{"id":68982,"date":"2021-08-23T18:09:00","date_gmt":"2021-08-23T12:39:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=68982"},"modified":"2021-08-31T18:28:53","modified_gmt":"2021-08-31T12:58:53","slug":"rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-4\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Exercise 2.4 Chapter 2 Function (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-117947\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.4.png\" alt=\"RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.4.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.4-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4:<\/strong> RD Sharma Solutions for Class 12, Maths Chapter 2, helps students who wish to achieve a good academic score in the exam.\u00a0<a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-functions\/\"><strong>RD Sharma Maths Solutions for Class 12 Chapter 2<\/strong><\/a> are expertly designed to increase students&#8217; confidence to understand the concepts covered in this chapter and how to solve problems in a short period of time.<\/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-6a04a88d1933c\" 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-6a04a88d1933c\"><\/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\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-4\/#download-rd-sharma-solutions-class-12-maths-chapter-2-exercise-24-pdf\" title=\"Download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4 PDF\">Download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4 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-12-maths-chapter-2-exercise-2-4\/#access-answers-to-rd-sharma-solutions-class-12-maths-chapter-2-exercise-24\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-4\/#is-rd-sharma-solutions-class-12-maths-chapter-2-exercise-24-for-free\" title=\"Is RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4 for free?\">Is RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4 for free?<\/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\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-4\/#where-can-i-download-rd-sharma-solutions-class-12-maths-chapter-2-exercise-24-free-pdf\" title=\"Where can I download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4 free PDF?\">Where can I download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4 free PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-12-maths-chapter-2-exercise-24-pdf\"><\/span>Download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4 PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\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-12-Maths-Chapter-2-Exercise-2.4.pdf\", \"#example1\");<\/script><\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-2-Exercise-2.4.pdf\">RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4<\/a><\/p>\n<p>Our experts prepare these materials based on the Class 12 CBSE syllabus, keeping in mind the types of questions asked in the <strong>RD Sharma solution<\/strong>. Chapter 2 Functions explains the function and domains of functions and functions. It has four practices. Students can easily get answers to problems in <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-12-solutions\/\"><strong>RD Sharma Solutions for Class 12<\/strong><\/a>.