{"id":13581,"date":"2023-03-07T16:40:21","date_gmt":"2023-03-07T11:10:21","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=13581"},"modified":"2025-10-09T16:52:52","modified_gmt":"2025-10-09T11:22:52","slug":"cbse-sample-papers-class-12-mathematics","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/","title":{"rendered":"Class 12 Maths sample papers 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-12-mathematics\/#_Class_12_Maths_Sample_Papers_for_2025\" >\u00a0Class 12 Maths Sample Papers for 2025<\/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-12-mathematics\/#Sample_Papers_of_Maths_Class_12_%E2%80%93_Free_PDF_Download\" >Sample Papers of Maths Class 12 &#8211; Free PDF Download<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#CBSE_Sample_Papers_Class_12_Mathematics_2025\" >CBSE Sample Papers Class 12 Mathematics 2025<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#Download_myCBSEguide_App_for_Effective_Exam_Preparation\" >Download myCBSEguide App for Effective Exam Preparation<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#Solution\" >Solution<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#Boost_Your_Exam_Preparation_with_the_myCBSEguide_App\" >Boost Your Exam Preparation with the myCBSEguide App<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#Sample_Papers_for_Class_12_2025\" >Sample Papers for Class 12 2025<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#Download_CBSE_Class_12_Sample_Papers_for_All_Subjects\" >Download CBSE Class 12 Sample Papers for All Subjects<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"_Class_12_Maths_Sample_Papers_for_2025\"><\/span><strong>\u00a0Class 12 Maths Sample Papers for 2025<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Looking for <strong>CBSE <a href=\"https:\/\/mycbseguide.com\/cbse-sample-papers-class-12.html\">Class 12<\/a> Mathematics sample papers<\/strong> for the <strong>2024-25 academic year<\/strong>? If you&#8217;re preparing for the <strong>Class 12 Maths exams in 2025<\/strong>, downloading <strong>Class 12 Maths sample papers 2025<\/strong> is an essential step in your exam preparation. These <strong>Class 12 Maths sample papers 2025<\/strong> are designed to align with the latest CBSE syllabus and exam pattern, providing you with a clear understanding of what to expect. By practicing these <strong>Class 12 Maths sample papers 2025<\/strong>, you can get a good grasp of key topics, improve problem-solving skills, and boost your confidence. Additionally, solving these sample papers will help you assess your readiness and identify areas that need more focus. Downloading and practicing <strong>Class 12 Maths sample papers 2025<\/strong> from reliable sources like the <strong>myCBSEguide app<\/strong> will ensure you&#8217;re fully prepared for the exam.Visit the <strong>myCBSEguide app<\/strong> to access <strong>high-quality sample papers<\/strong> with <strong>100% genuine solutions<\/strong> in <strong>PDF format<\/strong>. These sample papers are carefully designed to help <strong>Class 12 students<\/strong> practice and master important concepts, ensuring effective preparation for the <strong>2025 CBSE exam<\/strong>.<\/p>\n<p>If you&#8217;re searching for the <strong>best CBSE sample papers for Class 12 Mathematics 2025<\/strong>, the <strong>myCBSEguide student dashboard<\/strong> is your one-stop solution. It offers <strong>free downloadable sample papers<\/strong>, <strong>NCERT solutions<\/strong>, <strong>chapter-wise practice tests<\/strong>, <strong>revision notes<\/strong>, and much more to meet all your <strong>study needs<\/strong>.<\/p>\n<p>Start your exam preparation now by downloading the<a href=\"https:\/\/play.google.com\/store\/search?q=mycbseguide+app&amp;c=apps\"> <strong>myCBSEguide app<\/strong><\/a> or logging into the <strong>myCBSEguide <a href=\"https:\/\/mycbseguide.com\/dashboard\/home\">student dashboard<\/a><\/strong>. Get access to the <strong>best Class 12 Maths sample papers<\/strong> and prepare effectively for the <strong>2025 CBSE Mathematics exam<\/strong>.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"Sample_Papers_of_Maths_Class_12_%E2%80%93_Free_PDF_Download\"><\/span>Sample Papers of Maths Class 12 &#8211; Free PDF Download<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Download CBSE Class 12 Mathematics Sample Papers for 2025<\/strong><\/p>\n<p>The <strong>CBSE Class 12 Mathematics sample papers for 2025<\/strong> are now available! With the release of the new <strong>CBSE marking scheme<\/strong> and <strong>blueprint<\/strong> for the <strong>2025 Class 12 exams<\/strong>, students can now access <strong>high-quality sample papers<\/strong> to help them prepare effectively. These sample papers are available for <strong>free download<\/strong> in <strong>PDF format<\/strong> through the <strong>myCBSEguide app<\/strong> and website.<\/p>\n<p>To excel in the <strong>Class 12 Maths exams 2025<\/strong>, practicing with <strong>Class 12 Maths sample papers 2025<\/strong> is crucial for understanding the exam format and gaining confidence. These <strong>Class 12 Maths sample papers 2025<\/strong> are designed to give students a clear picture of the types of questions that will appear on the exam and help improve time management skills. By regularly solving these <strong>Class 12 Maths sample papers 2025<\/strong>, you can identify your strengths and weaknesses, allowing you to focus on areas that need improvement. Additionally, working through <strong>Class 12 Maths sample papers 2025<\/strong> will familiarize you with the marking scheme, ensuring that you\u2019re fully prepared for every section of the exam. To get started, download the <strong>Class 12 Maths sample papers 2025<\/strong> from trusted platforms like myCBSEguide, which offers updated papers according to the latest syllabus and exam pattern.<\/p>\n<p><strong>CBSE Sample Papers for Class 12 Mathematics<\/strong> come with <strong>solutions<\/strong> and are based on the latest <strong>CBSE syllabus<\/strong>, marking scheme, and exam blueprint. These sample papers provide a comprehensive preparation strategy and answer the most frequently asked question, <strong>&#8220;How to prepare for CBSE Class 12 board exams?&#8221;<\/strong> With these sample papers, you can practice and familiarize yourself with the exam format, ensuring you are fully prepared for the <strong>2025 CBSE board exams<\/strong>.<\/p>\n<p>Download the <strong>myCBSEguide app<\/strong> today to get the latest <strong>CBSE sample papers for Class 12 Mathematics<\/strong> with solutions. Start practicing now to enhance your exam preparation and perform well in your upcoming exams!<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg\" alt=\"Class 12 Maths Sample Papers\" width=\"600\" height=\"300\" \/><\/p>\n<p style=\"text-align: center;\"><strong>Sample Papers of Class 12 Maths 2025 with solution<\/strong><\/p>\n<p style=\"text-align: center;\"><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/1284\/type\/2\">Download as PDF<\/a><\/strong><\/p>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"CBSE_Sample_Papers_Class_12_Mathematics_2025\"><\/span><strong>CBSE Sample Papers Class 12 Mathematics 2025<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: center;\"><strong>Class 12 &#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 A = <span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 0 &amp; 1 \\\\ 0 &amp; 0 \\end{array}\\right]{\/tex}<\/span>, then A<sup>2023<\/sup> is equal to\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 2023 &amp; 0\\\\ 0 &amp; 2023 \\end{array}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 0 &amp; 1 \\\\ 0 &amp; 0 \\end{array}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 0 &amp; 2023 \\\\ 0 &amp; 0 \\end{array}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 0 &amp; 0 \\\\ 0 &amp; 0 \\end{array}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>If A is a 3<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>3 matrix and |A| = -2, then value of |A (adj A)| is\n<div style=\"margin-left: 20px;\">\n<p>a)-2<\/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)2<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)-8<\/p>\n<\/div>\n<\/li>\n<li>The system of linear equations<br \/>\n5x + ky = 5,<br \/>\n3x + 3y = 5;<br \/>\nwill be consistent if:<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)k = 5<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)k = -5<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)k\u00a0<span class=\"math-tex\">{tex}\\neq{\/tex}<\/span>\u00a0-3<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)k\u00a0<span class=\"math-tex\">{tex}\\neq{\/tex}<\/span>\u00a05<\/p>\n<\/div>\n<\/li>\n<li>If the matrix <span class=\"math-tex\">{tex}A=\\left[\\begin{array}{cc} 3-2 x &amp; x+1 \\\\ 2 &amp; 4 \\end{array}\\right]{\/tex}<\/span> is singular then x = ?.\n<div style=\"margin-left: 20px;\">\n<p>a)1<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)0<\/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>The direction ratios of a line parallel to z-axis are:\n<div style=\"margin-left: 20px;\">\n<p>a)&lt; 0,\u00a00, 0 &gt;<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)&lt; 1, 1, 0 &gt;<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)&lt; 1, 1, 1 &gt;<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)&lt; 0,\u00a00, 1 &gt;<\/p>\n<\/div>\n<\/li>\n<li>If y = e<sup>-x<\/sup> (A cos x + B sin x), then y is a solution of\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{d^{2} y}{d x^{2}}+2 \\frac{d y}{d x}+2 y=0{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{d^{2} y}{d x^{2}}-2 \\frac{d y}{d x}+2 y=0{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{d^{2} y}{d x^{2}}+2 y=0{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{d^{2} y}{d x^{2}}+2 \\frac{d y}{d x}=0{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>In an LPP, if the objective function z = ax + by has the same maximum value on two corner points of the feasible region, then the number of points at which z<sub>max <\/sub>occurs is:\n<div style=\"margin-left: 20px;\">\n<p>a)finite<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)infinite<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)2<\/p>\n<\/div>\n<\/li>\n<li>In <span class=\"math-tex\">{tex}\\triangle\\text {ABC}{\/tex}<\/span>, <span class=\"math-tex\">{tex}\\overrightarrow{\\mathrm{AB}}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\hat {\\text i} + \\hat {\\text j} + 2\\hat {\\text k}{\/tex}<\/span> and <span class=\"math-tex\">{tex}\\overrightarrow{\\mathrm{AC}}{\/tex}<\/span> = <span class=\"math-tex\">{tex}3\\hat {\\text i} &#8211; \\hat {\\text j} + 4\\hat {\\text k}{\/tex}<\/span>. If D is mid-point of BC, then vector <span class=\"math-tex\">{tex}\\overrightarrow{\\mathrm{AD}}{\/tex}<\/span> is equal to:\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\hat{\\text i}-\\hat{\\text j}+\\hat{\\text k}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}2 \\hat{\\text i}-2 \\hat{\\text j}+2 \\hat{\\text k}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}4 \\hat{\\text i}+6 \\hat{\\text k}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}2 \\hat{\\text i}+3 \\hat{\\text k}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li><span class=\"math-tex\">{tex}\\int\\limits_{0}^\\frac{\\pi}{8} {\/tex}<\/span>\u00a0tan<sup>2<\/sup>\u00a0(2x)\u00a0is equal to\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{4+\\pi}{8}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{4-\\pi}{8}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{4-\\pi}{2} {\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{4-\\pi}{4} {\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>If A = <span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 1 &amp; 0 \\\\ 0 &amp; 0 \\end{array}\\right]{\/tex}<\/span> and B = <span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 1 &amp; 1 \\\\ 0 &amp; 0 \\end{array}\\right]{\/tex}<\/span>, then B&#8217;A&#8217; is equal to:\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 1 &amp; 1 \\\\ 0 &amp; 0 \\end{array}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 1 &amp; 1 \\\\ 1 &amp; 1 \\end{array}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 1 &amp; 0 \\\\ 1 &amp; 0 \\end{array}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 0 &amp; 0 \\\\ 0 &amp; 0 \\end{array}\\right]{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>Objective function of an LPP is\n<div style=\"margin-left: 20px;\">\n<p>a)a function to be optimized<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)a function between the variables<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)a constraint<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)a relation between the variables<\/p>\n<\/div>\n<\/li>\n<li>If the angle between the vectors <span class=\"math-tex\">{tex}\\vec {\\text a}{\/tex}<\/span> and <span class=\"math-tex\">{tex}\\vec {\\text b}{\/tex}<\/span> is <span class=\"math-tex\">{tex}\\frac \\pi 4{\/tex}<\/span> and <span class=\"math-tex\">{tex}|\\vec{a} \\times \\vec{b}|{\/tex}<\/span> = 1, then <span class=\"math-tex\">{tex}\\vec{a} \\cdot \\vec{b}{\/tex}<\/span> is equal to\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}{\\sqrt{2}}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\sqrt{2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)1<\/p>\n<\/div>\n<\/li>\n<li>Find the area of the triangle with vertices (0,0), (4,2), and (1,1).\n<div style=\"margin-left: 20px;\">\n<p>a)1 sq.unit<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)2\u00a0sq.unit<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)0\u00a0sq.unit<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)5\u00a0sq.unit<\/p>\n<\/div>\n<\/li>\n<li>For any two events A and B, if <span class=\"math-tex\">{tex}\\text P(\\bar {\\text A}){\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac 12{\/tex}<\/span>, <span class=\"math-tex\">{tex}\\text P(\\bar {\\text B}){\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac 23{\/tex}<\/span> and P(A\u00a0<span class=\"math-tex\">{tex}\\cap{\/tex}<\/span> B) = <span class=\"math-tex\">{tex}\\frac 14{\/tex}<\/span>, then <span class=\"math-tex\">{tex}\\mathrm{P}\\left(\\frac{\\overline{\\mathrm{A}}}{\\overline{\\mathrm{B}}}\\right){\/tex}<\/span> equals:\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac 14{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac {3}{8}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac 89{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac 18{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>Which of the following transformations reduce the differential equation\u00a0<span class=\"math-tex\">{tex}\\frac{d z}{d x}+\\frac{z}{x} \\log z=\\frac{z}{x^{2}}(\\log z)^{2}{\/tex}<\/span>\u00a0into the form\u00a0<span class=\"math-tex\">{tex}\\frac{d u}{d x}+P(x) u=Q(x){\/tex}<\/span>\n<div style=\"margin-left: 20px;\">\n<p>a)u = e<sup>x<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)u = log x<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)u = (log z)<sup>2<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)u = (log z)<sup>-1<\/sup><\/p>\n<div class=\"flex-shrink-0 flex flex-col relative items-end\">\n<div>\n<div class=\"pt-0\">\n<div class=\"gizmo-bot-avatar flex h-8 w-8 items-center justify-center overflow-hidden rounded-full\">\n<div class=\"relative p-1 rounded-sm flex items-center justify-center bg-token-main-surface-primary text-token-text-primary h-8 w-8\"><strong>Download myCBSEguide App for Comprehensive Exam Preparation<\/strong><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"group\/conversation-turn relative flex w-full min-w-0 flex-col agent-turn\">\n<div class=\"flex-col gap-1 md:gap-3\">\n<div class=\"flex max-w-full flex-col flex-grow\">\n<div class=\"min-h-8 text-message flex w-full flex-col items-end gap-2 whitespace-normal break-words [.text-message+&amp;]:mt-5\" dir=\"auto\" data-message-author-role=\"assistant\" data-message-id=\"39267dde-462f-4131-afe1-10ee6cae1535\" data-message-model-slug=\"gpt-4o-mini\">\n<div class=\"flex w-full flex-col gap-1 empty:hidden first:pt-[3px]\">\n<div class=\"markdown prose w-full break-words dark:prose-invert dark\">\n<p>To practice more questions and boost your exam preparation, <strong>download the <a href=\"https:\/\/play.google.com\/store\/search?q=mycbseguide+app&amp;c=apps\">myCBSEguide<\/a> app<\/strong>. It provides <strong>complete study material<\/strong> for <strong>CBSE, NCERT<\/strong>, and competitive exams like <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong>. With access to <strong>sample papers<\/strong>, <strong>chapter-wise tests<\/strong>, <strong>NCERT solutions<\/strong>, and <strong>revision notes<\/strong>, myCBSEguide ensures that you\u2019re fully equipped to succeed in your exams.