{"id":30178,"date":"2023-03-28T21:28:01","date_gmt":"2023-03-28T15:58:01","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=30178"},"modified":"2025-10-09T16:55:02","modified_gmt":"2025-10-09T11:25:02","slug":"class-12-applied-maths-sample-paper","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/class-12-applied-maths-sample-paper\/","title":{"rendered":"Class 12 Applied Maths Sample Paper 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\/class-12-applied-maths-sample-paper\/#Class_12_Applied_Maths_Sample_Papers_2025\" >Class 12 Applied Maths Sample Papers 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\/class-12-applied-maths-sample-paper\/#Model_Papers_of_Applied_Maths_2024-25\" >Model Papers of Applied Maths 2024-25<\/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\/class-12-applied-maths-sample-paper\/#Applied_Maths_Sample_Question_Paper_2025\" >Applied Maths Sample Question Paper 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\/class-12-applied-maths-sample-paper\/#Solution\" >Solution<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/class-12-applied-maths-sample-paper\/#Class_12_Sample_Papers_2025\" >Class 12 Sample Papers 2025<\/a><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"Class_12_Applied_Maths_Sample_Papers_2025\"><\/span>Class 12 Applied Maths Sample Papers 2025<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Applied Maths<\/strong> is a newly introduced subject for <strong>commerce students<\/strong> in <strong>Class 12<\/strong>, offering an ideal choice for those who don\u2019t plan on pursuing higher studies in Mathematics. <strong>CBSE Class 12 Applied Maths Sample Papers 2025<\/strong> are now available for free download on the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide app<\/strong><\/a> and website in <strong>PDF format<\/strong>. Teachers can easily create customized exam papers or mock papers using the <a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\"><strong>Examin8 app<\/strong><\/a>, adding their name, logo, and branding. This app allows educators to design professional-quality question papers for practice or assessments. Enhance your teaching experience with <a href=\"https:\/\/examin8.com\/\"><strong>Examin8 Website<\/strong><\/a>\u00a0and streamline the paper creation process.<\/p>\n<p>These model question papers are designed to help students familiarize themselves with the exam format and improve their problem-solving skills. By practicing with these <strong>Class 12 Applied Maths sample papers<\/strong>, students can enhance their chances of scoring higher grades in the <strong>CBSE board exams<\/strong>.<\/p>\n<p><strong> Class 12 Sample papers <a class=\"button\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/4788\/type\/2\">Download as PDF<\/a><\/strong><\/p>\n<p>The <strong><a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\">CBSE Class 12<\/a> Applied Maths exam 2025<\/strong> will be conducted for <strong>80 marks<\/strong>, with the remaining <strong>20 marks<\/strong> allocated for internal assessments. These Mock Papers are a great tool for thorough preparation and boosting exam confidence.<br \/>\n<strong>Visit our Website <a class=\"button\" href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a><\/strong><\/p>\n<h2><span class=\"ez-toc-section\" id=\"Model_Papers_of_Applied_Maths_2024-25\"><\/span>Model Papers of Applied Maths 2024-25<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>The <strong>Class 12 Applied Maths<\/strong>\u00a0<strong>2025 session<\/strong> follows a well-defined structure with five sections. The <strong>question paper<\/strong> is designed to assess a range of skills, with a total of <strong>38 questions<\/strong> distributed across the sections:<\/p>\n<ul>\n<li><strong>Section A<\/strong>: This section consists of <strong>objective-type questions<\/strong> and is worth <strong>20 marks<\/strong>. It tests students\u2019 basic understanding of key concepts.<\/li>\n<li><strong>Section B<\/strong>: This section includes <strong>five very short answer questions<\/strong>, aimed at assessing quick recall and understanding of specific topics.<\/li>\n<li><strong>Section C<\/strong>: With <strong>6 three-mark questions<\/strong>, Section C covers a range of concepts and requires students to apply their knowledge in more detailed answers.<\/li>\n<li><strong>Section D<\/strong>: This section includes <strong>4 five-mark questions<\/strong>, focusing on more complex problem-solving and in-depth explanations.<\/li>\n<li><strong>Section E<\/strong>: The final section features <strong>3 case study-based questions<\/strong>, designed to test students\u2019 ability to apply mathematical concepts to real-world scenarios.<\/li>\n<\/ul>\n<p>In total, the <strong>Class 12 Applied Maths question paper<\/strong>\u00a0<strong>2025<\/strong> consists of <strong>38 questions<\/strong> spread across these five sections. Understanding this structure will help students plan their time and approach more effectively during the exam.<\/p>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Applied_Maths_Sample_Question_Paper_2025\"><\/span>Applied Maths Sample Question Paper 2025<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<hr \/>\n<p style=\"text-align: center;\"><strong>Class 12 &#8211; Applied Maths<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<p>Read the following instructions very carefully and strictly follow them:<\/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, 2 questions in Section C, 2 questions in Section D and one sub-part 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 is a square matrix of order 3 and |A| = 2, then the value of |-AA&#8217;| is\n<div style=\"margin-left: 20px;\">\n<p>a) 4<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b) -2<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c) -4<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d) 2<\/p>\n<\/div>\n<\/li>\n<li>A sample of 50 bulbs is taken at random. Out of 50 we found 15 bulbs are of Bajaj, 17 are of Surya and 18 are of Crompton. What is the point estimate of population proportion of Surya?\n<div style=\"margin-left: 20px;\">\n<p>a) 0.3<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b) 0.34<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c) 0.36<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d) 0.4<\/p>\n<\/div>\n<\/li>\n<li>The assumption in calculating annuity is that every payment is\n<div style=\"margin-left: 20px;\">\n<p>a) equal<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b) marginal<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c) nominal<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)different<\/p>\n<\/div>\n<\/li>\n<li>The maximum value of Z = 3x + 4y subject to the constraints: x + y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 4, x <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 0, y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 0 is:\n<div style=\"margin-left: 20px;\">\n<p>a)0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)18<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)16<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)12<\/p>\n<\/div>\n<\/li>\n<li>The matrix\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{ccc} 0 &amp; -5 &amp; 3 \\\\ 5 &amp; 0 &amp; -7 \\\\ -3 &amp; 7 &amp; 0 \\end{array}\\right]{\/tex}<\/span> is a\n<div style=\"margin-left: 20px;\">\n<p>a)symmetric matrix<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)skew-symmetric matrix<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)scalar matrix<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)diagonal matrix<\/p>\n<\/div>\n<\/li>\n<li>If the mean and the variance of a probability distribution are 4 and 2 respectively, then the probability of two successes is\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{219}{256}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{7}{64}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{37}{256}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>For the following probability distribution:<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>X<\/td>\n<td>-4<\/td>\n<td>-3<\/td>\n<td>-2<\/td>\n<td>-1<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td>P(X)<\/td>\n<td>0.1<\/td>\n<td>0.2<\/td>\n<td>0.3<\/td>\n<td>0.2<\/td>\n<td>0.2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The value of E(X) is:<\/p>\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)-1.8<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)-2<\/p>\n<\/div>\n<\/li>\n<li>Solution of the differential equation <span class=\"math-tex\">{tex}x \\frac{d y}{d x}+2 y{\/tex}<\/span>\u00a0=\u00a0x<sup>2<\/sup> is\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}y=\\frac{x^2}{4}+\\mathrm{C}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}y=\\frac{x^4+\\mathrm{C}}{4 x^2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}y=\\frac{x^2+C}{4 x^2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}y=\\frac{x^2+C}{x^2}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>A pipe fills <span class=\"math-tex\">{tex}\\frac{3}{7}^{\\text {th }}{\/tex}<\/span> part of a tank in 1 hour. The rest of the tank can be filled in\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{7}{3}{\/tex}<\/span> hours<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{3}{4}{\/tex}<\/span> hours<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{4}{3}{\/tex}<\/span> hours<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{7}{4}{\/tex}<\/span> hours<\/p>\n<\/div>\n<\/li>\n<li>If F(x) =\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} cos x &amp; sin x \\\\ -sin x &amp; cos x \\end{array}\\right]{\/tex}<\/span>,\u00a0then F(x) F(y) is equal to:\n<div style=\"margin-left: 20px;\">\n<p>a)F(x &#8211; y)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)F(xy)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)F(x)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)F(x + y)<\/p>\n<\/div>\n<\/li>\n<li>(09 : 30 + 16 : 40) in 24 hours clock is\n<div style=\"margin-left: 20px;\">\n<p>a)2 : 10<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)26 : 10<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)25 : 70<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)03 : 10<\/p>\n<\/div>\n<\/li>\n<li>The solution set of the inequation\u00a0|x + 2|\u00a0<span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a05\u00a0is\n<div style=\"margin-left: 20px;\">\n<p>a)(-7, 5)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)|x|\u00a0<span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a05<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)[-5, 