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-12-maths-chapter-2-exercise-24\"><\/span>Access answers to RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h4><span class=\"ez-toc-section\" id=\"exercise-24-page-no-268\"><\/span><strong>Exercise 2.4 Page No: 2.68<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p><strong>1. State with reason whether the following functions have inverse:<br \/>(i) f: {1, 2, 3, 4} \u2192 {10} with\u00a0f\u00a0= {(1, 10), (2, 10), (3, 10), (4, 10)}<\/strong><\/p>\n<p><strong>(ii) g: {5, 6, 7, 8} \u2192 {1, 2, 3, 4} with\u00a0g\u00a0= {(5, 4), (6, 3), (7, 4), (8, 2)}<\/strong><\/p>\n<p><strong>(iii) h: {2, 3, 4, 5} \u2192 {7, 9, 11, 13} with\u00a0h\u00a0= {(2, 7), (3, 9), (4, 11), (5, 13)}<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given f: {1, 2, 3, 4} \u2192 {10} with\u00a0f\u00a0= {(1, 10), (2, 10), (3, 10), (4, 10)}<br \/><br \/>We have:<br \/><br \/>f\u00a0(1) =\u00a0f\u00a0(2) =\u00a0f\u00a0(3) =\u00a0f\u00a0(4) = 10<\/p>\n<p>\u21d2\u00a0f\u00a0is not one-one.<\/p>\n<p>\u21d2\u00a0f is not a bijection.<br \/><br \/>So,\u00a0f\u00a0does not have an inverse.<\/p>\n<p>(ii) Given g: {5, 6, 7, 8} \u2192 {1, 2, 3, 4} with\u00a0g\u00a0= {(5, 4), (6, 3), (7, 4), (8, 2)}<br \/><br \/>from the question it is clear that g\u00a0(5) =\u00a0g\u00a0(7) = 4<\/p>\n<p>\u21d2 f\u00a0is not one-one.<\/p>\n<p>\u21d2 f\u00a0is not a bijection.<br \/><br \/>So,\u00a0f\u00a0does not have an inverse.<\/p>\n<p>(iii) Given h: {2, 3, 4, 5} \u2192 {7, 9, 11, 13} with\u00a0h\u00a0= {(2, 7), (3, 9), (4, 11), (5, 13)}<br \/><br \/>Here, different elements of the domain have different images in the co-domain.<br \/><br \/>\u21d2 h is one-one.<br \/><br \/>Also, each element in the co-domain has a pre-image in the domain.<\/p>\n<p>\u21d2\u00a0h is onto.<\/p>\n<p>\u21d2\u00a0h is a bijection.<\/p>\n<p>Therefore h inverse exists.<\/p>\n<p>\u21d2\u00a0h has an inverse and it is given by<br \/><br \/>h<sup>-1\u00a0<\/sup>= {(7, 2), (9, 3), (11, 4), (13, 5)}<\/p>\n<p><strong>2. Find\u00a0f\u00a0<sup>\u22121<\/sup>\u00a0if it exists: f:\u00a0A\u00a0\u2192\u00a0B,\u00a0where\u00a0<\/strong><\/p>\n<p><strong>(i) A\u00a0= {0, \u22121, \u22123, 2};\u00a0B\u00a0= {\u22129, \u22123, 0, 6} and\u00a0f(x) = 3\u00a0x.<\/strong><\/p>\n<p><strong>(ii) A\u00a0= {1, 3, 5, 7, 9};\u00a0B\u00a0= {0, 1, 9, 25, 49, 81} and\u00a0f(x) =\u00a0x<sup>2<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given A\u00a0= {0, \u22121, \u22123, 2};\u00a0B\u00a0= {\u22129, \u22123, 0, 6} and\u00a0f(x) = 3\u00a0x.<br \/><br \/>So,\u00a0f =\u00a0{(0, 0), (-1, -3), (-3, -9), (2, 6)}<\/p>\n<p>Here, different elements of the domain have different images in the co-domain.<br \/><br \/>Clearly, this is one-one.<br \/><br \/>Range of\u00a0f =\u00a0Range of\u00a0f\u00a0= B<br \/><br \/>so,\u00a0f\u00a0is a bijection and,<\/p>\n<p>Thus,\u00a0f<sup>\u00a0-1<\/sup>\u00a0exists. <br \/><br \/>Hence, f<sup>\u00a0-1<\/sup>=\u00a0{(0, 0), (-3, -1), (-9, -3), (6, 2)}<\/p>\n<p>(ii) Given A\u00a0= {1, 3, 5, 7, 9};\u00a0B\u00a0= {0, 1, 9, 25, 49, 81} and\u00a0f(x) =\u00a0x<sup>2<\/sup><br \/><br \/>So,\u00a0f =\u00a0{(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}<\/p>\n<p>Here, different elements of the domain have different images in the co-domain.<br \/><br \/>Clearly,\u00a0f\u00a0is one-one.<br \/><br \/>But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)<\/p>\n<p>\u21d2 f\u00a0is not a bijection.