin 2 \\beta}{5-\\cos 2 \\beta}{\/tex}<\/span><br \/>\nLHS\u00a0= RHS<br \/>\nHence Proved<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>LHS = <span class=\"math-tex\">{tex}cos 2x\\times\u00a0\u00a0cos{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac { x } { 2 }{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}- cos 3x\u00a0\\times\u00a0cos{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac { 9 x } { 2 }{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac { 1 } { 2 }{\/tex}<\/span>\u00a0[2 cos 2x\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0cos\u00a0<span class=\"math-tex\">{tex}\\frac { x } { 2 }{\/tex}<\/span>\u00a0 &#8211; 2 cos\u00a0<span class=\"math-tex\">{tex}\\frac { 9 x } { 2 }{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0cos 3x]\u00a0 [multiplying numerator and denominator by 2]\n=\u00a0<span class=\"math-tex\">{tex}\\frac { 1 } { 2 }{\/tex}<\/span>\u00a0[cos\u00a0<span class=\"math-tex\">{tex}\\left( 2 x + \\frac { x } { 2 } \\right){\/tex}<\/span>\u00a0+ cos\u00a0<span class=\"math-tex\">{tex}\\left( 2 x &#8211; \\frac { x } { 2 } \\right){\/tex}<\/span>\u00a0&#8211; cos\u00a0<span class=\"math-tex\">{tex}\\left( \\frac { 9 x } { 2 } + 3 x \\right){\/tex}<\/span>\u00a0&#8211; cos\u00a0<span class=\"math-tex\">{tex}\\left( \\frac { 9 x } { 2 } &#8211; 3 x \\right){\/tex}<\/span>]\u00a0 [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}2 cos\\ x\u00a0\\times\u00a0cos\\ y = cos (x + y) + cos (x &#8211; y){\/tex}<\/span>]\n=\u00a0<span class=\"math-tex\">{tex}\\frac { 1 } { 2 }{\/tex}<\/span>\u00a0[ cos\u00a0<span class=\"math-tex\">{tex}\\frac { 5 x } { 2 }{\/tex}<\/span>\u00a0+ cos\u00a0<span class=\"math-tex\">{tex}\\frac { 3 x } { 2 }{\/tex}<\/span>\u00a0&#8211; cos\u00a0<span class=\"math-tex\">{tex}\\frac { 15 x } { 2 }{\/tex}<\/span>\u00a0&#8211; cos\u00a0<span class=\"math-tex\">{tex}\\frac { 3 x } { 2 }{\/tex}<\/span>]\n=\u00a0<span class=\"math-tex\">{tex}\\frac { 1 } { 2 }{\/tex}<\/span>\u00a0[cos\u00a0<span class=\"math-tex\">{tex}\\frac { 5 x } { 2 }{\/tex}<\/span>\u00a0&#8211; cos\u00a0<span class=\"math-tex\">{tex}\\frac { 15 x } { 2 }{\/tex}<\/span>]\n=\u00a0<span class=\"math-tex\">{tex}\\frac { 1 } { 2 }{\/tex}<\/span>\u00a0[- 2 sin\u00a0<span class=\"math-tex\">{tex}\\left( \\frac { \\frac { 5 x } { 2 } + \\frac { 15 x } { 2 } } { 2 } \\right) \\sin \\left( \\frac { \\frac { 5 x } { 2 } &#8211; \\frac { 15 x } { 2 } } { 2 } \\right) ]{\/tex}<\/span>\u00a0 [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span><span class=\"math-tex\">{tex}\u00a0cos\\ x &#8211; cos\\ y = &#8211; 2sin{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left( \\frac { x + y } { 2 } \\right) \\cdot \\sin \\left( \\frac { x &#8211; y } { 2 } \\right) ]{\/tex}<\/span><br \/>\n= &#8211; sin 5x\u00a0sin\u00a0<span class=\"math-tex\">{tex}\\left( \\frac { &#8211; 5 x } { 2 } \\right){\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}= sin 5x\u00a0\\times\u00a0\u00a0sin{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac { 5 x } { 2 }{\/tex}<\/span>\u00a0[<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}sin (-\\theta\u00a0) = &#8211; sin{\/tex}<\/span><span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>]\n= RHS<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0LHS = RHS<br \/>\nHence proved.<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section E <\/b>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">The path traced by Javelin is parabola. A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane.<br \/>\ncompare x<sup>2<\/sup> = -16y with x<sup>2<\/sup> = -4ay<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> &#8211; 4a = -16<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> a = 4<br \/>\ncoordinates of focus for parabola x<sup>2<\/sup> = -4ay is (0, -a)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> coordinates of focus for given parabola is (0, -4)<\/li>\n<li style=\"text-align: left;\">compare x<sup>2<\/sup> = -16y with x<sup>2 <\/sup>= -4ay<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> -4a = -16<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> a = 4<br \/>\nEquation of directrix for parabola x<sup>2<\/sup> = -4ay is y = a<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> Equation of directrix for parabola