5]\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)[-7, 3]\n<\/div>\n<\/li>\n<li>In a 400 m race, A gives B a start of 5 seconds and beats him by 15 m. In another race of 400 m, A beats B by 7<span class=\"math-tex\">{tex}\\frac 17{\/tex}<\/span>seconds. Their respective speeds are:\n<div style=\"margin-left: 20px;\">\n<p>a)8 m\/sec, 7 m\/sec<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)5 m\/sec, 7 m\/sec<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)9 m\/sec, 7 m\/sec<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)6 m\/sec, 7 m\/sec<\/p>\n<\/div>\n<\/li>\n<li>Comer points of the feasible region for an LPP are : (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let z = 4x + 6y be the objective function. Then, Max. z &#8211; Min. z =\n<div style=\"margin-left: 20px;\">\n<p>a)48<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)42<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)60<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)18<\/p>\n<\/div>\n<\/li>\n<li>The objective function of an LPP is\n<div style=\"margin-left: 20px;\">\n<p>a)a relation between the variables<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)a function to be not optimized<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)a constrain<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)a function to be optimized<\/p>\n<\/div>\n<\/li>\n<li>For the purpose of t-test of significance, a random sample of size (n) 34 is drawn from a normal population, then the degree of freedom (v) is\n<div style=\"margin-left: 20px;\">\n<p>a)35<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{1}{34}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)33<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)34<\/p>\n<\/div>\n<\/li>\n<li>The value of <span class=\"math-tex\">{tex}\\int \\frac{1}{x+x \\log x} d x{\/tex}<\/span> is\n<div style=\"margin-left: 20px;\">\n<p>a)log (1 + log x)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)x + log x<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)x log (1 + log x)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)1 + log x<\/p>\n<\/div>\n<\/li>\n<li>A factory production is delayed for three weeks due to breakdown of a machine and unavailability of spare parts. Under which trend oscillation does this situation fall?\n<div style=\"margin-left: 20px;\">\n<p>a)Cyclical<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Seasonal<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)Secular<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)Irregular<\/p>\n<\/div>\n<\/li>\n<li>Let A and B be two square matrices of the same order.<br \/>\n<strong>Assertion (A):<\/strong> (A&#8217;BA) is symmetric if B is symmetric.<br \/>\n<strong>Reason (R):\u00a0<\/strong>(A&#8217;BA) is skew-symmetric if B is skew-symmetric.<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)Both A and R are true and R is the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Both A and R are true but R is not the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)A is true but R is false.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)A is false but R is true.<\/p>\n<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> The tangent at x = 1 to the curve y = x<sup>3<\/sup> &#8211; x<sup>2<\/sup> &#8211; x + 2 again meets the curve at x = -2.<br \/>\n<strong>Reason (R):<\/strong> When a equation of a tangent solved with the curve, repeated roots are obtained at point of tangency.<\/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<p><strong>Prepare for Exams with myCBSEguide App \u2013 Your Ultimate Study Companion<\/strong><\/p>\n<p>Visit our <a href=\"https:\/\/mycbseguide.com\/\"><strong>Website myCBSEGuide<\/strong><\/a>. To excel in your exams, download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide<\/strong><\/a> app today! This all-in-one app offers comprehensive study material for CBSE, NCERT, JEE (Main), NEET-UG, and NDA exams. Whether you&#8217;re preparing for school exams or competitive entrance tests, <strong>myCBSEguide<\/strong> provides a wealth of resources, including practice questions, solved papers, and more. With the <strong>Class 12 Applied Maths 2025<\/strong> syllabus in mind, we have curated a list of essential resources to help students excel in both theory and application-based questions.<\/p>\n<p>For <strong>teachers<\/strong>, the <a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\"><strong>Examin8<\/strong><\/a> app is the perfect tool to create customized practice papers with your branding, including your name and logo. Enhance your teaching experience and help students prepare better with personalized assessments.<\/p>\n<p><strong>Key Features of myCBSEguide:<\/strong><\/p>\n<ul>\n<li>Complete study material for <strong>CBSE, JEE (Main), NEET-UG,<\/strong> and <strong>NDA<\/strong>.<\/li>\n<li>Practice papers, solved examples, and previous years&#8217; question papers.<\/li>\n<li>Easy access to <strong>NCERT<\/strong> solutions and notes.<\/li>\n<li>User-friendly interface for quick learning.<\/li>\n<\/ul>\n<p><strong>Key Features of Examin8:<\/strong><\/p>\n<ul>\n<li>Create customized exams with your own name and logo.<\/li>\n<li>Personalize the content for students.<\/li>\n<li>Streamlined paper generation for teachers.<\/li>\n<\/ul>\n<p><strong>Download the myCBSEguide App<\/strong> now to start preparing smarter and stay ahead of the competition! You can access detailed NCERT solutions for every chapter of <strong>Class 12 Applied Maths 2025<\/strong> on our platform, ensuring you understand each concept thoroughly.<\/p>\n<p><strong>Explore Our Website for More Information &amp; Resources<\/strong><\/p>\n<p>We invite you to visit our website and explore a wide range of resources designed to help you succeed in your academic journey. Whether you&#8217;re preparing for <strong>CBSE<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, or <strong>NDA<\/strong> exams, our platform offers comprehensive study material, practice papers, and much more to help you prepare effectively.<\/p>\n<p><strong>Why visit our website <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a>?<\/strong><\/p>\n<ul>\n<li>Access <strong>high-quality study material<\/strong> and <strong>practice questions<\/strong>.<\/li>\n<li>Download useful apps like <strong>myCBSEguide<\/strong> and <a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\"><strong>Examin8<\/strong><\/a> for personalized learning.<\/li>\n<li>Stay updated with the latest resources and exam tips.<\/li>\n<\/ul>\n<p>Don&#8217;t miss out on the opportunity to enhance your preparation! Visit us today and take the next step towards achieving your goals.<\/p>\n<\/div>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section B<\/b><\/li>\n<li>Rahul borrowed \u20b9100000 from a co-operative society at the rate of 10%\u00a0p.a. for 2 years. Calculate his EMI using flat rate method.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>If money is worth 5% compare the present value of a perpetuity of \u20b92,000 payable at the end of each year with that of an ordinary annuity of \u20b92,000 per year for 100 years. (Given (1.05)<sup>-100<\/sup> = 0.0076)<\/li>\n<li>Assuming a four yearly cycle, calculate the trend by the method of moving averages from the following data:<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>Year<\/td>\n<td>1984<\/td>\n<td>1985<\/td>\n<td>1986<\/td>\n<td>1987<\/td>\n<td>1988<\/td>\n<td>1989<\/td>\n<td>1990<\/td>\n<td>1991<\/td>\n<td>1992<\/td>\n<td>1993<\/td>\n<\/tr>\n<tr>\n<td>Value<\/td>\n<td>12<\/td>\n<td>25<\/td>\n<td>39<\/td>\n<td>54<\/td>\n<td>70<\/td>\n<td>87<\/td>\n<td>105<\/td>\n<td>100<\/td>\n<td>82<\/td>\n<td>65<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Evaluate:\u00a0<span class=\"math-tex\">{tex}\\int_\\limits{1}^{2} \\frac{1}{x\\left(1+x^{2}\\right)}{\/tex}<\/span>dx<\/li>\n<li>If A is a non-singular square matrix of order 3 such that |adj A| = 225, find |A&#8217;|.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>If A =\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{r} 1 \\\\ -4 \\\\ 3 \\end{array}\\right]{\/tex}<\/span>\u00a0and B\u00a0= [-1 2 1], verify that (AB)&#8217; = B&#8217;A&#8217;.<\/li>\n<li>Find the unit digit in 183! + 3<sup>183<\/sup>.<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section C<\/b><\/li>\n<li>A bond has issued with the face (Par) value of \u20b9 1,000 at 10% coupon for three years The required rate of return is 8%. What is the value of the bond if the coupon amount is payable on half-yearly basis? Given (1.04)<sup>-6<\/sup> = 0.79031<\/li>\n<li>The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Show that the differential equation representing one parameter family of curves (x<sup>2<\/sup> &#8211; y<sup>2<\/sup>) = c(x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>)<sup>2<\/sup>\u00a0is (x<sup>2<\/sup>\u00a0&#8211; 3xy<sup>2<\/sup>) dx = (y<sup>2<\/sup>\u00a0&#8211; 3x<sup>2<\/sup>y) dy<\/li>\n<li>The marginal revenue of a company is given by MR = 100 + 20Q + 3Q<sup>2<\/sup>, where Q is the amount of units sold for a period. Find the total revenue function if at Q = 2 it is equal to 260.<\/li>\n<li>Fit a straight line trend by the method of least squares to the data given below:<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td><strong>Years<\/strong><\/td>\n<td>2012<\/td>\n<td>2013<\/td>\n<td>2014<\/td>\n<td>2015<\/td>\n<td>2016<\/td>\n<td>2017<\/td>\n<td>2018<\/td>\n<\/tr>\n<tr>\n<td>Sales (in tones)<\/td>\n<td>9<\/td>\n<td>11<\/td>\n<td>13<\/td>\n<td>12<\/td>\n<td>14<\/td>\n<td>15<\/td>\n<td>17<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Ten cartons are taken at random from an automatic filling machine. The mean net weight of the cartons is 11.8 kg and the standard deviation 0.15 kg. Does the sample mean differ significantly from the intended weight of 12 kg? [Given that for d.f. = 9, t<sub>0.05<\/sub> = 2.26]<\/li>\n<li>In a certain examination, the percentage of passes and distinction were 46 and 9 respectively. Estimate the average marks obtained by the candidate, the minimum pass and distinction marks being 40 and 75 respectively (assume the distribution of marks to be normal).\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>A factory produces bulbs. The probability that one bulb is defective is <span class=\"math-tex\">{tex}\\frac{1}{50}{\/tex}<\/span> and they are packed in boxes of 10. From a single box, find the probability that<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>none of the bulbs is defective<\/li>\n<li>exactly two bulbs are defective<\/li>\n<li>more than 8 bulbs work properly.