{"id":68980,"date":"2021-08-23T16:25:00","date_gmt":"2021-08-23T10:55:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=68980"},"modified":"2021-08-31T18:30:25","modified_gmt":"2021-08-31T13:00:25","slug":"rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-2\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Exercise 2.2 Chapter 2 Function (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-117941\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.2.png\" alt=\"RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.2.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.2-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2:<\/strong> RD Sharma Solutions for Class 12, Maths Chapter 2, helps students who wish to achieve a good academic score in the board exam. <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-12-solutions\/\"><strong>RD Sharma Class 12 Maths Solutions<\/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-69da49e307852\" 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-69da49e307852\"><\/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-2-2\/#download-rd-sharma-solutions-class-12-maths-chapter-2-exercise-22-pdf\" title=\"Download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2 PDF\">Download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2 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-2-2\/#access-answers-to-rd-sharma-solutions-class-12-maths-chapter-2-exercise-22\" title=\"Access answers to RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2\">Access answers to RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-2\/#exercise-22-page-no-246\" title=\"Exercise 2.2 Page No: 2.46\">Exercise 2.2 Page No: 2.46<\/a><\/li><\/ul><\/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\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-2\/#rd-sharma-solutions-class-12-maths-chapter-2-exercise-22-important-topics\" title=\"RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2: Important Topics\">RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2: Important Topics<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-2\/#faqs-on-rd-sharma-solutions-class-12-maths-chapter-2-exercise-22\" title=\"FAQs on RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2\">FAQs on RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-2\/#how-many-questions-are-there-in-rd-sharma-solutions-class-12-maths-chapter-2-exercise-22\" title=\"How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2?\">How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2?<\/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\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-2\/#is-rd-sharma-solutions-class-12-maths-chapter-2-exercise-22-for-free\" title=\"Is RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2 for free?