{"id":68981,"date":"2023-09-13T17:03:00","date_gmt":"2023-09-13T11:33:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=68981"},"modified":"2023-11-10T12:30:48","modified_gmt":"2023-11-10T07:00:48","slug":"rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-3\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Exercise 2.3 Chapter 2 Function (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-117944\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.3.png\" alt=\"RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.3.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.3-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3:&nbsp;<\/strong>In Chapter 2, Exercise 2.3 <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-12-solutions\/\"><strong>RD Sharma Class 12 Maths Solutions<\/strong><\/a>, you will solve problems concerning the composition of real functions. Class 12 is a turning point in a student&#8217;s life. It mainly prepares students to make important decisions about their future education and aspirations.&nbsp;You can refer to <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-functions\/\"><strong>RD Sharma Solutions Class 12 Maths Chapter 2 Functions<\/strong><\/a> Exercise 2.3 free PDF for a better understanding of the topics taught in this exercise.<\/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-69e76746ad665\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: 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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-2-3\/#where-can-i-download-rd-sharma-solutions-class-12-maths-chapter-2-exercise-23-free-pdf\" title=\"Where can I download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3 free PDF?\">Where can I download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3 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-23-pdf\"><\/span>Download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3 PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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-2Exercise2.3.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-2Exercise2.3.pdf\">RD Sharma Solutions for Class 12 Maths Chapter 2 Exercise 2.3<\/a><\/p>\n<p>Students&#8217; exam preparation is accelerated by solving these problems in a simple manner. <strong>RD Sharma Solutions<\/strong> mostly aids in revision and increases student confidence before appearing for the board exam.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-12-maths-chapter-2-exercise-23\"><\/span>Access answers to RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h4><span class=\"ez-toc-section\" id=\"exercise-23-page-no-254\"><\/span><strong>Exercise 2.3 Page No: 2.54<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p><strong>1. Find&nbsp;fog&nbsp;and&nbsp;gof,&nbsp;if <\/strong><\/p>\n<p><strong>(i) f&nbsp;(x)&nbsp;=&nbsp;e<sup>x<\/sup>,&nbsp;g (x)&nbsp;=&nbsp;log<sub>e<\/sub>&nbsp;x<\/strong><\/p>\n<p><strong>(ii) f&nbsp;(x)&nbsp;=&nbsp;x<sup>2<\/sup>,&nbsp;g (x)&nbsp;=&nbsp;cos&nbsp;x<\/strong><\/p>\n<p><strong>(iii) f&nbsp;(x)&nbsp;=&nbsp;|x|,&nbsp;g&nbsp;(x)&nbsp;=&nbsp;sin&nbsp;x<\/strong><\/p>\n<p><strong>(iv) f (x) = x+1, g(x) = e<sup>x<\/sup><\/strong><\/p>\n<p><strong>(v) f (x)&nbsp;=&nbsp;sin<sup>\u22121<\/sup>&nbsp;x,&nbsp;g(x)&nbsp;=&nbsp;x<sup>2<\/sup><\/strong><\/p>\n<p><strong>(vi) f&nbsp;(x)&nbsp;=&nbsp;x+1,&nbsp;g&nbsp;(x)&nbsp;=&nbsp;sin&nbsp;x<\/strong><\/p>\n<p><strong>(vii) f(x)=&nbsp;x&nbsp;+&nbsp;1,&nbsp;g&nbsp;(x)&nbsp;=&nbsp;2x&nbsp;+&nbsp;3<\/strong><\/p>\n<p><strong>(viii) f(x) = c, c \u2208 R, g(x) = sin x<sup>2<\/sup><\/strong><\/p>\n<p><strong>(ix) f(x) = x<sup>2<\/sup>&nbsp;+ 2 , g (x) = 1 \u2212 1\/ (1-x)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given f&nbsp;(x) = e<sup>x<\/sup>,&nbsp;g(x)&nbsp;=&nbsp;log<sub>e<\/sub>&nbsp;x<\/p>\n<p>Let f:&nbsp;R&nbsp;\u2192&nbsp;(0,&nbsp;\u221e);&nbsp;and g:&nbsp;(0,&nbsp;\u221e)&nbsp;\u2192&nbsp;R<\/p>\n<p>Now we have to calculate fog,<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>fog:&nbsp;(&nbsp;0,&nbsp;\u221e)&nbsp;\u2192&nbsp;R<\/p>\n<p>(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))<\/p>\n<p>=&nbsp;f&nbsp;(log<sub>e<\/sub>&nbsp;x)<\/p>\n<p>=&nbsp;log<sub>e<\/sub>&nbsp;e<sup>x<\/sup><\/p>\n<p>=&nbsp;x&nbsp;<\/p>\n<p>Now we have to calculate gof,<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;f&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;g.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R\u2192&nbsp;R<\/p>\n<p>(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))<\/p>\n<p>=&nbsp;g&nbsp;(e<sup>x<\/sup>)<\/p>\n<p>=&nbsp;log<sub>e<\/sub>&nbsp;e<sup>x<\/sup><\/p>\n<p>=&nbsp;x<\/p>\n<p>(ii) f&nbsp;(x)&nbsp;=&nbsp;x<sup>2<\/sup>,&nbsp;g(x)&nbsp;=&nbsp;cos&nbsp;x<\/p>\n<p>f:&nbsp;R\u2192&nbsp;[0,&nbsp;\u221e)&nbsp;;&nbsp;g:&nbsp;R\u2192[\u22121,&nbsp;1]<\/p>\n<p>Now we have to calculate fog,<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;not&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>\u21d2&nbsp;Domain&nbsp;(fog)&nbsp;=&nbsp;{x:&nbsp;x \u2208 domain&nbsp;of&nbsp;g&nbsp;and&nbsp;g&nbsp;(x)&nbsp;\u2208 domain&nbsp;of&nbsp;f}<\/p>\n<p>\u21d2&nbsp;Domain&nbsp;(fog)&nbsp;=&nbsp;x:&nbsp;x&nbsp;\u2208&nbsp;R&nbsp;and&nbsp;cos&nbsp;x&nbsp;\u2208&nbsp;R}<\/p>\n<p>\u21d2&nbsp;Domain&nbsp;of&nbsp;(fog)&nbsp;=&nbsp;R<\/p>\n<p>(fog):&nbsp;R\u2192&nbsp;R<\/p>\n<p>(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))<\/p>\n<p>=&nbsp;f&nbsp;(cos&nbsp;x)<\/p>\n<p>=&nbsp;cos<sup>2<\/sup>&nbsp;x<\/p>\n<p>Now we have to calculate gof,<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;f&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;g.