<\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section D<\/b><\/li>\n<li>A manufacturer produces two models of bikes-model X and model Y. Model X takes a 6 man-hours to make per unit, while model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and marketing costs are \u20b92000 and \u20b91000\u00a0per unit for Models X and Y respectively. The total funds available for these purposes are \u20b980000 per week. Profit per unit for models X and Yare noon and \u20b9 600 respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Form an L.P.P. and\u00a0solve it graphically using iso-profit\/iso-cost method.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Two tailors P and Q earn \u20b9 150 and \u20b9 200 per day respectively. P can stitch 6 shirts and 4 trousers a day, while Q can stitch 10 shirts and 4 trousers per day. How many days should each work to produce at least 60 shirts and 32 trousers at minimum labour cost?<\/li>\n<li>Let X denote the number of hours a Class 12 student studies during a randomly selected school day. The probability that X can take the values x<sub>i<\/sub>, for an unknown constant <strong>k<\/strong>:<br \/>\n<span class=\"math-tex\">{tex}P(X=k)=\\left\\{\\begin{array}{ccc} 0 \\cdot 1 &amp; \\text { if } &amp; x_i=0 \\\\ k x_i &amp; \\text { if } &amp; x_i=1 \\text { or } 2 \\\\ k\\left(5-x_i\\right) &amp; \\text { if } &amp; x_i=3 \\text { or } 4 \\end{array}\\right.{\/tex}<\/span><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Find the value of k.<\/li>\n<li>Determine the probability that the student studied for at least 2 hours.<\/li>\n<li>Determine the probability that the student studied for at most 2 hours.<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).<\/li>\n<li>Find the linear inequations for which the shaded area in figure\u00a0is the solution set.<br \/>\n<img decoding=\"async\" style=\"height: 240px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1699014636-cwymcr.jpg\" alt=\"\" data-imgur-src=\"kkpq4vU.png\" \/><\/li>\n<li>A machine costing <em>\u20b9\u00a0<\/em>200000 has an effective life of 7 years and its scrap value is <em>\u20b9\u00a0<\/em>30000. What amount should the company put into a sinking fund earning 5% per annum, so that it can replace the machine after its useful life? Assume that a new machine will cost <em>\u20b9\u00a0<\/em>300000 after 7 years.<\/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 \/>\nDerivative of y = f(x) w.r.t. x (if exists) is denoted by <span class=\"math-tex\">{tex}\\frac{d y}{d x}{\/tex}<\/span> or f'(x) and is called the first-order derivative of y. If we take the derivative of <span class=\"math-tex\">{tex}\\frac{d y}{d x}{\/tex}<\/span> again, then we get <span class=\"math-tex\">{tex}\\frac{d}{d x}\\left(\\frac{d y}{d x}\\right)=\\frac{d^{2} y}{d x^{2}}{\/tex}<\/span> or f&#8221;(x) and is called the second-order derivative of y. Similarly, <span class=\"math-tex\">{tex}\\frac{d}{d x}\\left(\\frac{d^{2} y}{d x^{2}}\\right){\/tex}<\/span> is denoted and defined as <span class=\"math-tex\">{tex}\\frac{d^{3} y}{d x^{3}}{\/tex}<\/span> or f'&#8221;(x) and is known as the third-order derivative of y and so on.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>If y = tan<sup>-1<\/sup><span class=\"math-tex\">{tex}\\left(\\frac{\\log \\left(e \/ x^{2}\\right)}{\\log \\left(e x^{2}\\right)}\\right){\/tex}<\/span> + tan<sup>-1<\/sup><span class=\"math-tex\">{tex}\\left(\\frac{3+2 \\log x}{1-6 \\log x}\\right){\/tex}<\/span>,then find the value of<span class=\"math-tex\">{tex}\\frac{d^{2}y}{d x^{2}}{\/tex}<\/span>. (1)<\/li>\n<li>If u = x<sup>2<\/sup> + y<sup>2<\/sup> and x = s + 3t, y = 2s &#8211; t, then find the value of <span class=\"math-tex\">{tex}\\frac{d^{2} u}{d s^{2}}{\/tex}<\/span>. (1)<\/li>\n<li>If f(x) = 2 log sin x, then find f'(x). (2)<br \/>\n<strong>OR<\/strong><br \/>\nIf f(x) = e<sup>x<\/sup>sinx, then find f&#8221;(x). (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\n<strong>What Is a Sinking Fund?<\/strong><br \/>\nA sinking fund contains money set aside or saved to pay off a debt or bond. A company that issues debt will need to pay that debt off in the future, and the sinking fund helps to soften the hardship of a large outlay of revenue. A sinking fund allows companies that have floated debt in the form of bonds gradually save money and avoid a large lump-sum payment at maturity.<br \/>\n<strong>Example:<\/strong><\/p>\n<ul>\n<li>Cost of Machine: \u20b92,00,000\/-<\/li>\n<li>Effective Life: 7 Years<\/li>\n<li>Scrap Value: \u20b930,000\/-<\/li>\n<li>Sinking Fund Earning Rate: 5%<\/li>\n<li>The Expected Cost of New Machine: \u20b93,00,000\/-<\/li>\n<\/ul>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>What is the money required for a new machine after 7 years? (1)<\/li>\n<li>What is the value of A, i and n here? (1)<\/li>\n<li>What formula will you use to get the requisite amount? (2)<br \/>\n<strong>OR<\/strong><br \/>\nWhat amount should the company put into a sinking fund earning 5% per annum to replace the machine after its useful life? (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nTwo schools P and Q decided to award their selected students for the values of discipline and honesty in the form of prizes at the rate of \u20b9 x and \u20b9 y respectively. School P decided to award respectively 3, 2 students a total prize money of \u20b9 2300 and school Q decided to award respectively 5, 3 students a total prize money of \u20b9 3700.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Write the matrix equation representing the above situation. (1)<\/li>\n<li>Find the value of the determinant of coefficients of x and y. (1)<\/li>\n<li>Find the values of x and y respectively (use Cramer&#8217;s rule). (2)<br \/>\n<strong>OR<\/strong><br \/>\nFind the inverse of matrix A. (2)<br \/>\nLooking to practice more questions and prepare effectively for your exams? Download the <strong>myCBSEguide App<\/strong> today! It offers a complete range of study materials for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. Whether you&#8217;re a student or a teacher, myCBSEguide provides you with the tools needed to ace your exams. Preparing for the <strong>Class 12 Applied Maths 2025<\/strong> exam requires a structured study plan, and we provide the necessary tools and resources to help you stay on track<strong>Why Choose myCBSEguide?<\/strong><\/p>\n<ul>\n<li>Access curated study notes, NCERT solutions, previous year papers, and more.<\/li>\n<li>Test your knowledge with a wide variety of practice questions to strengthen your understanding.<\/li>\n<li>Easy navigation for seamless learning on-the-go.<\/li>\n<\/ul>\n<p><strong>Examin8 App for Teachers<\/strong> Teachers can also benefit from the <strong>Examin8 App<\/strong> to create custom exam papers tailored with their name and logo. Personalize your practice papers to better serve your students and track their progress.<\/p>\n<p><strong>Key Features for Teachers:<\/strong><\/p>\n<ul>\n<li>Create <strong>customized exam papers<\/strong> with your branding.<\/li>\n<li>Generate <strong>personalized assessments<\/strong> to suit your teaching style.<\/li>\n<li>Track student progress efficiently.<\/li>\n<\/ul>\n<p>Start preparing smarter and enhance your exam readiness. <strong>Download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide App<\/a><\/strong> and the<a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\"> <strong>Examin8 App<\/strong><\/a> now! <strong>Class 12 Applied Maths 2025<\/strong> covers a wide range of topics, including matrices, determinants, and differential equations, all of which are critical for understanding advanced mathematics.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center; page-break-before: always;\"><strong>Class 12 &#8211; Applied Maths<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Solution\"><\/span><strong>Solution <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ol style=\"padding-left: 20px;\">\n<li style=\"text-align: center; display: block;\"><b>Section A <\/b><\/li>\n<li>(c)\n<p style=\"display: inline;\">-4<\/p>\n<p><b>Explanation: <\/b>Given |A|\u00a0= 2, order of A = 3.<br \/>\nSo, |-AA&#8217;| = |-A| <span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>\u00a0|A&#8217;| = (-1)<sup>3<\/sup>| A|<span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>|A|<br \/>\n= -1|A|<sup>2<\/sup>\u00a0= -1(2)<sup>2<\/sup>\u00a0= -4<\/li>\n<li>(b)\n<p style=\"display: inline;\">0.34<\/p>\n<p><b>Explanation: <\/b>0.34<\/li>\n<li>(a)\n<p style=\"display: inline;\">equal<\/p>\n<p><b>Explanation: <\/b>equal<\/li>\n<li>(c)\n<p style=\"display: inline;\">16<\/p>\n<p><b>Explanation: <\/b>16<\/li>\n<li>(b)\n<p style=\"display: inline;\">skew-symmetric matrix<\/p>\n<p><b>Explanation: <\/b>Let A =\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{rrr} 0 &amp; -5 &amp; 3 \\\\ 5 &amp; 0 &amp; -7 \\\\ -3 &amp; 7 &amp; 0 \\end{array}\\right]{\/tex}<\/span><br \/>\nSo, A&#8217; =\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{rrr} 0 &amp; 5 &amp; -3 \\\\ -5 &amp; 0 &amp; 7 \\\\ 3 &amp; -7 &amp; 0 \\end{array}\\right]{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}=-\\left[\\begin{array}{rrr} 0 &amp; -5 &amp; 3 \\\\ 5 &amp; 0 &amp; -7 \\\\ -3 &amp; 7 &amp; 0 \\end{array}\\right]{\/tex}<\/span>\u00a0= -A<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0A is a skew-symmetric matrix.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Option (skew-symmetric matrix) is the correct answer.