<\/p>\n<p>For <strong>teachers<\/strong>, the <strong><a href=\"https:\/\/play.google.com\/store\/search?q=Examin8&amp;c=apps\">Examin8.com<\/a> app<\/strong> offers a convenient platform to <strong>create customized question papers<\/strong> with your own name and logo. This tool is ideal for teachers who want to create unique practice papers for their students, based on their specific syllabus and requirements.<\/p>\n<p>Start your exam preparation today by downloading the <strong>myCBSEguide app<\/strong> or visiting the <strong><a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a> website and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a><\/strong> <strong>website. <\/strong>Whether you\u2019re a student or a teacher, myCBSEguide offers all the resources you need to excel.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>ABCD is a rhombus whose diagonals intersect at E. Then <span class=\"math-tex\">{tex}\\vec{\\mathrm{EA}}+\\vec{\\mathrm{EB}}+\\vec{\\mathrm{EC}}+ \\vec{\\mathrm{ED}}{\/tex}<\/span> equals<\/p>\n<p>a)<span class=\"math-tex\">{tex}\\vec 0{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\vec {\\text {AD}}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}2\\vec {\\text {AD}}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}2\\vec {\\text {BC}}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>If\u00a0<span class=\"math-tex\">{tex}y=\\sqrt{\\sin x+\\sqrt{\\sin x+\\sqrt{\\sin x+}}} \\ldots \\infty{\/tex}<\/span>\u00a0then\u00a0<span class=\"math-tex\">{tex}\\frac {dy }{dx}{\/tex}<\/span> = ?\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{\\sin x}{(2 y-1)}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{\\cos x}{(y-1)}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{\\cos x}{(2 y-1)}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{\\sin x}{(2 y+1)}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>The direction ratios of two lines are a, b, c\u00a0and (b &#8211;\u00a0c), (c &#8211; a), (a &#8211; b) respectively. The angle between these lines is\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{\\pi}{2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{\\pi}{4}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{\\pi}{3}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{3 \\pi}{4}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> If two positive numbers are such that sum is 16 and sum of their cubes is minimum, then numbers are 8, 8.<br \/>\n<strong>Reason (R): <\/strong>If f be a function defined on an interval I and c <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> l and let f be twice differentiable at c, then x = c is a point of local minima if f'(c) = 0 and f&#8221;(c) &gt; 0 and f(c) is local minimum value of f.<\/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> A function f: N\u00a0<span class=\"math-tex\">{tex}\\to{\/tex}<\/span> N be defined by\u00a0<span class=\"math-tex\">{tex}\\ f(n) = \\begin{cases} \\frac n2 &amp; \\quad \\text{if } n \\text{ is even}\\\\ \\frac {(n+1)}2 &amp; \\quad \\text{if } n \\text{ is odd} \\end{cases} {\/tex}<\/span>\u00a0 for all n <span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0N; is one-one.<br \/>\n<strong>Reason (R):<\/strong> A function f: A\u00a0<span class=\"math-tex\">{tex}\\to{\/tex}<\/span> B is said to be injective if a\u00a0<span class=\"math-tex\">{tex}\\not={\/tex}<\/span> b then\u00a0f(a)\u00a0<span class=\"math-tex\">{tex}\\not={\/tex}<\/span> f(b).<\/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>Find the value of <span class=\"math-tex\">{tex}\\cos ^{-1} \\frac{1}{2}+2 \\sin ^{-1} \\frac{1}{2}{\/tex}<\/span>.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Find the value of <span class=\"math-tex\">{tex}{\\tan ^{ &#8211; 1}}\\left( { &#8211; \\frac{1}{{\\sqrt 3 }}} \\right) + {\\cot ^{ &#8211; 1}}\\left( {\\frac{1}{{\\sqrt 3 }}} \\right) + {\\tan ^{ &#8211; 1}}\\left[ {\\sin \\left( {\\frac{{ &#8211; \\pi }}{2}} \\right)} \\right]{\/tex}<\/span>.<\/li>\n<li>Find the absolute maximum and minimum values of the function f given by f(x) = cos<sup>2<\/sup> x + sin x, x <span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0[0, <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>]<\/li>\n<li>Find the rate of change of the area of a circle with respect to its radius r\u00a0when\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>r = 3 cm<\/li>\n<li>r = 4 cm<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Find the intervals of function f(x) = (x &#8211; 1)(x &#8211; 2)<sup>2<\/sup>\u00a0is<\/p>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>increasing<\/li>\n<li>decreasing.<\/li>\n<\/ol>\n<\/li>\n<li>Evaluate:\u00a0<span class=\"math-tex\">{tex}\\int_{1}^{4} f(x) d x{\/tex}<\/span>, where f(x)\u00a0=|x &#8211; 1| + |x &#8211; 2| + |x &#8211; 3|<\/li>\n<li>A ladder 13 m long is leaning against a vertical wall. The bottom of the ladder is dragged away from the wall along the ground at the rate of 2 cm\/sec. How fast is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall?<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section C<\/b><\/li>\n<li>Evaluate\u00a0<span class=\"math-tex\">{tex} \\int \\frac { 2 x + 5 } { \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } } d x.{\/tex}<\/span><\/li>\n<li>There are three coins. One is a coin having tails on both faces, another is a biased coin that comes up tails 70% of the time and the third is an unbiased coin. One of the coins is chosen at random and tossed, it shows tail. Find the probability that it was a coin with tail on both the faces.<\/li>\n<li>Evaluate the definite integral\u00a0<span class=\"math-tex\">{tex} \\int _ { 1 } ^ { 2 } \\frac { 5 x ^ { 2 } } { x ^ { 2 } + 4 x + 3 }{\/tex}<\/span>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Evaluate: <span class=\"math-tex\">{tex}\\int \\limits_{-2}^1 \\sqrt{5-4 x-x^2} \\mathrm{~d} x{\/tex}<\/span><\/li>\n<li>Find the particular solution of the differential equation <span class=\"math-tex\">{tex}\\left[x \\sin ^{2}\\left(\\frac{y}{x}\\right)-y\\right]{\/tex}<\/span>dx + x dy = 0, given that y = <span class=\"math-tex\">{tex}\\frac{\\pi}{4}{\/tex}<\/span>when\u00a0x = 1\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Find the general solution\u00a0of\u00a0<span class=\"math-tex\">{tex}x \\log x \\frac{d y}{d x}+y=\\frac{2}{x} \\log x{\/tex}<\/span><\/li>\n<li>Find the Maximum and Minimum value of 2x + y<br \/>\nSubject to the constraints:<br \/>\nx + 3y <span class=\"math-tex\">{tex}\\ge{\/tex}<\/span> 6,<br \/>\nx &#8211; 3y <span class=\"math-tex\">{tex}\\ge{\/tex}<\/span> 3,<br \/>\n3x + 4y <span class=\"math-tex\">{tex}\\le{\/tex}<\/span> 24,<br \/>\n&#8211; 3x + 2y <span class=\"math-tex\">{tex}\\le{\/tex}<\/span>\u00a06,<br \/>\n5x + y <span class=\"math-tex\">{tex}\\ge{\/tex}<\/span>\u00a05,<br \/>\nwhere non-negative restrictions are x, y <span class=\"math-tex\">{tex}\\ge{\/tex}<\/span>\u00a00.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Solved the linear programming problem graphically:<br \/>\nMaximize Z = 60x + 15y<br \/>\nSubject to constraints<br \/>\nx + y <span class=\"math-tex\">{tex}\\le{\/tex}<\/span> 50<br \/>\n3x + y <span class=\"math-tex\">{tex}\\le{\/tex}<\/span>\u00a090<br \/>\nx, y <span class=\"math-tex\">{tex}\\ge{\/tex}<\/span>\u00a00<\/li>\n<li>Find the values of a and b so that the function f given by\u00a0<span class=\"math-tex\">{tex}f(x)=\\left\\{\\begin{aligned} 1, &amp; \\text { if } x \\leq 3 \\\\ a x+b, &amp; \\text { if } 3&lt;x&lt;5 \\\\ 7, &amp; \\text { if } x \\geq 5 \\end{aligned}\\right.{\/tex}<\/span>\u00a0is continuous at x = 3 and x = 5<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section D<\/b><\/li>\n<li>Using integration, find the area of the triangle formed by positive X-axis and tangent and normal to the circle x<sup>2<\/sup> + y<sup>2<\/sup> = 4 at (1, <span class=\"math-tex\">{tex}\\sqrt3{\/tex}<\/span>).<\/li>\n<li>Show that the function f : R <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span> {x <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> R : -1 &lt; x &lt; 1} defined by <span class=\"math-tex\">{tex}f(x) = \\frac{x}{{1 + |x|}}{\/tex}<\/span>, x <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> R is one-one and onto function.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>\u200bLet N be the set of all natural numbers and let R be a relation on N <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N, defined by<br \/>\n{a, b) R (c, d)\u00a0<span class=\"math-tex\">{tex}\\Leftrightarrow{\/tex}<\/span> ad = bc\u00a0for all (a, b), (c, d)\u00a0<span class=\"math-tex\">{tex}\\in{\/tex}<\/span> N <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N<br \/>\nShow that R is an equivalence relation on N\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0N. Also, find the equivalence class [(2, 6)]<\/li>\n<li>If <span class=\"math-tex\">{tex}A = \\left[ {\\begin{array}{*{20}{c}} 2&amp;{ &#8211; 3}&amp;5 \\\\ 3&amp;2&amp;{ &#8211; 4} \\\\ 1&amp;1&amp;{ &#8211; 2} \\end{array}} \\right]{\/tex}<\/span>, find A<sup>-1<\/sup>. Using A<sup>-1<\/sup> solve the system of equations 2x &#8211; 3y + 5z = 11; 3x + 2y &#8211; 4z = -5; x + y &#8211; 2z = -3<\/li>\n<li>Show that the straight lines whose direction cosines are given by the equations al + bm + cn = 0 and ul<sup>2<\/sup> + vm<sup>2<\/sup> + wn<sup>2<\/sup> = 0 are perpendicular, if<br \/>\na<sup>2\u00a0<\/sup>(v + w) + b<sup>2<\/sup><sub>\u00a0<\/sub>(u + w) + c<sup>2<\/sup><sub>\u00a0<\/sub>(u + v) = 0 and, parallel, if\u00a0<span class=\"math-tex\">{tex}\\frac{a^{2}}{u}+\\frac{b^{2}}{v}+\\frac{c^{2}}{w}{\/tex}<\/span>\u00a0= 0<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Find the shortest distance between the lines l<sub>1<\/sub> and l<sub>2<\/sub> whose vector equations\u00a0are<br \/>\n<span class=\"math-tex\">{tex}\\vec r = \\hat i + \\hat j + \\lambda (2\\hat i &#8211; \\hat j + \\hat k){\/tex}<\/span>\u00a0&#8230;(1)<br \/>\nand\u00a0<span class=\"math-tex\">{tex}\\vec r = 2\\hat i + \\hat j &#8211; \\hat k + \\mu (3\\hat i &#8211; 5\\hat j + 2\\hat k){\/tex}<\/span> &#8230;(2)<\/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 \/>\nAkshat and his friend Aditya were playing the snake and ladder game. They had their own dice to play the game. Akshat was having red dice whereas Aditya had black dice. In the beginning, they were using their own dice to play the game. But then they decided to make it faster and started playing with two dice together.<br \/>\n<img decoding=\"async\" style=\"height: 129px; width: 300px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1642671977-wpeagw.jpg\" alt=\"\" \/><br \/>\nAditya rolled down both\u00a0black and red die together.<br \/>\nFirst die is black and second is red.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (1)<\/li>\n<li>Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. (1)<\/li>\n<li>Find the conditional probability of obtaining the sum 10, given that the black\u00a0die resulted in even\u00a0number. (2)<br \/>\n<strong>OR<\/strong><br \/>\nFind the conditional probability of obtaining the doublet, given that the red die resulted in a number more\u00a0than 4. (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nThe slogans on chart papers are to be placed on a school bulletin board at the points A, B and C displaying A (follow Rules), B (Respect your elders) and C (Be a good human). The coordinates of these points are (1, 4, 2), (3, -3, -2) and (-2, 2, 6), respectively.<br \/>\n<img decoding=\"async\" style=\"height: 169px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1685421062-2vp289.jpg\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>If <span class=\"math-tex\">{tex}\\vec a{\/tex}<\/span>, <span class=\"math-tex\">{tex}\\vec b{\/tex}<\/span> and <span class=\"math-tex\">{tex}\\vec c{\/tex}<\/span> be the position vectors of points A, B, C, respectively, then find <span class=\"math-tex\">{tex}|\\vec{a}+\\vec{b}+\\vec{c}|{\/tex}<\/span>. (1)<\/li>\n<li>If <span class=\"math-tex\">{tex}\\vec{a}=4 \\hat{i}+6 \\hat{j}+12 \\hat{k}{\/tex}<\/span>, then find the unit vector in direction of <span class=\"math-tex\">{tex}\\vec a{\/tex}<\/span>. (1)<\/li>\n<li>Find area of <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABC. (2)<br \/>\n<strong>OR<\/strong><br \/>\nWrite the triangle law of addition for <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABC. Suppose, if the given slogans are to be placed on a straight line, then the value of <span class=\"math-tex\">{tex}|\\vec{a} \\times \\vec{b}+\\vec{b} \\times \\vec{c}+\\vec{c} \\times \\vec{a}|{\/tex}<\/span>. (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nA gardener wants to construct a rectangular\u00a0bed of garden in a circular patch of land. He takes the maximum perimeter of the rectangular region as possible. (Refer to the images given below for calculations)<br \/>\n<img decoding=\"async\" style=\"height: 145px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/7xU70FD.png\" alt=\"\" \/><img decoding=\"async\" style=\"height: 124px; width: 131px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/zCtJs7Q.png\" alt=\"\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Find the perimeter of rectangle in terms of any one side and radius of circle. (1)<\/li>\n<li>Find critical points to maximize the perimeter of rectangle? (1)<\/li>\n<li>Check for maximum or minimum value of perimeter at critical point. (2)<br \/>\n<strong>OR<\/strong><br \/>\nIf a rectangle of the maximum perimeter which can be inscribed in a circle of radius 10 cm is square, then the perimeter of region. (2)<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Download_myCBSEguide_App_for_Effective_Exam_Preparation\"><\/span><strong>Download myCBSEguide App for Effective Exam Preparation <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>For comprehensive exam preparation, <strong>download the <a href=\"https:\/\/play.google.com\/store\/search?q=mycbseguide+app&amp;c=apps\">myCBSEguide<\/a> app<\/strong>. It offers complete study material for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, and top competitive exams like <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong>. The app provides access to <strong>sample papers<\/strong>, <strong>chapter-wise practice tests<\/strong>, <strong>NCERT solutions<\/strong>, <strong>revision notes<\/strong>, and <strong>important questions<\/strong>, ensuring you are fully prepared for your exams.If you&#8217;re a <strong>teacher<\/strong>, the <strong><a href=\"https:\/\/play.google.com\/store\/search?q=Examin8&amp;c=apps\">Examin8<\/a> app<\/strong> is an excellent tool for creating customized question papers. With <strong>Examin8<\/strong>, you can design papers with your own name, logo, and content to suit your students&#8217; needs.