\">Is RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2 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-2-2\/#where-can-i-download-rd-sharma-solutions-class-12-maths-chapter-2-exercise-22-free-pdf\" title=\"Where can I download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2 free PDF?\">Where can I download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2 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-22-pdf\"><\/span>Download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2 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-Ex-2.2-1-1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-2-Ex-2.2-1-1.pdf\">RD Sharma Solutions for Class 12 Maths Chapter 2 Exercise 2.2<\/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-solutions-class-12-maths-chapter-2-functions\/\"><strong>RD Sharma Maths Solutions for Class 12 Chapter 2<\/strong><\/a> Exercise 2.2 from this article.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-12-maths-chapter-2-exercise-22\"><\/span>Access answers to RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"exercise-22-page-no-246\"><\/span>Exercise 2.2 Page No: 2.46<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Find\u00a0gof\u00a0and\u00a0fog\u00a0when\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0g\u00a0:\u00a0R\u00a0\u2192\u00a0R\u00a0is defined by\u00a0<\/strong><\/p>\n<p><strong>(i) f(x) = 2x\u00a0+ 3 and g(x) =\u00a0x<sup>2<\/sup>\u00a0+ 5.<\/strong><\/p>\n<p><strong>(ii) f(x) = 2x\u00a0+\u00a0x<sup>2<\/sup>\u00a0and g(x) =\u00a0x<sup>3<\/sup><\/strong><\/p>\n<p><strong>(iii) f\u00a0(x)\u00a0=\u00a0x<sup>2<\/sup>\u00a0+ 8\u00a0and g(x)\u00a0= 3<em>x<\/em><sup>3<\/sup>\u00a0+ 1<\/strong><\/p>\n<p><strong>(iv) f (x) =\u00a0x\u00a0and\u00a0g(x) = |x|\u00a0<\/strong><\/p>\n<p><strong>(v) f(x) =\u00a0x<sup>2<\/sup>\u00a0+ 2x\u00a0\u2212 3 and g(x) = 3x\u00a0\u2212 4\u00a0<\/strong><\/p>\n<p><strong>(vi) f(x) = 8x<sup>3<\/sup>\u00a0and g(x) =\u00a0x<sup>1\/3<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given,\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0g:\u00a0R\u00a0\u2192\u00a0R<br \/><br \/>So,\u00a0gof:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0fog:\u00a0R\u00a0\u2192\u00a0R<\/p>\n<p>Also given that f(x) = 2x\u00a0+ 3 and\u00a0g(x) =\u00a0x<sup>2<\/sup>\u00a0+ 5<br \/><br \/>Now, (gof) (x) = g (f (x))<br \/><br \/>= g\u00a0(2x\u00a0+3)<br \/><br \/>=\u00a0(2x\u00a0+ 3)<sup>2<\/sup>\u00a0+ 5<br \/><br \/>=\u00a04x<sup>2<\/sup>+ 9 + 12x\u00a0+5<br \/><br \/>=4x<sup>2<\/sup>+\u00a012x\u00a0+ 14<\/p>\n<p>Now, (fog) (x) = f (g (x))<br \/><br \/>= f (x<sup>2<\/sup>\u00a0+ 5)<br \/><br \/>=\u00a02\u00a0(x<sup>2<\/sup>\u00a0+ 5) +3<br \/><br \/>=\u00a02\u00a0x<sup>2<\/sup>+ 10 + 3<br \/><br \/>=\u00a02x<sup>2<\/sup>\u00a0+ 13<\/p>\n<p>(ii) Given,\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0g:\u00a0R\u00a0\u2192\u00a0R<br \/><br \/>so,\u00a0gof:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0fog:\u00a0R\u00a0\u2192\u00a0R<\/p>\n<p>f(x) = 2x\u00a0+\u00a0x<sup>2<\/sup>\u00a0and\u00a0g(x) =\u00a0x<sup>3<\/sup><\/p>\n<p>(gof)\u00a0(x)=\u00a0g\u00a0(f\u00a0(x))<\/p>\n<p>=\u00a0g\u00a0(2x+x<sup>2<\/sup>)<\/p>\n<p>=\u00a0(2x+x<sup>2<\/sup>)<sup>3<\/sup><\/p>\n<p>Now, (fog)\u00a0(x) =\u00a0f\u00a0(g\u00a0(x))<\/p>\n<p>=\u00a0f\u00a0(x<sup>3<\/sup>)<\/p>\n<p>=\u00a02\u00a0(x<sup>3<\/sup>) + (x<sup>3<\/sup>)<sup>2<\/sup><\/p>\n<p>= 2x<sup>3\u00a0<\/sup>+ x<sup>6<\/sup><\/p>\n<p>(iii) Given,\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0g:\u00a0R\u00a0\u2192\u00a0R<br \/><br \/>So,\u00a0gof:\u00a0R\u00a0\u2192\u00a0R and\u00a0fog:\u00a0R\u00a0\u2192\u00a0R<\/p>\n<p>f(x) =\u00a0x<sup>2<\/sup>\u00a0+ 8 and\u00a0g(x) = 3x<sup>3<\/sup>\u00a0+ 1<\/p>\n<p>(gof)\u00a0(x) =\u00a0g\u00a0(f\u00a0(x))<\/p>\n<p>=\u00a0g\u00a0(x<sup>2<\/sup>\u00a0+\u00a08)<\/p>\n<p>=\u00a03\u00a0(x<sup>2<\/sup>+8)<sup>3<\/sup>\u00a0+\u00a01<\/p>\n<p>Now, (fog)\u00a0(x) =\u00a0f\u00a0(g\u00a0(x))<\/p>\n<p>=\u00a0f\u00a0(3x<sup>3<\/sup>\u00a0+\u00a01)<\/p>\n<p>=\u00a0(3x<sup>3<\/sup>+1)<sup>2<\/sup>\u00a0+\u00a08<\/p>\n<p>=\u00a09x<sup>6<\/sup>\u00a0+\u00a06x<sup>3\u00a0<\/sup>+\u00a01 +\u00a08<\/p>\n<p>= 9x<sup>6\u00a0<\/sup>+ 6x<sup>3\u00a0<\/sup>+ 9<\/p>\n<p>(iv) Given,\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0g:\u00a0R\u00a0\u2192\u00a0R<br \/><br \/>So,\u00a0gof:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0fog:\u00a0R\u00a0\u2192\u00a0R<\/p>\n<p>f(x) =\u00a0x\u00a0and\u00a0g(x) = |x|<\/p>\n<p>(gof)\u00a0(x) =\u00a0g\u00a0(f\u00a0(x))<\/p>\n<p>=\u00a0g\u00a0(x)<\/p>\n<p>=\u00a0|x|<\/p>\n<p>Now (fog)\u00a0(x) =\u00a0f\u00a0(g\u00a0(x))<\/p>\n<p>=\u00a0f\u00a0(|x|)<\/p>\n<p>=\u00a0|x|<\/p>\n<p>(v) Given,\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0g:\u00a0R\u00a0\u2192\u00a0R<br \/><br \/>So,\u00a0gof:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0fog:\u00a0R\u00a0\u2192\u00a0R<\/p>\n<p>f(x) =\u00a0x<sup>2<\/sup>\u00a0+ 2x\u00a0\u2212 3 and\u00a0g(x) = 3x\u00a0\u2212 4<\/p>\n<p>(gof)\u00a0(x) =\u00a0g\u00a0(f(x))<\/p>\n<p>=\u00a0g\u00a0(x<sup>2\u00a0<\/sup>+ 2x \u2212 3)<\/p>\n<p>=\u00a03\u00a0(x<sup>2\u00a0<\/sup>+ 2x \u2212 3)\u00a0\u2212 4<\/p>\n<p>=\u00a03x<sup>2\u00a0<\/sup>+\u00a06x\u00a0\u2212\u00a09\u00a0\u2212\u00a04<\/p>\n<p>=\u00a03x<sup>2\u00a0<\/sup>+ 6x \u2212 13<\/p>\n<p>Now, (fog)\u00a0(x) =\u00a0f\u00a0(g\u00a0(x))<\/p>\n<p>=\u00a0f\u00a0(3x \u2212 4)<\/p>\n<p>=\u00a0(3x\u00a0\u2212\u00a04)<sup>2\u00a0<\/sup>+ 2\u00a0(3x\u00a0\u2212\u00a04)\u00a0\u22123<\/p>\n<p>=\u00a09x<sup>2\u00a0<\/sup>+ 16 \u2212 24x + 6x \u2013 8 \u2212 3<\/p>\n<p>=\u00a09x<sup>2\u00a0<\/sup>\u2212 18x\u00a0+\u00a05<\/p>\n<p>(vi) Given,\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0g:\u00a0R\u00a0\u2192\u00a0R<br \/><br \/>So,\u00a0gof:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0fog:\u00a0R\u00a0\u2192\u00a0R<\/p>\n<p>f(x) = 