<\/li>\n<li>(b)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{7}{64}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>Given np = 4 and npq = 2<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{n p q}{n p}=\\frac{2}{4} \\Rightarrow q=\\frac{1}{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> p = 1<span class=\"math-tex\">{tex}-\\frac{1}{2}=\\frac{1}{2}{\/tex}<\/span>, so\u00a0n <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> <span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0= 4 <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> n = 8<br \/>\nNow, P(X = 2) = <span class=\"math-tex\">{tex}{ }^8 C _2\\left(\\frac{1}{2}\\right)^2\\left(\\frac{1}{2}\\right)^6=\\frac{28}{256}=\\frac{7}{64}{\/tex}<\/span><\/li>\n<li>(c)\n<p style=\"display: inline;\">-1.8<\/p>\n<p><b>Explanation: <\/b>E(X) = <span class=\"math-tex\">{tex}\\sum {\/tex}<\/span>X P(X)<br \/>\n= -4\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0(0.1) + (-3 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a00.2) + (-2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a00.3) + (-1\u00a0 +0.2) + (0 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a00.2)<br \/>\n= -0.4 &#8211; 0.6 &#8211; 0.6 &#8211; 0.2 = -1.8<\/li>\n<li>(b)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}y=\\frac{x^4+\\mathrm{C}}{4 x^2}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\frac{d y}{d x}+\\frac{2}{x}{\/tex}<\/span>y = x <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0I.F. = <span class=\"math-tex\">{tex}e^{\\int \\frac{2}{x} d x}{\/tex}<\/span>\u00a0= <span class=\"math-tex\">{tex}e^{2 \\log x}{\/tex}<\/span>\u00a0= x<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Solution is y <span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span> x<sup>2<\/sup>\u00a0= <span class=\"math-tex\">{tex}\\int{\/tex}<\/span> x <span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span> x<sup>2<\/sup> dx + C<sub>1<\/sub><br \/>\ny <span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span> x<sup>2<\/sup>\u00a0= <span class=\"math-tex\">{tex}\\frac{x^4}{4}{\/tex}<\/span>\u00a0+ C<sub>1<\/sub> <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0y =\u00a0<span class=\"math-tex\">{tex}\\frac{x^4+C}{4 x^2}{\/tex}<\/span><\/li>\n<li>(c)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{4}{3}{\/tex}<\/span> hours<\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\frac {3}{7}{\/tex}<\/span>l\u00a0<span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span>\u00a01 min<br \/>\nremaining amount =\u00a0<span class=\"math-tex\">{tex}1 &#8211; \\frac {3}{7}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac {4}{7}{\/tex}<\/span>ltr<br \/>\nsince time taken to fill 1 litre<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac {3}{7}{\/tex}<\/span>min<br \/>\nSo time required to fill\u00a0<span class=\"math-tex\">{tex}\\frac {4}{7}{\/tex}<\/span>litres<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac {7}{3} \\times \\frac {4}{7}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac {4}{3}{\/tex}<\/span>min<\/li>\n<li>(d)\n<p style=\"display: inline;\">F(x + y)<\/p>\n<p><b>Explanation: <\/b>F(x + y)<\/li>\n<li>(a)\n<p style=\"display: inline;\">2 : 10<\/p>\n<p><b>Explanation: <\/b>(09: 30 + 16 : 40) (mod 24) = 26 : 10 (mod 24) = 2 : 10<\/li>\n<li>(d)\n<p style=\"display: inline;\">[-7, 3]\n<p><b>Explanation: <\/b>|x + 2| <span class=\"math-tex\">{tex}\\le{\/tex}<\/span> 5<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0-5 <span class=\"math-tex\">{tex}\\le{\/tex}<\/span> x + 2 <span class=\"math-tex\">{tex}\\le{\/tex}<\/span> 5<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0-7 <span class=\"math-tex\">{tex}\\le{\/tex}<\/span> x <span class=\"math-tex\">{tex}\\le{\/tex}<\/span> 3<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0 x\u00a0<span class=\"math-tex\">{tex}\\in{\/tex}<\/span>\u00a0[-7, -3]<\/li>\n<li>(a)\n<p style=\"display: inline;\">8 m\/sec, 7 m\/sec<\/p>\n<p><b>Explanation: <\/b>Suppose A covers 400 m in t seconds<br \/>\nThen, B covers 385 m in (t + 5) seconds<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0B covers 400 m =\u00a0<span class=\"math-tex\">{tex}\\left\\{\\frac{(t+5)}{385} \\times 400\\right\\}{\/tex}<\/span>sec<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{80(t+5)}{77}{\/tex}<\/span>\u00a0sec<br \/>\nAlso, B covers 400 m =\u00a0<span class=\"math-tex\">{tex}\\left(t+7 \\frac{1}{7}\\right){\/tex}<\/span>sec<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{(7 t+50)}{7}{\/tex}<\/span>sec<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{80(t+5)}{77}=\\frac{7 t+50}{7}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a080(t + 5) = 11(7t + 50)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(80t &#8211; 77t) = (550 &#8211; 400)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a03t = 150<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0t = 50<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0A&#8217;s speed<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{400}{50}{\/tex}<\/span>\u00a0m\/sec<br \/>\n= 8 m\/sec<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0B&#8217;s speed<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac {385}{55}{\/tex}<\/span>\u00a0m\/sec<br \/>\n= 7 m\/sec<\/li>\n<li>(c)\n<p style=\"display: inline;\">60<\/p>\n<p><b>Explanation: <\/b>Here the objective function is given by:<br \/>\nF = 4x + 6y<\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\"><strong>Corner points<\/strong><\/td>\n<td style=\"text-align: center;\"><strong>Z = 4x +6 y<\/strong><\/td>\n<\/tr>\n<tr>\n<td>(0, 2)<\/td>\n<td>12 (Min.)<\/td>\n<\/tr>\n<tr>\n<td>(3, 0)<\/td>\n<td>12 (Min.)<\/td>\n<\/tr>\n<tr>\n<td>(6, 0)<\/td>\n<td>24<\/td>\n<\/tr>\n<tr>\n<td>(6, 8)<\/td>\n<td>72 (Max.)<\/td>\n<\/tr>\n<tr>\n<td>(0, 5)<\/td>\n<td>30<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Maximum of F &#8211;\u00a0Minimum of F = 72 &#8211;\u00a012 = 60<\/li>\n<li>(d)\n<p style=\"display: inline;\">a function to be optimized<\/p>\n<p><b>Explanation: <\/b>A Linear programming problem is a linear function (also known as an objective function) subjected to certain constraints for which we need to find an optimal solution (i.e. either a maximum\/minimum value) depending on the requirement of the problem.<br \/>\nFrom the above definition, we can clearly say that the Linear programming problem&#8217;s objective is to either maximize\/minimize a given objective function, which means to optimize a function to get an optimum solution.<\/li>\n<li>(c)\n<p style=\"display: inline;\">33<\/p>\n<p><b>Explanation: <\/b>Given n = 34<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0degree of freedom (v) = 34 &#8211; 1 = 33<\/li>\n<li>(a)\n<p style=\"display: inline;\">log (1 + log x)<\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}I=\\int \\frac{1}{x+x \\log x} d x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}I=\\int \\frac{d x}{x(1+\\log x)}{\/tex}<\/span><br \/>\nPut 1 + log x = t<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{1}{x} d x=d t{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}I=\\int \\frac{1}{t} d t{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> \u00a0I = log |t| + C<br \/>\nI = log (1 + log x) + C<\/li>\n<li>(d)\n<p style=\"display: inline;\">Irregular<\/p>\n<p><b>Explanation: <\/b>Irregular<\/li>\n<li>(b)\n<p style=\"display: inline;\">Both A and R are true but R is not the correct explanation of A.<\/p>\n<p><b>Explanation: <\/b>Given A and B are square matrices of the same order, so, A&#8217;BA is defined.<br \/>\nLet B be symmetric, then B&#8217; = B.<br \/>\nNow, (A&#8217;BA)&#8217; = A&#8217;B'(A&#8217;)&#8217; = A&#8217;BA \u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0A&#8217;BA is symmetric.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Assertion is true.<br \/>\nLet B be skew-symmetric, then B&#8217; = -B.<br \/>\nNow, (A&#8217;BA)&#8217; = A&#8217;B'(A&#8217;)&#8217; = A'(-B)A = -A&#8217;BA<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0A&#8217;BA is skew-symmetric.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Both A and R are true but R is not the correct explanation of A.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Option (Both A and R are true but R is not the correct explanation of A) is the correct answer.<\/li>\n<li>(d)\n<p style=\"display: inline;\">A is false but R is true.<\/p>\n<p><b>Explanation: <\/b>When x = 1, then y = (1)<sup>3<\/sup> &#8211; (1)<sup>3<\/sup> &#8211; 1 + 2 = 1<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\frac {dy}{dx}{\/tex}<\/span> = 3x<sup>2<\/sup> &#8211; 2x &#8211; 1 <span class=\"math-tex\">{tex}\\left. \\Rightarrow \\frac {dy}{dx}\\right|_{x = 1}{\/tex}<\/span> = 0<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> Equation of tangent at point (1, 1) is<br \/>\ny &#8211; 1 = 0(x &#8211; 1) <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> y = 1<br \/>\nSolving with the curve, x<sup>3<\/sup> &#8211; x<sup>2<\/sup> &#8211; x + 2 = 1<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x<sup>3<\/sup> &#8211; x<sup>2<\/sup> &#8211; x + 1 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> (x &#8211; 1)(x<sup>2<\/sup> &#8211; 1) = 0 <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> = 1, 1, -1 [here, 1 is repeated root]\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> Tangent meets the curve again at x = -1<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> Assertion is false, Reason is true.<\/p>\n<p>Prepare smarter and score better by downloading the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide<\/strong> app<\/a> today! Whether you&#8217;re gearing up for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, or <strong>NDA exams<\/strong>, myCBSEguide offers comprehensive study resources, including practice questions, solved papers, NCERT solutions, and more. With expertly curated content and easy-to-use features, the app is designed to help you achieve top scores with confidence. As <strong>Class 12 Applied Maths 2025<\/strong> draws closer, students should prioritize understanding the application of mathematical concepts to real-world problems.<\/p>\n<p>For <strong>teachers<\/strong>, the <strong>Examin8<\/strong> app is an ideal solution to create custom exam papers. Personalize tests with your name, logo, and branding, making it easy to assess student progress and track performance.To excel in <strong>Class 12 Applied Maths 2025<\/strong>, it&#8217;s crucial to focus on problem-solving techniques that emphasize both theoretical knowledge and practical applications.<\/p>\n<p><strong>Why Choose myCBSEguide App?<\/strong><\/p>\n<ul>\n<li><strong>Complete study material<\/strong> for <strong>CBSE<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>.<\/li>\n<li>Access to <strong>NCERT solutions<\/strong>, practice papers, and exam preparation tips.<\/li>\n<li>User-friendly interface to enhance your learning experience anytime, anywhere.Our expert-created practice papers for <strong>Class 12 Applied Maths 2025<\/strong> will help you familiarize yourself with the exam format and the types of questions you\u2019ll encounter.<\/li>\n<\/ul>\n<p><strong>Key Benefits of <a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\">Examin8<\/a> for Teachers:<\/strong><\/p>\n<ul>\n<li>Easily create <strong>customized exam papers<\/strong> with your own name and logo.<\/li>\n<li>Design personalized assessments tailored to your students\u2019 needs.