x<sup>2<\/sup> = -16y is y = 4<br \/>\nLength of latus rectum is 4a = 4 \u00d7 4 = 16<\/li>\n<li style=\"text-align: left;\">Equation of parabola with axis along y &#8211; axis<br \/>\nx<sup>2<\/sup> = 4ay<br \/>\nwhich passes through (5, 2)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 25 = 4a <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 2<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 4a = <span class=\"math-tex\">{tex}\\frac{25}{2}{\/tex}<\/span><br \/>\nhence required equation of parabola is<br \/>\n<span class=\"math-tex\">{tex}x^2=\\frac{25}{2} y{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 2x<sup>2<\/sup> = 25y<br \/>\nEquation of directrix is y= -a<br \/>\nHence required equation of directrix is 8y + 25 = 0.<br \/>\n<strong>OR<\/strong><br \/>\nSince the focus (2,0) lies on the x-axis, the x-axis itself is the axis of the parabola.<br \/>\nHence the equation of the parabola is of the form either y<sup>2<\/sup> = 4ax or y<sup>2<\/sup> = -4ax.<br \/>\nSince the directrix is x = -2 and the focus is (2,0), the parabola is to be of the form y<sup>2<\/sup> = 4ax with a = 2.<br \/>\nHence the required equation is y<sup>2<\/sup> = 4(2)x = 8x<br \/>\nlength of latus rectum = 4a = 8<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">By using formula,<br \/>\n<span class=\"math-tex\">{tex}{\\sigma ^2} = \\frac{1}{N}\\left[ {\\sum\\limits_{i = 1}^n {{f_i}} {{\\left( {{x_i} &#8211; \\bar x} \\right)}^2}} \\right]{\/tex}<\/span><\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\">x<sub>i<\/sub><\/td>\n<td style=\"text-align: center;\">f<sub>i<\/sub><\/td>\n<td style=\"text-align: center;\">f<sub>i<\/sub>x<sub>i<\/sub><\/td>\n<td style=\"text-align: center;\">x<sub>i<\/sub> &#8211;\u00a0<span class=\"math-tex\">{tex}\\overline x{\/tex}<\/span><\/td>\n<td style=\"text-align: center;\">(x<sub>i<\/sub> &#8211;\u00a0<span class=\"math-tex\">{tex}\\overline x{\/tex}<\/span>)<sup>2<\/sup><\/td>\n<td style=\"text-align: center;\">f<sub>i<\/sub>(x<sub>i<\/sub> &#8211;\u00a0<span class=\"math-tex\">{tex}\\overline x{\/tex}<\/span>)<sup>2<\/sup><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">12<\/td>\n<td style=\"text-align: center;\">-10<\/td>\n<td style=\"text-align: center;\">100<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}300{\/tex}<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">8<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<td style=\"text-align: center;\">40<\/td>\n<td style=\"text-align: center;\">-6<\/td>\n<td style=\"text-align: center;\">36<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}180{\/tex}<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">11<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<td style=\"text-align: center;\">99<\/td>\n<td style=\"text-align: center;\">-3<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}81{\/tex}<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">17<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<td style=\"text-align: center;\">85<\/td>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}45{\/tex}<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">20<\/td>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">80<\/td>\n<td style=\"text-align: center;\">6<\/td>\n<td style=\"text-align: center;\">36<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}144{\/tex}<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">24<\/td>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">72<\/td>\n<td style=\"text-align: center;\">10<\/td>\n<td style=\"text-align: center;\">100<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}300{\/tex}<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">32<\/td>\n<td style=\"text-align: center;\">1<\/td>\n<td style=\"text-align: center;\">32<\/td>\n<td style=\"text-align: center;\">18<\/td>\n<td style=\"text-align: center;\">324<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}324{\/tex}<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Total<\/td>\n<td style=\"text-align: center;\">30<\/td>\n<td style=\"text-align: center;\">420<\/td>\n<td style=\"text-align: center;\"><\/td>\n<td style=\"text-align: center;\"><\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}1374{\/tex}<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Given,\u00a0N =\u00a0<span