{"id":67900,"date":"2023-09-09T16:01:00","date_gmt":"2023-09-09T10:31:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67900"},"modified":"2023-11-10T10:30:51","modified_gmt":"2023-11-10T05:00:51","slug":"rd-sharma-solutions-class-11-maths-chapter-1-exercise-1-7","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-1-exercise-1-7\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 1 Exercise 1.7 (Updated for 2024)"},"content":{"rendered":"\n<p><img src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-11-Solutions-Chapter-1-Exercise-1.7.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 1 Exercise 1.7\" width=\"1200\" height=\"675\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 1 Exercise 1.7:<\/strong> If you want to clear the concepts of Class 11 Maths then <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths<\/a> is the solution for you. Designed by subject matter experts, the solutions are reliable and easy to understand. Download the free PDF of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-1-sets\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 11 Solutions Chapter 1<\/a> Exercise 1.7 today.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 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\">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=\"ez-toc-toggle-icon-1\"><label for=\"item-69da200859ef9\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" 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 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-1-exercise-1-7\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-1-exercise-17\" title=\"How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 1 Exercise 1.7?\">How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 1 Exercise 1.7?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-for-class-11-maths-chapter-1-exercise-17-pdf\"><\/span>Download RD Sharma Solutions for Class 11 Maths Chapter 1 Exercise 1.7 PDF:<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/1.7.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 11 Solutions Chapter 1 Exercise 1.7<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/1.7.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-for-class-11-maths-chapter-1-exercise-17\"><\/span>Access RD Sharma Solutions for Class 11 Maths Chapter 1 Exercise 1.7<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-for-class-11-maths-exercise-17-chapter-1-%e2%80%93-sets\"><\/span>Access answers to RD Sharma Solutions for Class 11 Maths Exercise 1.7 Chapter 1 \u2013 Sets<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. For any two sets A and B, prove that: A<\/strong>\u2018<strong>\u00a0\u2013 B<\/strong>\u2018<strong>\u00a0= B \u2013 A<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>To prove, A\u2019 \u2013 B\u2019 = B \u2013 A<\/p>\n<p>Firstly we need to show<\/p>\n<p>A\u2019 \u2013 B\u2019\u00a0\u2286\u00a0B \u2013 A<\/p>\n<p>Let x \u2208 A\u2019 \u2013 B\u2019<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0A\u2019 and x\u00a0\u2209\u00a0B\u2019<\/p>\n<p>\u21d2\u00a0x\u00a0\u2209\u00a0A and x\u00a0\u2208\u00a0B (since, A\u00a0\u2229\u00a0A\u2019 = \u03d5 )<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0B \u2013 A<\/p>\n<p>It is true for all x\u00a0\u2208\u00a0A\u2019 \u2013 B\u2019<\/p>\n<p>\u2234\u00a0A\u2019 \u2013 B\u2019 = B \u2013 A<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>2. For any two sets A and B, prove the following:<br \/>(i) A\u00a0\u2229\u00a0(A<\/strong>\u2018<strong>\u00a0\u222a\u00a0B) = A\u00a0\u2229\u00a0B<\/strong><\/p>\n<p><strong>(ii) A \u2013 (A \u2013 B) = A\u00a0\u2229\u00a0B<\/strong><\/p>\n<p><strong>(iii) A\u00a0\u2229\u00a0(A\u00a0\u222a\u00a0B\u2019) =\u00a0\u03d5<\/strong><\/p>\n<p><strong>(iv) A \u2013 B = A \u0394 (A\u00a0\u2229\u00a0B)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>A\u00a0\u2229\u00a0(A\u2019\u00a0\u222a\u00a0B) = A\u00a0\u2229\u00a0B<\/p>\n<p>Let us consider LHS A\u00a0\u2229\u00a0(A\u2019\u00a0\u222a\u00a0B)<\/p>\n<p>Expanding<\/p>\n<p>(A\u00a0\u2229\u00a0A\u2019)\u00a0\u222a\u00a0(A \u2229\u00a0B)<\/p>\n<p>We know (A \u2229 A\u2019) =\u03d5<\/p>\n<p>\u21d2\u00a0\u03d5\u00a0\u222a\u00a0(A\u2229\u00a0B)<\/p>\n<p>\u21d2\u00a0(A \u2229\u00a0B)<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(ii)\u00a0<\/strong>A \u2013 (A \u2013 B) = A\u00a0\u2229\u00a0B<\/p>\n<p>For any sets A and B, we have De-Morgan\u2019s law<\/p>\n<p>(A\u00a0\u222a B)\u2019 = A\u2019 \u2229 B\u2019, (A\u00a0\u2229\u00a0B) \u2018 = A\u2019\u00a0\u222a\u00a0B\u2019<\/p>\n<p>Consider LHS<\/p>\n<p>= A \u2013 (A\u2013B)<\/p>\n<p>= A\u00a0\u2229\u00a0(A\u2013B)\u2019<\/p>\n<p>= A \u2229 (A\u2229B\u2019)\u2019<\/p>\n<p>= A \u2229 (A\u2019 \u222a B\u2019)\u2019) (since (B\u2019)\u2019 = B)<\/p>\n<p>= A\u00a0\u2229\u00a0(A\u2019 \u222a B)<\/p>\n<p>= (A\u00a0\u2229\u00a0A\u2019)\u00a0\u222a\u00a0(A\u00a0\u2229\u00a0B)<\/p>\n<p>= \u03d5 \u222a (A \u2229 B) (since A \u2229 A\u2019 = \u03d5)<\/p>\n<p>= (A\u00a0\u2229\u00a0B) (since, \u03d5\u00a0\u222a\u00a0x = x, for any set)<\/p>\n<p>= RHS<\/p>\n<p>\u2234\u00a0LHS=RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(iii)\u00a0<\/strong>A\u00a0\u2229\u00a0(A\u00a0\u222a\u00a0B\u2019) =\u00a0\u03d5<\/p>\n<p>Let us consider LHS A\u00a0\u2229\u00a0(A\u00a0\u222a\u00a0B\u2019)<\/p>\n<p>= A \u2229\u00a0(A\u00a0\u222a\u00a0B\u2019)<\/p>\n<p>= A \u2229 (A\u2019\u2229\u00a0B\u2019) (By De\u2013Morgan\u2019s law)<\/p>\n<p>= (A \u2229 A\u2019) \u2229 B\u2019 (since A \u2229 A\u2019 = \u03d5)<\/p>\n<p>=\u00a0\u03d5\u00a0\u2229\u00a0B\u2019<\/p>\n<p>=\u00a0\u03d5 (since, \u03d5\u00a0\u2229\u00a0B\u2019 = \u03d5)<\/p>\n<p>= RHS<\/p>\n<p>\u2234\u00a0LHS=RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(iv)\u00a0<\/strong>A \u2013 B = A \u0394 (A\u00a0\u2229\u00a0B)<\/p>\n<p>Let us consider RHS A \u0394 (A\u00a0\u2229\u00a0B)<\/p>\n<p>A \u0394 (A \u2229 B) (since E \u0394 F = (E\u2013F) \u222a (F\u2013E))<\/p>\n<p>= (A \u2013 (A \u2229 B)) \u222a (A \u2229 B \u2013A) (since E \u2013 F = E \u2229 F\u2019)<\/p>\n<p>= (A\u00a0\u2229\u00a0(A\u00a0\u2229\u00a0B)\u2019)\u00a0\u222a\u00a0(A \u2229 B \u2229 A\u2019)<\/p>\n<p>= (A\u00a0\u2229\u00a0(A\u2019 \u222a B\u2019))\u00a0\u222a\u00a0(A \u2229 A\u2019 \u2229 B) (by using De-Morgan\u2019s law and associative law)<\/p>\n<p>= (A\u00a0\u2229\u00a0A\u2019) \u222a (A \u2229 B\u2019) \u222a (\u03d5 \u2229 B) (by using distributive law)<\/p>\n<p>=\u00a0\u03d5\u00a0\u222a\u00a0(A\u00a0\u2229\u00a0B\u2019)\u00a0\u222a\u00a0\u03d5<\/p>\n<p>= A\u00a0\u2229\u00a0B\u2019 (since, A\u00a0\u2229\u00a0B\u2019 = A\u2013B)<\/p>\n<p>= A \u2013 B<\/p>\n<p>= LHS<\/p>\n<p>\u2234\u00a0LHS=RHS<\/p>\n<p>Hence Proved<\/p>\n<p><strong>3. If A, B, C are three sets such that A\u00a0\u2282\u00a0B, then prove that C \u2013 B\u00a0\u2282\u00a0C \u2013 A.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, ACB<\/p>\n<p>To prove: C \u2013 B\u00a0\u2282\u00a0C \u2013 A<\/p>\n<p>Let us consider, x\u00a0\u2208\u00a0C\u2013B<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0C and x \u2209\u00a0B<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0C and x \u2209\u00a0A<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0C \u2013 A<\/p>\n<p>Thus, x\u00a0\u2208\u00a0C\u2013B\u00a0\u21d2\u00a0x\u00a0\u2208\u00a0C \u2013 A<\/p>\n<p>This is true for all x\u00a0\u2208\u00a0C\u2013B<\/p>\n<p>\u2234\u00a0C \u2013 B\u00a0\u2282\u00a0C \u2013 A<\/p>\n<p>Hence proved.