<strong>Download the myCBSEguide app<\/strong> now and start preparing for success in your <strong>CBSE<\/strong>, <strong>JEE<\/strong>, <strong>NEET<\/strong>, or <strong>NDA exams<\/strong>. For teachers,\u00a0 visit <strong><a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a> Website and <\/strong><strong><a href=\"https:\/\/play.google.com\/store\/search?q=Examin8&amp;c=apps\">Examin8<\/a> Website<\/strong>\u00a0provides an easy and efficient way to generate tailored question papers.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center; page-break-before: always;\"><strong>Class 12 &#8211; Mathematics<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Solution\"><\/span><strong>Solution <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ol style=\"padding-left: 20px;\">\n<li style=\"list-style-type: none;\">\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}\\left[\\begin{array}{ll} 0 &amp; 0 \\\\ 0 &amp; 0 \\end{array}\\right]{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 0 &amp; 0 \\\\ 0 &amp; 0 \\end{array}\\right]{\/tex}<\/span><\/li>\n<li>(d)\n<p style=\"display: inline;\">-8<\/p>\n<p><b>Explanation: <\/b>-8<\/li>\n<li>(d)\n<p style=\"display: inline;\">k\u00a0<span class=\"math-tex\">{tex}\\neq{\/tex}<\/span>\u00a05<\/p>\n<p><b>Explanation: <\/b>We have, 5x + ky &#8211; 5 = 0<br \/>\nand\u00a03x + 3y &#8211; 5 = 0<br \/>\nFor consistent system<br \/>\n<span class=\"math-tex\">{tex}\\begin{equation} \\frac{5}{3} \\neq \\frac{k}{3} \\end{equation}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0k\u00a0<span class=\"math-tex\">{tex}\\neq{\/tex}<\/span>\u00a05<\/li>\n<li>(a)\n<p style=\"display: inline;\">1<\/p>\n<p><b>Explanation: <\/b>When a given matrix is singular then the given matrix determinant is 0.<br \/>\n|A| = 0<br \/>\nGiven, <span class=\"math-tex\">{tex}A=\\left(\\begin{array}{cc} 3-2 x &amp; x+1 \\\\ 2 &amp; 4 \\end{array}\\right) {\/tex}<\/span><br \/>\n|A| = 0<br \/>\n4(3 &#8211; 2x) &#8211; 2(x + 1) = 0<br \/>\n12 &#8211; 8x &#8211; 2x &#8211; 2 = 0<br \/>\n10 &#8211; 10x = 0<br \/>\n10(1 &#8211; x) = 0<br \/>\nx = 1<\/li>\n<li>(d)\n<p style=\"display: inline;\">&lt; 0,\u00a00, 1 &gt;<\/p>\n<p><b>Explanation: <\/b>&lt; 0,\u00a00, 1 &gt;<\/li>\n<li>(a)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{d^{2} y}{d x^{2}}+2 \\frac{d y}{d x}+2 y=0{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>Given that, y = e<sup>-x<\/sup> (A cos x + sin x)<br \/>\nDifferentiating both sides w.r.t. x we get<br \/>\n<span class=\"math-tex\">{tex}\\frac{dy}{dx}{\/tex}<\/span>\u00a0= -e<sup>-x<\/sup> (A cos x + B sin x) + e<sup>-x<\/sup> (-A sin x + B cos x)<br \/>\n<span class=\"math-tex\">{tex}\\frac{dy}{dx}{\/tex}<\/span>\u00a0= -y + e<sup>x<\/sup> (-A sin x + B cos x)<br \/>\nAgain, differentiating both sides w.r.t. x, we get<br \/>\n<span class=\"math-tex\">{tex}\\frac{d^{2} y}{d x^{2}}=-\\frac{d y}{d x}{\/tex}<\/span>\u00a0+ e<sup>-x<\/sup> (-A cos x &#8211; B sin x) &#8211; e<sup>-x<\/sup>(-A sin x + b cos x)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{d^{2} y}{d x^{2}}=-\\frac{d y}{d x}-y-\\left[\\frac{d y}{d x}+y\\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{d^{2} y}{d x^{2}}=-2 \\frac{d y}{d x}-2 y{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{d^{2} y}{d x^{2}}+2 \\frac{d y}{d x}+2 y=0{\/tex}<\/span><\/li>\n<li>(c)\n<p style=\"display: inline;\">infinite<\/p>\n<p><b>Explanation: <\/b>In a LPP, if the objective f<sup>n<\/sup>\u00a0Z = ax + by has the maximum value on two corner point of the feasible region then every point on the line segment joining these two points gives the same maximum value.<br \/>\nhence, Z<sub>max<\/sub>\u00a0occurs at infinite no of times.<\/li>\n<li>(d)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}2 \\hat{\\text i}+3 \\hat{\\text k}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}2 \\hat{\\text i}+3 \\hat{\\text k}{\/tex}<\/span><\/li>\n<li>(b)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{4-\\pi}{8}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>I = <span class=\"math-tex\">{tex}\\int_{0}^\\frac{\\pi}{8} {\/tex}<\/span>\u00a0tan<sup>2<\/sup>(2 x)\u00a0dx<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\int_{0}^\\frac{\\pi}{8} {\/tex}<\/span>[sec<sup>2<\/sup> (2x) &#8211; 1]dx<br \/>\n=\u00a0<span class=\"math-tex\">{tex}[\\frac {\\tan 2x}2 &#8211; x]^{\\frac {\\pi}8}_0{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\left[\\frac {\\tan \\frac {\\pi}4}2 &#8211; \\frac {\\pi}8-\\frac {\\tan 0}2 &#8211; 0\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac 12-\\frac {\\pi}8{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{4-\\pi}{8}{\/tex}<\/span><\/li>\n<li>(c)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 1 &amp; 0 \\\\ 1 &amp; 0 \\end{array}\\right]{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 1 &amp; 0 \\\\ 1 &amp; 0 \\end{array}\\right]{\/tex}<\/span><\/li>\n<li>(a)\n<p style=\"display: inline;\">a function to be optimized<\/p>\n<p><b>Explanation: <\/b>a function to be optimized<br \/>\nThe objective function of a linear programming problem is either to be maximized or minimized i.e. objective function is to be optimized.<\/li>\n<li>(d)\n<p style=\"display: inline;\">1<\/p>\n<p><b>Explanation: <\/b>1<\/li>\n<li>(a)\n<p style=\"display: inline;\">1 sq.unit<\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\frac{1}{2}\\left| {\\begin{array}{*{20}{c}} 0&amp;0&amp;1 \\\\ 4&amp;2&amp;1 \\\\ 1&amp;1&amp;1 \\end{array}} \\right| = \\frac{1}{2}[ 1(4 &#8211; 2)] = \\frac{1}{2}(2) = 1{\/tex}<\/span><\/li>\n<li>(b)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac {3}{8}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\frac {3}{8}{\/tex}<\/span><\/li>\n<li>(d)\n<p style=\"display: inline;\">u = (log z)<sup>-1<\/sup><\/p>\n<p><b>Explanation: <\/b>We have ,<\/p>\n<p><span class=\"math-tex\">{tex}\\frac{d z}{d x}+\\frac{z}{x} \\log z=\\frac{z}{x^{2}}(\\log z)^{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{d z}{d x}=\\frac{z}{x^{2}}(\\log z)^{2}-\\frac{z}{x} \\log z{\/tex}<\/span>\u00a0&#8230;.(i)<br \/>\nPut v\u00a0= (logz)<sup>-1<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\frac{d v}{d x}=\\frac{-1}{(\\log z)^{2}} \\frac{1}{z} \\frac{d z}{d x}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{d z}{d x}=-z(\\log z)^{2} \\frac{d v}{d x}{\/tex}<\/span>\u00a0&#8230;.(ii)<br \/>\nFrom (i) and (ii)<br \/>\n<span class=\"math-tex\">{tex}-z(\\log z)^{2} \\frac{d v}{d x}=\\frac{z}{x^{2}}(\\log z)^{2}-\\frac{z}{x} \\log z{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}(\\log z) \\frac{d v}{d x}=-\\frac{1}{x^{2}}(\\log z)+\\frac{1}{x}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{d v}{d x}=-\\frac{1}{x^{2}}+\\frac{u}{x}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{d v}{d x}-\\frac{u}{x}=-\\frac{1}{x^{2}}{\/tex}<\/span><br \/>\nP(x) =\u00a0<span class=\"math-tex\">{tex}-1 \\over x{\/tex}<\/span>, q (x) =\u00a0<span class=\"math-tex\">{tex}-1 \\over x^2{\/tex}<\/span><br \/>\nGiven differential equation can be reduced.<br \/>\nusing v = (log z)<sup>-1<\/sup><\/li>\n<li>(a)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\vec 0{\/tex}<\/span><\/p>\n<p><b>Explanation:<\/b>ABCD is a rhombus and its diagonal intersects at E.<br \/>\n<img decoding=\"async\" style=\"height: 96px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/O8e3Not.png\" alt=\"\" \/><br \/>\nAccording to the properties of rhombus,<br \/>\n<span class=\"math-tex\">{tex}|\\vec{\\mathrm{AB}}|=|\\vec{\\mathrm{BC}}|=|\\vec{\\mathrm{CD}}|=|\\vec{\\mathrm{DA}}|{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\vec{\\mathrm{EA}}=-\\vec{\\mathrm{EC}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\vec{\\mathrm{ED}}=-\\vec{\\mathrm{EB}}{\/tex}<\/span><br \/>\nConsider the given expression <span class=\"math-tex\">{tex}\\vec{\\mathrm{EA}}+\\vec{\\mathrm{EB}}+\\vec{\\mathrm{EC}}+\\vec{\\mathrm{ED}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=-\\vec{\\mathrm{EC}}+\\vec{\\mathrm{EB}}+\\vec{\\mathrm{EC}}-\\vec{\\mathrm{EB}}{\/tex}<\/span>\u00a0&#8230;&#8230;(<span class=\"math-tex\">{tex}\\because\\vec{\\mathrm{EA}}=-\\vec{\\mathrm{EC}}{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\vec{\\mathrm{ED}}=-\\vec{\\mathrm{EB}}{\/tex}<\/span>)<br \/>\n= 0<br \/>\nHence, the correct answer is 0.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Boost_Your_Exam_Preparation_with_the_myCBSEguide_App\"><\/span><strong>Boost Your Exam Preparation with the myCBSEguide App<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Ready to practice more and prepare efficiently for your exams? <strong>Download the <a href=\"https:\/\/play.google.com\/store\/search?q=mycbseguide+app&amp;c=apps\">myCBSEguide app<\/a><\/strong> today for access to comprehensive <strong>study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, and key entrance exams such as <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong>. Whether you\u2019re looking for <strong>sample papers<\/strong>, <strong>chapter-wise practice tests<\/strong>, <strong>NCERT solutions<\/strong>, or <strong>important question papers<\/strong>, the myCBSEguide app offers everything you need to excel in your exams. If you&#8217;re aiming for success in the <strong>Class 12 Maths exams 2025<\/strong>, practicing with <strong>Class 12 Maths sample papers 2025<\/strong> is essential. By regularly solving <strong>Class 12 Maths sample papers 2025<\/strong>, students can improve their speed and accuracy, which is crucial for scoring well. Moreover, <strong>Class 12 Maths sample papers 2025<\/strong> are an excellent tool for identifying your strengths and weaknesses, allowing you to tailor your study plan effectively. To make your preparation even more comprehensive, download the best <strong>Class 12 Maths sample papers 2025<\/strong> from trusted platforms like the myCBSEguide app, which provides high-quality papers with solutions to ensure thorough revision.<\/p>\n<p>For <strong>teachers<\/strong>, the <strong><a href=\"https:\/\/play.google.com\/store\/search?q=Examin8&amp;c=apps\">Examin8.com<\/a> app<\/strong> allows you to easily create customized question papers with your name and logo. This app is perfect for crafting personalized practice papers and assessments for your students.<\/p>\n<p><strong>Start preparing smarter today<\/strong>\u2014download the <strong>myCBSEguide app<\/strong> and explore a wide range of resources to help you succeed in your <strong>CBSE<\/strong>, <strong>NEET<\/strong>, <strong>JEE<\/strong>, or <strong>NDA exams<\/strong>. Teachers can also use <strong><a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>\u00a0to create tailored exam papers that fit their specific syllabus and requirements.<\/p>\n<p>(c)<\/li>\n<li>\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{\\cos x}{(2 y-1)}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>Given:<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow y=\\sqrt{\\sin x+\\sqrt{\\sin x+\\sqrt{\\sin x+}}}{\/tex}<\/span><br \/>\nWe can write it as<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow y=\\sqrt{\\sin x+y}{\/tex}<\/span><br \/>\nSquaring we get<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0y<sup>2<\/sup>\u00a0= sin x + y<br \/>\nDifferentiating with respect to x,we get<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow 2 y \\frac{d y}{d x}=\\cos x+\\frac{d y}{d x}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{d y}{d x}=\\frac{\\cos x}{(2 y-1)}{\/tex}<\/span><\/li>\n<li>(a)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{\\pi}{2}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>\u00a0Let&#8217;s\u00a0consider the first parallel vector to be\u00a0<span class=\"math-tex\">{tex}\\vec{a}=a \\hat{i}+b \\hat{j}+c \\hat{k}{\/tex}<\/span>\u00a0and second parallel vector be\u00a0<span class=\"math-tex\">{tex}\\overrightarrow{\\mathrm{b}}=(\\mathrm{b}-\\mathrm{c}) \\hat{\\mathrm{i}}+(\\mathrm{c}-\\mathrm{a}) \\hat{\\mathrm{j}}+(\\mathrm{a}-\\mathrm{b}) \\hat{\\mathrm{k}}{\/tex}<\/span><br \/>\nFor the angle, we can use the formula\u00a0<span class=\"math-tex\">{tex}\\cos \\alpha=\\frac{\\vec{a} \\cdot \\vec{b}}{|\\vec{a}| \\times|\\vec{b}|}{\/tex}<\/span><br \/>\nFor that, we need to find the magnitude of these vectors<br \/>\n<span class=\"math-tex\">{tex}|\\vec{a}|=\\sqrt{a^{2}+b^{2}+(c)^{2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{a^{2}+b^{2}+c^{2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}|\\overrightarrow{\\mathrm{b}}|=\\sqrt{(\\mathrm{b}-\\mathrm{c})^{2}+(\\mathrm{c}-\\mathrm{a})^{2}+(\\mathrm{a}-\\mathrm{b})^{2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{2\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\cos \\alpha=\\frac{(a \\hat{\\imath}+b \\hat{\\jmath}+c \\hat{k}) \\cdot((b-c) \\hat{\\imath}+(c-a) \\hat{\\jmath}+(a-b) \\hat{k})}{\\sqrt{2\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)} \\times \\sqrt{a^{2}+b^{2}+c^{2}}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\cos \\alpha=\\frac{a b-a c+b c-b a+c a-c b}{\\sqrt{2\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)} \\times \\sqrt{a^{2}+b^{2}+c^{2}}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\cos \\alpha=\\frac{0}{\\sqrt{2\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)} \\times \\sqrt{a^{2}+b^{2}+c^{2}}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\alpha=\\cos ^{-1}(0){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\alpha=\\frac{\\pi}{2}{\/tex}<\/span><\/li>\n<li>(a)\n<p style=\"display: inline;\">Both A and R are true and R is the correct explanation of A.<\/p>\n<p><b>Explanation: <\/b>Let one number be x, then the other number will be (16 &#8211; x).<br \/>\nLet the sum of the cubes of these numbers be denoted by S.<br \/>\nThen, S = x<sup>3 <\/sup>+ (16 &#8211; x)<sup>3<\/sup><br \/>\nOn differentiating w.r.t. x, we get<br \/>\n<span class=\"math-tex\">{tex}\\frac{d S}{d x}{\/tex}<\/span> = 3x<sup>2 <\/sup>+ 3(16 &#8211; x)<sup>2<\/sup>(-1)<br \/>\n= 3x<sup>2 <\/sup>&#8211; 3(16 &#8211; x)<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow\u00a0\\frac{d^2 S}{d x^2}{\/tex}<\/span> = 6x + 6(16 &#8211; x) = 96<br \/>\nFor minima put <span class=\"math-tex\">{tex}\\frac {dS}{dx}{\/tex}<\/span> = 0.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> 3x<sup>2 <\/sup>&#8211; 3(16 &#8211; x)<sup>2 <\/sup>= 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x<sup>2 <\/sup>&#8211; (256 + x<sup>2 <\/sup>&#8211; 32x) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 32x = 256<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x = 8<br \/>\nAt x = 8, <span class=\"math-tex\">{tex}\\left(\\frac{d^2 S}{d x^2}\\right)_{x=8}{\/tex}<\/span> = 96 &gt; 0<br \/>\nBy second derivative test, x = 8 is the point of local minima of S.<br \/>\nThus, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 &#8211; 8 = 8<br \/>\nHence, the required numbers are 8 and 8.<\/li>\n<li>(d)\n<p style=\"display: inline;\">A is false but R is true.<\/p>\n<p><b>Explanation: <\/b><strong>Assertion <\/strong>is false because distinct elements in N has equal images.