{"id":68111,"date":"2021-08-31T15:49:00","date_gmt":"2021-08-31T10:19:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=68111"},"modified":"2021-08-31T17:35:14","modified_gmt":"2021-08-31T12:05:14","slug":"rd-sharma-solutions-class-11-maths-chapter-31-exercise-31-6","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-31-exercise-31-6\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 31 Exercise 31.6 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-121616\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-31-Exercise-31.6.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-31-Exercise-31.6.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-31-Exercise-31.6-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6:\u00a0<\/strong>We&#8217;ll look at the validity of statements in this exercise. Students who want to do well in their board exams should look over RD Sharma Class 11 Solutions. The <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\"><strong>RD Sharma Class 11 Solutions<\/strong><\/a> covers the methods in detail, without skipping any important aspects of the problem-solving process. The pdf of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-31-mathematical-reasoning\/\"><strong>RD Sharma Solutions Class 11 Maths Chapter 31<\/strong><\/a> Exercise 31.6 can be downloaded by students using the links provided below.<\/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-69d3fa9b1030f\" 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 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-31-exercise-31-6\/#download-rd-sharma-class-11-solutions-chapter-31-exercise-316-free-pdf\" title=\"Download RD Sharma Class 11 Solutions Chapter 31 Exercise 31.6 Free PDF\">Download RD Sharma Class 11 Solutions Chapter 31 Exercise 31.6 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-31-exercise-31-6\/#access-answers-to-rd-sharma-solutions-class-11-maths-chapter-31-exercise-316-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6- Important Question with Answers\">Access answers to RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6- Important Question with Answers<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-31-exercise-31-6\/#where-can-i-download-rd-sharma-solutions-class-11-maths-chapter-31-exercise-316-free-pdf\" title=\"Where can I download RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 free PDF?\">Where can I download RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 free PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-31-exercise-31-6\/#what-are-the-benefits-of-using-kopykitab%e2%80%99s-rd-sharma-solutions-class-11-maths-chapter-31-exercise-316\" title=\"What are the benefits of using Kopykitab\u2019s RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6?\">What are the benefits of using Kopykitab\u2019s RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-11-solutions-chapter-31-exercise-316-free-pdf\"><\/span>Download RD Sharma Class 11 Solutions Chapter 31 Exercise 31.6 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-31-Ex-31.6-1.pdf\">RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6<\/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\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-31-Ex-31.6-1.pdf\", \"#example1\");<\/script><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-11-maths-chapter-31-exercise-316-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Check the validity of the following statements:<br \/>(i) p: 100 is a multiple of 4 and 5.<br \/>(ii) q: 125 is a multiple of 5 and 7.<br \/>(iii) r: 60 is a multiple of 3 or 5.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a0p: 100 is a multiple of 4 and 5.<\/p>\n<p>We know that 100 is a multiple of 4 as well as 5. So, the given statement is true.<\/p>\n<p>Hence, the statement is true.<\/p>\n<p><br \/><strong>(ii)<\/strong>\u00a0q: 125 is a multiple of 5 and 7<\/p>\n<p>We know that\u00a0125 is a multiple of 5 and not a multiple of 7. So, the given statement is false.<\/p>\n<p>Hence, the statement is false.<\/p>\n<p><br \/><strong>(iii)<\/strong>\u00a0r: 60 is a multiple of 3 or 5.<\/p>\n<p>We know that 60 is a multiple of 3 as well as 5. So, the given statement is true.<\/p>\n<p>Hence, the statement is true.<\/p>\n<p><strong>2. Check whether the following statement is true or not:<br \/>(i) p: If x and y are odd integers, then x + y is an even integer.<\/strong><\/p>\n<p><strong>(ii) q : if x, y are integer such that xy is even, then at least one of x and y is an even integer.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>p: If x and y are odd integers, then x + y is an even integer.<\/p>\n<p>Let us assume that \u2018p\u2019 and \u2018q\u2019 be the statements given by<\/p>\n<p>p: x and y are odd integers.<\/p>\n<p>q: x + y is an even integer<\/p>\n<p>the given statement can be written as :<\/p>\n<p>if p, then q.