<\/li>\n<li>Track student performance and streamline exam management.<\/li>\n<\/ul>\n<p>Don\u2019t wait \u2013 <strong>download <a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a><\/strong> now and start your journey towards exam success!<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section B <\/b><\/li>\n<li style=\"clear: both;\">P = \u20b9100000, i = <span class=\"math-tex\">{tex}\\frac{10}{12 \\times 100}=\\frac{1}{120}{\/tex}<\/span>, n = 2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>12 = 24.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0EMI\u00a0= <span class=\"math-tex\">{tex}\\frac{\\mathrm{P} \\ \\ + \\ \\ \\mathrm{P} n i}{n}{\/tex}<\/span>\u00a0= <span class=\"math-tex\">{tex}\\frac{100000 \\ \\ + \\ \\ 100000 \\ \\ \\times \\ \\ \\frac{1}{120} \\ \\ \\times \\ \\ 24}{24} {\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{100000 \\ \\ + \\ \\ 20000}{24}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{120000}{24}{\/tex}<\/span>\u00a0= \u20b95000<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Let P be the present value of a perpetuity of \u20b92,000 payable at the end of each year when money is worth 5%. It is given that<br \/>\ni =\u00a0<span class=\"math-tex\">{tex}\\frac{5}{100}{\/tex}<\/span>\u00a0= 0.05 and R = 2,000<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0P =\u00a0<span class=\"math-tex\">{tex}\\frac{R}{i} \\Rightarrow{\/tex}<\/span>\u00a0P =\u00a0\u20b9<span class=\"math-tex\">{tex}\\frac{2,000}{0.05}{\/tex}<\/span>\u00a0=\u00a0\u20b940,000<br \/>\nLet P<sub>1<\/sub> be the present value of an ordinary annuity of \u20b92,000 per year for 100 years. Then,<br \/>\nP<sub>1<\/sub>\u00a0= R<span class=\"math-tex\">{tex}\\left\\{\\frac{1-(1-i)^{-n}}{i}\\right\\}{\/tex}<\/span><br \/>\nWe have, R = 2,000, i =\u00a0<span class=\"math-tex\">{tex}\\frac{5}{100}{\/tex}<\/span>\u00a0= 0.05 and n = 100<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0P<sub>1<\/sub>\u00a0=\u00a0\u20b92,000\u00a0<span class=\"math-tex\">{tex}\\left\\{\\frac{1-(1.05)^{-100}}{0.05}\\right\\}{\/tex}<\/span>\u00a0=\u00a0\u20b92,000<span class=\"math-tex\">{tex}\\left(\\frac{1-0.0076}{0.05}\\right){\/tex}<\/span>\u00a0=\u00a0\u20b9<span class=\"math-tex\">{tex}\\frac{2,000 \\times 0.9924}{0.05}{\/tex}<\/span> =\u00a0\u20b939,696<br \/>\nWe observe that the present value of the perpetuity is more than that of ordinary annuity.<\/li>\n<li><img decoding=\"async\" style=\"height: 289px; width: 400px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1707737011-6jcvd3.jpg\" \/><\/li>\n<li>Let\u00a0<span class=\"math-tex\">{tex}\\frac{1}{x\\left(1+x^{2}\\right)}=\\frac{A}{x}+\\frac{B x+C}{1+x^{2}}{\/tex}<\/span>\u00a0&#8230;(i)<br \/>\nThen, 1 = A(1 + x<sup>2<\/sup>) + (Bx + C)x &#8230;(ii)<br \/>\nPutting x = 0 in (ii), we get A = 1<br \/>\nComparing the coefficients of x<sup>2<\/sup> and x in (ii), we get<br \/>\nA + B = 0 and C = 0\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0B = -1 and C = 0 [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0A = 1]\nSubstituting the values of A, B and C in (i), we get<br \/>\n<span class=\"math-tex\">{tex}\\frac{1}{x\\left(1+x^{2}\\right)}=\\frac{1}{x}-\\frac{x}{1+x^{2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0I =\u00a0<span class=\"math-tex\">{tex}\\int_\\limits{1}^{2} \\frac{1}{x\\left(1+x^{2}\\right)}{\/tex}<\/span>\u00a0dx =\u00a0<span class=\"math-tex\">{tex}\\int_\\limits{1}^{2} \\frac{1}{x}{\/tex}<\/span>\u00a0dx &#8211;\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2} \\int_\\limits{1}^{2} \\frac{2 x}{1+x^{2}}{\/tex}<\/span>\u00a0dx =\u00a0<span class=\"math-tex\">{tex}[\\log x]_{1}^{2}-\\frac{1}{2}\\left[\\log \\left(1+x^{2}\\right)\\right]_{1}^{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0I = (log 2 &#8211; log 1) &#8211;\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0(log 5 &#8211; log 2) = log 2 &#8211;\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0log 5 +\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0log 2 =\u00a0<span class=\"math-tex\">{tex}\\frac{3}{2}{\/tex}<\/span>\u00a0log 2 &#8211;\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0log 5<\/li>\n<li style=\"clear: both;\">As A is a non-singular square matrix of order 3, |adj A| = |A|<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 225 = |A|<sup>2<\/sup> <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> |A| = 15, -15<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> |A&#8217;| = 15, -15 (<span class=\"math-tex\">{tex}\\because{\/tex}<\/span> |A&#8217;| = |A|)<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>AB =\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{r} 1 \\\\ -4 \\\\ 3 \\end{array}\\right]{\/tex}<\/span>\u00a0[-1 2 1 ] =\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{rrr} -1 &amp; 2 &amp; 1 \\\\ 4 &amp; -8 &amp; -4 \\\\ -3 &amp; 6 &amp; 3 \\end{array}\\right]{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(AB)&#8217; =\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{rrr} -1 &amp; 4 &amp; -3 \\\\ 2 &amp; -8 &amp; 6 \\\\ 1 &amp; -4 &amp; 3 \\end{array}\\right]{\/tex}<\/span>\u00a0and\u00a0 B&#8217;A&#8217; = [-1 2 1]&#8217;\u00a0<span class=\"math-tex\">{tex}\\left[\\begin{array}{r} 1 \\\\ -4 \\\\ 3 \\end{array}\\right]^{\\prime}=\\left[\\begin{array}{r} -1 \\\\ 2 \\\\ 1 \\end{array}\\right]{\/tex}<\/span>\u00a0[1 -4 3]\u00a0\u00a0<span class=\"math-tex\">{tex}= \\left[\\begin{array}{rrr} -1 &amp; 4 &amp; -3 \\\\ 2 &amp; -8 &amp; 6 \\\\ 1 &amp; -4 &amp; 3 \\end{array}\\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> (AB)&#8217; = B&#8217;A&#8217;.<\/li>\n<li>We have, 5! = 120, 6! = 720, 7! =5040 etc.<br \/>\nIn fact units digit in n! for n <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a05 is 0. Therefore, units digit in 183! is 0. Consequently, units digit in 183! + 3<sup>183<\/sup> is same as the units digit in 3<sup>183<\/sup>.<br \/>\nNow, 3<sup>2<\/sup> = 9 <span class=\"math-tex\">{tex}\\equiv{\/tex}<\/span> -1 (mod 10) and, 3<sub>183<\/sub> = (3<sup>2<\/sup>)<sup>91<\/sup> <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a03<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a03<sup>2<\/sup> <span class=\"math-tex\">{tex}\\equiv{\/tex}<\/span> -1 (mod 10)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(3<sup>2<\/sup>)<sup>91\u00a0<\/sup><span class=\"math-tex\">{tex}\\equiv{\/tex}<\/span> (-1)<sup>91<\/sup> (mod 10)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(3<sup>2<\/sup>)<sup>91<\/sup> <span class=\"math-tex\">{tex}\\equiv{\/tex}<\/span> -1\u00a0(mod 10)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(3<sup>2<\/sup>)<sup>91<\/sup> <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a03 <span class=\"math-tex\">{tex}\\equiv{\/tex}<\/span> -3 (mod 10)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(3<sup>2<\/sup>)<sup>91<\/sup>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 3 <span class=\"math-tex\">{tex}\\equiv{\/tex}<\/span>\u00a07 (mod 10)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a03<sup>183<\/sup> <span class=\"math-tex\">{tex}\\equiv{\/tex}<\/span> 7 (mod 10)<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Units digit in 3<sup>183<\/sup> is 7.<br \/>\nHence, the units digit in 183! + 3<sup>183<\/sup> is 7<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section C <\/b><\/li>\n<li>Given, P = \u20b9 1,000 Annual Coupon Payment<br \/>\n= \u20b9 1,000 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 10% = \u20b9 100<br \/>\nSemi-annual Coupon Payment,<br \/>\nC = \u20b9 100 + 2 = \u20b9 50<br \/>\nr = 8% + 2<br \/>\n= 4%<br \/>\n= 0.04<br \/>\nN = 3 year <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 2 = 6 periods for semi-annual coupon payments<br \/>\nBond Value = C <span class=\"math-tex\">{tex}\\times \\frac{1-(1+r)^{-N}}{r}+\\frac{P}{(1+r)^{-N}}{\/tex}<\/span><br \/>\n= 50 <span class=\"math-tex\">{tex}\\times \\frac{1-(1+0.04)^{-6}}{0.04}{\/tex}<\/span><span class=\"math-tex\">{tex}+\\frac{1000}{(1+0.04)^{6}}{\/tex}<\/span><br \/>\n= 50 <span class=\"math-tex\">{tex}\\times \\frac{1-0.79031}{0.04}{\/tex}<\/span>\u00a0+ 1000 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 0.79031<br \/>\n= \u20b9 262.1125 + 790.31<br \/>\n= \u20b9 1053.42<\/li>\n<li style=\"clear: both;\">Let A be the quantity of bacteria present in culture at any time t and initial quantity of bacteria is A<sub>0<\/sub><br \/>\n<span class=\"math-tex\">{tex}\\frac{d A}{d A} \\propto A{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{d A}{d t}=\\lambda A {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{d A}{A}=\\lambda d t {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\int \\frac{d A}{A}= \\int \\lambda d t {\/tex}<\/span><br \/>\nlog A = <span class=\"math-tex\">{tex}\\lambda{\/tex}<\/span>t + c &#8230;(i)<br \/>\nInitially, A = A<sub>0<\/sub>, t = 0<br \/>\nlog A<sub>0<\/sub> = 0 + c<br \/>\nlog A<sub>0<\/sub> =\u00a0c<br \/>\nNow equation (i) becomes,<br \/>\nlog A = <span class=\"math-tex\">{tex}\\lambda{\/tex}<\/span>t + log A<sub>0<\/sub><br \/>\n<span class=\"math-tex\">{tex}\\log \\left(\\frac{A}{A_{0}}\\right) = \\lambda t{\/tex}<\/span>\u00a0&#8230;(ii)<br \/>\nGiven A = 2 A<sub>0<\/sub>\u00a0when t = 6 hours<br \/>\n<span class=\"math-tex\">{tex}\\log \\left(\\frac{A}{A_{0}}\\right) = 6 \\lambda{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{\\log 2}{6} = \\lambda{\/tex}<\/span><br \/>\nNow equation (ii) becomes,<br \/>\n<span class=\"math-tex\">{tex}\\log \\left(\\frac{A}{A_{0}}\\right) = \\frac{\\log 2}{6} t{\/tex}<\/span><br \/>\nNow, A = 8 A<sub>0<\/sub><br \/>\nso, <span class=\"math-tex\">{tex}\\log \\left(\\frac{8 A_{0}}{A_{0}}\\right) = \\frac{\\log 2}{6} t{\/tex}<\/span><br \/>\nlog 2<sup>3<\/sup> =\u00a0<span class=\"math-tex\">{tex}\\frac{\\log 2}{6} t {\/tex}<\/span><br \/>\n3 log 2 =\u00a0<span class=\"math-tex\">{tex}\\frac{\\log 2}{6} t{\/tex}<\/span><br \/>\n18 = t<br \/>\nHence, Bacteria becomes 8 times in 18 hours.