{"id":92657,"date":"2021-10-07T13:19:00","date_gmt":"2021-10-07T07:49:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=92657"},"modified":"2021-10-14T16:43:39","modified_gmt":"2021-10-14T11:13:39","slug":"ncert-solutions-class-12-maths-chapter-6-exercise-6-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-class-12-maths-chapter-6-exercise-6-4\/","title":{"rendered":"NCERT Solutions Class 12 Maths Chapter 6 Application of Derivatives\u00a0Exercise 6.4 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-93019\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/05\/6.4.jpg\" alt=\"NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/05\/6.4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/05\/6.4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4: <\/strong>The NCERT Solutions <a href=\"https:\/\/www.kopykitab.com\/Class-12-Maths\">Class 12 Maths<\/a> Chapter 6 Exercise 6.4 comprises all the questions of Chapter 6 &#8211; Application of derivates exercise 6.4.\u00a0<\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-for-class-12-maths-chapter-6-application-of-derivatives\/\" target=\"_blank\" rel=\"noopener\"><strong>NCERT Solutions For Class 12 Maths Chapter 6: Application of Derivatives\u00a0<\/strong><\/a><\/li>\n<\/ul>\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-69db2225beaf5\" 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 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" 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><\/label><input  type=\"checkbox\" id=\"item-69db2225beaf5\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-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:\/\/www.kopykitab.com\/blog\/ncert-solutions-class-12-maths-chapter-6-exercise-6-4\/#ncert-solutions-class-12-maths-chapter-6-exercise-64\" title=\"NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4\">NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4<\/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\/ncert-solutions-class-12-maths-chapter-6-exercise-6-4\/#download-ncert-solutions-class-12-maths-chapter-6-exercise-64-pdf\" title=\"Download NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 PDF:\">Download NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 PDF:<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-class-12-maths-chapter-6-exercise-6-4\/#access-chapter-6-exercise-64-class-12-maths-question-answer\" title=\"Access Chapter 6 Exercise 6.4 Class 12 Maths Question Answer\">Access Chapter 6 Exercise 6.4 Class 12 Maths Question Answer<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-class-12-maths-chapter-6-exercise-6-4\/#faqs-on-the-ncert-solutions-for-class-12-maths-chapter-6-exercise-64\" title=\"FAQs on the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4\">FAQs on the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-class-12-maths-chapter-6-exercise-6-4\/#how-much-does-the-downloading-of-ncert-solutions-class-12-maths-chapter-6-exercise-64-pdf-costs\" title=\"How much does the downloading of NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 PDF costs?\">How much does the downloading of NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 PDF costs?<\/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\/ncert-solutions-class-12-maths-chapter-6-exercise-6-4\/#from-where-can-we-download-the-ncert-solutions-class-12-maths-chapter-6-exercise-64-pdf\" title=\"From where can we download the NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 PDF?\">From where can we download the NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-class-12-maths-chapter-6-exercise-6-4\/#is-there-a-step-by-step-explanation-of-exercise-64-class-12-questions\" title=\"Is there a step-by-step explanation of Exercise 6.4 Class 12 questions?\">Is there a step-by-step explanation of Exercise 6.4 Class 12 questions?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-class-12-maths-chapter-6-exercise-6-4\/#what-is-the-total-weightage-of-chapter-6-applications-of-derivatives\" title=\"What is the total weightage of Chapter 6, Applications of derivatives?\">What is the total weightage of Chapter 6, Applications of derivatives?