<\/p>\n<p>Let p be true. Then, x and y are odd integers<\/p>\n<p>x = 2m+1, y = 2n+1 for some integers m, n<\/p>\n<p>x + y = (2m+1) + (2n+1)<\/p>\n<p>x + y = (2m+2n+2)<\/p>\n<p>x + y = 2(m+n+1)<\/p>\n<p>x + y is an integer<\/p>\n<p>q is true.<\/p>\n<p>So, p is true\u00a0and\u00a0q is true.<\/p>\n<p>Hence, \u201cif p, then q \u201cis a true statement.\u201d<\/p>\n<p><strong>(ii)\u00a0<\/strong>q: if x, y are integer such that xy is even, then at least one of x and y is an even integer.<\/p>\n<p>Let us assume that p and q be the statements given by<\/p>\n<p>p: x and y are integers and xy is an even integer.<\/p>\n<p>q: At least one of x and y is even.<\/p>\n<p>Let p be true, and then xy is an even integer.<\/p>\n<p>So,<\/p>\n<p>xy = 2(n + 1)<\/p>\n<p>Now,<\/p>\n<p>Let x = 2(k + 1)<\/p>\n<p>Since, x is an even integer, xy = 2(k + 1). y is also an even integer.<\/p>\n<p><br \/>Now take x = 2(k + 1) and y = 2(m + 1)<\/p>\n<p>xy = 2(k + 1).2(m + 1) = 2.2(k + 1)(m + 1)<\/p>\n<p>So, it is also true.<\/p>\n<p>Hence, the statement is true.<\/p>\n<p><strong>3. Show that the statement<br \/>p : \u201cIf x is a real number such that x<sup>3<\/sup>\u00a0+ x = 0, then x is 0\u201d is true by<br \/>(i) Direct method<br \/>(ii) method of Contrapositive<br \/>(iii) method of contradiction<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a0Direct Method:<\/p>\n<p>Let us assume that \u2018q\u2019 and \u2018r\u2019 be the statements given by<\/p>\n<p>q: x is a real number such that x<sup>3\u00a0<\/sup>+ x=0.<\/p>\n<p>r: x is 0.<\/p>\n<p>The given statement can be written as:<\/p>\n<p>if q, then r.<\/p>\n<p>Let q be true. Then, x is a real number such that x<sup>3\u00a0<\/sup>+ x = 0<\/p>\n<p>x is a real number such that x(x<sup>2\u00a0<\/sup>+ 1) = 0<\/p>\n<p>x = 0<\/p>\n<p>r is true<\/p>\n<p>Thus, q is true<\/p>\n<p>Therefore, q is true\u00a0and\u00a0r is true.<\/p>\n<p>Hence, p is true.<\/p>\n<p><strong>(ii)<\/strong>\u00a0Method of Contrapositive:<\/p>\n<p>Let r be false. Then,<\/p>\n<p>R is not true<\/p>\n<p>x \u2260 0, x\u2208R<\/p>\n<p>x(x<sup>2<\/sup>+1)\u22600, x\u2208R<\/p>\n<p>q is not true<\/p>\n<p>Thus, -r = -q<\/p>\n<p>Hence, p : q\u00a0and\u00a0r is true<\/p>\n<p><br \/><strong>(iii)<\/strong>\u00a0Method of Contradiction:<\/p>\n<p>If possible, let p be false. Then,<\/p>\n<p>P is not true<\/p>\n<p>-p is true<\/p>\n<p>-p (p =&gt; r) is true<\/p>\n<p>q and \u2013r is true<\/p>\n<p>x is a real number such that x<sup>3<\/sup>+x = 0and x\u2260 0<\/p>\n<p>x =0 and x\u22600<\/p>\n<p>This is a contradiction.<\/p>\n<p>Hence, p is true.<\/p>\n<p><strong>4. Show that the following statement is true by the method of the contrapositive<br \/>p: \u201cIf x is an integer and x<sup>2<\/sup>\u00a0is odd, then x is also odd.\u201d<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us assume that \u2018q\u2019 and \u2018r\u2019 be the statements given<\/p>\n<p>q: x is an integer and x<sup>2<\/sup>\u00a0is odd.<\/p>\n<p>r: x is an odd integer.<\/p>\n<p>The given statement can be written as:<\/p>\n<p>p: if q, then r.<\/p>\n<p>Let r be false. Then,<\/p>\n<p>x is not an odd integer, then x is an even integer<\/p>\n<p>x = (2n) for some integer n<\/p>\n<p>x<sup>2<\/sup>\u00a0= 4n<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0is an even integer<\/p>\n<p>Thus, q is False<\/p>\n<p>Therefore, r is false\u00a0and\u00a0q is false<\/p>\n<p>Hence, p: \u201c if q, then r\u201d is a true statement.<\/p>\n<p><strong>5.<\/strong>\u00a0<strong>Show that the following statement is true<br \/>\u201cThe integer n is even if and only if n<sup>2<\/sup>\u00a0is even\u201d<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the statements,<\/p>\n<p>p: Integer n is even<\/p>\n<p>q: If n<sup>2<\/sup>\u00a0is even<\/p>\n<p>Let p be true. Then,<\/p>\n<p>Let n = 2k<\/p>\n<p>Squaring both the sides, we get,<\/p>\n<p>n<sup>2<\/sup>\u00a0= 4k<sup>2<\/sup><\/p>\n<p>n<sup>2<\/sup>\u00a0= 2.2k<sup>2<\/sup><\/p>\n<p>n<sup>2<\/sup>\u00a0is an even number.<\/p>\n<p>So, q is true when p is true.<\/p>\n<p>Hence, the given statement is true.<\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 Q6.<\/strong><\/p>\n<p><strong>6. By giving a counter example, show that the following statement is not true.<br \/>p: \u201cIf all the angles of a triangle are equal, then the triangle is an obtuse-angled triangle.\u201d<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider a triangle ABC with all angles equal.<\/p>\n<p>Then, each angle of the triangle is equal to 60.<\/p>\n<p>So, ABC is not an obtuse angle triangle.<\/p>\n<p>Hence, the statement \u201cp: If all the angles of a triangle are equal, then the triangle is an obtuse-angled triangle\u201d is False.<\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 Q7.<\/strong><\/p>\n<p><strong>7. Which of the following statements are true and which are false? In each case give a valid reason for saying so<br \/>(i) p: Each radius of a circle is a chord of the circle.<br \/>(ii) q: The centre of a circle bisects each chord of the circle.