8x<sup>3<\/sup>\u00a0and\u00a0g(x) =\u00a0x<sup>1\/3<\/sup><\/p>\n<p>(gof)\u00a0(x) =\u00a0g\u00a0(f\u00a0(x))<\/p>\n<p>=\u00a0g\u00a0(8x<sup>3<\/sup>)<\/p>\n<p>= (8x<sup>3<\/sup>)<sup>1\/3<\/sup><\/p>\n<p>= [(2x)<sup>3<\/sup>]<sup>1\/3<\/sup><\/p>\n<p>=\u00a02x<\/p>\n<p>Now, (fog)\u00a0(x) =\u00a0f\u00a0(g\u00a0(x))<\/p>\n<p>= f (x<sup>1\/3<\/sup>)<\/p>\n<p>= 8 (x<sup>1\/3<\/sup>)<sup>3<\/sup><\/p>\n<p>=\u00a08x<\/p>\n<p><strong>2. Let\u00a0f\u00a0= {(3, 1), (9, 3), (12, 4)} and\u00a0g\u00a0= {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that\u00a0gof\u00a0and\u00a0fog are both defined. Also, find\u00a0fog\u00a0and\u00a0gof.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f = {(3, 1), (9, 3), (12, 4)} and\u00a0g\u00a0= {(1, 3), (3, 3) (4, 9) (5, 9)}<br \/><br \/>f\u00a0: {3, 9, 12} \u2192 {1, 3, 4} and\u00a0g\u00a0: {1, 3, 4, 5} \u2192 {3, 9}<\/p>\n<p>Co-domain of\u00a0f\u00a0is a subset of the domain of\u00a0g.<br \/><br \/>So,\u00a0gof\u00a0exists and\u00a0gof: {3, 9, 12} \u2192 {3, 9}<\/p>\n<p>(gof)\u00a0(3) = g\u00a0(f\u00a0(3)) = g\u00a0(1)\u00a0= 3<\/p>\n<p>(gof)\u00a0(9) = g\u00a0(f\u00a0(9)) = g\u00a0(3) = 3<\/p>\n<p>(gof)\u00a0(12) = g\u00a0(f\u00a0(12)) = g\u00a0(4) = 9<\/p>\n<p>\u21d2\u00a0gof\u00a0= {(3,\u00a03),\u00a0(9,\u00a03),\u00a0(12,\u00a09)}<\/p>\n<p>Co-domain of\u00a0g\u00a0is a subset of the domain of\u00a0f.<br \/><br \/>So,\u00a0fog\u00a0exists and\u00a0fog: {1, 3, 4, 5} \u2192 {3, 9, 12}<\/p>\n<p>(fog)\u00a0(1) = f\u00a0(g\u00a0(1)) = f\u00a0(3) = 1<\/p>\n<p>(fog)\u00a0(3) = f\u00a0(g\u00a0(3)) = f\u00a0(3) = 1<\/p>\n<p>(fog)\u00a0(4) = f\u00a0(g\u00a0(4)) = f\u00a0(9) = 3<\/p>\n<p>(fog)\u00a0(5) = f\u00a0(g\u00a0(5)) = f\u00a0(9) = 3<\/p>\n<p>\u21d2\u00a0fog = {(1,\u00a01),\u00a0(3,\u00a01),\u00a0(4,\u00a03),\u00a0(5,\u00a03)}<\/p>\n<p><strong>3. Let\u00a0f\u00a0= {(1, \u22121), (4, \u22122), (9, \u22123), (16, 4)} and\u00a0g\u00a0= {(\u22121, \u22122), (\u22122, \u22124), (\u22123, \u22126), (4, 8)}. Show that\u00a0gof\u00a0is defined while\u00a0fog\u00a0is not defined. Also, find\u00a0gof.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f\u00a0= {(1, \u22121), (4, \u22122), (9, \u22123), (16, 4)} and\u00a0g\u00a0= {(\u22121, \u22122), (\u22122, \u22124), (\u22123, \u22126), (4, 8)}<br \/><br \/>f: {1, 4, 9, 16} \u2192 {-1, -2, -3, 4} and\u00a0g: {-1, -2, -3, 4} \u2192 {-2, -4, -6, 8}<\/p>\n<p>Co-domain of\u00a0f\u00a0= domain of\u00a0g<br \/><br \/>So,\u00a0gof\u00a0exists and\u00a0gof: {1, 4, 9, 16} \u2192 {-2, -4, -6, 8}<\/p>\n<p>(gof)\u00a0(1)\u00a0=\u00a0g\u00a0(f\u00a0(1))\u00a0=\u00a0g\u00a0(\u22121)\u00a0=\u00a0\u22122<\/p>\n<p>(gof)\u00a0(4)\u00a0=\u00a0g\u00a0(f\u00a0(4)) = g\u00a0(\u22122)\u00a0=\u00a0\u22124<\/p>\n<p>(gof)\u00a0(9)\u00a0=\u00a0g\u00a0(f\u00a0(9))\u00a0=\u00a0g\u00a0(\u22123)\u00a0=\u00a0\u22126<\/p>\n<p>(gof)\u00a0(16)\u00a0= g\u00a0(f\u00a0(16))\u00a0= g\u00a0(4)\u00a0=\u00a08<\/p>\n<p>So,\u00a0gof\u00a0=\u00a0{(1,\u00a0\u22122),\u00a0(4,\u00a0\u22124),\u00a0(9,\u00a0\u22126),\u00a0(16,\u00a08)}<\/p>\n<p>But the co-domain of\u00a0g\u00a0is not same as the domain of\u00a0f.