<br \/>\nfor example f(1) = <span class=\"math-tex\">{tex}\\frac {(1+1)}{2}{\/tex}<\/span>\u00a0= 1<br \/>\nf(2) = <span class=\"math-tex\">{tex}\\frac {2}{2}{\/tex}<\/span>\u00a0= 1<br \/>\n<strong>Reason <\/strong>is true because for injective function if elements are not equal then their images should be unequal.<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section B <\/b><\/li>\n<li style=\"clear: both;\">Let <span class=\"math-tex\">{tex}\\cos ^{-1}\\left(\\frac{1}{2}\\right)=x{\/tex}<\/span>. Then,\u00a0<span class=\"math-tex\">{tex}\\cos x=\\frac{1}{2}=\\cos \\left(\\frac{\\pi}{3}\\right){\/tex}<\/span>.<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\cos ^{-1}\\left(\\frac{1}{2}\\right)=\\frac{\\pi}{3}{\/tex}<\/span><br \/>\nLet <span class=\"math-tex\">{tex}\\sin ^{-1}\\left(\\frac{1}{2}\\right)=y{\/tex}<\/span>. Then, <span class=\"math-tex\">{tex}\\sin y=\\frac{1}{2}=\\sin \\left(\\frac{\\pi}{6}\\right){\/tex}<\/span>.<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\sin ^{-1}\\left(\\frac{1}{2}\\right)=\\frac{\\pi}{6}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\cos ^{-1}\\left(\\frac{1}{2}\\right)+2 \\sin ^{-1}\\left(\\frac{1}{2}\\right){\/tex}<\/span> <span class=\"math-tex\">{tex}=\\frac{\\pi}{3}+\\frac{2 \\pi}{6}=\\frac{\\pi}{3}+\\frac{\\pi}{3}=\\frac{2 \\pi}{3}{\/tex}<\/span><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>We have, <span class=\"math-tex\">{tex}{\\tan ^{ &#8211; 1}}\\left( { &#8211; \\frac{1}{{\\sqrt 3 }}} \\right) + {\\cot ^{ &#8211; 1}}\\left( {\\frac{1}{{\\sqrt 3 }}} \\right) + {\\tan ^{ &#8211; 1}}\\left[ {\\sin \\left( {\\frac{{ &#8211; \\pi }}{2}} \\right)} \\right]{\/tex}<\/span>.<br \/>\n<span class=\"math-tex\">{tex} = {\\tan ^{ &#8211; 1}}\\left( {\\tan \\frac{{5\\pi }}{6}} \\right) + {\\cot ^{ &#8211; 1}}\\left( {\\cot \\frac{\\pi }{3}} \\right) + {\\tan ^{ &#8211; 1}}( &#8211; 1){\/tex}<\/span>.<br \/>\n<span class=\"math-tex\">{tex}= {\\tan ^{ &#8211; 1}}\\left[ {\\tan \\left( {\\pi &#8211; \\frac{\\pi }{6}} \\right)} \\right] + {\\cot ^{ &#8211; 1}}\\left[ {\\cot \\left( {\\frac{\\pi }{3}} \\right)} \\right]{\/tex}<\/span><span class=\"math-tex\">{tex}+ ta{n^{ &#8211; 1}}\\left[ {\\tan \\left( {\\pi &#8211; \\frac{\\pi }{4}} \\right)} \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= {\\tan ^{ &#8211; 1}}\\left( { &#8211; \\tan \\frac{\\pi }{6}} \\right) + {\\cot ^{ &#8211; 1}}\\left( {\\cot \\frac{\\pi }{3}} \\right) + {\\tan ^{ &#8211; 1}}{\/tex}<\/span><span class=\"math-tex\">{tex}\\left( { &#8211; \\tan \\frac{\\pi }{4}} \\right){\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left[ \\begin{gathered} \\because {\\tan ^{ &#8211; 1}}\\left( {\\tan x} \\right) = x,x \\in \\left( { &#8211; \\frac{\\pi }{2},\\frac{\\pi }{2}} \\right) \\hfill \\\\ {\\cot ^{ &#8211; 1}}(\\cot x) = x,x \\in (0,\\;\\pi ) \\hfill \\\\ and\\;{\\tan ^{ &#8211; 1}}( &#8211; x) = &#8211; {\\tan ^{ &#8211; 1}}x \\hfill \\\\ \\end{gathered} \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= &#8211; \\frac{\\pi }{6} + \\frac{\\pi }{3} &#8211; \\frac{\\pi }{4} = \\frac{{ &#8211; 2\\pi + 4\\pi &#8211; 3\\pi }}{{12}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{ &#8211; 5\\pi + 4\\pi }}{{12}} = \\frac{{ &#8211; \\pi }}{{12}}{\/tex}<\/span><\/li>\n<li>It is given that f(x) = cos<sup>2<\/sup> x + sin x,\u00a0<span class=\"math-tex\">{tex}x \\in[0, \\pi]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}f^\\prime{\/tex}<\/span>(x) = 2 cos x (-sin x) + cos x<br \/>\n= -2 sin x cos x + cos x<br \/>\nNow, if <span class=\"math-tex\">{tex}f^\\prime{\/tex}<\/span>(x) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02 sin x cos x = cos x<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0cos x (2 sin x &#8211; 1) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0sin x =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0or cos x = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}x=\\frac{\\pi}{6},\\frac{5\\pi}{6}{\/tex}<\/span>\u00a0or\u00a0<span class=\"math-tex\">{tex}\\frac{\\pi}{2}{\/tex}<\/span>\u00a0as\u00a0<span class=\"math-tex\">{tex}\\mathrm{x} \\in[0, \\pi]{\/tex}<\/span><br \/>\nNext, evaluating the value of f at critical points\u00a0<span class=\"math-tex\">{tex}x=\\frac{\\pi}{2}{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}x=\\frac{\\pi}{6}{\/tex}<\/span>\u00a0and at the end points of the interval\u00a0<span class=\"math-tex\">{tex}[0, \\pi]{\/tex}<\/span>, (i.e. at x = 0 and\u00a0<span class=\"math-tex\">{tex}x=\\pi{\/tex}<\/span>\u00a0), we get,<br \/>\n<span class=\"math-tex\">{tex}f\\left(\\frac{\\pi}{6}\\right)=\\cos ^{2} \\frac{\\pi}{6}+\\sin \\frac{\\pi}{6}=\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}+\\frac{1}{2}=\\frac{5}{4}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}f(\\frac{5\\pi}{6}) = cos^2(\\frac{5\\pi}{6})+ sin(\\frac{5\\pi}{6}) = cos^2(\\pi-\\frac\\pi6)+sin(\\pi-\\frac\\pi6)=cos^2\\frac\\pi6-sin\\frac\\pi6=\\frac54{\/tex}<\/span><br \/>\nf(0) = cos<sup>2<\/sup> 0 + sin 0 = 1 + 0 = 1<br \/>\n<span class=\"math-tex\">{tex}f(\\pi)=\\cos ^{2} \\pi+\\sin \\pi{\/tex}<\/span>\u00a0= (-1)<sup>2<\/sup> + 0 = 1<br \/>\n<span class=\"math-tex\">{tex}f\\left(\\frac{\\pi}{2}\\right)=\\cos ^{2} \\frac{\\pi}{2}+\\sin \\frac{\\pi}{2}=0+1=1{\/tex}<\/span><br \/>\nTherefore, the absolute maximum value of f is\u00a0<span class=\"math-tex\">{tex}\\frac{5}{4}{\/tex}<\/span>\u00a0occurring at x =\u00a0<span class=\"math-tex\">{tex}\\frac{\\pi}{6}{\/tex}<\/span>\u00a0and the absolute minimum value of f is 1 occurring at x = 1,\u00a0<span class=\"math-tex\">{tex}\\frac{\\pi}{2}{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>.<\/li>\n<li style=\"clear: both;\">Let x\u00a0denote the area of the circle of variable radius r.<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>Area of circle <span class=\"math-tex\">{tex}\\left( x \\right) = \\pi {r^2}{\/tex}<\/span><span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span>\u00a0Rate of change of area x\u00a0w.r.t. r<span class=\"math-tex\">{tex}\\frac{{dx}}{{dr}} = \\pi \\left( {2r} \\right) = 2\\pi r{\/tex}<\/span>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>When r = 3\u00a0cm, then<br \/>\n<span class=\"math-tex\">{tex}\\frac{{dx}}{{dr}} = 2\\pi \\left( 3 \\right) = 6\\pi c{m^2}\/\\sec {\/tex}<\/span><\/li>\n<li>When r = 4 cm, then<br \/>\n<span class=\"math-tex\">{tex}\\frac{{dx}}{{dr}} = 2\\pi \\left( 4 \\right) = 8\\pi c{m^2}\/\\sec {\/tex}<\/span><\/li>\n<\/ol>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Given function is f(x) = (x &#8211; 1)(x &#8211; 2)<sup>2\u00a0<\/sup>= x<sup>2\u00a0<\/sup>&#8211; 4x + 4 (x &#8211; 1)<br \/>\n=\u00a0x<sup>3\u00a0<\/sup>&#8211; 4x<sup>2\u00a0<\/sup>+ 4x &#8211; x<sup>2\u00a0<\/sup>+ 4x &#8211; 4<br \/>\nf(x) = x<sup>3\u00a0<\/sup>&#8211;\u00a05x<sup>2\u00a0<\/sup>+ 8x &#8211; 4<br \/>\nf\u2019(x) = 3x<sup>2\u00a0<\/sup>&#8211; 10x + 8<br \/>\nf\u2019(x) = 3x<sup>2\u00a0<\/sup>&#8211; 6x &#8211; 4x + 8<br \/>\n= 3x(x &#8211; 2) -4(x &#8211; 2)<br \/>\n= (3x &#8211; 4)(x &#8211; 2)<br \/>\n<img decoding=\"async\" style=\"width: 200px; height: 43px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/e6F6GHm.png\" alt=\"\" data-imgur-src=\"e6F6GHm.png\" \/><br \/>\nFunction f(x) is decreasing for x<span class=\"math-tex\">{tex}\\in{\/tex}<\/span>[4\/3, 2] and increasing in\u00a0x<span class=\"math-tex\">{tex}\\in{\/tex}<\/span><span class=\"math-tex\">{tex}(-\\infty, 4 \/ 3) \\cup(2, \\infty){\/tex}<\/span>.<\/li>\n<li>Let y =\u00a0<span class=\"math-tex\">{tex}\\int_{1}^{4} f(x) d x{\/tex}<\/span>, then we have<br \/>\n<span class=\"math-tex\">{tex}y=\\int_{1}^{2} f(x) d x+\\int_{2}^{3} f(x) d x+\\int_{3}^{4} f(x) d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\int_{1}^{2}[(x-1)-(x-2)-(x-3)] d x{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}+\\int_{2}^{3}\\{(x-1)+(x-2)-(x-3)\\} d x{\/tex}<\/span><span class=\"math-tex\">{tex}+\\int_{3}^{4}\\{(x-1)+(x-2)+(x-3)\\} d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\int_{1}^{2}(-x+4) d x+\\int_{2}^{3} x d x+\\int_{3}^{4}(3 x-6) d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\left[\\frac{-x^{2}}{2}+4 x\\right]_{1}^{2}+\\left[\\frac{x^{2}}{2}\\right]_{2}^{3}+\\left[\\frac{3 x^{2}}{2}-6 x\\right]_{3}^{4}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}=\\left(\\frac{5}{2}+\\frac{5}{2}+\\frac{9}{2}\\right)=\\frac{19}{2}{\/tex}<\/span><\/li>\n<li>Let AB be the ladder &amp; length of ladder is 5m<br \/>\ni..e, AB = 5<br \/>\n&amp; OB be the wall &amp; OA be the ground.<br \/>\n<img decoding=\"async\" style=\"height: 152px; width: 130px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/KZ9DgO9.png\" alt=\"\" \/><br \/>\nSuppose OA = x &amp; OB = y<br \/>\nGiven that<br \/>\nThe bottom of the ladder is pulled along the ground, away the wall at the rate of 2cm\/s<br \/>\ni.e.,\u00a0<span class=\"math-tex\">{tex}\\frac{d x}{d t}{\/tex}<\/span>\u00a0= 2cm\/sec &#8230;.. (i)<br \/>\nWe need to calculate at which rate height of ladder on the wall.<br \/>\nDecreasing when foot of the ladder is 4 m away from the wall<br \/>\ni.e. we need to calculate <span class=\"math-tex\">{tex}\\frac{d y}{d t}{\/tex}<\/span>\u00a0 when x = 4 cm<br \/>\nWall OB is perpendicular to the ground OA<br \/>\n<img decoding=\"async\" style=\"height: 132px; width: 100px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/c5nFLHg.png\" alt=\"\" \/><br \/>\nUsing Pythagoras theorem,we get<br \/>\n(OB)<sup>2<\/sup> +\u00a0(OA)<sup>2<\/sup> = (AB)<sup>2<\/sup><br \/>\ny<sup>2<\/sup> + x<sup>2<\/sup> = (5)<sup>2<\/sup><br \/>\ny<sup>2<\/sup> + x<sup>2<\/sup> = 24 &#8230;. (ii)<br \/>\nDifferentiating w.r.t. time,we get<br \/>\n<span class=\"math-tex\">{tex}\\frac{d\\left(y^{2}+x^{2}\\right)}{d t}=\\frac{d(25)}{d t}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{d\\left(y^{2}\\right)}{d t}+\\frac{d\\left(x^{2}\\right)}{d t}=0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{d\\left(y^{2}\\right)}{d t} \\times \\frac{d y}{d y}+\\frac{d\\left(x^{2}\\right)}{d t} \\times \\frac{d x}{d x}=0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}2 y \\times \\frac{d y}{d t}+2 x \\times \\frac{d x}{d t}=0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}2 y \\times \\frac{d y}{d x}+2 x \\times(2)=0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}2 y \\frac{d y}{d t}+4 x=0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}2 y \\frac{d y}{d t}=-4 x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{d y}{d t}=\\frac{-4 x}{2 y}{\/tex}<\/span><br \/>\nWe need to find\u00a0<span class=\"math-tex\">{tex}\\frac{d y}{d t}{\/tex}<\/span>\u00a0when x = 4cm<br \/>\n<span class=\"math-tex\">{tex}\\left.\\frac{d y}{d t}\\right|_{x=4}=\\frac{-4 \\times 4}{2 y}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\left.\\frac{d y}{d t}\\right|_{x=4}=\\frac{-16}{2 y}{\/tex}<\/span>\u00a0&#8230;.. (iii)<br \/>\nFinding value of y<br \/>\nFrom (ii)<br \/>\nx<sup>2<\/sup> + y<sup>2<\/sup> = 25<br \/>\nPutting x = 4<br \/>\n(4)<sup>2<\/sup> + y<sup>2<\/sup> = 25<br \/>\ny<sup>2<\/sup> = 9<br \/>\ny = 3<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section C <\/b><\/li>\n<li>According to the question,\u00a0\u00a0<span class=\"math-tex\">{tex} I = \\int \\frac { 2 x + 5 } { \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } } d x{\/tex}<\/span><br \/>\nAbove integral can be written as :<span class=\"math-tex\">{tex} I = \\int \\frac { 2 x + 5+1-1 } { \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } } d x{\/tex}<\/span><span class=\"math-tex\">{tex} I = \\int \\frac { 2 x + 6-1 } { \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } } d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} I = \\int \\frac { &#8211; ( &#8211; 2 x &#8211; 6 ) &#8211; 1 } { \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } } d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = &#8211; \\int \\frac { &#8211; 2 x &#8211; 6 } { \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } } d x &#8211; \\int \\frac { d x } { \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} let \\ I = &#8211; I _ { 1 } &#8211; I _ { 2 }{\/tex}<\/span>\u00a0&#8230;(i)<br \/>\n<span class=\"math-tex\">{tex} I _ { 1 } = \\int \\frac { &#8211; 2 x &#8211; 6 } { \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } } d x{\/tex}<\/span><br \/>\nPut <span class=\"math-tex\">{tex}7 &#8211; 6x &#8211; x^2 = t{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}(-6 &#8211; 2x) dx = dt{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\therefore \\quad I _ { 1 } = \\int \\frac { d t } { \\sqrt { t } } = \\int t ^ { &#8211; 1 \/ 2 } d t= 2 \\sqrt { t } + C_1{\/tex}<\/span><span class=\"math-tex\">{tex} t = 7-6x &#8211; x^2{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 2 \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } + C _ { 1 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} I _ { 2 } = \\int \\frac { d x } { \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } }{\/tex}<\/span><span class=\"math-tex\">{tex}=\\int \\frac { d x } { \\sqrt { -(-7 + 6 x + x ^ { 2 } )} }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\int \\frac { d x } { \\sqrt { &#8211; \\left( &#8211; 7 + 6 x + x ^ { 2 } + 9 &#8211; 9 \\right) } }{\/tex}<\/span><span class=\"math-tex\">{tex} = \\int \\frac { d x } { \\sqrt { &#8211; \\left( 6 x + x ^ { 2 } + 9 &#8211; 16 \\right) } }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\int \\frac { d x } { \\sqrt { &#8211; \\left[ ( x + 3 ) ^ { 2 } &#8211; 16 \\right] } }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\quad I _ { 2 } = \\int \\frac { d x } { \\sqrt { ( 4 ) ^ { 2 } &#8211; ( x + 3 ) ^ { 2 } } }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\sin ^ { &#8211; 1 } \\left( \\frac { x + 3 } { 4 } \\right) + C _ { 2 }{\/tex}<\/span><span class=\"math-tex\">{tex} \\left[ \\because \\int \\frac { d x } { \\sqrt { a ^ { 2 } &#8211; x ^ { 2 } } } = \\sin ^ { &#8211; 1 } \\frac { x } { a } + c \\right]{\/tex}<\/span><br \/>\nPutting the values of I<sub>1<\/sub> and I<sub>2<\/sub> in\u00a0Equation (i),<br \/>\nwe get<span class=\"math-tex\">{tex}I= &#8211; 2 \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } }-C_1- \\sin ^ { &#8211; 1 } \\left( \\frac { x + 3 } { 4 } \\right) -C_2{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}I= &#8211; 2 \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } &#8211; \\sin ^ { &#8211; 1 } \\left( \\frac { x + 3 } { 4 } \\right) -C_1-C_2{\/tex}<\/span><\/p>\n<p><span class=\"math-tex\">{tex}I= &#8211; 2 \\sqrt { 7 &#8211; 6 x &#8211; x ^ { 2 } } &#8211; \\sin ^ { &#8211; 1 } \\left( \\frac { x + 3 } { 4 } \\right) + C \\ [\\because C = -C_1 &#8211; C_2]{\/tex}<\/span><\/li>\n<li>E<sub>1<\/sub>: Selected coin has tail on both faces<br \/>\nE<sub>2<\/sub>: Selected coin is biased<br \/>\nE<sub>3<\/sub>: Selected coin is unbiased<br \/>\nA: Tail comes up<br \/>\nP(E<sub>1<\/sub>) = P(E<sub>2<\/sub>) = P(E<sub>3<\/sub>) = <span class=\"math-tex\">{tex}\\frac {1} {3}{\/tex}<\/span><br \/>\nP(A|E<sub>1<\/sub>) = 1, <span class=\"math-tex\">{tex}\\mathrm{P}\\left(\\mathrm{AlE}_2\\right)=\\frac{7}{10}, \\mathrm{P}\\left(\\mathrm{AlE}_3\\right)=\\frac{1}{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\mathrm{P}\\left(\\mathrm{E}_1 \\mid \\mathrm{A}\\right)=\\frac{\\mathrm{P}\\left(\\mathrm{E}_1\\right) \\mathrm{P}\\left(\\mathrm{A} \\mid E_1\\right)}{\\mathrm{P}\\left(\\mathrm{E}_1\\right) \\mathrm{P}\\left(\\mathrm{A} \\mid E_1\\right)+\\mathrm{P}\\left(\\mathrm{E}_2\\right) \\mathrm{P}\\left(\\mathrm{A} \\mid