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"ncert-solutions-class-12-maths-chapter-6-exercise-64\"><\/span>NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/05\/6.4-1.pdf\" target=\"_blank\" rel=\"noopener\">NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4 pdf<\/a><\/p>\n<p>\u00a0<\/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\/05\/6.4-1.pdf\", \"#example1\");<\/script><\/p>\n<h3><span class=\"ez-toc-section\" id=\"download-ncert-solutions-class-12-maths-chapter-6-exercise-64-pdf\"><\/span>Download NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 PDF:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: center;\"><a style=\"width: auto; padding: 18px; cursor: pointer; font-weight: bold; border-radius: 40px; color: #ffffff; background: #ff4500;\" href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/05\/6.4-1.pdf\" target=\"_blank\" rel=\"noopener\">Download Exercise 6.4 Class 12 NCERT Solution Free PDF<\/a><\/p>\n<p><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;NCERT Solutions Class 12 maths Chapter 6 Exercise 6.3&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 has various questions and apart from that, there are other practice questions with a detailed explanation. All these solutions are designed with precision with the help of subject matter experts.\u00a0<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-chapter-6-exercise-64-class-12-maths-question-answer\"><\/span>Access Chapter 6 Exercise 6.4 Class 12 Maths Question Answer<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Ex 6.4 Class 12 Maths\u00a0Question 1.<\/strong><br \/>Using differentials, find the approximate value of each of the following up to 3 places of decimal.<br \/>(i) (25.3)^(1\/2)<br \/><strong>Solution:<\/strong><br \/>(i) y + \u2206y = 25.3^(1\/2)<br \/>= 25+0.3^(1\/2)<br \/>= x+\u0394x^(1\/2)<br \/>\u2234 x = 25<br \/>\u2206x = 0.3<br \/>\u21d2 y = \u221ax<\/p>\n<p><strong>Ex 6.4 Class 12 Maths\u00a0Question 2.<\/strong><br \/>Find the approximate value of f (2.01), where f (x) = 4x\u00b2 + 5x + 2<br \/><strong>Solution:<\/strong><br \/>f(x+\u2206x) = f(2.01), f(x) = f (2) = 4.2\u00b2 + 5.2 + 2 = 28,<br \/>f\u2019 (x) = 8x + 5 Now, f(x + \u2206x) = f(x) + \u2206f(x)<br \/>= f(x) + f\u2019 (x) \u2022 \u2206x = 28 + (8x + 5) \u2206x<br \/>= 28 + (16 + 5) x 0.01<br \/>= 28 + 21 x 0.01<br \/>= 28 + 0.21<br \/>Hence,f(2 x 01)<br \/>= 28 x 21.<\/p>\n<p><strong>Ex 6.4 Class 12 Maths\u00a0Question 3.<\/strong><br \/>Find the approximate value of f (5.001), where f(x) = x<sup>3<\/sup>\u00a0\u2013 7x<sup>2<\/sup>\u00a0+15.<br \/><strong>Solution:<\/strong><br \/>Let x + \u2206x = 5.001, x = 5 and \u2206x = 0.001,<br \/>f(x) = f(5) = \u2013 35<br \/>f(x + \u2206x) = f(x) + \u2206f(x) = f(x) + f'(x).\u2206x<br \/>= (x<sup>3<\/sup>\u00a0\u2013 7x\u00b2 + 15) + (3x\u00b2 \u2013 14x) \u00d7 \u2206x<br \/>f(5.001) = \u2013 35 + (3 \u00d7 5\u00b2 \u2013 14 \u00d7 5) \u00d7 0.001<br \/>\u21d2 f (5.001) = \u2013 35 + 0.005<br \/>= \u2013 34.995.<\/p>\n<p><strong>Ex 6.4 Class 12 Maths\u00a0Question 4.<\/strong><br \/>Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.<br \/><strong>Solution:<\/strong><br \/>The side of the cube = x meters.<br \/>Increase in side = 1% = 0.01 \u00d7 x = 0.01 x<br \/>Volume of cube V= x<sup>3<\/sup><br \/>\u2234 \u2206v =dvdx\u00a0\u00d7 \u2206x<br \/>= 3x\u00b2 \u00d7 0.01 x<br \/>= 0.03 x<sup>3<\/sup>\u00a0m<sup>3<\/sup><\/p>\n<p><strong>Ex 6.4 Class 12 Maths\u00a0Question 5.<\/strong><br \/>Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.<br \/><strong>Solution:<\/strong><br \/>The side of the cube = x m;<br \/>Decrease in side = 1% = 0.01 x<br \/>Increase in side = \u2206x = \u2013 0.01 x<br \/>Surface area of cube = 6x\u00b2 m\u00b2 = S<br \/>\u2234\u00a0dsdx\u00a0\u00d7 \u2206x = 12x \u00d7 (- 0.01 x)<br \/>= \u2013 0.12 x\u00b2 m\u00b2.<\/p>\n<p><strong>Ex 6.4 Class 12 Maths\u00a0Question 6.<\/strong><br \/>If the radius of a sphere is measured as 7m with an error of 0.02 m, then find the approximate error in calculating its volume.<br \/><strong>Solution:<\/strong><br \/>Radius of the sphere = 7m : \u2206r = 0.02 m.<br \/>Volume of the sphere V =\u00a043\u03c0r3<br \/>\u0394V=dVdr\u00d7\u0394r=43.\u03c0.3r2\u00d7\u0394r<br \/>= 4\u03c0 \u00d7 7\u00b2 \u00d7 0.02<br \/>= 3.92 \u03c0m\u00b3<\/p>\n<p><strong>Ex 6.4 Class 12 Maths\u00a0Question 7.<\/strong><br \/>If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.<br \/><strong>Solution:<\/strong><br \/>Radius of the sphere = 9 m: \u2206r = 0.03m<br \/>Surface area of sphere S = 4\u03c0r\u00b2<br \/>\u2206s =\u00a0dsdr\u00a0\u00d7 \u2206r<br \/>= 8\u03c0r \u00d7 \u2206r<br \/>= 8\u03c0 \u00d7 9 \u00d7 0.03<br \/>= 2.16 \u03c0m\u00b2.<\/p>\n<p><strong>Ex 6.4 Class 12 Maths\u00a0Question 8.