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R\u2192R<\/p>\n<p>(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))<\/p>\n<p>=&nbsp;g&nbsp;(x<sup>2<\/sup>)<\/p>\n<p>=&nbsp;cos&nbsp;x<sup>2<\/sup><\/p>\n<p>(iii) Given f&nbsp;(x)&nbsp;=&nbsp;|x|,&nbsp;g(x)&nbsp;=&nbsp;sin&nbsp;x<\/p>\n<p>f:&nbsp;R&nbsp;\u2192&nbsp;(0,&nbsp;\u221e)&nbsp;;&nbsp;g&nbsp;:&nbsp;R\u2192[\u22121,&nbsp;1]<\/p>\n<p>Now we have to calculate fog,<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R\u2192R<\/p>\n<p>(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))<\/p>\n<p>=&nbsp;f&nbsp;(sin&nbsp;x)<\/p>\n<p>=&nbsp;|sin&nbsp;x|<\/p>\n<p>Now we have to calculate gof,<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;f&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;g.<\/p>\n<p>\u21d2&nbsp;fog&nbsp;:&nbsp;R\u2192&nbsp;R<\/p>\n<p>(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))<\/p>\n<p>=&nbsp;g&nbsp;(|x|)<\/p>\n<p>=&nbsp;sin&nbsp;|x|<\/p>\n<p>(iv) Given f&nbsp;(x)&nbsp;=&nbsp;x + 1,&nbsp;g(x)&nbsp;=&nbsp;e<sup>x<\/sup><\/p>\n<p>f:&nbsp;R\u2192R&nbsp;;&nbsp;g:&nbsp;R&nbsp;\u2192&nbsp;[&nbsp;1,&nbsp;\u221e)<\/p>\n<p>Now we have calculate fog:<\/p>\n<p>Clearly,&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R\u2192R<\/p>\n<p>(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))<\/p>\n<p>=&nbsp;f&nbsp;(e<sup>x<\/sup>)<\/p>\n<p>=&nbsp;e<sup>x&nbsp;<\/sup>+ 1<\/p>\n<p>Now we have to compute gof,<\/p>\n<p>Clearly,&nbsp;range&nbsp;of&nbsp;f&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;domain&nbsp;of&nbsp;g.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R\u2192R<\/p>\n<p>(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))<\/p>\n<p>=&nbsp;g&nbsp;(x+1)<\/p>\n<p>=&nbsp;e<sup>x+1<\/sup><\/p>\n<p>(v) Given f&nbsp;(x)&nbsp;=&nbsp;sin&nbsp;<sup>\u22121<\/sup>&nbsp;x,&nbsp;g(x)&nbsp;=&nbsp;x<sup>2<\/sup><\/p>\n<p>f:&nbsp;[\u22121,1]\u2192 [(-\u03c0)\/2 ,\u03c0\/2];&nbsp;g&nbsp;:&nbsp;R&nbsp;\u2192&nbsp;[0,&nbsp;\u221e)&nbsp;<\/p>\n<p>Now we have to compute&nbsp;fog:<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;not&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>Domain&nbsp;(fog)&nbsp;=&nbsp;{x:&nbsp;x&nbsp;\u2208&nbsp;domain&nbsp;of&nbsp;g&nbsp;and&nbsp;g&nbsp;(x)&nbsp;\u2208&nbsp;domain&nbsp;of&nbsp;f}<\/p>\n<p>Domain&nbsp;(fog) = {x:&nbsp;x&nbsp;\u2208&nbsp;R&nbsp;and&nbsp;x<sup>2<\/sup>&nbsp;\u2208&nbsp;[\u22121, 1]}<\/p>\n<p>Domain&nbsp;(fog) = {x:&nbsp;x&nbsp;\u2208&nbsp;R&nbsp;and&nbsp;x&nbsp;\u2208&nbsp;[\u22121, 1]}<\/p>\n<p>Domain&nbsp;of&nbsp;(fog) =&nbsp;[\u22121, 1]<\/p>\n<p>fog:&nbsp;[\u22121,1]&nbsp;\u2192&nbsp;R&nbsp;<\/p>\n<p>(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))<\/p>\n<p>=&nbsp;f&nbsp;(x<sup>2<\/sup>)<\/p>\n<p>=&nbsp;sin<sup>\u22121<\/sup>&nbsp;(x<sup>2<\/sup>)<\/p>\n<p>Now we have to compute&nbsp;gof:<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;f&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;g.