<br \/><br \/>So,\u00a0f<sup>\u00a0-1<\/sup>does not exist.<\/p>\n<p><strong>3. Consider\u00a0f: {1, 2, 3} \u2192 {a,\u00a0b,\u00a0c} and\u00a0g: {a,\u00a0b,\u00a0c} \u2192 {apple, ball, cat} defined as\u00a0f\u00a0(1) =\u00a0a,\u00a0f\u00a0(2) =\u00a0b,\u00a0f\u00a0(3) =\u00a0c,\u00a0g\u00a0(a) = apple,\u00a0g\u00a0(b) = ball and\u00a0g\u00a0(c) = cat. Show that\u00a0f,\u00a0g\u00a0and\u00a0gof\u00a0are invertible. Find\u00a0f<sup>\u22121<\/sup>,\u00a0g<sup>\u22121<\/sup>\u00a0and\u00a0gof<sup>\u22121<\/sup>and show that (gof)<sup>\u22121<\/sup>\u00a0=\u00a0f\u00a0<sup>\u22121<\/sup>o\u00a0g<sup>\u22121<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f = {(1, a), (2, b), (c , 3)} and g = {(a , apple) , (b , ball) , (c , cat)} Clearly , f and g are bijections.<\/p>\n<p>So, f and g are invertible.\u00a0<\/p>\n<p>Now,<\/p>\n<p>f\u00a0<sup>-1<\/sup>\u00a0= {(a ,1) , (b , 2) , (3,c)} and g<sup>-1<\/sup>\u00a0= {(apple, a), (ball , b), (cat , c)}<\/p>\n<p>So, f<sup>-1<\/sup>\u00a0o g<sup>-1<\/sup>= {apple, 1), (ball, 2), (cat, 3)}\u2026\u2026\u2026 (1)<\/p>\n<p>f: {1,2,3,}\u00a0\u2192 {a, b, c} and g: {a, b, c}\u00a0\u2192 {apple, ball, cat}<\/p>\n<p>So, gof: {1, 2, 3}\u00a0\u2192 {apple, ball, cat}<\/p>\n<p>\u21d2 (gof) (1) = g (f (1)) = g (a) = apple<\/p>\n<p>(gof) (2) = g (f (2))<\/p>\n<p>= g (b)<\/p>\n<p>= ball,<\/p>\n<p>And (gof) (3) = g (f (3))<\/p>\n<p>= g (c)<\/p>\n<p>= cat\u00a0<\/p>\n<p>\u2234 gof = {(1, apple), (2, ball), (3, cat)}<\/p>\n<p>Clearly, gof is a bijection.<\/p>\n<p>So, gof is invertible.\u00a0<\/p>\n<p>(gof)<sup>-1\u00a0<\/sup>=\u00a0{(apple, 1), (ball, 2), (cat, 3)}\u2026\u2026. (2)<\/p>\n<p>Form (1) and (2), we get<\/p>\n<p>(gof)<sup>-1<\/sup>\u00a0= f<sup>-1<\/sup>\u00a0o g\u00a0<sup>-1<\/sup><\/p>\n<p><strong>4. Let\u00a0A\u00a0= {1, 2, 3, 4};\u00a0B\u00a0= {3, 5, 7, 9};\u00a0C\u00a0= {7, 23, 47, 79} and\u00a0f:\u00a0A\u00a0\u2192\u00a0B,\u00a0g:\u00a0B\u00a0\u2192\u00a0C\u00a0be defined as\u00a0f(x) = 2x\u00a0+ 1 and\u00a0g(x) =\u00a0x<sup>2<\/sup>\u00a0\u2212 2. Express (gof)<sup>\u22121<\/sup>\u00a0and\u00a0f<sup>\u22121<\/sup>\u00a0og<sup>\u22121<\/sup>\u00a0as the sets of ordered pairs and verify that (gof)<sup>\u22121<\/sup>\u00a0=\u00a0f<sup>\u22121<\/sup>\u00a0og<sup>\u22121<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that f (x) = 2x + 1<\/p>\n<p>\u21d2\u00a0f=\u00a0{(1,\u00a02(1) + 1),\u00a0(2,\u00a02(2) + 1),\u00a0(3,\u00a02(3) + 1),\u00a0(4,\u00a02(4) + 1)}<\/p>\n<p>= {(1,\u00a03),\u00a0(2,\u00a05),\u00a0(3,\u00a07),\u00a0(4,\u00a09)}<\/p>\n<p>Also given that g(x) = x<sup>2<\/sup>\u22122<\/p>\n<p>\u21d2\u00a0g =\u00a0{(3,\u00a03<sup>2<\/sup>\u22122),\u00a0(5,\u00a05<sup>2<\/sup>\u22122),\u00a0(7,\u00a07<sup>2<\/sup>\u22122),\u00a0(9,\u00a09<sup>2<\/sup>\u22122)}<\/p>\n<p>= {(3,\u00a07),\u00a0(5,\u00a023),\u00a0(7,\u00a047),\u00a0(9,\u00a079)}<\/p>\n<p>Clearly\u00a0f\u00a0and\u00a0g\u00a0are\u00a0bijections\u00a0and,\u00a0hence,\u00a0f<sup>\u22121<\/sup>:\u00a0B\u2192 A\u00a0and\u00a0g<sup>\u22121<\/sup>:\u00a0C\u2192 B\u00a0exist.<\/p>\n<p>So,\u00a0f<sup>\u22121<\/sup>=\u00a0{(3,\u00a01),\u00a0(5,\u00a02),\u00a0(7,\u00a03),\u00a0(9,\u00a04)}\u00a0<\/p>\n<p>And\u00a0g<sup>\u22121<\/sup>=\u00a0{(7,\u00a03),\u00a0(23,\u00a05),\u00a0(47,\u00a07),\u00a0(79,\u00a09)}<\/p>\n<p>Now,\u00a0(f<sup>\u22121\u00a0<\/sup>o\u00a0g<sup>\u22121<\/sup>):\u00a0C\u2192 A<\/p>\n<p>f<sup>\u22121<\/sup>\u00a0o\u00a0g<sup>\u22121\u00a0<\/sup>= {(7,\u00a01),\u00a0(23,\u00a02),\u00a0(47,\u00a03),\u00a0(79,\u00a04)}\u2026\u2026\u2026.