class=\"math-tex\">{tex}\\sum f_i{\/tex}<\/span>\u00a0= 30,\u00a0<span class=\"math-tex\">{tex}\\sum f_ix_i{\/tex}<\/span>\u00a0= 420 and\u00a0<span class=\"math-tex\">{tex}\\sum f_{i}\\left(x_{i}-\\overline{x}\\right)^{2}{\/tex}<\/span>\u00a0= 1374<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\bar x = \\frac{{\\sum\\limits_{i = 1}^7 {{f_i}} {x_i}}}{N}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{420}{30}{\/tex}<\/span>\u00a0= 14<br \/>\nVariance (<span class=\"math-tex\">{tex}\\sigma^2{\/tex}<\/span>) =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{N}\\sum\\limits_{i = 1}^7 {{f_i}} {\\left( {{x_i} &#8211; \\bar x} \\right)^2}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{30}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}1374 = 45.8{\/tex}<\/span><br \/>\nStandard deviation,\u00a0<span class=\"math-tex\">{tex}\\sigma{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\sqrt{\\sigma^2}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\sqrt{45.8}{\/tex}<\/span> = 6.77<\/li>\n<li style=\"text-align: left;\">Variance (<span class=\"math-tex\">{tex}\\sigma^2{\/tex}<\/span>) =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{N}\\sum\\limits_{i = 1}^7 {{f_i}} {\\left( {{x_i} &#8211; \\bar x} \\right)^2}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{30}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}1374 = 45.8{\/tex}<\/span><\/li>\n<li style=\"text-align: left;\">Given,\u00a0N =\u00a0<span class=\"math-tex\">{tex}\\sum f_i{\/tex}<\/span>\u00a0= 30,\u00a0<span class=\"math-tex\">{tex}\\sum f_ix_i{\/tex}<\/span>\u00a0= 420 and\u00a0<span class=\"math-tex\">{tex}\\sum f_{i}\\left(x_{i}-\\overline{x}\\right)^{2}{\/tex}<\/span>\u00a0= 1374<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\bar x = \\frac{{\\sum\\limits_{i = 1}^7 {{f_i}} {x_i}}}{N}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{420}{30}{\/tex}<\/span>\u00a0= 14<br \/>\n<strong>OR<br \/>\n<span class=\"math-tex\">{tex}\\sigma^{2}=\\frac{1}{N} \\Sigma\\left(x_{i}-\\bar{x}\\right){\/tex}<\/span><\/strong><\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">Since, at least 3 questions from each part have to be selected<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\">Part I<\/td>\n<td style=\"text-align: center;\">Part II<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So number of ways are<br \/>\n3 questions from part I and 5 questions from part II can be selected in <span class=\"math-tex\">{tex}n^8 C_3 \\times{ }^7 C_5{\/tex}<\/span> ways<br \/>\n4 questions from part I and 4 questions from part II can be selected in <span class=\"math-tex\">{tex}{ }^8 C_4 \\times{ }^7 C_4{\/tex}<\/span> ways<br \/>\n5 questions from part I and 3 questions from part II can be selected in <span class=\"math-tex\">{tex}{ }^8 C_5 \\times{ }^7 C_3{\/tex}<\/span> ways<br \/>\nSo required number of ways are<br \/>\n<span class=\"math-tex\">{tex}{ }^8 C_3 \\times{ }^7 C_5+{ }^8 C_4 \\times{ }^7 C_4+{ }^8 C_5 \\times{ }^7 C_3{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{8 !}{5 ! \\times 3 !} \\times \\frac{7 !}{5 ! \\times 2 !}+\\frac{8 !}{4 ! \\times 4 !} \\times \\frac{7 !}{4 ! \\times 3 !}+\\frac{8 !}{5 ! \\times 3 !} \\times \\frac{7 !}{4 ! \\times 3 !}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{8 \\times 7 \\times 6}{3 \\times 2 \\times 1} \\times \\frac{7 \\times 6}{2 \\times 1}+\\frac{8 \\times 7 \\times 6 \\times 5}{4 \\times 3 \\times 2 \\times 1} \\times \\frac{7 \\times 6 \\times 5}{3 \\times 2 \\times 1}+\\frac{8 \\times 7 \\times 6}{3 \\times 2 \\times 1} \\times \\frac{7 \\times 6 \\times 5 \\times 4}{4 \\times 3 \\times 2 \\times 1}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 56 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 21 + 70 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 35 + 56 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 35<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 1176\u00a0+ 2450 + 1960<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 5586<\/li>\n<li style=\"text-align: left;\">Ashish is selecting 3 questions from part I so he has to select remaining 5 questions from part II<br \/>\nThe number of ways of selection is<br \/>\n3 questions from part I and 5 questions from part II can be selected in\u00a0<span class=\"math-tex\">{tex}{ }^8 C_3 \\times{ }^7 C_5{\/tex}<\/span>\u00a0ways<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{ }^8 C_3 \\times{ }^7 C_5{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{8 !