<br \/><br \/>So,\u00a0fog\u00a0does not exist.<\/p>\n<p><strong>4. Let\u00a0A\u00a0= {a,\u00a0b,\u00a0c},\u00a0B\u00a0= {u,\u00a0v,\u00a0w} and let\u00a0f\u00a0and\u00a0g\u00a0be two functions from\u00a0A\u00a0to\u00a0B\u00a0and from\u00a0B\u00a0to\u00a0A,\u00a0respectively, defined as: f\u00a0= {(a,\u00a0v), (b,\u00a0u), (c,\u00a0w)},\u00a0g\u00a0= {(u,\u00a0b), (v,\u00a0a), (w,\u00a0c)}.<br \/>Show that\u00a0f\u00a0and\u00a0g\u00a0both are bijections and find\u00a0fog\u00a0and\u00a0gof.<\/strong><\/p>\n<p>\u00a0<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f\u00a0= {(a,\u00a0v), (b,\u00a0u), (c,\u00a0w)},\u00a0g\u00a0= {(u,\u00a0b), (v,\u00a0a), (w,\u00a0c)}.<br \/><br \/>Also given that A\u00a0= {a,\u00a0b,\u00a0c},\u00a0B\u00a0= {u,\u00a0v,\u00a0w}<\/p>\n<p>Now we have to show f and g both are bijective.<\/p>\n<p>Consider f\u00a0= {(a,\u00a0v), (b,\u00a0u), (c,\u00a0w)} and\u00a0f:\u00a0A \u2192 B<br \/><br \/>Injectivity of\u00a0f: No two elements of\u00a0A\u00a0have the same image in B.<br \/><br \/>So,\u00a0f\u00a0is one-one.<br \/><br \/>Surjectivity of\u00a0f: Co-domain of\u00a0f\u00a0= {u,\u00a0v,\u00a0w}<br \/><br \/>Range of\u00a0f\u00a0= {u,\u00a0v,\u00a0w}<br \/><br \/>Both are same.<br \/><br \/>So,\u00a0f\u00a0is onto.<br \/><br \/>Hence,\u00a0f\u00a0is a bijection.<\/p>\n<p>Now consider g\u00a0= {(u,\u00a0b), (v,\u00a0a), (w,\u00a0c)} and\u00a0g:\u00a0B \u2192 A<br \/><br \/>Injectivity of\u00a0g: No two elements of\u00a0B have the same image in A.<br \/><br \/>So,\u00a0g\u00a0is one-one.<br \/><br \/>Surjectivity of g: Co-domain of\u00a0g\u00a0= {a,\u00a0b,\u00a0c}<br \/><br \/>Range of\u00a0g\u00a0= {a,\u00a0b,\u00a0c}<br \/><br \/>Both are the same.<br \/><br \/>So,\u00a0g\u00a0is onto.<br \/><br \/>Hence,\u00a0g\u00a0is a bijection.<\/p>\n<p>Now we have to find fog,<br \/><br \/>we know that Co-domain of\u00a0g\u00a0is same as the domain of\u00a0f.<br \/><br \/>So,\u00a0fog\u00a0exists and\u00a0fog: {u\u00a0v,\u00a0w}\u00a0\u2192\u00a0{u,\u00a0v,\u00a0w}<\/p>\n<p>(fog)\u00a0(u)\u00a0=\u00a0f\u00a0(g\u00a0(u))\u00a0=\u00a0f\u00a0(b)\u00a0=\u00a0u<\/p>\n<p>(fog)\u00a0(v)\u00a0=\u00a0f\u00a0(g\u00a0(v))\u00a0=\u00a0f\u00a0(a)\u00a0=\u00a0v<\/p>\n<p>(fog)\u00a0(w)\u00a0=\u00a0f\u00a0(g\u00a0(w))\u00a0=\u00a0f\u00a0(c)\u00a0=\u00a0w<\/p>\n<p>So,\u00a0fog\u00a0=\u00a0{(u,\u00a0u),\u00a0(v,\u00a0v),\u00a0(w,\u00a0w)}<\/p>\n<p>Now we have to find gof,<br \/><br \/>Co-domain of\u00a0f\u00a0is same as the domain of\u00a0g.