E_2\\right)+\\mathrm{P}\\left(\\mathrm{E}_3\\right) \\mathrm{P}\\left(\\mathrm{AlE}_3\\right)}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{\\frac{1}{3} \\times 1}{\\frac{1}{3} \\times 1+\\frac{1}{3} \\times \\frac{7}{10}+\\frac{1}{3} \\times \\frac{1}{2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{10}{22}{\/tex}<\/span> or <span class=\"math-tex\">{tex}\\frac{5}{11}{\/tex}<\/span><\/li>\n<li style=\"clear: both;\">According to the question,\u00a0<span class=\"math-tex\">{tex} I = \\int _ { 1 } ^ { 2 } \\frac { 5 x ^ { 2 } } { x ^ { 2 } + 4 x + 3 } d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} I = 5 \\int _ { 1 } ^ { 2 } \\left( 1 + \\frac { &#8211; 4 x &#8211; 3 } { x ^ { 2 } + 4 x + 3 } \\right) d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 5 \\int _ { 1 } ^ { 2 } d x &#8211; 5 \\int _ { 1 } ^ { 2 } \\frac { 4 x + 3 } { x ^ { 2 } + 4 x + 3 } d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\quad I = 5 [ x ] _ { 1 } ^ { 2 } &#8211; 5 \\int _ { 1 } ^ { 2 } \\frac { 4 x + 3 } { ( x + 3 ) ( x + 1 ) } d x{\/tex}<\/span><br \/>\nUsing partial fraction,<br \/>\nlet\u00a0<span class=\"math-tex\">{tex} \\frac { 4 x + 3 } { ( x + 3 ) ( x + 1 ) } = \\frac { A } { x + 3 } + \\frac { B } { x + 1 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\quad \\frac { 4 x + 3 } { ( x + 3 ) ( x + 1 ) } = \\frac { A ( x + 1 ) + B ( x + 3 ) } { ( x + 3 ) ( x + 1 ) }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}4x + 3 = A(x + 1) + B(x + 3){\/tex}<\/span><br \/>\nComparing the coefficients of like from both sides,<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow A + B = 4 \\Rightarrow A = 4 &#8211; B{\/tex}<\/span><br \/>\nand A + 3B\u00a0 = 3<span class=\"math-tex\">{tex} \\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}4 &#8211; B + 3B = 3{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\quad B = &#8211; \\frac { 1 } { 2 },{\/tex}<\/span>then\u00a0<span class=\"math-tex\">{tex} A = 4 + \\frac { 1 } { 2 } = \\frac { 9 } { 2 }{\/tex}<\/span><br \/>\nNow, from Equation\u00a0(i), we get<br \/>\n<span class=\"math-tex\">{tex} I = 5 ( 2 &#8211; 1 ) &#8211; 5 \\int _ { 1 } ^ { 2 } \\left( \\frac { 9 \/ 2 } { x + 3 } + \\frac { &#8211; 1 \/ 2 } { x + 1 } \\right) d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 5 &#8211; 5 \\left[ \\frac { 9 } { 2 } \\log | x + 3 | &#8211; \\frac { 1 } { 2 } \\log | x + 1 | \\right] _ { 1 } ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 5 &#8211; 5\\left. {\\left[ {\\left( {\\frac{9}{2}\\log 5 &#8211; \\frac{1}{2}\\log 3} \\right)} \\right. &#8211; \\left( {\\frac{9}{2}\\log 4 &#8211; \\frac{1}{2}\\log 2} \\right)} \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 5 &#8211; 5 \\left[ \\frac { 9 } { 2 } ( \\log 5 &#8211; \\log 4 ) &#8211; \\frac { 1 } { 2 } ( \\log 3 &#8211; \\log 4 ]\\right.{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 5 &#8211; 5 \\left[ \\frac { 9 } { 2 } \\log \\frac { 5 } { 4 } &#8211; \\frac { 1 } { 2 } \\log \\frac { 3 } { 2 } \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 5 &#8211; \\frac { 45 } { 2 } \\log \\frac { 5 } { 4 } + \\frac { 5 } { 2 } \\log \\frac { 3 } { 2 }{\/tex}<\/span><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p><span class=\"math-tex\">{tex}I=\\int\\sqrt{5-4x-x^2}\\;dx{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}I=\\int\\sqrt{-\\left(x^2+4x+4-4-5\\right)}\\;dx{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}I=\\int\\sqrt{-\\left(x+2\\right)^2+9}\\;dx{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}I=\\int\\sqrt{\\left(3\\right)^2-\\left(x+2\\right)^2}\\;dx{\/tex}<\/span><br \/>\nAs, <span class=\"math-tex\">{tex}\\int\\sqrt{\\left(a\\right)^2-\\left(x\\right)^2}\\;dx{\/tex}<\/span> <span class=\"math-tex\">{tex}=\\frac x2\\sqrt{a^2-x^2}+\\frac{a^2}2\\sin^{-1}{\\textstyle\\frac xa}+C{\/tex}<\/span><br \/>\nSo, <span class=\"math-tex\">{tex}\\int\\sqrt{\\left(3\\right)^2-\\left(x+2\\right)^2}\\;dx{\/tex}<\/span> <span class=\"math-tex\">{tex}=\\frac{x+2}2\\sqrt{5-4x-x^2}+\\frac92\\sin^{-1}{\\textstyle\\frac{x+2}3}+C{\/tex}<\/span><\/li>\n<li style=\"clear: both;\">We can rewrite the given differential equation as,<br \/>\n<span class=\"math-tex\">{tex}\\frac{d y}{d x}=\\frac{y}{x}-\\sin ^{2}\\left(\\frac{y}{x}\\right){\/tex}<\/span><br \/>\nThis is of the form\u00a0<span class=\"math-tex\">{tex}\\frac{d y}{d x}=f\\left(\\frac{y}{x}\\right){\/tex}<\/span>\u00a0So, it is homogeneous.<br \/>\nPutting y = vx and\u00a0<span class=\"math-tex\">{tex}\\frac{d y}{d x}=v+x \\frac{d v}{d x}{\/tex}<\/span>\u00a0in (i), we get<br \/>\n<span class=\"math-tex\">{tex}v+x \\frac{d v}{d x}{\/tex}<\/span>\u00a0= v -sin<sup>2<\/sup> v<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad x \\frac{d v}{d x}{\/tex}<\/span>\u00a0= -sin<sup>2<\/sup> v<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow-cosec ^{2} v d v=\\frac{1}{x} d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\int\\left(-cosec ^{2} v\\right) d v=\\int \\frac{1}{x} d x{\/tex}<\/span>\u00a0[on integrating both sides]\n<span class=\"math-tex\">{tex}\\Rightarrow {\/tex}<\/span>\u00a0cot v = log |x| + C,\u00a0where C\u00a0is an arbitrary constant<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow {\/tex}<\/span>\u00a0cot <span class=\"math-tex\">{tex}\\frac{y}{x}{\/tex}<\/span>\u00a0= log |x| + C &#8230;(ii) <span class=\"math-tex\">{tex}\\left[\\because v=\\frac{y}{x}\\right]{\/tex}<\/span><br \/>\nPutting x = 1 and y =\u00a0<span class=\"math-tex\">{tex}\\frac{\\pi}{4}{\/tex}<\/span>\u00a0in (ii), we get C = 1.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0cot\u00a0<span class=\"math-tex\">{tex}\\frac{y}{x}{\/tex}<\/span>= log |x| + 1 is the desired solution.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>It is given that\u00a0<span class=\"math-tex\">{tex}x \\log x \\frac{d y}{d x}+y=\\frac{2}{x} \\log x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{d y}{d x}+\\frac{y}{x \\log x}=\\frac{2}{x^{2}}{\/tex}<\/span>,<br \/>\nThis is equation in the form of\u00a0<span class=\"math-tex\">{tex}\\frac{d y}{d x}+p y=Q{\/tex}<\/span>\u00a0(where, p =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{x \\log x}{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}Q=\\frac{2}{x^{2}}{\/tex}<\/span>\u00a0)<br \/>\nNow, I.F. =\u00a0<span class=\"math-tex\">{tex}\\mathrm{e}^{\\int \\mathrm{pdx}}=\\mathrm{e}^{\\int \\frac{1}{\\mathrm{xlog} \\mathrm{x}} \\mathrm{dx}}=\\mathrm{e}^{\\log |\\log \\mathrm{x}|}=\\log \\mathrm{x}{\/tex}<\/span><br \/>\nThus, the solution of the given differential equation is given by the relation:<br \/>\n<span class=\"math-tex\">{tex}\\mathrm{y}(\\mathrm{I}. \\mathrm{F} )=\\int(\\mathrm{Q} \\times \\mathrm{I.F}) \\mathrm{d} \\mathrm{x}+\\mathrm{C}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\mathrm{y} . \\log \\mathrm{x}=\\int\\left[\\frac{2}{\\mathrm{x}^{2}} \\cdot \\log \\mathrm{x}\\right] \\mathrm{d} \\mathrm{x}+\\mathrm{C}{\/tex}<\/span>\u00a0&#8230;.(i)<br \/>\nNow,\u00a0<span class=\"math-tex\">{tex}\\int\\left[\\frac{2}{x^{2}} \\cdot \\log x\\right] d x=2 \\int\\left(\\log x \\cdot \\frac{1}{x^{2}}\\right) d x{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}2\\left[\\log x . \\int \\frac{1}{x^{2}} d x-\\int\\left\\{\\frac{d}{d x}(\\log x) \\cdot \\int \\frac{1}{x^{2}} d x\\right\\} d x\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}2\\left[\\log _{\\mathrm{X}}\\left(-\\frac{1}{\\mathrm{x}}\\right)-\\int\\left(\\frac{1}{\\mathrm{x}} \\cdot\\left(-\\frac{1}{\\mathrm{x}}\\right)\\right) \\mathrm{d} \\mathrm{x}\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}2\\left[-\\frac{\\log x}{x}+\\int \\frac{1}{x^{2}} d x\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}2\\left[-\\frac{\\log x}{x}-\\frac{1}{x}\\right]{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}-\\frac{2}{x}(1+\\log x){\/tex}<\/span><br \/>\nNow, substituting the value in (i), we get,<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\mathrm{y} \\cdot \\log \\mathrm{x}=-\\frac{2}{\\mathrm{x}}(1+\\log \\mathrm{x})+\\mathrm{C}{\/tex}<\/span><br \/>\nTherefore, the required general solution of the given differential equation is<br \/>\n<span class=\"math-tex\">{tex}\\text { y. } \\log x=-\\frac{2}{x}(1+\\log x)+C{\/tex}<\/span><\/li>\n<li style=\"clear: both;\">First, we will convert the given inequations into equations, we obtain the following equations:<br \/>\nx + y = 4,<br \/>\nx + y = 3,<br \/>\nx &#8211; 2y = 2,<br \/>\nx = 0 and y = 0<br \/>\nThe line x + 3y = 6 meets the coordinate axis at A(6, 0) and B(0, 2). Join these points to obtain the line x + 3y = 6<br \/>\nClearly, (0,0) does not satisfies the inequation x + 3y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 6 . So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.<br \/>\nThe line x &#8211; 3y = 3 meets the coordinate axis at C(3,0) and D(0, &#8211; 1). Join these points to obtain the line x &#8211; 3y = 3. Clearly, (0,0) satisfies the inequation x &#8211; 3y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 3.50, the region in xy-plane that contains the origin represents the solution set of the given equation.<br \/>\nThe line 3x + 4 y = 24 meets the coordinate axis at E(8,0) and F(0,6). Join these points to obtain the line 3 x + 4 y = 24. Clearly, (0, 0) satisfies the inequation 3 x + 4 y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 24. So, the region in xy-plane that contains the origin represents the solution set of the given equation.<br \/>\nThe line -3x + 2y = 6 meets the coordinate axis at G( &#8211; 2,0) and H(0,3). Join these points to obtain the line &#8211; 3 x + 2y = 6 Clearly, (0,0) satisfies the inequation &#8211; 3 x + 2y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 6 . So, the region in xy-plane that contains the origin represents the solution set of the given equation.<br \/>\nThe line 5x + y = 5 meets the coordinate axis at x(1,0) and y(0,5) ) . Join these points to obtain line 5x + y = 5<br \/>\nClearly, (0,0) does not satisfies the inequation 5 x + y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 5 . So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.<br \/>\nThe region represented by x <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 0 and y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 0 (non negative restrictions)since every point in the first quadrant satisfies these inequations. So, the\u00a0 graph will be in first quadrant.These lines are drawn using a suitable scale.<br \/>\n<img decoding=\"async\" style=\"width: 250px; height: 186px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/YYu6rE7.png\" alt=\"\" data-imgur-src=\"YYu6rE7.png\" \/><br \/>\nThe corner points of the feasible region are <span class=\"math-tex\">{tex}P\\left(\\frac{4}{13}, \\frac{45}{13}\\right){\/tex}<\/span> <span class=\"math-tex\">{tex}\\mathrm{k}\\left(\\frac{4}{3}, 5\\right){\/tex}<\/span>, <span class=\"math-tex\">{tex}L \\left(\\frac{84}{13}, \\frac{15}{13}\\right){\/tex}<\/span>, <span class=\"math-tex\">{tex}\\mathrm{M}\\left(\\frac{9}{2}, \\frac{1}{2}\\right){\/tex}<\/span>, <span class=\"math-tex\">{tex}\\mathrm{N}\\left(\\frac{9}{14}, \\frac{25}{14}\\right){\/tex}<\/span><br \/>\nThe value of objective function Z = 2x + y at the corner point are as follows:<br \/>\n<span class=\"math-tex\">{tex}P\\left(\\frac{4}{13}, \\frac{45}{13}\\right){\/tex}<\/span>\u00a0:\u00a0<span class=\"math-tex\">{tex}2 \\times \\frac{4}{13}+\\frac{45}{13}=\\frac{53}{13}{\/tex}<\/span><br \/>\n<s><span class=\"math-tex\">{tex}\\mathrm{K}\\left(\\frac{4}{3}, 5\\right){\/tex}<\/span><\/s>:\u00a0<span class=\"math-tex\">{tex}2 \\times \\frac{4}{3}+5=\\frac{23}{3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}L\\left(\\frac{84}{13}, \\frac{15}{13}\\right){\/tex}<\/span>\u00a0;\u00a0<span class=\"math-tex\">{tex}2 \\times \\frac{84}{13}+\\frac{15}{13}=\\frac{183}{13}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}M\\left(\\frac{9}{2}, \\frac{1}{2}\\right){\/tex}<\/span>\u00a0:\u00a0<span class=\"math-tex\">{tex}2 \\times \\frac{9}{2}+\\frac{1}{2}=\\frac{19}{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\mathrm{N}\\left(\\frac{9}{14}, \\frac{25}{14}\\right){\/tex}<\/span>\u00a0:\u00a0<span class=\"math-tex\">{tex}2 \\times \\frac{9}{14}+\\frac{25}{14}=\\frac{43}{14}{\/tex}<\/span><br \/>\nWe see that the minimum value of the objective function Z is <span class=\"math-tex\">{tex}\\frac{43}{14}{\/tex}<\/span> which is at <span class=\"math-tex\">{tex}\\mathrm{N}\\left(\\frac{9}{14}, \\frac{25}{14}\\right){\/tex}<\/span>, and maximum value of the objective function is <span class=\"math-tex\">{tex}\\frac{183}{13}{\/tex}<\/span> which is at <span class=\"math-tex\">{tex}L\\left(\\frac{84}{13}, \\frac{15}{13}\\right){\/tex}<\/span>.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>We have to maximize Z= 60 x + 15y First, we will convert the given inequations into equations, we obtain the following equations:<br \/>\nx + y = 50, 3x + y = 90, x = 0 and y = 0<br \/>\nRegion represented by x + y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 50 :<br \/>\nThe line x + y = 50 meets the coordinate axes at A(50, 0) and B(0, 50) respectively. By joining these points we obtain the line 3x + 5 y = 15 Clearly (0, 0) satisfies the inequation x + y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 50 . Therefore, the region containing the origin represents the solution<br \/>\nset of the inequation x + y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 50<br \/>\nRegion represented by 3x + y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 90 :<br \/>\nThe line 3x + y = 90 meets the coordinate axes at C(30, 0) and D(0, 90) respectively. By joining these points we obtain the line 3x + y = 90 Clearly (0, 0) satisfies the inequation 3x + y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 90 . Therefore, the region containing the origin represents the solution set of the inequation 3x + y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 90<br \/>\nRegion represented by x <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 0 and y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 0 :<br \/>\nsince, every point in the first quadrant satisfies these inequations. Therefore, the first quadrant is the region represented by the inequations x <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 0, and y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 0.<br \/>\nThe feasible region is given by\u00a0<span class=\"math-tex\">{tex}{\/tex}<\/span><span class=\"math-tex\">{tex}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}{\/tex}<\/span><span class=\"math-tex\">{tex}{\/tex}<\/span><br \/>\n<img decoding=\"async\" style=\"width: 250px; height: 328px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/Fqi4Bnl.png\" alt=\"\" data-imgur-src=\"Fqi4Bnl.png\" \/><br \/>\nThe corner points of the feasible region are O(0, 0), C(30, 0) E(20, 30) and B(0, 50)<br \/>\nThe values of Z\u00a0 at these corner points are as follows given by<br \/>\nCorner point Z = 60 x + 15 y<br \/>\nO(0, 0) : <span class=\"math-tex\">{tex}60 \\times 0+15 \\times 0=0{\/tex}<\/span><br \/>\nC(30, 0) : <span class=\"math-tex\">{tex}60 \\times 30+15 \\times 0=1800{\/tex}<\/span><br \/>\nE(20, 30) : <span class=\"math-tex\">{tex}60 \\times 20+15 \\times 30=1650{\/tex}<\/span><br \/>\nB(0, 50) : <span class=\"math-tex\">{tex}60 \\times 0+15 \\times 50=750{\/tex}<\/span><br \/>\nTherefore, the maximum value of Z is 1800 at the point (30, 0) Hence, x = 30 and y = 0 is the optimal solution of the given LPP. Thus, the optimal value of Z is 1800.This is the required solution.