<br \/>(iii) r: Circle is a particular case of an ellipse.<br \/>(iv) s: If x and y are integers such that x &gt; y, then \u2013 x &lt; \u2013 y.<br \/>(v) t: \u221a11\u00a0is a rational number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>p: Each radius of a circle is a chord of the circle.<\/p>\n<p>The Radius of the circle is not its chord.<\/p>\n<p>Hence, this statement is False.<\/p>\n<p><strong>(ii)<\/strong> q: The centre of a circle bisects each chord of the circle.<\/p>\n<p>A chord does not have to pass through the center.<\/p>\n<p>Hence, this statement is False.<\/p>\n<p><strong>(iii)<\/strong>\u00a0r: Circle is a particular case of an ellipse.<\/p>\n<p>A circle can be an ellipse in a particular case when the circle has equal axes.<\/p>\n<p>Hence, this statement is true.<\/p>\n<p><br \/><strong>(iv)<\/strong>\u00a0s: If x and y are integers such that x &gt; y, then \u2013 x &lt; \u2013 y.<\/p>\n<p>For any two integers, if x \u2013 y id positive then \u2013(x-y) is negative.<\/p>\n<p>Hence, this statement is true.<\/p>\n<p><br \/><strong>(v)<\/strong>\u00a0t: \u221a11\u00a0is a rational number.<\/p>\n<p>The square root of prime numbers is irrational numbers.<\/p>\n<p>Hence, this statement is False.<\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 Q8.<\/strong><\/p>\n<p><strong>8. Determine whether the argument used to check the validity of the following statement is correct:<br \/>p: \u201cIf x<sup>2<\/sup>\u00a0is irrational, then x is rational.\u201d<br \/>The statement is true because the number x<sup>2<\/sup>\u00a0= \u03c0<sup>2<\/sup>\u00a0is irrational, therefore x = \u03c0 is irrational.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Argument Used: x<sup>2<\/sup>\u00a0= \u03c0<sup>2<\/sup>\u00a0is irrational, therefore x = \u03c0 is irrational.<\/p>\n<p>p: \u201cIf x<sup>2<\/sup>\u00a0is irrational, then x is rational.\u201d<\/p>\n<p><br \/>Let us take an irrational number given by x = \u221ak, where k is a rational number.<\/p>\n<p>Squaring both sides, we get,<\/p>\n<p>x<sup>2<\/sup>\u00a0= k<\/p>\n<p>x<sup>2<\/sup>\u00a0is a rational number and contradicts our statement.<\/p>\n<p>Hence, the given argument is wrong.<\/p>\n<p>We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 11, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-11-maths-chapter-31-exercise-316\"><\/span>FAQs on RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6<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-1630397497996\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-class-11-maths-chapter-31-exercise-316\"><\/span>How many questions are there in RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 8 questions in RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630397604631\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-solutions-class-11-maths-chapter-31-exercise-316-free-pdf\"><\/span>Where can I download RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630397605856\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-using-kopykitab%e2%80%99s-rd-sharma-solutions-class-11-maths-chapter-31-exercise-316\"><\/span>What are the benefits of using Kopykitab\u2019s RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>The following are some of the benefits of using Kopykitab\u2019s RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6:<br \/>1. Correct answers according to the last CBSE guidelines and syllabus.<br \/>2. The RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 is written in simple language to assist students in their board examination, &amp; competitive examination preparation.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6:\u00a0We&#8217;ll look at the validity of statements in this exercise. Students who want to do well in their board exams should look over RD Sharma Class 11 Solutions. The RD Sharma Class 11 Solutions covers the methods in detail, without skipping any important aspects of the &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 31 Exercise 31.6 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-31-exercise-31-6\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 31 Exercise 31.6 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":121616,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334,73564,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68111"}],"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\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=68111"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68111\/revisions"}],"predecessor-version":[{"id":121707,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68111\/revisions\/121707"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/121616"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=68111"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=68111"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=68111"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}