(1)<\/p>\n<p>Also,\u00a0f:\u00a0A\u2192B\u00a0and\u00a0g:\u00a0B\u00a0\u2192\u00a0C,<\/p>\n<p>\u21d2\u00a0gof:\u00a0A\u00a0\u2192\u00a0C,\u00a0(gof)\u00a0<sup>\u22121<\/sup>\u00a0:\u00a0C\u2192 A<\/p>\n<p>So,\u00a0f<sup>\u22121<\/sup>\u00a0o\u00a0g<sup>\u22121<\/sup>and\u00a0(gof)<sup>\u22121\u00a0<\/sup>have\u00a0same\u00a0domains.<\/p>\n<p>(gof) (x) = g\u00a0(f\u00a0(x))<\/p>\n<p>=g\u00a0(2x + 1)<\/p>\n<p>=(2x +1 )<sup>2<\/sup>\u2212 2<\/p>\n<p>\u21d2\u00a0(gof)\u00a0(x)\u00a0=\u00a04x<sup>2\u00a0<\/sup>+\u00a04x\u00a0+ 1 \u2212 2<\/p>\n<p>\u21d2\u00a0(gof)\u00a0(x)\u00a0=\u00a04x<sup>2<\/sup>+\u00a04x\u00a0\u22121<\/p>\n<p>Then,\u00a0(gof)\u00a0(1)\u00a0=\u00a0g\u00a0(f\u00a0(1))\u00a0<\/p>\n<p>=\u00a04 + 4 \u2212 1\u00a0<\/p>\n<p>=7,<\/p>\n<p>(gof) (2) = g\u00a0(f\u00a0(2))<\/p>\n<p>= 4(2)<sup>2<\/sup>\u00a0+ 4(2) \u2013 1 = 23,<\/p>\n<p>(gof) (3) = g\u00a0(f\u00a0(3))<\/p>\n<p>= 4(3)<sup>2<\/sup>\u00a0+ 4(3) \u2013 1 = 47\u00a0and\u00a0<\/p>\n<p>(gof) (4) = g\u00a0(f\u00a0(4))<\/p>\n<p>= 4(4)<sup>2<\/sup>\u00a0+ 4(4) \u2212 1 = 79<\/p>\n<p>So,\u00a0gof = {(1,\u00a07),\u00a0(2,\u00a023),\u00a0(3,\u00a047),\u00a0(4,\u00a079)}<\/p>\n<p>\u21d2 (gof)<sup>\u2013 1<\/sup>\u00a0= {(7,\u00a01),\u00a0(23,\u00a02),\u00a0(47,\u00a03),\u00a0(79,\u00a04)}\u2026\u2026 (2)<\/p>\n<p>From\u00a0(1)\u00a0and\u00a0(2),\u00a0we\u00a0get:<\/p>\n<p>(gof)<sup>\u22121\u00a0<\/sup>=\u00a0f<sup>\u22121<\/sup>\u00a0o\u00a0g<sup>\u22121<\/sup><\/p>\n<p><strong>5. Show that the function\u00a0f:\u00a0Q\u00a0\u2192\u00a0Q,\u00a0defined by\u00a0f(x) = 3x\u00a0+ 5, is invertible. Also, find\u00a0f<sup>\u22121<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given function\u00a0f:\u00a0Q\u00a0\u2192\u00a0Q,\u00a0defined by\u00a0f(x) = 3x\u00a0+ 5<\/p>\n<p>Now we have to show that the given function is invertible.<\/p>\n<p>Injection of f:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be two elements of the domain (Q),<\/p>\n<p>Such that f(x) = f(y)<\/p>\n<p>\u21d2 3x\u00a0+ 5 = 3y\u00a0+ 5<\/p>\n<p>\u21d2 3x\u00a0= 3y<\/p>\n<p>\u21d2 x = y<br \/><br \/>so,\u00a0f is one-one.<\/p>\n<p>Surjection of\u00a0f:<br \/><br \/>Let\u00a0y\u00a0be in the co-domain (Q),<\/p>\n<p>Such that\u00a0f(x) = y<\/p>\n<p>\u21d2 3x +5 = y\u00a0<\/p>\n<p>\u21d2 3x = y \u2013 5<\/p>\n<p>\u21d2 x = (y -5)\/3 belongs to Q domain<\/p>\n<p>\u21d2 f\u00a0is onto.<br \/><br \/>So,\u00a0f\u00a0is a bijection and, hence, it is invertible.<\/p>\n<p>Now we have to find\u00a0f<sup>-1<\/sup>:<\/p>\n<p>Let\u00a0f<sup>-1<\/sup>(x) = y\u2026\u2026 (1)<\/p>\n<p>\u21d2\u00a0x = f(y)<\/p>\n<p>\u21d2\u00a0x = 3y + 5<\/p>\n<p>\u21d2\u00a0x\u00a0\u22125\u00a0=\u00a03y<\/p>\n<p>\u21d2 y = (x \u2013 5)\/3<\/p>\n<p>Now substituting this value in (1) we get<\/p>\n<p>So, f<sup>-1<\/sup>(x) = (x \u2013 5)\/3<\/p>\n<p><strong>6. Consider\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0given by\u00a0f(x) = 4x\u00a0+ 3. Show that\u00a0f\u00a0is invertible. Find the inverse of\u00a0f.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f:\u00a0R\u00a0\u2192\u00a0R\u00a0given by\u00a0f(x) = 4x\u00a0+ 3<\/p>\n<p>Now we have to show that the given function is invertible.