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>The given equation of one parameter family of curves is<br \/>\nx<sup>2<\/sup> &#8211; y<sup>2<\/sup> c(x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>)<sup>2<\/sup>\u00a0&#8230;(i)<br \/>\nDifferentiating (i) with respect to x, we get<br \/>\n2x &#8211; 2y<span class=\"math-tex\">{tex}\\frac {dy}{dx}{\/tex}<\/span>\u00a0= 2c(x<sup>2<\/sup> + y<sup>2<\/sup>)(2x + 2y<span class=\"math-tex\">{tex}\\frac {dy}{dx}{\/tex}<\/span>)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(x\u00a0&#8211; y<span class=\"math-tex\">{tex}\\frac {dy}{dx}{\/tex}<\/span>) =\u00a02c(x<sup>2<\/sup> + y<sup>2<\/sup>)(x + y<span class=\"math-tex\">{tex}\\frac {dy}{dx}{\/tex}<\/span>) &#8230;(ii)<br \/>\nOn substituting the value of c obtained from (i) in (ii), we get,<br \/>\n<span class=\"math-tex\">{tex}\\left(x-y \\frac{d y}{d x}\\right)=\\frac{2\\left(x^{2}-y^{2}\\right)\\left(x^{2}+y^{2}\\right)}{\\left(x^{2}+y^{2}\\right)^{2}}\\left(x+y \\frac{d y}{d x}\\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(x<sup>2<\/sup> + y<sup>2<\/sup>)(x\u00a0&#8211; y<span class=\"math-tex\">{tex}\\frac {dy}{dx}{\/tex}<\/span>) =\u00a02(x<sup>2<\/sup>\u00a0&#8211;\u00a0y<sup>2<\/sup>)(x + y<span class=\"math-tex\">{tex}\\frac {dy}{dx}{\/tex}<\/span>)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0{x(x<sup>2<\/sup> + y<sup>2<\/sup>) &#8211; 2x(x<sup>2<\/sup>\u00a0&#8211;\u00a0y<sup>2<\/sup>)} =\u00a0<span class=\"math-tex\">{tex}\\frac {dy}{dx}{\/tex}<\/span>{2y(x<sup>2<\/sup>\u00a0&#8211;\u00a0y<sup>2<\/sup>) +\u00a0y(x<sup>2<\/sup>\u00a0+\u00a0y<sup>2<\/sup>)}<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(3xy<sup>2<\/sup> &#8211; x<sup>3<\/sup>) =\u00a0<span class=\"math-tex\">{tex}\\frac {dy}{dx}{\/tex}<\/span>(3x<sup>2<\/sup>y &#8211; y<sup>3<\/sup>)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(x<sup>3<\/sup> &#8211; 3 xy<sup>2<\/sup>) dx = (y<sup>3<\/sup> &#8211; 3x<sup>2<\/sup>y) dy, which is the given differential equation.<\/li>\n<li>We find the total revenue function TR by integrating the marginal revenue function MR:<br \/>\nTR (Q) = <span class=\"math-tex\">{tex}\\int{\/tex}<\/span> MR (Q) dQ<br \/>\n= <span class=\"math-tex\">{tex}\\int{\/tex}<\/span> (100 + 20Q + 3Q<sup>2<\/sup>) dQ<br \/>\n= 100Q + 10Q<sup>2 <\/sup> + Q<sup>3 <\/sup> + C<br \/>\nThe constant of integration C can be determined using the initial condition TR (Q=2) = 260. Hence,<br \/>\n200 + 40 + 8 + C = 260<br \/>\nC = 12<br \/>\nSo, the total revenue function is given by<br \/>\nTR (Q) = 100Q + 10Q<sup>2 <\/sup> + Q<sup>3 <\/sup> + 12<\/li>\n<li>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\">Years (t<sub>i<\/sub>)<\/th>\n<th scope=\"col\">Sales (y<sub>i<\/sub>)<\/th>\n<th scope=\"col\">x<sub>i<\/sub> = t<sub>i<\/sub> &#8211; 2015<\/th>\n<th scope=\"col\"><span class=\"math-tex\">{tex}x ^2_i{\/tex}<\/span><\/th>\n<th scope=\"col\">x<sub>i<\/sub>y<sub>i<\/sub><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>2012<\/td>\n<td>9<\/td>\n<td>-3<\/td>\n<td>9<\/td>\n<td>-27<\/td>\n<\/tr>\n<tr>\n<td>2013<\/td>\n<td>11<\/td>\n<td>-2<\/td>\n<td>4<\/td>\n<td>-22<\/td>\n<\/tr>\n<tr>\n<td>2014<\/td>\n<td>13<\/td>\n<td>-1<\/td>\n<td>1<\/td>\n<td>-13<\/td>\n<\/tr>\n<tr>\n<td>2015<\/td>\n<td>12<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td>2016<\/td>\n<td>14<\/td>\n<td>1<\/td>\n<td>1<\/td>\n<td>14<\/td>\n<\/tr>\n<tr>\n<td>2017<\/td>\n<td>15<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>30<\/td>\n<\/tr>\n<tr>\n<td>2018<\/td>\n<td>17<\/td>\n<td>3<\/td>\n<td>9<\/td>\n<td>51<\/td>\n<\/tr>\n<tr>\n<td>n = 7<\/td>\n<td><span class=\"math-tex\">{tex}\\sum y_i{\/tex}<\/span>\u00a0= 91<\/td>\n<td><span class=\"math-tex\">{tex}\\sum x_i{\/tex}<\/span>\u00a0= 0<\/td>\n<td><span class=\"math-tex\">{tex}\\sum x_i^2{\/tex}<\/span>\u00a0= 28<\/td>\n<td><span class=\"math-tex\">{tex}\\sum x_i y_i{\/tex}<\/span>\u00a0= 33<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span class=\"math-tex\">{tex}a = \\frac {\\sum y _i}{n} = \\frac {91}{7}{\/tex}<\/span>\u00a0=13<br \/>\n<span class=\"math-tex\">{tex}b = \\frac {\\sum x_i y _i}{\\sum x_i^2} = \\frac {33}{28}{\/tex}<\/span>\u00a0= 1.179<br \/>\nThe equation of the straight line trend is<br \/>\ny = ax + b<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> y = 13x + 1.179<\/li>\n<li><span class=\"math-tex\">{tex}\\mu ={\/tex}<\/span> Population mean = 12 Kg<br \/>\n<span class=\"math-tex\">{tex}\\overline{X}{\/tex}<\/span> = Sample mean = 11.8 Kg<br \/>\nn = 10<br \/>\nSample standard deviation = s = 0.15<br \/>\nNull Hypothesis H<sub>0 <\/sub>= There is no significance between the sample mean<br \/>\n<span class=\"math-tex\">{tex}\\overline{X}{\/tex}<\/span> and the population mean \u200b<span class=\"math-tex\">{tex}\\mu {\/tex}<\/span>.<br \/>\nAlternate Hypothesis H<sub>1<\/sub> = There is significance between the sample mean <span class=\"math-tex\">{tex}\\overline{X}{\/tex}<\/span> and the population mean \u200b<span class=\"math-tex\">{tex}\\mu {\/tex}<\/span><br \/>\nLet t be the test statistic given by<br \/>\n<span class=\"math-tex\">{tex}t = \\frac{\\overline{X}-\\mu}{\\frac{s}{\\sqrt{n-1}}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}t = \\left(\\frac{11.8-12}{0.15}\\right)\\times 3{\/tex}<\/span><br \/>\n= -4<br \/>\nThe test statistic t follows student t-distribution with (10-1)=9 degrees of freedom<br \/>\nIt is given that t<sub>0.05<\/sub> = 2.26<br \/>\nWe observe that,<br \/>\n|t| = 4&gt;2.26<br \/>\n<sub><span class=\"math-tex\">{tex}\\implies{\/tex}<\/span><\/sub>Calculate |t| &gt; tabulated t<sub>9<\/sub>(0.05)<br \/>\nSo, the null hypothesis is rejected at a 5% level of significance.<br \/>\nHence there is a significance between the sample mean <span class=\"math-tex\">{tex}\\overline{X}{\/tex}<\/span> and the population mean\u200b <span class=\"math-tex\">{tex}\\mu{\/tex}<\/span>\u200b\u200b\u200b\u200b\u200b.<\/li>\n<li style=\"clear: both;\">Let X denote the marks obtained by the candidates. Let <span class=\"math-tex\">{tex}\\mu{\/tex}<\/span>\u00a0be mean and <span class=\"math-tex\">{tex}\\sigma {\/tex}<\/span>\u00a0be the standard deviation of the normal distribution.<br \/>\nLet Z be the standard normal variate. Then,<br \/>\nZ =\u00a0<span class=\"math-tex\">{tex}\\frac{X-\\mu}{\\sigma}{\/tex}<\/span><br \/>\nNow, X = 40\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0Z =\u00a0<span class=\"math-tex\">{tex}\\frac{40-\\mu}{\\sigma}{\/tex}<\/span>\u00a0= Z<sub>1<\/sub>\u00a0(say) and, X = 75\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0Z =\u00a0\u00a0<span class=\"math-tex\">{tex}\\frac{75-\\mu}{\\sigma}{\/tex}<\/span>\u00a0= Z<sub>2<\/sub>\u00a0(say)<br \/>\nNow,<br \/>\nP(X <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span> 40) = 0.46 [Given]\n= P(Z <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a0Z<sub>1<\/sub>) = 0.46<br \/>\n=\u00a0P(Z <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a00) &#8211; P(0 <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z<sub>1<\/sub>) = 0.46 [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0P(Z <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z<sub>1<\/sub>) &lt; 0.5\u00a0<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Z<sub>1<\/sub>\u00a0&gt; 0]\n= 0.5 &#8211; P(0 <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z<sub>1<\/sub>) = 0.46<br \/>\n= P(0 <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z<sub>1<\/sub>) = 0.04<br \/>\n= Z<sub>1<\/sub> = 0.1<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{40-\\mu}{\\sigma}{\/tex}<\/span>\u00a0= 0.1<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\mu{\/tex}<\/span>\u00a0+ 0.1\u00a0<span class=\"math-tex\">{tex}\\sigma {\/tex}<\/span>\u00a0= 40 &#8230;(i)<br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 58px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1623066672-qnh2qx.jpg\" alt=\"\" data-imgur-src=\"i5l1ErS.png\" \/><br \/>\nAnd,<br \/>\nP(X <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a075) = 0.09\u00a0[Given]\n= P(Z <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a0Z<sub>2<\/sub>) = 0.09<br \/>\n=\u00a0P(Z <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a00) &#8211; P(0 <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z<sub>2<\/sub>) = 0.09\u00a0[<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0P(Z <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z<sub>2<\/sub>) &lt; 0.5\u00a0<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Z<sub>2<\/sub>\u00a0&gt; 0]\n= 0.5 &#8211; P(0 <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z<sub>2<\/sub>) = 0.09<br \/>\n= P(0 <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0Z<sub>2<\/sub>) = 0.41<br \/>\n= Z<sub>2<\/sub> = 1.34<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{75-\\mu}{\\sigma}{\/tex}<\/span>\u00a0= 1.34<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\mu{\/tex}<\/span>\u00a0+ 1.34\u00a0<span class=\"math-tex\">{tex}\\sigma {\/tex}<\/span>\u00a0= 75 &#8230;(ii)<br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 67px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1623066675-vubza6.jpg\" alt=\"\" data-imgur-src=\"GJeTSho.png\" \/><br \/>\nSolving (i) and (ii), we get <span class=\"math-tex\">{tex}\\mu{\/tex}<\/span>\u00a0= 37.18 and <span class=\"math-tex\">{tex}\\sigma {\/tex}<\/span>\u00a0= 28.22.<br \/>\nThus, the average mark obtained by the candidates is 37<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Let X is the random variable that denotes that a bulb is defective.<br \/>\nAlso, n = 10, <span class=\"math-tex\">{tex}p = \\frac{1}{{50}}{\/tex}<\/span> and <span class=\"math-tex\">{tex}q = \\frac{{49}}{{50}}{\/tex}<\/span> and P(X = r) <span class=\"math-tex\">{tex}{ = ^n}{C_r }{p^r}{q^{n &#8211; r }}{\/tex}<\/span><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>None of the bulbs are defective i.e., r = 0<br \/>\n<span class=\"math-tex\">{tex}\\therefore P(X = r) = {P_{(0)}}{ = ^{10}}{C_0}{\\left( {\\frac{1}{{50}}} \\right)^0}{\\left( {\\frac{{49}}{{50}}} \\right)^{10 &#8211; 0}} = {\\left( {\\frac{{49}}{{50}}} \\right)^{10}}{\/tex}<\/span><\/li>\n<li>Exactly two bulbs are defective i.e., r = 2<br \/>\n<span class=\"math-tex\">{tex}\\therefore P(X = r) = {P_{(2)}}{ = ^{10}}{C_2}{\\left( {\\frac{1}{{50}}} \\right)^2}{\\left( {\\frac{{49}}{{50}}} \\right)^8}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{10!}}{{8!2!}}{\\left( {\\frac{1}{{50}}} \\right)^2} \\cdot {\\left( {\\frac{{49}}{{50}}} \\right)^8} = 45 \\times {\\left( {\\frac{1}{{50}}} \\right)^{10}} \\times {(49)^8}{\/tex}<\/span><\/li>\n<li>More than 8 bulbs work properly i.e., there are less than 2 bulbs that are defective.<br \/>\nSo, r &lt; 2 <span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> r = 0,1<br \/>\n<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span> P(X = r) = P(r &lt; 2) = P(0) + P(1)<br \/>\n<span class=\"math-tex\">{tex}{ = ^{10}}{C_0}{\\left( {\\frac{1}{{50}}} \\right)^0}{\\left( {\\frac{{49}}{{50}}} \\right)^{10 &#8211; 0}}{ + ^{10}}{C_1}{\\left( {\\frac{1}{{50}}} \\right)^1}{\\left( {\\frac{{49}}{{50}}} \\right)^{10 &#8211; 1}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = {\\left( {\\frac{{49}}{{50}}} \\right)^{10}} + \\frac{{10!