<\/li>\n<li>Given function is:\u00a0<span class=\"math-tex\">{tex}f(x)=\\left\\{\\begin{array}{l} 1, \\text { if } x \\leq 3 \\\\ a x+b, \\text { if } 3&lt;x&lt;5 \\\\ 7, \\text { if } x \\geq 5 \\end{array}\\right.{\/tex}<\/span><br \/>\nWe have,<br \/>\n(LHL at x = 3) =\u00a0<span class=\"math-tex\">{tex}\\lim _\\limits{x \\rightarrow 3^{-}} f(x)=\\lim _\\limits{h \\rightarrow 0} f(3-h){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\lim _\\limits{h \\rightarrow 0}(1)=1{\/tex}<\/span><br \/>\n(RHL at x = 3) =\u00a0<span class=\"math-tex\">{tex}\\lim _\\limits{x \\rightarrow 3^{+}} f(x)=\\lim _\\limits{h \\rightarrow 0} f(3+h){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\lim _\\limits{h \\rightarrow 0} a(3+h)+b=3 a+b{\/tex}<\/span><br \/>\n(LHL at x = 5) =\u00a0<span class=\"math-tex\">{tex}\\lim _\\limits{x \\rightarrow 5^{-}} f(x)=\\lim _\\limits{h \\rightarrow 0} f(5-h){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\lim _\\limits{h \\rightarrow 0}(a(5-h)+b)=5 a+b{\/tex}<\/span><br \/>\n(RHL at x = 5) =\u00a0<span class=\"math-tex\">{tex}\\lim _\\limits{x \\rightarrow 5^{+}} f(x)=\\lim _\\limits{h \\rightarrow 0} f(5+h){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\lim _\\limits{h \\rightarrow 0}{\/tex}<\/span>\u00a07 = 7<br \/>\nIf f(x) is continuous at x = 3 and 5, then<br \/>\n<span class=\"math-tex\">{tex}\\lim _\\limits{x \\rightarrow 3^{-}} f(x)=\\lim _\\limits{x \\rightarrow 3^{+}} f(x) \\text { and } \\lim _\\limits{x \\rightarrow 5^{-}} f(x)=\\lim _\\limits{x \\rightarrow 5^{+}} f(x){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a01 = 3a + b &#8230;(i)<br \/>\nand 5a + b = 7 &#8230;(ii)<br \/>\nOn solving eqs. (i) and (ii), we get,<br \/>\na = 3 and b = -8<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section D <\/b><\/li>\n<li>According to the question ,Given equation of circle is <span class=\"math-tex\">{tex}x^2 + y^2 = 4{\/tex}<\/span>&#8230;&#8230;(i)<br \/>\nOn differentiating Eq. (i) w.r.t. x , we get<br \/>\n<span class=\"math-tex\">{tex}2x + 2y\\frac{dy}{dx}{\/tex}<\/span>= 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad x + y \\frac { d } { d x } = 0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\frac { d y } { d x } = &#8211; \\frac { x } { y }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\left( \\frac { d y } { d x } \\right) _ { ( 1 , \\sqrt { 3 } ) } = &#8211; \\frac { 1 } { \\sqrt { 3 } }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore m = &#8211; \\frac { 1 } { \\sqrt { 3 } }{\/tex}<\/span>\u00a0( &#8216;m&#8217; is slope of tangent )<br \/>\nNow, equation of tangent at point (1, <span class=\"math-tex\">{tex}\\sqrt3{\/tex}<\/span>) is<br \/>\n<span class=\"math-tex\">{tex}( y &#8211; \\sqrt { 3 } ) = &#8211; \\frac { 1 } { \\sqrt { 3 } } ( x &#8211; 1 ){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\sqrt3 y &#8211; 3 = -x + 1{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad x + \\sqrt { 3 } y = 4{\/tex}<\/span>\u00a0&#8230;&#8230;.(ii)<br \/>\nequation of normal passing through\u00a0point (1, <span class=\"math-tex\">{tex}\\sqrt3{\/tex}<\/span>) andslope of normal =\u00a0<span class=\"math-tex\">{tex}\\frac {-1}{m_{(tangent )}} = \\sqrt3{\/tex}<\/span><span class=\"math-tex\">{tex}(y &#8211;\u00a0\\sqrt3) =\u00a0\\sqrt3 (x &#8211; 1){\/tex}<\/span>(y &#8211;\u00a0<span class=\"math-tex\">{tex}\\sqrt3{\/tex}<\/span>) =\u00a0<span class=\"math-tex\">{tex}\\sqrt3{\/tex}<\/span>x &#8211;\u00a0<span class=\"math-tex\">{tex}\\sqrt3{\/tex}<\/span>\u00a0&#8230;&#8230;..(iii)<br \/>\nNow, the Eqs. (i) and (ii) can be represented in the graph as shown below:<br \/>\n<img decoding=\"async\" style=\"width: 200px; height: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/OiW6vtR.png\" alt=\"\" data-imgur-src=\"OiW6vtR.png\" \/><br \/>\nOn putting y = 0 in Eq. (i), we get<br \/>\nx + 0 = 4<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x = 4<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> the tangent line x + <span class=\"math-tex\">{tex}\\sqrt3{\/tex}<\/span>y = 4 cuts the X-axis at A(4,0).<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Required area = Area of shaded region OAB<br \/>\n<span class=\"math-tex\">{tex}= \\int _ { 0 } ^ { 1 } y _ { ( \\text { equation of normal) } } d x{\/tex}<\/span>+\u00a0<span class=\"math-tex\">{tex}\\int _ { 1 } ^ { 4 } y _\\text{ (equation of tangent) } d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\int _ { 0 } ^ { 1 } \\sqrt { 3 } x d x + \\int _ { 1 } ^ { 4 } \\left( \\frac { 4 &#8211; x } { \\sqrt { 3 } } \\right) d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\sqrt { 3 } \\left[ \\frac { x ^ { 2 } } { 2 } \\right] _ { 0 } ^ { 1 } + \\frac { 1 } { \\sqrt { 3 } } \\left[ 4 x &#8211; \\frac { x ^ { 2 } } { 2 } \\right] _ { 1 } ^ { 4 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\frac { \\sqrt { 3 } } { 2 } + \\frac { 1 } { \\sqrt { 3 } } \\left[ 16 &#8211; \\frac { 16 } { 2 } &#8211; 4 + \\frac { 1 } { 2 } \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\frac { \\sqrt { 3 } } { 2 } + \\frac { 1 } { \\sqrt { 3 } } \\left[ 12 &#8211; \\frac { 15 } { 2 } \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\frac { \\sqrt { 3 } } { 2 } + \\frac { 1 } { \\sqrt { 3 } } \\left[ \\frac { 9 } { 2 } \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\frac { \\sqrt { 3 } } { 2 } + \\frac { 3 \\sqrt { 3 } } { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\frac { 4 \\sqrt { 3 } } { 2 } {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= 2 \\sqrt { 3 }{\/tex}<\/span>sq units.<\/li>\n<li style=\"clear: both;\">f is one-one: For any x, y <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> R, we have f(x) : f(y)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{x}{{1 + |x|}} = \\frac{y}{{|y| + 1}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> xy + x = xy + y<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x = y<br \/>\nTherefore, f is one-one function.<br \/>\nIf f is one-one, let y = R \u2013 {1}, then f(x) = y<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{x}{{x + 1}} = y{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow x = \\frac{y}{{1 &#8211; y}}{\/tex}<\/span><br \/>\nIt is clear that x <span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0R for all y = R \u2013 {1}, also x <span class=\"math-tex\">{tex}\\ne{\/tex}<\/span>=-1<br \/>\nBecause x = -1<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{y}{{1 &#8211; y}} = &#8211; 1{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> y = -1 + y which is not possible.<br \/>\nThus for each R \u2013 {1} there exists <span class=\"math-tex\">{tex}x = \\frac{y}{{1 &#8211; y}} \\in{\/tex}<\/span> R \u2013 {1} such that<br \/>\n<span class=\"math-tex\">{tex}f(x) = \\frac{x}{{x + 1}} = \\frac{{\\frac{y}{{1 &#8211; y}}}}{{\\frac{y}{{1 &#8211; y}} + 1}} = y{\/tex}<\/span><br \/>\nTherefore f is onto function.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>We observe the following properties of relation R.<br \/>\nReflexivity: Let (a, b) be an arbitrary element of N <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N. Then,<br \/>\n(a, b)\u00a0<span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0N\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0N<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a, b\u00a0<span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0N<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0ab = ba [By commutativity of multiplication on N]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(a, b) R (a, b)<br \/>\nThus, (a, b) R (a, b) for all (a, b) <span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0N\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N.<br \/>\nSo, R is reflexive on N <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N.<br \/>\nSymmetry: Let (a, b), (c, d)\u00a0<span class=\"math-tex\">{tex}\\in{\/tex}<\/span> N <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N be such that (a, b) R (c, d). Then,<br \/>\n(a, b) R (c, d)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0ad = bc<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0cb =\u00a0da [By commutativity of multiplication on N]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(c,d) R (a, b)<br \/>\nThus, (a, b) R (c, d)\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> (c, d) R (a, b) for all (a, b), (c, d ) <span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0N <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N .<br \/>\nSo, R is symmetric on N <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N.<br \/>\nTransitivity: Let (a, b), (c, d), (e, f)\u00a0<span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0N\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>N such that (a, b) R (c, d) and (c, d) R (e, f). Then,<br \/>\n<span class=\"math-tex\">{tex}\\begin{array}{l}{(a, b) R(c, d) \\Rightarrow a d=b c} \\\\ {(c, d) R(e, f) \\Rightarrow c f=d e}\\end{array} \\} \\Rightarrow(a d)(c f){\/tex}<\/span>=\u00a0(bc) (de)\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0af = be\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(a, b) R (e, f)<br \/>\nThus, (a, b) R (c, d) and (c, d) R (e, f) <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> (a, b) R (e, f) for all (a, b), (c, d), (e, f) <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> N<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N.<br \/>\nSo, R is transitive on N <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N.<br \/>\nHence, R being reflexive, symmetric and transitive, is an equivalence relation on N<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N.<br \/>\n[(2, 6)] = {(x, y) <span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0N<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N :(x, y) R (2, 6)}<br \/>\n= {( x, y) <span class=\"math-tex\">{tex}\\in{\/tex}<\/span> N <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> N :3x = y}<br \/>\n= {(x,\u00a03x ) : x\u00a0<span class=\"math-tex\">{tex}\\in{\/tex}<\/span> N ] ={(1, 3), (2, 6), (3, 9), (4,12),..}<\/li>\n<li>Given: Matrix <span class=\"math-tex\">{tex}A = \\left[ {\\begin{array}{*{20}{c}} 2&amp;{ &#8211; 3}&amp;5 \\\\ 3&amp;2&amp;{ &#8211; 4} \\\\ 1&amp;1&amp;{ &#8211; 2} \\end{array}} \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0|A|\u00a0=\u00a0<span class=\"math-tex\">{tex}\\left| {\\begin{array}{*{20}{c}} 2&amp;{ &#8211; 3}&amp;5 \\\\ 3&amp;2&amp;{ &#8211; 4} \\\\ 1&amp;1&amp;{ &#8211; 2} \\end{array}} \\right|{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> |A| = 2(-4 + 4) &#8211; (-3)(-6 + 4) + 5(3 &#8211; 2) = 0 &#8211; 6 + 5 = -1 <span class=\"math-tex\">{tex} \\ne{\/tex}<\/span> 0<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0A<sup>-1<\/sup> exists and A<sup>-1<\/sup>\u00a0=\u00a0<span class=\"math-tex\">{tex} \\frac{1}{{\\left| A \\right|}}{\/tex}<\/span>\u00a0(adj. A)&#8230;(i)<br \/>\nNow, A<sub>11<\/sub> = 0, A<sub>12<\/sub> = 2, A<sub>13<\/sub> = 1 and A<sub>21<\/sub> = -1, A<sub>22<\/sub> = -9, A<sub>23<\/sub> = -5 and A<sub>31<\/sub> = 2, A<sub>32<\/sub> = 23, A<sub>33<\/sub> = 13<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0adj.\u00a0A =<span class=\"math-tex\">{tex} \\left[ {\\begin{array}{*{20}{c}} 0&amp;2&amp;1 \\\\ { &#8211; 1}&amp;{ &#8211; 9}&amp;{ &#8211; 5} \\\\ 2&amp;{23}&amp;{13} \\end{array}} \\right]&#8217; = \\left[ {\\begin{array}{*{20}{c}} 0&amp;{ &#8211; 1}&amp;2 \\\\ 2&amp;{ &#8211; 9}&amp;{23} \\\\ 1&amp;{ &#8211; 5}&amp;{13} \\end{array}} \\right]{\/tex}<\/span><br \/>\nFrom eq. (i),<br \/>\nA<sup>-1\u00a0<\/sup>=\u00a0<span class=\"math-tex\">{tex} \\frac{1}{{ &#8211; 1}}\\left[ {\\begin{array}{*{20}{c}} 0&amp;{ &#8211; 1}&amp;2 \\\\ 2&amp;{ &#8211; 9}&amp;{23} \\\\ 1&amp;{ &#8211; 5}&amp;{13} \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 0&amp;1&amp;{ &#8211; 2} \\\\ { &#8211; 2}&amp;9&amp;{ &#8211; 23} \\\\ { &#8211; 1}&amp;5&amp;{ &#8211; 13} \\end{array}} \\right]{\/tex}<\/span><br \/>\nNow, Matrix form of given equations is AX = B<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\left[ {\\begin{array}{*{20}{c}} 2&amp;{ &#8211; 3}&amp;5 \\\\ 3&amp;2&amp;{ &#8211; 4} \\\\ 1&amp;1&amp;{ &#8211; 2} \\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}} x \\\\ y \\\\ z \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} {11} \\\\ { &#8211; 5} \\\\ { &#8211; 3} \\end{array}} \\right]{\/tex}<\/span><br \/>\nHere A =<span class=\"math-tex\">{tex} \\left[ {\\begin{array}{*{20}{c}} 2&amp;{ &#8211; 3}&amp;5 \\\\ 3&amp;2&amp;{ &#8211; 4} \\\\ 1&amp;1&amp;{ &#8211; 2} \\end{array}} \\right]{\/tex}<\/span>, X = <span class=\"math-tex\">{tex}\\left[\\begin{array}{l} x \\\\ y \\\\ z \\end{array}\\right]{\/tex}<\/span>\u00a0and B =<span class=\"math-tex\">{tex} \\left[ {\\begin{array}{*{20}{c}} {11} \\\\ { &#8211; 5} \\\\ { &#8211; 3} \\end{array}} \\right]{\/tex}<\/span><br \/>\nTherefore, solution is unique and X = A<sup>-1<\/sup>B<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\left[ {\\begin{array}{*{20}{c}} x \\\\ y \\\\ z \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 0&amp;1&amp;{ &#8211; 2} \\\\ { &#8211; 2}&amp;9&amp;{ &#8211; 23} \\\\ { &#8211; 1}&amp;5&amp;{ &#8211; 13} \\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}} {11} \\\\ { &#8211; 5} \\\\ { &#8211; 3} \\end{array}} \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\left[ {\\begin{array}{*{20}{c}} {0 &#8211; 5 + 6} \\\\ { &#8211; 22 &#8211; 45 + 69} \\\\ { &#8211; 11 &#8211; 25 + 39} \\end{array}} \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\left[ {\\begin{array}{*{20}{c}} 1 \\\\ 2 \\\\ 3 \\end{array}} \\right]{\/tex}<\/span><br \/>\nTherefore, x = 1, y = 2 and z = 3<\/li>\n<li style=\"clear: both;\">The given equations are<br \/>\nal + bm + cn = 0 &#8230;.(i)<br \/>\nand, ul<sup>2<\/sup> + vm<sup>2<\/sup> + wn<sup>2<\/sup> = 0 ..(ii)<br \/>\nFrom (i), we get<br \/>\nn = &#8211;<span class=\"math-tex\">{tex}\\left(\\frac{a l+b m}{c}\\right){\/tex}<\/span><br \/>\nSubstituting n = &#8211;<span class=\"math-tex\">{tex}\\left(\\frac{a l+b m}{c}\\right){\/tex}<\/span>\u00a0in (ii), we get<br \/>\nul<sup>2<\/sup> + vm<sup>2<\/sup> +\u00a0w\u00a0<span class=\"math-tex\">{tex}\\frac{(a l+b m)^{2}}{c^{2}}{\/tex}<\/span>\u00a0= 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(c<sup>2<\/sup>u + a<sup>2<\/sup>w)l<sup>2\u00a0<\/sup>+ 2abwlm + (c<sup>2<\/sup>v + b<sup>2<\/sup>w)m<sup>2\u00a0<\/sup>= 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left(a^{2} w+c^{2} u\\right)\\left(\\frac{l}{m}\\right)^{2}+2 a b w\\left(\\frac{l}{m}\\right)+\\left(b^{2} w+c^{2} v\\right)=0{\/tex}<\/span>\u00a0&#8230;.(iii)<br \/>\nThis is a quadratic equation in\u00a0<span class=\"math-tex\">{tex}\\frac{l}{m}{\/tex}<\/span>.\u00a0So, it gives two values of\u00a0<span class=\"math-tex\">{tex}\\frac{l}{m}{\/tex}<\/span>. Suppose the two values be\u00a0<span class=\"math-tex\">{tex}\\frac{l_{1}}{m_{1}}{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\frac{l_{2}}{m_{2}}{\/tex}<\/span>.<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\quad \\frac{l_{1}}{m_{1}}, \\frac{l_{2}}{m_{2}}=\\frac{b^{2} w+c^{2} v}{a^{2} w+c^{2} u} \\Rightarrow \\frac{l_{1} l_{2}}{b^{2} w+c^{2} v}=\\frac{m_{1} m_{2}}{a^{2} w+c^{2} u}{\/tex}<\/span>\u00a0&#8230;..(iv)<br \/>\nSimilarly, by making a quadratic equation in\u00a0<span class=\"math-tex\">{tex}\\frac{m}{n}{\/tex}<\/span>, we obtain<br \/>\n<span class=\"math-tex\">{tex}\\frac{m_{1} m_{2}}{a^{2} w+c^{2} u}=\\frac{n_{1} n_{2}}{a^{2} v+b^{2} u}{\/tex}<\/span>\u00a0&#8230;.