<br \/><br \/>So,\u00a0fog\u00a0exists and\u00a0gof: {a,\u00a0b,\u00a0c}\u00a0\u2192\u00a0{a,\u00a0b,\u00a0c}<\/p>\n<p>(gof)\u00a0(a)\u00a0=\u00a0g\u00a0(f\u00a0(a))\u00a0=\u00a0g\u00a0(v)\u00a0=\u00a0a<\/p>\n<p>(gof)\u00a0(b)\u00a0=\u00a0g\u00a0(f\u00a0(b))\u00a0=\u00a0g\u00a0(u)\u00a0=\u00a0b<\/p>\n<p>(gof)\u00a0(c)\u00a0=\u00a0g\u00a0(f\u00a0(c))\u00a0=\u00a0g\u00a0(w)\u00a0=\u00a0c<\/p>\n<p>So,\u00a0gof\u00a0=\u00a0{(a,\u00a0a),\u00a0(b,\u00a0b),\u00a0(c,\u00a0c)}<\/p>\n<p><strong>5. Find\u00a0fog\u00a0(2) and\u00a0gof\u00a0(1) when f:\u00a0R\u00a0\u2192\u00a0R;\u00a0f(x) =\u00a0x<sup>2<\/sup>\u00a0+ 8 and\u00a0g:\u00a0R\u00a0\u2192\u00a0R;\u00a0g(x) = 3x<sup>3<\/sup>\u00a0+ 1.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f:\u00a0R\u00a0\u2192\u00a0R;\u00a0f(x) =\u00a0x<sup>2<\/sup>\u00a0+ 8 and\u00a0g:\u00a0R\u00a0\u2192\u00a0R;\u00a0g(x) = 3x<sup>3<\/sup>\u00a0+ 1.<\/p>\n<p>Consider (fog)\u00a0(2)\u00a0=\u00a0f\u00a0(g\u00a0(2))\u00a0<\/p>\n<p>=\u00a0f\u00a0(3 \u00d7 2<sup>3\u00a0<\/sup>+ 1)\u00a0<\/p>\n<p>=\u00a0f(3 \u00d7 8 + 1)<\/p>\n<p>= f (25)<\/p>\n<p>= 25<sup>2<\/sup>\u00a0+ 8<\/p>\n<p>=\u00a0633<\/p>\n<p>(gof)\u00a0(1)\u00a0=\u00a0g\u00a0(f\u00a0(1))\u00a0<\/p>\n<p>=\u00a0g\u00a0(1<sup>2\u00a0<\/sup>+ 8)\u00a0<\/p>\n<p>=\u00a0g\u00a0(9)\u00a0<\/p>\n<p>=\u00a03 \u00d7 9<sup>3\u00a0<\/sup>+ 1\u00a0<\/p>\n<p>=\u00a02188<\/p>\n<p><strong>6. Let\u00a0R<sup>+<\/sup>\u00a0be the set of all non-negative real numbers. If\u00a0f:\u00a0R<sup>+<\/sup>\u00a0\u2192\u00a0R<sup>+<\/sup>\u00a0and\u00a0g\u00a0:\u00a0R<sup>+<\/sup>\u00a0\u2192\u00a0R<sup>+<\/sup>\u00a0are defined as f(x)=x<sup>2<\/sup>\u00a0and g(x)=+ \u221ax, find fog\u00a0and\u00a0gof. Are they equal functions.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f:\u00a0R<sup>+<\/sup>\u00a0\u2192\u00a0R<sup>+<\/sup>\u00a0and\u00a0g:\u00a0R<sup>+<\/sup>\u00a0\u2192\u00a0R<sup>+<\/sup><br \/><br \/>So,\u00a0fog:\u00a0R<sup>+<\/sup>\u00a0\u2192\u00a0R<sup>+<\/sup>\u00a0and\u00a0gof:\u00a0R<sup>+<\/sup>\u00a0\u2192\u00a0R<sup>+<\/sup><br \/><br \/>Domains of\u00a0fog\u00a0and\u00a0gof\u00a0are the same.<\/p>\n<p>Now we have to find fog and gof also we have to check whether they are equal or not,<\/p>\n<p>Consider (fog)\u00a0(x)\u00a0=\u00a0f\u00a0(g (x))<\/p>\n<p>= f (\u221ax)<\/p>\n<p>= \u221ax<sup>2<\/sup><\/p>\n<p>= x<\/p>\n<p>Now consider (gof)\u00a0(x)\u00a0= g (f (x))<\/p>\n<p>= g (x<sup>2<\/sup>)<\/p>\n<p>= \u221ax<sup>2<\/sup><\/p>\n<p>= x<\/p>\n<p>So,\u00a0(fog)\u00a0(x)\u00a0=\u00a0(gof)\u00a0(x),\u00a0\u2200x\u00a0\u2208 R<sup>+<\/sup><\/p>\n<p>Hence,\u00a0fog\u00a0=\u00a0gof<\/p>\n<p><strong>7. Let\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0g:\u00a0R\u00a0\u2192\u00a0R\u00a0be defined by\u00a0f(x) =\u00a0x<sup>2<\/sup>\u00a0and\u00a0g(x) =\u00a0x\u00a0+ 1. Show that\u00a0fog\u00a0\u2260\u00a0gof.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f:\u00a0R\u00a0\u2192\u00a0R\u00a0and\u00a0g:\u00a0R\u00a0\u2192 R.<br \/><br \/>So, the domains of\u00a0f\u00a0and\u00a0g\u00a0are the same.