}}{{1!9!}} \\cdot \\frac{1}{{50}} \\cdot {\\left( {\\frac{{49}}{{50}}} \\right)^9}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = {\\left( {\\frac{{49}}{{50}}} \\right)^{10}} + \\frac{1}{5} \\cdot {\\left( {\\frac{{49}}{{50}}} \\right)^9} = {\\left( {\\frac{{49}}{{50}}} \\right)^9}\\left( {\\frac{{49}}{{50}} + \\frac{1}{5}} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = {\\left( {\\frac{{49}}{{50}}} \\right)^9}\\left( {\\frac{{59}}{{50}}} \\right) = \\frac{{59{{(49)}^9}}}{{{{(50)}^{10}}}}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<ol style=\"padding-left: 20px;\">\n<li style=\"text-align: center; display: block;\"><b>Section D<\/b><\/li>\n<li style=\"clear: both;\">Let x and y be the number of bikes of model X and Y respectively, then the problem can be formulated as an L.P.P.as follows:<br \/>\nMaximize Z = 1000x + 600y subject to constraints<br \/>\n6x + 10y\u00a0<span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0450\u00a0(man hours constraint)<br \/>\ni.e. 3x + 5y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a0225<br \/>\n2000x + 1000\u00a0<span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 80000\u00a0(handling and marketing constraints)<br \/>\ni.e. 2x + y\u00a0<span class=\"math-tex\">{tex}\\leq{\/tex}<\/span> 80<br \/>\nx\u00a0<span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a00, y\u00a0<span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a00\u00a0(non-negativity constraints)<br \/>\nDraw the lines 3x + 5y = 225 and 2x + y = 80 and shade the region satisfied by the above inequalities.<br \/>\n<img decoding=\"async\" style=\"height: 248px; width: 210px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1707280471-db67d6.jpg\" \/><br \/>\nThe feasible region OABC is convex and bounded.<br \/>\nThe corner points are O(0, 0), A(40, 0), B(25, 30) and C(0, 45).<br \/>\nNow, let us give some value to Z say 12000 and draw a dotted line 1000x + 600y = 12000 which is iso-profit line. Move this line parallel to itself over the feasible region.<br \/>\nIt passes through all corner points one by one. The farthest corner point it crosses is B(25, 30) which gives us the optimal solution.<br \/>\nZ = 1000 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a025 + 600\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 30 i.e. Z = 43000.<br \/>\nHence, 25 bikes of model X and 30 bikes of model X should be manufactured to obtain maximum profit of \u20b9 43000<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Let the tailor P work for x days and the tailor Q work for y days respectively.<br \/>\nHere, the problem can be formulated as an L.P.P. as follows:<br \/>\nMinimize Z = 150x + 200y<br \/>\nSubject to the constraints:<br \/>\n6x + 10y\u00a0<span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a060<br \/>\nor 3x + 5y\u00a0<span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a030 &#8230;(i)<br \/>\n4x + 4y\u00a0<span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a032<br \/>\nor x + y\u00a0<span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a08 &#8230;(ii)<br \/>\nand x\u00a0<span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a00, y\u00a0<span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a00<br \/>\nConverting them into equations we obtain the following equations:<br \/>\n3x + 5y = 30, x + y = 8<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0y =\u00a0<span class=\"math-tex\">{tex}\\frac{30-3 x}{5}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0y = 8 &#8211; x<\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>x<\/td>\n<td>0<\/td>\n<td>10<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td>y<\/td>\n<td>6<\/td>\n<td>0<\/td>\n<td>3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>x<\/td>\n<td>0<\/td>\n<td>8<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td>y<\/td>\n<td>8<\/td>\n<td>0<\/td>\n<td>3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img decoding=\"async\" style=\"height: 287px; width: 300px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1646645850-xedskh.jpg\" \/><br \/>\nThe shaded region in the diagram represent the feasible region.<br \/>\nThe corner points are A(10, 0), B(5, 3) and C(0, 8)<br \/>\nAt the corner point the value of Z = 150x + 200y<br \/>\nAt (10, 0) Z = 1500<br \/>\nAt B( 5, 3) Z = 150\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a05 + 200\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a03<br \/>\n= 750 + 600 = 1350<br \/>\nAt C(0, 8) Z = 1600<br \/>\nAs the feasible region is unbounded, we draw the graph of the half-plane.<br \/>\n150x + 200y &lt; 1350<br \/>\n3x + 4y &lt; 27<br \/>\nThere is no point common with the feasible region, therefore, Z has minimum value.<br \/>\nMinimum value of Z is \u20b9\u00a01350 and it occurs at the point B(5, 3).<br \/>\nHence, the labour cost is \u20b9\u00a01350 when P works for 5 days and Q works for 3 days.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>0.1 + k + 2k + 2k + k = 1<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a00.1 + 6k = 1<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow {k}=\\frac{3}{20}{\/tex}<\/span><\/li>\n<li><span class=\"math-tex\">{tex}{P}({X} \\geq {2}){\/tex}<\/span>\u00a0= P(2) + P(3) + P(4)<br \/>\n= 2k + 2k + k<br \/>\n= 5k =\u00a0<span class=\"math-tex\">{tex}\\frac{3}{4}{\/tex}<\/span><\/li>\n<li><span class=\"math-tex\">{tex}{P}({X} \\leq {2}){\/tex}<\/span>\u00a0= P(0) + P(1) + P(2)<br \/>\n= 0.1 + k + 2k<br \/>\n<span class=\"math-tex\">{tex}=\\frac{1}{10}+\\frac{9}{20}=\\frac{11}{20}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n<li style=\"clear: both;\">\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Given: first six positive integers.<br \/>\nTwo numbers can be selected at random (without replacement) from the first six positive integer in 6 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a05 = 30 ways.<br \/>\nX denotes the larger of the two numbers obtained. Hence, X can take any value of 2, 3, 4, 5 or 6.<br \/>\nFor X = 2, the possible observations are (1, 2) and (2, 1)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow P(X)=\\frac{2}{30}=\\frac{1}{15}{\/tex}<\/span><br \/>\nFor X = 3, the possible observations are (1, 3), (3, 1), (2, 3) and (3, 2).<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow P(X)=\\frac{4}{30}=\\frac{2}{15}{\/tex}<\/span><br \/>\nFor X = 4, the possible observations are (1, 4), (4, 1), (2, 4), (4, 2), (3, 4) and (4, 3).<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow P(X)=\\frac{6}{30}=\\frac{1}{5}{\/tex}<\/span><br \/>\nFor X = 5, the possible observations are (1, 5), (5, 1), (2, 5), (5, 2), (3, 5), (5, 3), (5, 4) and (4, 5).<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow P(X)=\\frac{8}{30}=\\frac{4}{15}{\/tex}<\/span><br \/>\nFor X = 6, the possible observations are (1, 6), (6, 1), (2, 6), (6, 2), (3, 6), (6, 3) (6, 4), (4, 6), (5, 6) and (6, 5).<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow P(X)=\\frac{10}{30}=\\frac{1}{3}{\/tex}<\/span><br \/>\nHence, the required probability distribution is,<\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>X<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>6<\/td>\n<\/tr>\n<tr>\n<td>P(X)<\/td>\n<td><span class=\"math-tex\">{tex}\\frac{1}{15}{\/tex}<\/span><\/td>\n<td><span class=\"math-tex\">{tex}\\frac{2}{15}{\/tex}<\/span><\/td>\n<td><span class=\"math-tex\">{tex}\\frac{1}{5}{\/tex}<\/span><\/td>\n<td><span class=\"math-tex\">{tex}\\frac{4}{15}{\/tex}<\/span><\/td>\n<td><span class=\"math-tex\">{tex}\\frac{1}{3}{\/tex}<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Therefore E(X) =\u00a0<span class=\"math-tex\">{tex}2 \\times \\frac{1}{15}+3 \\times \\frac{2}{15}+4 \\times \\frac{1}{5}+5 \\times \\frac{4}{15}+6 \\times \\frac{1}{3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow E(X)=\\frac{14}{3}{\/tex}<\/span><\/li>\n<li>Consider the line x + 2y = 8. We observe that the shaded region and the origin are on the same side of the line x + 2y = 8 and (0, 0) satisfies the linear constraint x + 2y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a08. So, we must have one inequations as x + 2y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a08.<br \/>\n<img decoding=\"async\" style=\"height: 240px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1699014637-ajrv4k.jpg\" alt=\"\" data-imgur-src=\"kkpq4vU.png\" \/><br \/>\nNow, consider the line 2x + y = 2. We find that the shaded region and the origin are on the opposite sides of the line 2x + y = 2 and (0, 0) does not satisfy the inequation 2x + y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a02. So, the second inequations is 2x + y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a02.<br \/>\nFinally, consider the line x &#8211; y = 1. We observe that the shaded region and the origin are on the same side of line x &#8211; y = 1. We observe that the shaded region and the origin are on the same side of line x &#8211; y = 1 and (0, 0) satisfies x &#8211; y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a01. So, the third constraint is x &#8211; y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a01.<br \/>\nWe also notice that the shaded region is above x-axis and is on the right side of y-axis. So, we must have x <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a00 and y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a00. Thus, the linear inequations corresponding to the given solution set are<br \/>\nx + 2y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a08, 2x + y <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a02, x &#8211; y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a01, x <span class=\"math-tex\">{tex}\\geq{\/tex}<\/span>\u00a00, y <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a00<\/li>\n<li>Cost of new machine = \u20b9 300000<br \/>\nScrap value of old machine = \u20b9 30000<br \/>\nHence, the money required for new machine after 7 years<br \/>\n= \u20b9 300000 &#8211; \u20b9 30000 = \u20b9 270000<br \/>\nSo, we have A = \u20b9 270000, i = <span class=\"math-tex\">{tex}\\frac{5}{100}{\/tex}<\/span>\u00a0= 0.05, n = 7<br \/>\nUsing formula, A = R<span class=\"math-tex\">{tex}\\left[\\frac{(1+i)^{n}-1}{i}\\right]{\/tex}<\/span>, we get<br \/>\n270000 = R<span class=\"math-tex\">{tex}\\left[\\frac{(1.05)^{7}-1}{0.05}\\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0R =\u00a0<span class=\"math-tex\">{tex}\\frac{270000 \\times 0.05}{(1.05)^{7}-1}{\/tex}<\/span>\u00a0[Let x = (1.05)<sup>7\u00a0<\/sup><span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x = 7 log 1.05 = 7 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 0.0212 = 0.1484\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x = antilog 0.1484\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x = 1.407]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0R\u00a0= <span class=\"math-tex\">{tex}\\frac{13500}{1.407-1}=\\frac{13500}{0.407}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0R\u00a0= 33169.53<br \/>\nHence, the company should deposit \u20b9 33169.53 at the end of each year for 7 years.