<\/p>\n<p>Consider injection of\u00a0f:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be two elements of domain (R),<\/p>\n<p>Such that f(x) = f(y)<br \/><br \/>\u21d2 4x\u00a0+ 3 = 4y\u00a0+ 3<br \/><br \/>\u21d2 4x\u00a0= 4y<br \/><br \/>\u21d2 x = y<br \/><br \/>So,\u00a0f\u00a0is one-one.<\/p>\n<p>Now surjection of\u00a0f: <br \/><br \/>Let\u00a0y\u00a0be in the co-domain (R),<\/p>\n<p>Such that\u00a0f(x) = y.<\/p>\n<p>\u21d2 4x + 3 = y\u00a0<\/p>\n<p>\u21d2 4x = y -3<\/p>\n<p>\u21d2 x = (y-3)\/4 in R\u00a0(domain)<\/p>\n<p>\u21d2 f\u00a0is onto.<br \/><br \/>So,\u00a0f\u00a0is a bijection and, hence, it is invertible.<\/p>\n<p>Now we have to find\u00a0f<sup>\u00a0-1<\/sup><\/p>\n<p>Let f<sup>-1<\/sup>(x) = y\u2026\u2026. (1)<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0f\u00a0(y)<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a04y\u00a0+\u00a03<\/p>\n<p>\u21d2\u00a0x\u00a0\u2212\u00a03\u00a0=\u00a04y<\/p>\n<p>\u21d2 y = (x -3)\/4<\/p>\n<p>Now substituting this value in (1) we get<\/p>\n<p>So, f<sup>-1<\/sup>(x) = (x-3)\/4<\/p>\n<p><strong>7. Consider\u00a0f:\u00a0R\u00a0\u2192\u00a0R<sup>+<\/sup>\u00a0\u2192 [4, \u221e) given by\u00a0f(x) =\u00a0x<sup>2<\/sup>\u00a0+ 4. Show that\u00a0f\u00a0is invertible with inverse\u00a0f<sup>\u22121<\/sup>\u00a0of\u00a0f\u00a0given by f<sup>\u22121<\/sup>(x) = \u221a (x-4) where R<sup>+<\/sup>\u00a0is the set of all non-negative real numbers.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f:\u00a0R\u00a0\u2192\u00a0R<sup>+<\/sup>\u00a0\u2192 [4, \u221e) given by\u00a0f(x) =\u00a0x<sup>2<\/sup>\u00a0+ 4.<\/p>\n<p>Now we have to show that f is invertible,<\/p>\n<p>Consider injection of\u00a0f:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be two elements of the domain (Q),<\/p>\n<p>Such that f(x) = f(y)\u00a0<\/p>\n<p>\u21d2\u00a0x<sup>2\u00a0<\/sup>+ 4 = y<sup>2\u00a0<\/sup>+ 4<\/p>\n<p>\u21d2\u00a0x<sup>2\u00a0<\/sup>= y<sup>2<\/sup><\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0y (as\u00a0co-domain\u00a0as\u00a0R+)<\/p>\n<p>So,\u00a0f\u00a0is one-one<\/p>\n<p>Now surjection of\u00a0f:<br \/><br \/>Let\u00a0y\u00a0be in the co-domain (Q),<\/p>\n<p>Such that\u00a0f(x) = y<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0+ 4 = y<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0= y \u2013 4<\/p>\n<p>\u21d2 x = \u221a (y-4) in R<\/p>\n<p>\u21d2\u00a0f\u00a0is onto.<br \/><br \/>So,\u00a0f\u00a0is a bijection and, hence, it is invertible.<\/p>\n<p>Now we have to find\u00a0f<sup>-1<\/sup>:<\/p>\n<p>Let\u00a0f<sup>\u22121<\/sup>\u00a0(x)\u00a0=\u00a0y\u2026\u2026 (1)<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0f\u00a0(y)<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0y<sup>2<\/sup>\u00a0+\u00a04<\/p>\n<p>\u21d2\u00a0x\u00a0\u2212\u00a04\u00a0=\u00a0y<sup>2<\/sup><\/p>\n<p>\u21d2 y = \u221a (x-4)<\/p>\n<p>So, f<sup>-1<\/sup>(x) = \u221a (x-4)<\/p>\n<p>Now substituting this value in (1) we get,<\/p>\n<p>So, f<sup>-1<\/sup>(x) = \u221a (x-4)<\/p>\n<p><strong>8. If f(x) = (4x + 3)\/ (6x \u2013 4), x \u2260 (2\/3)\u00a0show that\u00a0fof(x) =\u00a0x, for all x\u00a0\u2260 (2\/3). What is the inverse of\u00a0f?