(v)<br \/>\nFrom (iv) and (v), we get<br \/>\n<span class=\"math-tex\">{tex}\\frac{l_{1} l_{2}}{b^{2} w+c^{2} v}=\\frac{m_{1} m_{2}}{a^{2} w+c^{2} u}=\\frac{n_{1} n_{2}}{a^{2} v+b^{2} u}=\\lambda{\/tex}<\/span>\u00a0(say)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}l_{1} l_{2}=\\lambda\\left(b^{2} w+c^{2} v\\right), m_{1} m_{2}=\\lambda\\left(a^{2} w+c^{2} u\\right), n_{1} n_{2}=\\lambda\\left(a^{2} v+b^{2} u\\right){\/tex}<\/span><br \/>\nFor the given lines to be perpendicular, we must have<br \/>\nl<sub>1<\/sub> l<sub>2<\/sub> + m<sub>1<\/sub> m<sub>2<\/sub> + n<sub>1<\/sub> n<sub>2<\/sub> = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\lambda\\left(b^{2} w+c^{2} v\\right)+\\lambda\\left(a^{2} w+c^{2} u\\right)+\\lambda\\left(a^{2} v+b^{2} u\\right)=0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a<sup>2<\/sup> (v + w) + b<sup>2<\/sup> (u + w) + c<sup>2<\/sup> (u + v) = 0<br \/>\nFor the given lines to be parallel, the direction cosines must be equal and so the roots of the equation (iii) must be equal.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a04a<sup>2\u00a0<\/sup>b<sup>2\u00a0<\/sup>w<sup>2\u00a0<\/sup>\u2212 4(a<sup>2<\/sup>w + c<sup>2<\/sup>u) (b<sup>2<\/sup>w + c<sup>2<\/sup>v) = 0\u00a0[On equating discriminant to zero]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a<sup>2<\/sup> c<sup>2<\/sup> vw + b<sup>2<\/sup>\u00a0c<sup>2<\/sup> uw + c<sup>4<\/sup> uv = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a<sup>2<\/sup> vw + b<sup>2<\/sup>\u00a0c<sup>2<\/sup> uw + c<sup><span style=\"font-size: 10.8333px;\">2<\/span><\/sup><span style=\"font-size: 10.8333px;\">u<\/span>v = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{a^{2}}{u}+\\frac{b^{2}}{v}+\\frac{c^{2}}{w}{\/tex}<\/span>\u00a0= 0\u00a0[Dividing throughout by uvw] Hence the required result is proved<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p><span class=\"math-tex\">{tex}{\\vec a_1} = \\hat i + \\hat j,{\\vec b_1} = 2\\hat i &#8211; \\hat j + \\hat k{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}{\\vec a_2} = 2\\hat i + \\hat j &#8211; \\hat k,{\\vec b_2} = 3\\hat i &#8211; 5\\hat j + 2\\hat k{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}{\\vec a_2} &#8211; {\\vec a_1} = \\hat i &#8211; \\hat k{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}{\\vec b_1} \\times {\\vec b_2} = \\left| {\\begin{array}{*{20}{c}} {\\hat i}&amp;{\\hat j}&amp;{\\hat k} \\\\ 2&amp;{ &#8211; 1}&amp;1 \\\\ 3&amp;{ &#8211; 5}&amp;2 \\end{array}} \\right|{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\hat i (-2+5)-\\hat j(4-3)+\\hat k(-10+3){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 3\\hat i &#8211; \\hat j &#8211; 7\\hat k{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\left| {{{\\vec b}_1} \\times {{\\vec b}_2}} \\right|=\\sqrt {9+1+49}= \\sqrt {59} {\/tex}<\/span><br \/>\nAlso,\u00a0<span class=\"math-tex\">{tex}(\\vec b_1\u00d7\\vec b_2).(\\vec a_2-\\vec a_1)=(3\\hat i-\\hat j-7 \\hat k)(\\hat i-\\hat k)=3+7+0=10{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}d=\\left|\\frac{(\\vec b_1\u00d7\\vec b_2).(\\vec a_2-\\vec a_1)}{|\\vec b_1\u00d7\\vec b_2|}\\right|=\\frac{10}{\\sqrt{59}}{\/tex}<\/span><\/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;\">Let A represents obtaining a sum greater than 9 and B represents black die resulted in a 5.<br \/>\nn(S) = 36<br \/>\nn(A) = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}\u00a0=\u00a06<br \/>\nn(B) = {(5, 1),\u00a0(5, 2),\u00a0(5, 3),\u00a0(5, 4),\u00a0(5, 5)} = 6<br \/>\nn(A <span class=\"math-tex\">{tex}\\cap{\/tex}<\/span> B) = {(5, 5),\u00a0(5, 6)} = 2<br \/>\nP(A\/B) = <span class=\"math-tex\">{tex}\\frac{{P\\left( {A \\cap B} \\right)}}{{P\\left( B \\right)}}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac{{\\frac{{n\\left( {A \\cap B} \\right)}}{{n\\left( S \\right)}}}}{{\\frac{{n\\left( B \\right)}}{{n\\left( S \\right)}}}}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac{{\\frac{2}{{36}}}}{{\\frac{6}{{36}}}} = \\frac{1}{3}{\/tex}<\/span><\/li>\n<li style=\"text-align: left;\">Let A represents obtaining a sum\u00a08\u00a0and B represents red\u00a0die resulted in number less than 4.<br \/>\nn(S) = 36<br \/>\nn(A) =\u00a0{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\u00a0= 5<br \/>\nn(B) = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)} = 18<br \/>\nn(A <span class=\"math-tex\">{tex}\\cap{\/tex}<\/span> B) = {(5, 3), (6, 2)} = 2<br \/>\nP(A\/B) = <span class=\"math-tex\">{tex}\\frac{{P\\left( {A \\cap B} \\right)}}{{P\\left( B \\right)}}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac{{\\frac{{n\\left( {A \\cap B} \\right)}}{{n\\left( S \\right)}}}}{{\\frac{{n\\left( B \\right)}}{{n\\left( S \\right)}}}}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac{{\\frac{2}{{36}}}}{{\\frac{{18}}{{36}}}} = \\frac{1}{9}{\/tex}<\/span><\/li>\n<li style=\"text-align: left;\">Let A represents obtaining a sum 10\u00a0and B represents black die resulted in even number.<br \/>\nn(S) = 36<br \/>\nn(A) = {(4, 6), (6, 4), (5, 5)} = 3<br \/>\nn(B) =\u00a0{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}\u00a0= 18<br \/>\nn(A <span class=\"math-tex\">{tex}\\cap{\/tex}<\/span> B) = {(4, 6), (6, 4)} = 2<br \/>\nP(A\/B) = <span class=\"math-tex\">{tex}\\frac{{P\\left( {A \\cap B} \\right)}}{{P\\left( B \\right)}} = \\frac{{\\frac{{n\\left( {A \\cap B} \\right)}}{{n\\left( S \\right)}}}}{{\\frac{{n\\left( B \\right)}}{{n\\left( S \\right)}}}}{\/tex}<\/span> =\u00a0<span class=\"math-tex\">{tex}\\frac{{\\frac{2}{{36}}}}{{\\frac{{18}}{{36}}}} = \\frac{1}{9}{\/tex}<\/span><br \/>\n<strong>OR<\/strong><br \/>\nLet A represents getting doublet\u00a0and B represents red\u00a0die resulted in number greater than 4.<br \/>\nn(S) = 36<br \/>\nn(A) = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}\u00a0= 6<br \/>\nn(B) =\u00a0{(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 5), (5, 6), (6, 5), (6, 6)}\u00a0= 12<br \/>\nn(A <span class=\"math-tex\">{tex}\\cap{\/tex}<\/span> B) = {(4, 4), (5, 5), (6, 6)}\u00a0= 3<br \/>\nP(A\/B) = <span class=\"math-tex\">{tex}\\frac{{P\\left( {A \\cap B} \\right)}}{{P\\left( B \\right)}} = \\frac{{\\frac{{n\\left( {A \\cap B} \\right)}}{{n\\left( S \\right)}}}}{{\\frac{{n\\left( B \\right)}}{{n\\left( S \\right)}}}}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac{{\\frac{3}{{36}}}}{{\\frac{{12}}{{36}}}} = \\frac{1}{4}{\/tex}<\/span><\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>\n<p style=\"text-align: left;\">Here,<br \/>\nPosition vector of A is <span class=\"math-tex\">{tex}\\vec{a}=\\hat{i}+4 \\hat{j}+2 \\hat{k} {\/tex}<\/span><br \/>\nPosition vector of B is <span class=\"math-tex\">{tex}\\vec{b}=3 \\hat{i}-3 \\hat{j}-2 \\hat{k}{\/tex}<\/span><br \/>\nPosition vector of C is <span class=\"math-tex\">{tex}\\vec{c}=-2 \\hat{i}+2 \\hat{j}+6 \\hat{k}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\vec{a}+\\vec{b}+\\vec{c}{\/tex}<\/span> = (1 + 3 &#8211; 2)<span class=\"math-tex\">{tex}\\hat i{\/tex}<\/span> + (4 &#8211; 3 + 2)<span class=\"math-tex\">{tex}\\hat j{\/tex}<\/span> + (2 &#8211; 2 + 6)<span class=\"math-tex\">{tex}\\hat k{\/tex}<\/span><br \/>\n= 2<span class=\"math-tex\">{tex}\\hat i{\/tex}<\/span> + 3<span class=\"math-tex\">{tex}\\hat j{\/tex}<\/span> + 6<span class=\"math-tex\">{tex}\\hat k{\/tex}<\/span><br \/>\nThus, <span class=\"math-tex\">{tex}|\\vec{a}+\\vec{b}+\\vec{c}|{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\left|\\sqrt{(2)^2+(3)^2+(6)^2}\\right|{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}|\\sqrt{4+9+16}|{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\sqrt {29}{\/tex}<\/span><\/p>\n<\/li>\n<li>\n<p style=\"text-align: left;\">Given, <span class=\"math-tex\">{tex}\\vec{a}=4 \\hat{i}+6 \\hat{j}+12 \\hat{k}{\/tex}<\/span>,<br \/>\n<span class=\"math-tex\">{tex}|\\vec{a}|=\\sqrt{4^2+6^2+12^2}{\/tex}<\/span> = 14<br \/>\nTherefore, the unit vector in direction of <span class=\"math-tex\">{tex}\\vec a{\/tex}<\/span> is given by<br \/>\n<span class=\"math-tex\">{tex}\\hat{a} =\\frac{\\vec{a}}{|\\vec{a}|}=\\frac{4 \\hat{i}+6 \\hat{j}+12 \\hat{k}}{14}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{4}{14} \\hat{i}+\\frac{6}{14} \\hat{j}+\\frac{12}{14} \\hat{k}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{2}{7} \\hat{i}+\\frac{3}{7} \\hat{j}+\\frac{6}{7} \\hat{k}{\/tex}<\/span><\/p>\n<\/li>\n<li>\n<p style=\"text-align: left;\">We have, A(1, 4, 2), B(3, -3, -2) and C(-2, 2, 6)<br \/>\nNow, <span class=\"math-tex\">{tex}\\overrightarrow{A B}=\\vec{b}-\\vec{a}=2 \\hat{i}-7 \\hat{j}-4 \\hat{k}{\/tex}<\/span><br \/>\nand <span class=\"math-tex\">{tex}\\overrightarrow{A C}=\\vec{c}-\\vec{a}=-3 \\hat{i}-2 \\hat{j}+4 \\hat{k}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> <span class=\"math-tex\">{tex}\\overrightarrow{A B} \\times \\overrightarrow{A C}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\left|\\begin{array}{ccc} \\hat{i} &amp; \\hat{j} &amp; \\hat{k} \\\\ 2 &amp; -7 &amp; -4 \\\\ -3 &amp; -2 &amp; 4 \\end{array}\\right|{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\hat{i}(-28-8)-\\hat{j}(8-12)+\\hat{k}(-4-21){\/tex}<\/span><br \/>\n= &#8211; <span class=\"math-tex\">{tex}36 \\hat{i}+4 \\hat{j}-25 \\hat{k}{\/tex}<\/span><br \/>\nNow, <span class=\"math-tex\">{tex}|\\overrightarrow{A B} \\times \\overrightarrow{A C}|=\\sqrt{(-36)^2+4^2+(-25)^2}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}|\\sqrt{1296+16+625}|=\\sqrt{1937}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> Area of <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABC = <span class=\"math-tex\">{tex}\\frac{1}{2}|\\overrightarrow{A B} \\times \\overrightarrow{A C}|{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{1}{2} \\sqrt{1937}{\/tex}<\/span> sq. units<br \/>\n<strong>OR<\/strong><br \/>\nTriangle law of addition for <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABC is given by<br \/>\n<span class=\"math-tex\">{tex}\\overrightarrow{A B}+\\overrightarrow{B C}+\\overrightarrow{C A}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\vec 0{\/tex}<\/span><br \/>\nIf the given points lie on the straight line, then the points will be collinear and so area of <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABC = 0<br \/>\nThen, <span class=\"math-tex\">{tex}|\\vec{a} \\times \\vec{b}+\\vec{b} \\times \\vec{c}+\\vec{c} \\times \\vec{a}|{\/tex}<\/span> = 0.<br \/>\nAlso, if a, b, c are the position vector of the three vertices A, B and C of <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABC, then area of triangle is <span class=\"math-tex\">{tex}\\frac{1}{2}|\\vec{a} \\times \\vec{b}+\\vec{b} \\times \\vec{c}+\\vec{c} \\times \\vec{a}|{\/tex}<\/span>.<\/p>\n<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">Let \u2018y\u2019 be the breadth and \u2018x\u2019 be the length of rectangle and\u00a0\u2018a\u2019 is radius of given circle.<br \/>\nFrom fig 4a<sup>2<\/sup> = x<sup>2<\/sup> + y<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0y<sup>2<\/sup> = 4a<sup>2<\/sup> &#8211; x<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0y =\u00a0<span class=\"math-tex\">{tex}\\sqrt {4{a^2}\\, &#8211; {x^2}}{\/tex}<\/span><br \/>\nPerimeter (P) = 2x + 2y =\u00a0<span class=\"math-tex\">{tex}2\\left( {x + \\sqrt {4{a^2}\\, &#8211; {x^2}} } \\right){\/tex}<\/span><\/li>\n<li style=\"text-align: left;\">We know that P =\u00a0<span class=\"math-tex\">{tex}2\\left(x+\\sqrt{4 a^2-x^2}\\right){\/tex}<\/span><br \/>\nCritical points to maximize perimeter <span class=\"math-tex\">{tex}\\frac{dP}{dx}{\/tex}<\/span>\u00a0= 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{d p}{d x}=2\\left(1+\\frac{1}{2 \\sqrt{4 a^2-x^2}}(-2 x)\\right){\/tex}<\/span>\u00a0= 0<br \/>\n<span class=\"math-tex\">{tex}2\\left(\\frac{\\sqrt{4 a^2-x^2}-x}{\\sqrt{4 a^2-x^2}}\\right){\/tex}<\/span>\u00a0= 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\sqrt{4 a^2-x^2}{\/tex}<\/span>\u00a0= x<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a04a<sup>2<\/sup> &#8211; x<sup>2<\/sup> = x<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02a<sup>2<\/sup> = x<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x =\u00a0<span class=\"math-tex\">{tex}\\pm \\sqrt{2a}{\/tex}<\/span><br \/>\nwhen x =\u00a0<span class=\"math-tex\">{tex}\\sqrt{2a}{\/tex}<\/span>, y =\u00a0<span class=\"math-tex\">{tex}\\sqrt{2a}{\/tex}<\/span><br \/>\nwhen x =\u00a0<span class=\"math-tex\">{tex}-\\sqrt{2a}{\/tex}<\/span>\u00a0not possible as &#8216;x&#8217; is length critical point is (<span class=\"math-tex\">{tex}\\sqrt{2a}{\/tex}<\/span>,\u00a0<span class=\"math-tex\">{tex}\\sqrt{2a}{\/tex}<\/span>)<\/li>\n<li style=\"text-align: left;\"><span class=\"math-tex\">{tex}\\frac{d p}{d x}=2\\left(1+\\frac{1}{2 \\sqrt{4 a^2-x^2}}(-2 x)\\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{d^2 P}{d x^2}=-2\\left(\\frac{\\sqrt{4 a^2-x^2}-(x)\\left(\\frac{-2 x}{2 \\sqrt{4^2-x^2}}\\right)}{\\left(4 a^2-x^2\\right)}\\right){\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}-2\\left(\\frac{\\left(4 a^2-x^2\\right)+x^2}{\\left(4 a^2-x^2\\right)^{3 \/ 2}}\\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\left.\\Rightarrow \\frac{d^2 P}{d x^2}\\right]_{x=a \\sqrt{2}}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}-2\\left(\\frac{4 a^2}{\\left(4 a^2-2 a^2\\right)^{3 \/ 2}}\\right)=\\frac{-2}{(2 \\sqrt{2}) a}&lt;0{\/tex}<\/span><br \/>\nPerimeter is maximum at a critical point.<br \/>\n<strong>OR<\/strong><br \/>\nFrom the above results know that x = y =\u00a0<span class=\"math-tex\">{tex}\\sqrt{2}a{\/tex}<\/span><br \/>\na = radius<br \/>\nHere, x = y =\u00a0<span class=\"math-tex\">{tex}10\\sqrt{2}{\/tex}<\/span><br \/>\nPerimeter = P = 4\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0side =\u00a0<span class=\"math-tex\">{tex}40\\sqrt{2}{\/tex}<\/span>\u00a0cm<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p><strong>Download myCBSEguide App for Complete Exam Preparation<\/strong><\/p>\n<p>To practice more questions and prepare effectively for your exams, <strong>download the <a href=\"https:\/\/play.google.com\/store\/search?q=mycbseguide+app&amp;c=apps\">myCBSEguide app<\/a><\/strong>. It offers comprehensive study material for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. The app provides access to <strong>sample papers<\/strong>, <strong>previous year questions<\/strong>, <strong>chapter-wise practice tests<\/strong>, and more, helping you get ready for exams in a structured manner.<\/p>\n<p>For <strong>teachers<\/strong>, the <strong><a href=\"https:\/\/play.google.com\/store\/search?q=Examin8&amp;c=apps\">Examin8.com<\/a> app<\/strong> allows you to easily create customized papers with your own name and logo, making exam preparation more efficient for your students.<\/p>\n<p><strong>Note<\/strong>: These are just the questions! To view and download the <strong>complete question paper with solutions<\/strong>, install the <strong>myCBSEguide app<\/strong> from the <strong>Google Play Store<\/strong> or log in to our <strong>Student Dashboard<\/strong> on the <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide website<\/strong><\/a>.<\/p>\n<p>Get started today and ensure you are fully prepared for your exams with <strong>myCBSEguide<\/strong>!