<\/p>\n<p>Consider (fog)\u00a0(x)\u00a0=\u00a0f\u00a0(g\u00a0(x))\u00a0<\/p>\n<p>=\u00a0f\u00a0(x + 1) =\u00a0(x + 1)<sup>2<\/sup>\u00a0<\/p>\n<p>=\u00a0x<sup>2\u00a0<\/sup>+ 1 + 2x<\/p>\n<p>Again consider (gof)\u00a0(x)\u00a0=\u00a0g\u00a0(f\u00a0(x))\u00a0<\/p>\n<p>=\u00a0g\u00a0(x<sup>2<\/sup>) = x<sup>2\u00a0<\/sup>+ 1<\/p>\n<p>So,\u00a0fog \u2260 gof<\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-12-maths-chapter-2-exercise-22-important-topics\"><\/span>RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2: 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<div>\n<p>We have provided complete details of RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2. 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-22\"><\/span>FAQs on RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<\/div>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1629723983397\" 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-22\"><\/span>How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 7 questions in RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629724018380\" 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-22-for-free\"><\/span>Is RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2 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 2.2 for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629724042190\" 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-22-free-pdf\"><\/span>Where can I download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2 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 for Class 12 Maths Chapter 2 Exercise 2.2 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 2.2: RD Sharma Solutions for Class 12, Maths Chapter 2, helps students who wish to achieve a good academic score in the board exam. RD Sharma Class 12 Maths Solutions are expertly designed to increase students&#8217; confidence to understand the concepts covered in this chapter and &#8230; <a title=\"RD Sharma Solutions for Class 12 Maths Exercise 2.2 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-2-2\/\" aria-label=\"More on RD Sharma Solutions for Class 12 Maths Exercise 2.2 Chapter 2 Function (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":117941,"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\/68980"}],"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=68980"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68980\/revisions"}],"predecessor-version":[{"id":121955,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68980\/revisions\/121955"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/117941"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=68980"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=68980"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=68980"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}