}{5 ! \\times 3 !} \\times \\frac{7 !}{5 ! \\times 2 !}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{8 \\times 7 \\times 6}{3 \\times 2 \\times 1} \\times \\frac{7 \\times 6}{2 \\times 1}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a056\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a021<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a01176<\/li>\n<li style=\"text-align: left;\">4 questions from part I\u00a0and 4 questions from part II can be selected<br \/>\n<span class=\"math-tex\">{tex}{ }^8 C_4 \\times{ }^7 C_4{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{8 !}{4 ! \\times 4 !} \\times \\frac{7 !}{4 ! \\times 3 !}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{8 \\times 7 \\times 6 \\times 5}{4 \\times 3 \\times 2 \\times 1} \\times \\frac{7 \\times 6 \\times 5}{3 \\times 2 \\times 1}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 70\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a035<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 2450<br \/>\n<strong>OR<\/strong><br \/>\n6 questions from part I\u00a0and 2 questions from part II can be selected or<br \/>\n2 questions from part I and 6 questions from part II can be selected<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{ }^8 C_6 \\times{ }^7 C_2+{ }^8 C_2 \\times{ }^7 C_6{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{8 !}{6 ! \\times 2 !} \\times \\frac{7 !}{2 ! \\times 5 !}+\\frac{8 !}{6 ! \\times 2 !} \\times \\frac{7 !}{1 ! \\times 6 !}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{8 \\times 7}{2 \\times 1} \\times \\frac{7 \\times 6}{2 \\times 1}+\\frac{8 \\times 7}{2 \\times 1} \\times 7{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a028\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a021 + 28\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a07<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0588 + 196 = 784<br \/>\n<strong>Download myCBSEguide App for Effective Exam Preparation and Practice<\/strong>For comprehensive <strong>exam preparation<\/strong>, download the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app<\/strong>. It provides <strong>complete study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. With <strong>sample papers<\/strong>, <strong>NCERT solutions<\/strong>, <strong>practice tests<\/strong>, and <strong>revision notes<\/strong>, the app ensures that students are well-equipped to excel in their exams. Whether you are struggling with specific topics or looking to reinforce your understanding, these sample papers offer a range of problems to solve.Additionally, <strong>teachers<\/strong> can use the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> app<\/strong> to create <strong>customized exam papers<\/strong> with their <strong>own name and logo<\/strong>, allowing for personalized assessments. Whether you&#8217;re a student looking to improve your exam readiness or a teacher crafting custom exams, both <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a> Website and <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a> Website offer the perfect tools for success. <strong>CBSE Sample Papers Class 11 Mathematics<\/strong> help students assess their preparation level and identify areas that need more focus.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Sample_Papers_for_Class_11_2025\"><\/span><strong>CBSE Sample Papers for Class 11 2025<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-physics\/1340\/\"><strong>Physics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-chemistry\/1356\/\"><strong>Chemistry<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-mathematics\/1371\/\"><strong>Mathematics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-biology\/1388\/\"><strong>Biology<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-accountancy\/1411\/\"><strong>Accountancy<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-economics\/1423\/\"><strong>Economics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-business-studies\/1740\/\"><strong>Business Studies<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-computer-science\/1852\/\"><strong>Computer Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-informatics-practices\/1874\/\"><strong>Informatics