<\/strong><br \/>If f (x) = 3x\u00b2 + 15x + 5, then the approximate value of f (3.02) is<br \/>(a) 47.66<br \/>(b) 57.66<br \/>(c) 67.66<br \/>(d) 77.66<br \/><strong>Solution:<\/strong><br \/>(d) x + \u2206x = 3.02, where x=30, \u2206x=.02,<br \/>\u2206f(x) = f(x + \u2206x) \u2013 f(x)<br \/>\u21d2 f(x + \u2206x) = f(x) + \u2206f(x) = f(x) + f\u2019 (x)\u2206x<br \/>Now f(x) = 3\u00d72 + 15x + 5; f(3) = 77, f\u2019 (x) = 6x + 15<br \/>f\u2019 (3) = 33<br \/>\u2234 f (3.02) = 87 + 33 x 0 02 = 77.66<\/p>\n<p><strong>Ex 6.4 Class 12 Maths\u00a0Question 9.<\/strong><br \/>The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is<br \/>(a) 0.06 x\u00b3 m\u00b3<br \/>(b) 0.6 x\u00b3 m\u00b3<br \/>(c) 0.09 x\u00b3 m\u00b3<br \/>(d) 0.9 x\u00b3 m\u00b3<br \/><strong>Solution:<\/strong><br \/>(c) Side of a cube = x meters<br \/>Volume of cube = x\u00b3,<br \/>for \u2206x. \u21d2 3% of x = 0.03 x<br \/>Let \u2206v be the change in v0l. \u2206v =\u00a0dvdx\u00a0x \u2206x = 3x\u00b2 \u00d7 \u2206x<br \/>But, \u2206x = 0.03 x<br \/>\u21d2 \u2206v = 3x\u00b2 x 0.03 x<br \/>= 0.09 x\u00b3m\u00b3<\/p>\n<p>All the best to the students appearing for the Class 12th board exam. Here is the detailed blog of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4. For further queries regarding the Class 12th <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> exam, you can ask in the comment box.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-the-ncert-solutions-for-class-12-maths-chapter-6-exercise-64\"><\/span>FAQs on the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4<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-1622201854458\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-the-downloading-of-ncert-solutions-class-12-maths-chapter-6-exercise-64-pdf-costs\"><\/span>How much does the downloading of NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 PDF costs?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>It is free of cost.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1622201886007\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-we-download-the-ncert-solutions-class-12-maths-chapter-6-exercise-64-pdf\"><\/span>From where can we download the NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can easily download it from Kopykitab.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1622201904796\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-there-a-step-by-step-explanation-of-exercise-64-class-12-questions\"><\/span>Is there a step-by-step explanation of Exercise 6.4 Class 12 questions?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, all the steps are in an elaborate manner.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1622202026662\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-is-the-total-weightage-of-chapter-6-applications-of-derivatives\"><\/span>What is the total weightage of Chapter 6, Applications of derivatives?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Chapter 6 has a total weightage of 35 marks.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4: The NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 comprises all the questions of Chapter 6 &#8211; Application of derivates exercise 6.4.\u00a0 NCERT Solutions For Class 12 Maths Chapter 6: Application of Derivatives\u00a0 NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.4 NCERT Solutions for &#8230; <a title=\"NCERT Solutions Class 12 Maths Chapter 6 Application of Derivatives\u00a0Exercise 6.4 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-class-12-maths-chapter-6-exercise-6-4\/\" aria-label=\"More on NCERT Solutions Class 12 Maths Chapter 6 Application of Derivatives\u00a0Exercise 6.4 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":93019,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73413,2917,73693],"tags":[73226,3570,75043],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/92657"}],"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\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=92657"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/92657\/revisions"}],"predecessor-version":[{"id":139287,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/92657\/revisions\/139287"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/93019"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=92657"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=92657"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=92657"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}