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>It is given that f(x) = (4x + 3)\/ (6x \u2013 4), x \u2260 2\/3<\/p>\n<p>Now we have to show fof(x) =\u00a0x<\/p>\n<p>(fof)(x) = f (f(x))<\/p>\n<p>= f ((4x+ 3)\/ (6x \u2013 4))<\/p>\n<p>= (4((4x + 3)\/ (6x -4)) + 3)\/ (6 ((4x +3)\/ (6x \u2013 4)) \u2013 4)<\/p>\n<p>= (16x + 12 + 18x \u2013 12)\/ (24x + 18 \u2013 24x + 16)<\/p>\n<p>= (34x)\/ (34)<\/p>\n<p>= x<\/p>\n<p>Therefore, fof(x) = x for all x \u2260 2\/3<\/p>\n<p>=&gt; fof\u00a0= 1<\/p>\n<p>Hence, the given function\u00a0f\u00a0is invertible and the inverse of\u00a0f\u00a0is\u00a0f\u00a0itself.<\/p>\n<p><strong>9. Consider\u00a0f:\u00a0R<sup>+<\/sup>\u00a0\u2192 [\u22125, \u221e) given by\u00a0f(x) = 9x<sup>2<\/sup>\u00a0+ 6x\u00a0\u2212 5. Show that\u00a0f\u00a0is invertible with<\/strong><\/p>\n<p><strong>f<sup>-1<\/sup>(x) = (\u221a(x +6)-1)\/3\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f:\u00a0R<sup>+<\/sup>\u00a0\u2192 [\u22125, \u221e) given by\u00a0f(x) = 9x<sup>2<\/sup>\u00a0+ 6x\u00a0\u2013 5<\/p>\n<p>We have to show that f is invertible.<\/p>\n<p>Injectivity of\u00a0f:<br \/><br \/>Let x and y be two elements of domain (R<sup>+<\/sup>),<\/p>\n<p>Such that f(x) = f(y)<\/p>\n<p>\u21d2\u00a09x<sup>2\u00a0<\/sup>+ 6x \u2013 5 = 9y<sup>2\u00a0<\/sup>+\u00a06y\u00a0\u2212\u00a05<\/p>\n<p>\u21d2\u00a09x<sup>2\u00a0<\/sup>+ 6x = 9y<sup>2\u00a0<\/sup>+ 6y<\/p>\n<p>\u21d2 x = y (As, x, y \u2208 R<sup>+<\/sup>)<\/p>\n<p>So,\u00a0f\u00a0is one-one.<\/p>\n<p>Surjectivity of\u00a0f:<br \/><br \/>Let\u00a0y\u00a0is in the co domain (Q)<\/p>\n<p>Such that\u00a0f(x) = y<\/p>\n<p>\u21d2 9x<sup>2<\/sup>\u00a0+ 6x \u2013 5 = y<\/p>\n<p>\u21d2 9x<sup>2<\/sup>\u00a0+ 6x = y + 5<\/p>\n<p>\u21d2 9x<sup>2<\/sup>\u00a0+ 6x +1 = y + 6 (By adding 1 on both sides)<\/p>\n<p>\u21d2 (3x + 1)<sup>2<\/sup>\u00a0= y + 6<\/p>\n<p>\u21d2 3x + 1 = \u221a(y + 6)<\/p>\n<p>\u21d2 3x = \u221a (y + 6) \u2013 1<\/p>\n<p>\u21d2 x = (\u221a (y + 6)-1)\/3 in R<sup>+<\/sup>\u00a0(domain)<\/p>\n<p>f\u00a0is onto.<br \/><br \/>So,\u00a0f\u00a0is a bijection and hence, it is invertible.<\/p>\n<p>Now we have to find f<sup>-1<\/sup><\/p>\n<p>Let\u00a0f<sup>\u22121<\/sup>(x)\u00a0=\u00a0y\u2026.. (1)<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0f\u00a0(y)<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a09y<sup>2\u00a0<\/sup>+\u00a06y\u00a0\u2212\u00a05<\/p>\n<p>\u21d2\u00a0x\u00a0+\u00a05\u00a0=\u00a09y<sup>2\u00a0<\/sup>+ 6y<\/p>\n<p>\u21d2\u00a0x\u00a0+\u00a06 =\u00a09y<sup>2<\/sup>+\u00a06y\u00a0+\u00a01 (adding\u00a01\u00a0on\u00a0both\u00a0sides)<\/p>\n<p>\u21d2\u00a0x\u00a0+\u00a06\u00a0=\u00a0(3y\u00a0+\u00a01)<sup>2<\/sup><\/p>\n<p>\u21d2 3y + 1 = \u221a (x + 6)<\/p>\n<p>\u21d2\u00a03y =\u221a(x +6) -1<\/p>\n<p>\u21d2 y = (\u221a (x+6)-1)\/3<\/p>\n<p>Now substituting this value in (1) we get,<\/p>\n<p>So, f<sup>-1<\/sup>(x)\u00a0= (\u221a (x+6)-1)\/3\u00a0<\/p>\n<p><strong>10. If\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0be defined by\u00a0f(x) =\u00a0x<sup>3<\/sup>\u00a0\u22123, then prove that\u00a0f<sup>\u22121<\/sup>\u00a0exists and find a formula for\u00a0f<sup>\u22121<\/sup>. Hence, find\u00a0f<sup>\u22121\u00a0<\/sup>(24) and\u00a0f<sup>\u22121<\/sup>\u00a0(5).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f:\u00a0R\u00a0\u2192\u00a0R\u00a0be defined by\u00a0f(x) =\u00a0x<sup>3<\/sup>\u00a0\u22123<\/p>\n<p>Now we have to prove that f<sup>\u22121<\/sup>\u00a0exists<\/p>\n<p>Injectivity of\u00a0f:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be two elements in domain (R),<\/p>\n<p>Such\u00a0that,\u00a0x<sup>3<\/sup>\u00a0\u2212\u00a03\u00a0=\u00a0y<sup>3<\/sup>\u00a0\u2212\u00a03<\/p>\n<p>\u21d2\u00a0x<sup>3<\/sup>\u00a0=\u00a0y<sup>3<\/sup><\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is one-one.