Practices<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-english-core\/1856\/\"><strong>English Core<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%95%E0%A5%8B%E0%A4%B0\/1866\/\"><strong>Hindi Core<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%90%E0%A4%9A%E0%A5%8D%E0%A4%9B%E0%A4%BF%E0%A4%95\/1868\/\"><strong>Hindi Elective<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-history\/1870\/\"><strong>History<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-political-science\/1880\/\"><strong>Political Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-geography\/1864\/\"><strong>Geography<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-sociology\/1882\/\"><strong>Sociology<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-physical-education\/1878\/\"><strong>Physical Education<\/strong><\/a><\/li>\n<li><strong><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11\/1339\/\">Other Subjects<\/a><\/strong><\/li>\n<\/ul>\n<h3><span class=\"ez-toc-section\" id=\"Download_Class_11_Sample_Papers_for_All_Subjects_from_myCBSEguide\"><\/span><strong>Download <a href=\"https:\/\/mycbseguide.com\/cbse-sample-papers-class-11.html\">Class 11 Sample Papers for All Subjects<\/a> from <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a><\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>For comprehensive <strong>exam preparation<\/strong>, download <strong>Class 11 sample papers<\/strong> for <strong>Physics<\/strong>, <strong>Chemistry<\/strong>, <strong>Biology<\/strong>, <strong>History<\/strong>, <strong>Political Science<\/strong>, <strong>Economics<\/strong>, <strong>Geography<\/strong>, <strong>Computer Science<\/strong>, <strong>Home Science<\/strong>, <strong>Accountancy<\/strong>, and <strong>Business Studies<\/strong> from the <strong>myCBSEguide app<\/strong> or website. <strong>myCBSEguide<\/strong> provides <strong>sample papers with solutions<\/strong>, <strong>chapter-wise test papers<\/strong>, <strong>NCERT solutions<\/strong>, <strong>NCERT Exemplar solutions<\/strong>, and <strong>quick revision notes<\/strong> to help students prepare effectively for their exams.<\/p>\n<p>Additionally, <strong>CBSE guess papers<\/strong> and <strong>important question papers<\/strong> are available for free download to assist students in their revision. All these study resources are accessible through the <strong>myCBSEguide app<\/strong>, the <strong>best app for CBSE students<\/strong>, and our user-friendly website. Start your preparation today with <strong>myCBSEguide<\/strong> for the latest and most accurate study materials. These <strong>CBSE Sample Papers Class 11 Mathematics<\/strong> are designed to mirror the actual exam format, helping students become familiar with the types of questions they will face.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>CBSE Sample Paper for Class 11 Mathematics Download Free Class 11 Mathematics Sample Papers 2024-25 from myCBSEguide As per the newly released CBSE marking scheme and blueprint for Class 11, myCBSEguide provides updated Class 11 Mathematics sample papers 2024-25. These sample papers are designed to match the latest exam pattern and are available for free &#8230; <a title=\"CBSE Sample Papers Class 11 Mathematics 2025\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-mathematics\/\" aria-label=\"More on CBSE Sample Papers Class 11 Mathematics 2025\">Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1339,1999,1414,2030],"tags":[163,1527,12,1959,1967,1030,1340],"class_list":["post-13680","post","type-post","status-publish","format-standard","hentry","category-cbse-sample-papers","category-class-11-sample-papers","category-mathematics-cbse-class-11","category-maths-sample-papers-class-11-sample-papers","tag-cbse-class-11","tag-cbse-question-paper","tag-cbse-sample-papers","tag-cbse-sample-papers-2024","tag-cbse-sample-papers-2025","tag-class-11-mathematics","tag-model-question-papers"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>CBSE Sample Papers Class 11 Mathematics 2025 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"CBSE Sample Papers Class 11 Mathematics\u00a0myCBSEguide 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