<\/p>\n<p><strong>4.<\/strong>\u00a0<strong>For any two sets A and B, prove that<br \/>(i) (A\u00a0\u222a\u00a0B) \u2013 B = A \u2013 B<\/strong><\/p>\n<p><strong>(ii) A \u2013 (A\u00a0\u2229\u00a0B) = A \u2013 B<\/strong><\/p>\n<p><strong>(iii) A \u2013 (A \u2013 B) = A\u00a0\u2229\u00a0B<\/strong><\/p>\n<p><strong>(iv) A\u00a0\u222a\u00a0(B \u2013 A) = A\u00a0\u222a\u00a0B<\/strong><\/p>\n<p><strong>(v) (A \u2013 B)\u00a0\u222a\u00a0(A\u00a0\u2229\u00a0B) = A<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>(A\u00a0\u222a\u00a0B) \u2013 B = A \u2013 B<\/p>\n<p>Let us consider LHS (A\u00a0\u222a\u00a0B) \u2013 B<\/p>\n<p>= (A\u2013B)\u00a0\u222a\u00a0(B\u2013B)<\/p>\n<p>= (A\u2013B) \u222a \u03d5 (since B\u2013B = \u03d5)<\/p>\n<p>= A\u2013B (since, x\u00a0\u222a\u00a0\u03d5 = x for any set)<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(ii)\u00a0<\/strong>A \u2013 (A\u00a0\u2229\u00a0B) = A \u2013 B<\/p>\n<p>Let us consider LHS A \u2013 (A\u00a0\u2229\u00a0B)<\/p>\n<p>= (A\u2013A)\u00a0\u2229 (A\u2013B)<\/p>\n<p>= \u03d5 \u2229 (A \u2013 B) (since A-A = \u03d5)<\/p>\n<p>= A \u2013 B<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(iii)\u00a0<\/strong>A \u2013 (A \u2013 B) = A\u00a0\u2229\u00a0B<\/p>\n<p>Let us consider LHS A \u2013 (A \u2013 B)<\/p>\n<p>Let, x\u00a0\u2208\u00a0A \u2013 (A\u2013B)\u00a0=\u00a0x\u00a0\u2208\u00a0A and x\u00a0\u2209\u00a0(A\u2013B)<\/p>\n<p>x\u00a0\u2208\u00a0A and x\u00a0\u2209\u00a0(A\u00a0\u2229\u00a0B)<\/p>\n<p>=\u00a0x\u00a0\u2208\u00a0A \u2229 (A\u00a0\u2229\u00a0B)<\/p>\n<p>= x\u00a0\u2208\u00a0(A\u00a0\u2229\u00a0B)<\/p>\n<p>=\u00a0(A\u00a0\u2229\u00a0B)<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(iv)\u00a0<\/strong>A\u00a0\u222a\u00a0(B \u2013 A) = A\u00a0\u222a\u00a0B<\/p>\n<p>Let us consider LHS A\u00a0\u222a\u00a0(B \u2013 A)<\/p>\n<p>Let, x\u00a0\u2208\u00a0A\u00a0\u222a\u00a0(B \u2013A)\u00a0\u21d2\u00a0x\u00a0\u2208\u00a0A or x\u00a0\u2208\u00a0(B \u2013 A)<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0A or x\u00a0\u2208\u00a0B and x\u00a0\u2209\u00a0A<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0B<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0(A\u00a0\u222a\u00a0B) (since,\u00a0B\u00a0\u2282\u00a0(A\u00a0\u222a\u00a0B))<\/p>\n<p>This is true for all x\u00a0\u2208\u00a0A\u00a0\u222a\u00a0(B\u2013A)<\/p>\n<p>\u2234\u00a0A \u222a (B\u2013A) \u2282 (A \u222a B)\u2026\u2026 (1)<\/p>\n<p>Conversely,<\/p>\n<p>Let x\u00a0\u2208\u00a0(A\u00a0\u222a\u00a0B)\u00a0\u21d2\u00a0x\u00a0\u2208\u00a0A or x\u00a0\u2208\u00a0B<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0A or x\u00a0\u2208\u00a0(B\u2013A) (since,\u00a0B\u00a0\u2282\u00a0(A\u00a0\u222a\u00a0B))<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0A\u00a0\u222a\u00a0(B\u2013A)<\/p>\n<p>\u2234\u00a0(A \u222a B) \u2282\u00a0A \u222a (B\u2013A)\u2026\u2026 (2)<\/p>\n<p>From 1 and 2, we get,<\/p>\n<p>A\u00a0\u222a\u00a0(B \u2013 A) = A\u00a0\u222a\u00a0B<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(v)\u00a0<\/strong>(A \u2013 B)\u00a0\u222a\u00a0(A\u00a0\u2229\u00a0B) = A<\/p>\n<p>Let us consider LHS (A \u2013 B)\u00a0\u222a\u00a0(A\u00a0\u2229\u00a0B)<\/p>\n<p>Let, x\u00a0\u2208\u00a0A<\/p>\n<p>Then either x\u00a0\u2208\u00a0(A\u2013B) or