<\/p>\n<p>Surjectivity of\u00a0f:<br \/><br \/>Let\u00a0y\u00a0be in the co-domain (R)<\/p>\n<p>Such that\u00a0f(x) = y<\/p>\n<p>\u21d2 x<sup>3<\/sup>\u00a0\u2013 3 = y<\/p>\n<p>\u21d2\u00a0x<sup>3<\/sup>\u00a0= y + 3\u00a0<\/p>\n<p>\u21d2 x = \u221b(y+3) in R<\/p>\n<p>\u21d2\u00a0f\u00a0is onto.<br \/><br \/>So,\u00a0f\u00a0is a bijection and, hence, it is invertible.<\/p>\n<p>Finding\u00a0f\u00a0<sup>-1<\/sup>:<\/p>\n<p>Let\u00a0f<sup>-1<\/sup>(x)\u00a0=\u00a0y\u2026\u2026.. (1)<\/p>\n<p>\u21d2\u00a0x=\u00a0f(y)<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0y<sup>3\u00a0<\/sup>\u2212 3<\/p>\n<p>\u21d2\u00a0x\u00a0+\u00a03\u00a0=\u00a0y<sup>3<\/sup><\/p>\n<p>\u21d2\u00a0y = \u221b(x + 3)\u00a0= f<sup>-1<\/sup>(x) [from\u00a0(1)]<\/p>\n<p>So, f<sup>-1<\/sup>(x) = \u221b(x + 3)<\/p>\n<p>Now, f<sup>-1<\/sup>(24) = \u221b (24 + 3)<\/p>\n<p>= \u221b27<\/p>\n<p>= \u221b3<sup>3<\/sup><\/p>\n<p>= 3<\/p>\n<p>And f<sup>-1<\/sup>(5) =\u221b (5 + 3)<\/p>\n<p>= \u221b8<\/p>\n<p>= \u221b2<sup>3<\/sup><\/p>\n<p>= 2<\/p>\n<p><strong>11. A function\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0is defined as\u00a0f(x) =\u00a0x<sup>3<\/sup>\u00a0+ 4. Is it a bijection or not? In case it is a bijection, find\u00a0f<sup>\u22121<\/sup>\u00a0(3).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that f:\u00a0R\u00a0\u2192\u00a0R\u00a0is defined as\u00a0f(x) =\u00a0x<sup>3<\/sup>\u00a0+ 4<\/p>\n<p>Injectivity of\u00a0f:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be two elements of domain (R),<\/p>\n<p>Such that f\u00a0(x)\u00a0=\u00a0f\u00a0(y)<\/p>\n<p>\u21d2\u00a0x<sup>3<\/sup>\u00a0+\u00a04\u00a0=\u00a0y<sup>3<\/sup>\u00a0+\u00a04<\/p>\n<p>\u21d2\u00a0x<sup>3<\/sup>\u00a0=\u00a0y<sup>3<\/sup><\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is one-one.<\/p>\n<p>Surjectivity of\u00a0f:<br \/><br \/>Let\u00a0y\u00a0be in the co-domain (R),<\/p>\n<p>Such that\u00a0f(x) = y.<\/p>\n<p>\u21d2\u00a0x<sup>3<\/sup>\u00a0+ 4 = y\u00a0<\/p>\n<p>\u21d2 x<sup>3<\/sup>\u00a0= y \u2013 4<\/p>\n<p>\u21d2 x = \u221b (y \u2013 4) in R (domain)<\/p>\n<p>\u21d2 f\u00a0is onto.<br \/><br \/>So,\u00a0f\u00a0is a bijection and, hence, it is invertible.<\/p>\n<p>Finding\u00a0f<sup>-1<\/sup>:<\/p>\n<p>Let\u00a0f<sup>\u22121<\/sup>\u00a0(x)\u00a0=\u00a0y\u2026\u2026 (1)<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0f\u00a0(y)<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0y<sup>3\u00a0<\/sup>+\u00a04<\/p>\n<p>\u21d2\u00a0x\u00a0\u2212\u00a04\u00a0=\u00a0y<sup>3<\/sup><\/p>\n<p>\u21d2 y =\u221b (x-4)<\/p>\n<p>So,\u00a0f<sup>-1<\/sup>(x)\u00a0=\u221b (x-4) [from\u00a0(1)]<\/p>\n<p>f<sup>-1\u00a0<\/sup>(3) = \u221b(3 \u2013 4)<\/p>\n<p>= \u221b-1<\/p>\n<p>= -1<\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-12-maths-chapter-2-exercise-24-important-topics\"><\/span>RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4: Important Topics<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Let us have a look at some of the important concepts that are discussed in this chapter.