<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<ol style=\"padding-left: 20px;\">\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;\">Given, y = tan<sup>-1<\/sup><span class=\"math-tex\">{tex}\\left(\\frac{\\log \\left(\\frac{e}{x^{2}}\\right)}{\\log e x^{2}}\\right){\/tex}<\/span> + tan<sup>-1<\/sup><span class=\"math-tex\">{tex}\\left(\\frac{3+2 \\log x}{1-6 \\log x}\\right){\/tex}<\/span><br \/>\n= tan<sup>-1<\/sup><span class=\"math-tex\">{tex}\\left(\\frac{1-\\log x^{2}}{1+\\log x^{2}}\\right){\/tex}<\/span>+ tan<sup>-1<\/sup><span class=\"math-tex\">{tex}\\left(\\frac{3+2 \\log x}{1-6 \\log x}\\right){\/tex}<\/span><br \/>\n= tan<sup>-1<\/sup>(1) &#8211; tan<sup>-1<\/sup>(2log x) + tan<sup>-1<\/sup>(3) + tan<sup>-1<\/sup>(2log x)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> y = tan<sup>-1<\/sup>(1) + tan<sup>-1<\/sup>(3)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> <span class=\"math-tex\">{tex}\\frac{d y}{d x}{\/tex}<\/span> = 0 <span class=\"math-tex\">{tex}\\Rightarrow \\frac{d^{2} y}{d x^{2}}{\/tex}<\/span> = 0<\/li>\n<li style=\"text-align: left;\">Given, x = s + 3t, y = 2s &#8211; t <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> <span class=\"math-tex\">{tex}\\frac{d x}{d s}=1{\/tex}<\/span>,<span class=\"math-tex\">{tex}\\frac{d y}{d s}=2 {\/tex}<\/span><br \/>\nNow, u = x<sup>2<\/sup> + y<sup>2<\/sup> <span class=\"math-tex\">{tex}\\frac{d u}{d s}=2 x \\frac{d x}{d s}+2 y \\frac{d y}{d s}{\/tex}<\/span> = 2x + 4y<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{d^{2} u}{d s^{2}}=2\\left(\\frac{d x}{d s}\\right)+4\\left(\\frac{d y}{d s}\\right) \\Rightarrow \\frac{d^{2} u}{d s^{2}}{\/tex}<\/span> = 2(1) + 4(2) = 10<\/li>\n<li style=\"text-align: left;\">We have,f{x) = 2 log sin x<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> f'(x) = 2<span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{1}{\\sin x}{\/tex}<\/span><span class=\"math-tex\">{tex}\\cdot{\/tex}<\/span>cos x = 2 cos x<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> <span class=\"math-tex\">{tex}f^{\\prime \\prime}{\/tex}<\/span>(x) = -2 cosec<sup>2<\/sup>x<br \/>\n<strong>OR<\/strong><br \/>\nWe have, f(x) = e<sup>x<\/sup>sinx<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> f'(x) = e<sup>x<\/sup>cosx + e<sup>x<\/sup>sinx = e<sup>x<\/sup>(cosx + sinx)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> e<sup>x<\/sup>(x) = e<sup>x<\/sup>(cosx &#8211; sinx) + e<sup>x<\/sup>(cosx + sinx) = 2e<sup>x<\/sup> cosx<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> f&#8221;(x) = 2[e<sup>x<\/sup>cosx &#8211; e<sup>x<\/sup>sinx] = 2e<sup>x<\/sup>[cosx &#8211; sinx]<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">Cost of new machine = \u20b9 300000<br \/>\nScrap value of old machine = \u20b9 30000<br \/>\nHence, the money required for new machine after 7 years<br \/>\n= \u20b9 300000 &#8211; \u20b9 30000 = \u20b9 270000<\/li>\n<li style=\"text-align: left;\">A = \u20b9 270000, i =\u00a0<span class=\"math-tex\">{tex}\\frac {5}{100}{\/tex}<\/span> = 0.05, n =\u00a07<\/li>\n<li style=\"text-align: left;\">A =\u00a0<span class=\"math-tex\">{tex}R\\left[\\frac{(1+i)^n-1}{i}\\right]{\/tex}<\/span><br \/>\n<strong>OR<\/strong><br \/>\nCost of new machine = \u20b9300000<br \/>\nScrap value of old machine = \u20b930000<br \/>\nHence, the money required for new machine after 7 years<br \/>\n= \u20b9300000 &#8211; \u20b930000 = \u20b9270000<br \/>\nSo, we have A = \u20b9270000, i =\u00a0<span class=\"math-tex\">{tex}\\frac {5}{100}{\/tex}<\/span> = 0.05, n =\u00a07<br \/>\nUsing formula,\u00a0A =\u00a0<span class=\"math-tex\">{tex}R\\left[\\frac{(1+i)^n-1}{i}\\right]{\/tex}<\/span>, we get<br \/>\n270000 =\u00a0<span class=\"math-tex\">{tex}\\mathrm{R}\\left[\\frac{(1.05)^7-1}{0.05}\\right]{\/tex}<\/span><br \/>\n[Let x = (1.05)<sup>7<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0log x = 7 log 1.05 = 7 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a00.0212 = 0.1484<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x = antilog 0.1484<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x = 1.407<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow R=\\frac{270000 \\times 0.05}{(1.05)^7-1}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow R=\\frac{13500}{1.407-1}=\\frac{13500}{0.407}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0R = 33169.53<br \/>\nHence, the company should deposit \u20b933169.53 at the end of each year for 7 years.<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">The linear equations representing the given situation are<br \/>\n3x + 2y = 2300<br \/>\n5x + 3y = 3700.<br \/>\nThe matrix equation representing these equations is<br \/>\n<span class=\"math-tex\">{tex}\\left[\\begin{array}{ll} 3 &amp; 2 \\\\ 5 &amp; 3 \\end{array}\\right]{\/tex}<\/span><span class=\"math-tex\">{tex}\\left[\\begin{array}{l} x \\\\ y \\end{array}\\right]{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\left[\\begin{array}{l} 2300 \\\\ 3700 \\end{array}\\right]{\/tex}<\/span><\/li>\n<li style=\"text-align: left;\">The determinant of the coefficients of x and y is <span class=\"math-tex\">{tex}\\left|\\begin{array}{ll}3 &amp; 2 \\\\ 5 &amp; 3\\end{array}\\right|{\/tex}<\/span>.<br \/>\nIts value = 3 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 3 &#8211; 5 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 2 = -1.<\/li>\n<li style=\"text-align: left;\">D<sub>1<\/sub> = <span class=\"math-tex\">{tex}\\left|\\begin{array}{ll} 2300 &amp; 2 \\\\ 3700 &amp; 3 \\end{array}\\right|{\/tex}<\/span> = 2300 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 3 &#8211; 3700 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 2 = -500,<br \/>\nD<sub>2<\/sub> = <span class=\"math-tex\">{tex}\\left|\\begin{array}{ll} 3 &amp; 2300 \\\\ 5 &amp; 3700 \\end{array}\\right|{\/tex}<\/span> = 3 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 3700 &#8211; 5 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 2300 = -400.<br \/>\nx = <span class=\"math-tex\">{tex}\\frac{\\mathrm{D}_1}{\\mathrm{D}}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac{-500}{-1}{\/tex}<\/span> = 500, y = <span class=\"math-tex\">{tex}\\frac{\\mathrm{D}_2}{\\mathrm{D}}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac{-400}{-1}{\/tex}<\/span> = 400.<br \/>\n<strong>OR<\/strong><br \/>\n|A| = <span class=\"math-tex\">{tex}\\left|\\begin{array}{ll} 3 &amp; 2 \\\\ 5 &amp; 3 \\end{array}\\right|{\/tex}<\/span> = 3 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 3 &#8211; 2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 5 = -1.<br \/>\nA<sup>-1<\/sup> = <span class=\"math-tex\">{tex}\\frac{1}{|\\mathrm{~A}|}{\/tex}<\/span> adj A = <span class=\"math-tex\">{tex}\\frac{1}{-1}{\/tex}<\/span><span class=\"math-tex\">{tex}\\left[\\begin{array}{rr} 3 &amp; -2 \\\\ -5 &amp; 3 \\end{array}\\right]{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\left[\\begin{array}{rr} -3 &amp; 2 \\\\ 5 &amp; -3 \\end{array}\\right]{\/tex}<\/span>.<\/p>\n<p style=\"text-align: left;\"><strong>Download the myCBSEguide App for Comprehensive Exam Preparation <\/strong>Prepare effectively for your exams by downloading the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide app<\/strong><\/a>, offering complete <strong>study material for CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. With access to <strong>practice questions<\/strong>, <strong>sample papers<\/strong>, and <strong>detailed solutions<\/strong>, the app is designed to help you improve your skills and boost your exam performance. <strong>Class 12 Applied Maths 2025<\/strong> students can benefit from our interactive video lessons that break down complex problems into easy-to-understand steps. To perform well in the <strong>Class 12 Applied Maths 2025<\/strong> exam, it\u2019s essential to practice regularly and solve as many problems as possible.<\/p>\n<p style=\"text-align: left;\"><strong>Examin8 App: Create Customized Exam Papers for Teachers<\/strong><br \/>\nTeachers can streamline their exam preparation process with the <a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\"><strong>Examin8 app<\/strong><\/a>. This tool allows educators to easily create <strong>customized exam papers<\/strong> with their own name, logo, and branding, offering a professional and efficient way to generate assessments tailored to their students&#8217; needs. Stay ahead in your studies with our regularly updated content for <strong>Class 12 Applied Maths 2025<\/strong>, keeping you aligned with the latest exam trends and syllabus changes.<\/p>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h2><span class=\"ez-toc-section\" id=\"Class_12_Sample_Papers_2025\"><\/span>Class 12 Sample Papers 2025<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>You can download sample papers for other subjects as well. These model papers are available for download on <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide app<\/strong><\/a>. Please browse through the resources provided below to access CBSE class 12th Model question papers for all the subjects. We offer downloadable practice papers designed specifically for <strong>Class 12 Applied Maths 2025<\/strong>, giving you the perfect opportunity to test your knowledge.<\/p>\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","protected":false},"excerpt":{"rendered":"<p>Class 12 Applied Maths Sample Papers 2025 Applied Maths is a newly introduced subject for commerce students in Class 12, offering an ideal choice for those who don\u2019t plan on pursuing higher studies in Mathematics. CBSE Class 12 Applied Maths Sample Papers 2025 are now available for free download on the myCBSEguide app and website &#8230; <a title=\"Class 12 Applied Maths Sample Paper 2025\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/class-12-applied-maths-sample-paper\/\" aria-label=\"More on Class 12 Applied Maths Sample Paper 2025\">Read 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