x\u00a0\u2208\u00a0(A\u00a0\u2229\u00a0B)<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0(A\u2013B)\u00a0\u222a\u00a0(A\u00a0\u2229\u00a0B)<\/p>\n<p>\u2234\u00a0A\u00a0\u2282\u00a0(A \u2013 B)\u00a0\u222a\u00a0(A\u00a0\u2229\u00a0B)\u2026. (1)<\/p>\n<p>Conversely,<\/p>\n<p>Let x\u00a0\u2208\u00a0(A\u2013B)\u00a0\u222a\u00a0(A\u00a0\u2229\u00a0B)<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0(A\u2013B) or x\u00a0\u2208\u00a0(A\u00a0\u2229\u00a0B)<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0A and x\u00a0\u2209\u00a0B or x\u00a0\u2208\u00a0B<\/p>\n<p>\u21d2\u00a0x\u00a0\u2208\u00a0A<\/p>\n<p>(A\u2013B)\u00a0\u222a\u00a0(A\u00a0\u2229\u00a0B)\u00a0\u2282\u00a0A\u2026\u2026\u2026. (2)<\/p>\n<p>\u2234\u00a0From (1) and (2), We get<\/p>\n<p>(A\u2013B)\u00a0\u222a\u00a0(A\u00a0\u2229\u00a0B) = A<\/p>\n<p>Hence proved.<\/p>\n<p>You will be able to come up with an effective strategy after you examine the <a href=\"https:\/\/rdsharmasolution.com\/\" target=\"_blank\" rel=\"noopener\">RD Sharma<\/a> solutions in Chapter 1 on a regular basis. You will be able to appear for <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 11 Maths Exam with a better plan if you keep the important points from Chapter 1 solutions in your mind.\u00a0<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-11-maths-chapter-1-exercise-17\"><\/span><strong>FAQs on <\/strong>RD Sharma Solutions Class 11 Maths Chapter 1 Exercise 1.7<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630341541582\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-even-beneficial-to-study-rd-sharma-solutions-class-11-maths-chapter-1-exercise-17\"><\/span><strong>Is it even beneficial to study RD Sharma Solutions Class 11 Maths Chapter 1 Exercise 1.7?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, your preparation will be strengthened with this amazing help book. All your questions will be answered by this book.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630341555486\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-class-11-maths-chapter-1-exercise-17-pdf-offline\"><\/span><strong>Can I access the RD Sharma Solutions Class 11 Maths Chapter 1 Exercise 1.7 PDF offline?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630341569532\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-1-exercise-17\"><\/span><strong>How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 1 Exercise 1.7?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 1 Exercise 1.7: If you want to clear the concepts of Class 11 Maths then RD Sharma Solutions Class 11 Maths is the solution for you. Designed by subject matter experts, the solutions are reliable and easy to understand. Download the free PDF of RD Sharma Class 11 &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 1 Exercise 1.7 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-1-exercise-1-7\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 1 Exercise 1.7 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":249,"featured_media":123589,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67900"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/249"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=67900"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67900\/revisions"}],"predecessor-version":[{"id":505686,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67900\/revisions\/505686"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/123589"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67900"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67900"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67900"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}