<\/p>\n<ul>\n<li>Classification of functions\n<ul>\n<li>Types of functions\n<ul>\n<li>Constant function<\/li>\n<li>Identity function<\/li>\n<li>Modulus function<\/li>\n<li>Integer function<\/li>\n<li>Exponential function<\/li>\n<li>Logarithmic function<\/li>\n<li>Reciprocal function<\/li>\n<li>Square root function<\/li>\n<\/ul>\n<\/li>\n<li>Operations on real functions<\/li>\n<li>Kinds of functions\n<ul>\n<li>One-one function<\/li>\n<li>On-to function<\/li>\n<li>Many one function<\/li>\n<li>In to function<\/li>\n<li>Bijection<\/li>\n<\/ul>\n<\/li>\n<li>Composition of functions<\/li>\n<li>Properties of the composition of functions<\/li>\n<li>Composition of real function<\/li>\n<li>Inverse of a function<\/li>\n<li>Inverse of an element<\/li>\n<li>Relation between graphs of a function and its inverse<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>We have provided complete details of RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 12 Exam, 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-12-maths-chapter-2-exercise-24\"><\/span>FAQs on RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4\u00a0 \u00a0\u00a0<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-1629727427464\" 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-12-maths-chapter-2-exercise-24\"><\/span>How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 11 questions in RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629727464131\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-solutions-class-12-maths-chapter-2-exercise-24-for-free\"><\/span>Is RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4 for free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, You can get RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4 for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629727474906\" 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-12-maths-chapter-2-exercise-24-free-pdf\"><\/span>Where can I download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4 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 12 Maths Chapter 2 Exercise 2.4 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4: RD Sharma Solutions for Class 12, Maths Chapter 2, helps students who wish to achieve a good academic score in the exam.\u00a0RD Sharma Maths Solutions for Class 12 Chapter 2 are expertly designed to increase students&#8217; confidence to understand the concepts covered in this chapter &#8230; <a title=\"RD Sharma Solutions for Class 12 Maths Exercise 2.4 Chapter 2 Function (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-4\/\" aria-label=\"More on RD Sharma Solutions for Class 12 Maths Exercise 2.4 Chapter 2 Function (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":117947,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3429,73223,73664],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68982"}],"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=68982"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68982\/revisions"}],"predecessor-version":[{"id":121951,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68982\/revisions\/121951"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/117947"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=68982"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=68982"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=68982"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}