<\/p>\n<p style=\"text-align: center;\"><b><strong><a class=\"button\" href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\">Download myCBSEguide App<\/a><\/strong><\/b><\/p>\n<h2><span class=\"ez-toc-section\" id=\"Sample_Papers_for_Class_12_2025\"><\/span>Sample Papers for Class 12 2025<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-physics\/1251\/\"><strong>Physics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-chemistry\/1267\/\" target=\"_blank\" rel=\"noopener\"><strong>Chemistry<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-mathematics\/1284\/\" target=\"_blank\" rel=\"noopener\"><strong>Mathematics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-biology\/1298\/\" target=\"_blank\" rel=\"noopener\"><strong>Biology<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-english-core\/1855\/\" target=\"_blank\" rel=\"noopener\"><strong>English Core<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-business-studies\/1727\/\" target=\"_blank\" rel=\"noopener\"><strong>Business Studies<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-economics\/1327\/\" target=\"_blank\" rel=\"noopener\"><strong>Economics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-accountancy\/1315\/\" target=\"_blank\" rel=\"noopener\"><strong>Accountancy<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-computer-science\/1851\/\" target=\"_blank\" rel=\"noopener\"><strong>Computer Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-informatics-practices\/1873\/\" target=\"_blank\" rel=\"noopener\"><strong>Informatics Practices<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%95%E0%A5%8B%E0%A4%B0\/1865\/\" target=\"_blank\" rel=\"noopener\"><strong>Hindi Core<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-%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\/1867\/\" target=\"_blank\" rel=\"noopener\"><strong>Hindi Elective<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-history\/1869\/\" target=\"_blank\" rel=\"noopener\"><strong>History<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-political-science\/1879\/\" target=\"_blank\" rel=\"noopener\"><strong>Political Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-geography\/1863\/\" target=\"_blank\" rel=\"noopener\"><strong>Geography<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-home-science\/1871\/\" target=\"_blank\" rel=\"noopener\"><strong>Home Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12-physical-education\/1877\/\" target=\"_blank\" rel=\"noopener\"><strong>Physical Education<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-12\/1250\/\" target=\"_blank\" rel=\"noopener\"><strong>Other Subjects<\/strong><\/a><\/li>\n<\/ul>\n<h3><span class=\"ez-toc-section\" id=\"Download_CBSE_Class_12_Sample_Papers_for_All_Subjects\"><\/span><strong>Download CBSE Class 12 Sample Papers for All Subjects<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Looking for <strong>Class 12 sample papers<\/strong> for <strong>Physics, Chemistry, Biology, History, Political Science, Economics, Geography, Computer Science, Home Science, Accountancy, Business Studies,<\/strong> and more? Visit the <strong><a href=\"https:\/\/play.google.com\/store\/search?q=mycbseguide+app&amp;c=apps\">myCBSEguide<\/a> app<\/strong> or the <strong><a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a> website<\/strong> to download <strong>free sample papers<\/strong> with solutions in <strong>PDF format<\/strong>. For students preparing for the <strong>Class 12 Maths exams 2025<\/strong>, solving <strong>Class 12 Maths sample papers 2025<\/strong> is one of the most effective ways to practice. By regularly working through <strong>Class 12 Maths sample papers 2025<\/strong>, students can familiarize themselves with key concepts and improve their problem-solving speed. Additionally, <strong>Class 12 Maths sample papers 2025<\/strong> help students assess their preparation levels and pinpoint areas that need further revision. Download the <strong>Class 12 Maths sample papers 2025<\/strong> from reliable sources like the myCBSEguide app, which provides updated papers based on the latest syllabus and exam guidelines.<\/p>\n<p>myCBSEguide provides <strong>sample papers with solutions<\/strong>, <strong>chapter-wise test papers<\/strong>, <strong>NCERT solutions<\/strong>, <strong>NCERT Exemplar solutions<\/strong>, <strong>quick revision notes<\/strong>, <strong>CBSE guess papers<\/strong>, and <strong>important question papers<\/strong>. These resources are designed to help you prepare efficiently for your <strong>CBSE Class 12 exams<\/strong>.<\/p>\n<p>All sample papers are updated according to the latest <strong>CBSE syllabus<\/strong> and <strong>marking scheme<\/strong>. With the <strong>best app for CBSE students<\/strong> and the <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide website<\/strong><\/a>, you can access comprehensive study material and enhance your exam preparation. Start practicing today and boost your chances of success in your Class 12 exams!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0Class 12 Maths Sample Papers for 2025 Looking for CBSE Class 12 Mathematics sample papers for the 2024-25 academic year? If you&#8217;re preparing for the Class 12 Maths exams in 2025, downloading Class 12 Maths sample papers 2025 is an essential step in your exam preparation. These Class 12 Maths sample papers 2025 are designed &#8230; <a title=\"Class 12 Maths sample papers 2025\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/\" aria-label=\"More on Class 12 Maths sample papers 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":[1339,2000,1432,2048],"tags":[147,1867,1527,12,439,1933,1959,1967,321,1340],"class_list":["post-13581","post","type-post","status-publish","format-standard","hentry","category-cbse-sample-papers","category-class-12-sample-papers","category-mathematics-cbse-class-12","category-maths-sample-papers-class-12-sample-papers","tag-cbse-class-12","tag-cbse-class-12-mathematics","tag-cbse-question-paper","tag-cbse-sample-papers","tag-cbse-sample-papers-2017-18","tag-cbse-sample-papers-2023","tag-cbse-sample-papers-2024","tag-cbse-sample-papers-2025","tag-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>Class 12 Maths sample papers 2025 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"2025 Class 12 Maths sample papers for the academic year 2024-25 with a complete solution are available here in PDF file format for free download.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Class 12 Maths sample papers 2025 | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"2025 Class 12 Maths sample papers for the academic year 2024-25 with a complete solution are available here in PDF file format for free download.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/\" \/>\n<meta property=\"og:site_name\" content=\"myCBSEguide\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/mycbseguide\/\" \/>\n<meta property=\"article:published_time\" content=\"2023-03-07T11:10:21+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2025-10-09T11:22:52+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg\" \/>\n<meta name=\"author\" content=\"myCBSEguide\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:site\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"myCBSEguide\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"48 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/\"},\"author\":{\"name\":\"myCBSEguide\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/f67796d5f5c5a468e8c680aaaad21519\"},\"headline\":\"Class 12 Maths sample papers 2025\",\"datePublished\":\"2023-03-07T11:10:21+00:00\",\"dateModified\":\"2025-10-09T11:22:52+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/\"},\"wordCount\":11297,\"publisher\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg\",\"keywords\":[\"CBSE Class 12\",\"CBSE Class 12 Mathematics\",\"CBSE Question Paper\",\"CBSE Sample Papers\",\"CBSE sample papers 2017-18\",\"CBSE Sample Papers 2023\",\"CBSE Sample Papers 2024\",\"CBSE Sample Papers 2025\",\"Mathematics\",\"Model Question Papers\"],\"articleSection\":[\"CBSE Sample Papers\",\"Class 12 Sample Papers\",\"Mathematics\",\"Maths Sample Papers\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/\",\"url\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/\",\"name\":\"Class 12 Maths sample papers 2025 | myCBSEguide\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#primaryimage\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg\",\"datePublished\":\"2023-03-07T11:10:21+00:00\",\"dateModified\":\"2025-10-09T11:22:52+00:00\",\"description\":\"2025 Class 12 Maths sample papers for the academic year 2024-25 with a complete solution are available here in PDF file format for free download.\",\"breadcrumb\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#primaryimage\",\"url\":\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg\",\"contentUrl\":\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/mycbseguide.com\/blog\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"CBSE\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/cbse\/\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Class 12\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/cbse\/cbse-class-12\/\"},{\"@type\":\"ListItem\",\"position\":4,\"name\":\"Mathematics\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/cbse\/cbse-class-12\/mathematics-cbse-class-12\/\"},{\"@type\":\"ListItem\",\"position\":5,\"name\":\"Class 12 Maths sample papers 2025\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#website\",\"url\":\"https:\/\/mycbseguide.com\/blog\/\",\"name\":\"myCBSEguide\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/mycbseguide.com\/blog\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\",\"name\":\"myCBSEguide\",\"url\":\"https:\/\/mycbseguide.com\/blog\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png\",\"contentUrl\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png\",\"width\":180,\"height\":180,\"caption\":\"myCBSEguide\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/\"},\"sameAs\":[\"https:\/\/www.facebook.com\/mycbseguide\/\",\"https:\/\/x.com\/mycbseguide\",\"https:\/\/www.linkedin.com\/company\/mycbseguide\/\",\"http:\/\/in.pinterest.com\/mycbseguide\/\",\"https:\/\/www.youtube.com\/channel\/UCxuqSnnygFzwJG0pwogCNEQ\"]},{\"@type\":\"Person\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/f67796d5f5c5a468e8c680aaaad21519\",\"name\":\"myCBSEguide\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"Class 12 Maths sample papers 2025 | myCBSEguide","description":"2025 Class 12 Maths sample papers for the academic year 2024-25 with a complete solution are available here in PDF file format for free download.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/","og_locale":"en_US","og_type":"article","og_title":"Class 12 Maths sample papers 2025 | myCBSEguide","og_description":"2025 Class 12 Maths sample papers for the academic year 2024-25 with a complete solution are available here in PDF file format for free download.","og_url":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/","og_site_name":"myCBSEguide","article_publisher":"https:\/\/www.facebook.com\/mycbseguide\/","article_published_time":"2023-03-07T11:10:21+00:00","article_modified_time":"2025-10-09T11:22:52+00:00","og_image":[{"url":"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg","type":"","width":"","height":""}],"author":"myCBSEguide","twitter_card":"summary_large_image","twitter_creator":"@mycbseguide","twitter_site":"@mycbseguide","twitter_misc":{"Written by":"myCBSEguide","Est. reading time":"48 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#article","isPartOf":{"@id":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/"},"author":{"name":"myCBSEguide","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/f67796d5f5c5a468e8c680aaaad21519"},"headline":"Class 12 Maths sample papers 2025","datePublished":"2023-03-07T11:10:21+00:00","dateModified":"2025-10-09T11:22:52+00:00","mainEntityOfPage":{"@id":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/"},"wordCount":11297,"publisher":{"@id":"https:\/\/mycbseguide.com\/blog\/#organization"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#primaryimage"},"thumbnailUrl":"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg","keywords":["CBSE Class 12","CBSE Class 12 Mathematics","CBSE Question Paper","CBSE Sample Papers","CBSE sample papers 2017-18","CBSE Sample Papers 2023","CBSE Sample Papers 2024","CBSE Sample Papers 2025","Mathematics","Model Question Papers"],"articleSection":["CBSE Sample Papers","Class 12 Sample Papers","Mathematics","Maths Sample Papers"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/","url":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/","name":"Class 12 Maths sample papers 2025 | myCBSEguide","isPartOf":{"@id":"https:\/\/mycbseguide.com\/blog\/#website"},"primaryImageOfPage":{"@id":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#primaryimage"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#primaryimage"},"thumbnailUrl":"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg","datePublished":"2023-03-07T11:10:21+00:00","dateModified":"2025-10-09T11:22:52+00:00","description":"2025 Class 12 Maths sample papers for the academic year 2024-25 with a complete solution are available here in PDF file format for free download.","breadcrumb":{"@id":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#primaryimage","url":"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg","contentUrl":"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_maths_sample_paper.jpg"},{"@type":"BreadcrumbList","@id":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-mathematics\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/mycbseguide.com\/blog\/"},{"@type":"ListItem","position":2,"name":"CBSE","item":"https:\/\/mycbseguide.com\/blog\/category\/cbse\/"},{"@type":"ListItem","position":3,"name":"Class 12","item":"https:\/\/mycbseguide.com\/blog\/category\/cbse\/cbse-class-12\/"},{"@type":"ListItem","position":4,"name":"Mathematics","item":"https:\/\/mycbseguide.com\/blog\/category\/cbse\/cbse-class-12\/mathematics-cbse-class-12\/"},{"@type":"ListItem","position":5,"name":"Class 12 Maths sample papers 2025"}]},{"@type":"WebSite","@id":"https:\/\/mycbseguide.com\/blog\/#website","url":"https:\/\/mycbseguide.com\/blog\/","name":"myCBSEguide","description":"","publisher":{"@id":"https:\/\/mycbseguide.com\/blog\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/mycbseguide.com\/blog\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/mycbseguide.com\/blog\/#organization","name":"myCBSEguide","url":"https:\/\/mycbseguide.com\/blog\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/","url":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png","contentUrl":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png","width":180,"height":180,"caption":"myCBSEguide"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/mycbseguide\/","https:\/\/x.com\/mycbseguide","https:\/\/www.linkedin.com\/company\/mycbseguide\/","http:\/\/in.pinterest.com\/mycbseguide\/","https:\/\/www.youtube.com\/channel\/UCxuqSnnygFzwJG0pwogCNEQ"]},{"@type":"Person","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/f67796d5f5c5a468e8c680aaaad21519","name":"myCBSEguide"}]}},"_links":{"self":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/13581","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/comments?post=13581"}],"version-history":[{"count":32,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/13581\/revisions"}],"predecessor-version":[{"id":30967,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/13581\/revisions\/30967"}],"wp:attachment":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/media?parent=13581"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/categories?post=13581"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/tags?post=13581"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}