<\/p>\n<p>fog:&nbsp;[\u22121, 1]&nbsp;\u2192&nbsp;R<\/p>\n<p>(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))<\/p>\n<p>=&nbsp;g&nbsp;(sin<sup>\u22121<\/sup>&nbsp;x)<\/p>\n<p>=&nbsp;(sin<sup>\u22121<\/sup>&nbsp;x)<sup>2<\/sup><\/p>\n<p>(vi) Given f(x)&nbsp;=&nbsp;x+1,&nbsp;g(x)&nbsp;=&nbsp;sin&nbsp;x<\/p>\n<p>f:&nbsp;R\u2192R&nbsp;;&nbsp;g:&nbsp;R\u2192[\u22121,&nbsp;1]<\/p>\n<p>Now we have to compute&nbsp;fog<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>Set&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R\u2192&nbsp;R<\/p>\n<p>(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))<\/p>\n<p>=&nbsp;f&nbsp;(sin&nbsp;x)<\/p>\n<p>=&nbsp;sin&nbsp;x&nbsp;+&nbsp;1<\/p>\n<p>Now we have to compute gof,<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;f&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;g.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R&nbsp;\u2192&nbsp;R<\/p>\n<p>(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))<\/p>\n<p>=&nbsp;g&nbsp;(x+1)<\/p>\n<p>=&nbsp;sin&nbsp;(x+1)<\/p>\n<p>(vii) Given f&nbsp;(x)&nbsp;=&nbsp;x+1,&nbsp;g&nbsp;(x)&nbsp;=&nbsp;2x&nbsp;+&nbsp;3<\/p>\n<p>f:&nbsp;R\u2192R&nbsp;;&nbsp;g:&nbsp;R&nbsp;\u2192&nbsp;R<\/p>\n<p>Now we have to compute&nbsp;fog&nbsp;<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R\u2192&nbsp;R<\/p>\n<p>(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))<\/p>\n<p>=&nbsp;f&nbsp;(2x+3)<\/p>\n<p>=&nbsp;2x&nbsp;+&nbsp;3&nbsp;+&nbsp;1<\/p>\n<p>=&nbsp;2x&nbsp;+&nbsp;4<\/p>\n<p>Now we have to compute&nbsp;gof<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;f&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;g.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R&nbsp;\u2192&nbsp;R<\/p>\n<p>(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))<\/p>\n<p>=&nbsp;g&nbsp;(x+1)<\/p>\n<p>=&nbsp;2&nbsp;(x&nbsp;+&nbsp;1)&nbsp;+&nbsp;3<\/p>\n<p>=&nbsp;2x&nbsp;+&nbsp;5<\/p>\n<p>(viii) Given f&nbsp;(x)&nbsp;=&nbsp;c,&nbsp;g&nbsp;(x)&nbsp;=&nbsp;sin&nbsp;x<sup>2<\/sup><\/p>\n<p>f:&nbsp;R&nbsp;\u2192&nbsp;{c}&nbsp;;&nbsp;g:&nbsp;R\u2192&nbsp;[&nbsp;0,&nbsp;1&nbsp;]<\/p>\n<p>Now we have to compute&nbsp;fog<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>fog:&nbsp;R\u2192R<\/p>\n<p>(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))<\/p>\n<p>=&nbsp;f&nbsp;(sin&nbsp;x<sup>2<\/sup>)<\/p>\n<p>=&nbsp;c<\/p>\n<p>Now we have to compute gof,<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;f&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;g.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R\u2192&nbsp;R<\/p>\n<p>(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))<\/p>\n<p>=&nbsp;g&nbsp;(c)<\/p>\n<p>=&nbsp;sin&nbsp;c<sup>2<\/sup><\/p>\n<p>(ix) Given f&nbsp;(x)&nbsp;=&nbsp;x<sup>2<\/sup>+&nbsp;2 and g&nbsp;(x)&nbsp;=&nbsp;1 \u2013 1 \/ (1 \u2013 x)<\/p>\n<p>f:&nbsp;R&nbsp;\u2192&nbsp;[&nbsp;2,&nbsp;\u221e&nbsp;)<\/p>\n<p>For&nbsp;domain&nbsp;of&nbsp;g:&nbsp;1\u2212&nbsp;x&nbsp;\u2260&nbsp;0&nbsp;<\/p>\n<p>\u21d2&nbsp;x&nbsp;\u2260&nbsp;1<\/p>\n<p>\u21d2&nbsp;Domain&nbsp;of&nbsp;g =&nbsp;R \u2212 {1}<\/p>\n<p>g&nbsp;(x&nbsp;)= 1 \u2013 [1\/(1 \u2013 x)] = (1 \u2013 x \u2013 1)\/ (1 \u2013 x) = -x\/(1 \u2013 x)<\/p>\n<p>For&nbsp;range&nbsp;of&nbsp;g<\/p>\n<p>y = (- x)\/ (1 \u2013 x)<\/p>\n<p>\u21d2&nbsp;y&nbsp;\u2013&nbsp;x y&nbsp;=&nbsp;\u2212&nbsp;x<\/p>\n<p>\u21d2&nbsp;y&nbsp;=&nbsp;x y&nbsp;\u2212&nbsp;x<\/p>\n<p>\u21d2&nbsp;y&nbsp;=&nbsp;x&nbsp;(y\u22121)<\/p>\n<p>\u21d2 x = y\/(y \u2013 1)<\/p>\n<p>Range&nbsp;of&nbsp;g&nbsp;= R \u2212 {1}<\/p>\n<p>So,&nbsp;g:&nbsp;R \u2212 {1} \u2192 R \u2212 {1}<\/p>\n<p>Now we have to compute&nbsp;fog<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>\u21d2&nbsp;fog:&nbsp;R&nbsp;\u2212&nbsp;{1} \u2192&nbsp;R<\/p>\n<p>(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))<\/p>\n<p>= f (-x\/ (1 \u2013 x))<\/p>\n<p>= ((-x)\/ (1 \u2013 x))<sup>2<\/sup>&nbsp;+ 2<\/p>\n<p>= (x<sup>2<\/sup>&nbsp;+ 2x<sup>2<\/sup>&nbsp;+ 2 \u2013 4x) \/ (1 \u2013 x)<sup>2<\/sup><\/p>\n<p>= (3x<sup>2&nbsp;<\/sup>\u2013 4x + 2)\/ (1 \u2013 x)<sup>2<\/sup><\/p>\n<p>Now we have to compute&nbsp;gof<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;f&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;g.<\/p>\n<p>\u21d2&nbsp;gof:&nbsp;R\u2192R<\/p>\n<p>(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))<\/p>\n<p>=&nbsp;g&nbsp;(x<sup>2<\/sup>&nbsp;+&nbsp;2)<\/p>\n<p>= 1 \u2013 1 \/ (1 \u2013 (x<sup>2<\/sup>&nbsp;+ 2))<\/p>\n<p>= \u2013 1\/ (1 \u2013 (x<sup>2<\/sup>&nbsp;+ 2))<\/p>\n<p>= (x<sup>2<\/sup>&nbsp;+ 2)\/ (x<sup>2<\/sup>&nbsp;+ 1)<\/p>\n<p><strong>2. Let&nbsp;f(x) =&nbsp;x<sup>2<\/sup>&nbsp;+&nbsp;x&nbsp;+ 1 and&nbsp;g(x) = sin&nbsp;x. Show that&nbsp;fog&nbsp;\u2260&nbsp;gof.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f(x) =&nbsp;x<sup>2<\/sup>&nbsp;+&nbsp;x&nbsp;+ 1 and&nbsp;g(x) = sin&nbsp;x<\/p>\n<p>Now we have to prove fog&nbsp;\u2260&nbsp;gof<\/p>\n<p>(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))&nbsp;<\/p>\n<p>=&nbsp;f (sin&nbsp;x)&nbsp;<\/p>\n<p>=&nbsp;sin<sup>2&nbsp;<\/sup>x&nbsp;+&nbsp;sin&nbsp;x&nbsp;+&nbsp;1<\/p>\n<p>And&nbsp;(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))&nbsp;<\/p>\n<p>=&nbsp;g&nbsp;(x<sup>2<\/sup>+&nbsp;x&nbsp;+&nbsp;1)&nbsp;<\/p>\n<p>=&nbsp;sin&nbsp;(x<sup>2<\/sup>+&nbsp;x&nbsp;+&nbsp;1)<\/p>\n<p>So,&nbsp;fog&nbsp;\u2260&nbsp;gof.<\/p>\n<p><strong>3. If&nbsp;f(x) = |x|, prove that&nbsp;fof&nbsp;=&nbsp;f.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f(x) = |x|,<\/p>\n<p>Now we have to prove that&nbsp;fof&nbsp;=&nbsp;f.<\/p>\n<p>Consider (fof)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(f&nbsp;(x))&nbsp;<\/p>\n<p>=&nbsp;f&nbsp;(|x|)&nbsp;<\/p>\n<p>=&nbsp;||x||&nbsp;<\/p>\n<p>=&nbsp;|x|&nbsp;<\/p>\n<p>=&nbsp;f&nbsp;(x)<\/p>\n<p>So,<\/p>\n<p>(fof)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(x),&nbsp;\u2200x&nbsp;\u2208&nbsp;R<\/p>\n<p>Hence,&nbsp;fof&nbsp;=&nbsp;f<\/p>\n<p><strong>4. If&nbsp;f(x) = 2x&nbsp;+ 5 and&nbsp;g(x) =&nbsp;x<sup>2<\/sup>&nbsp;+ 1 be two real functions, then describe each of the following functions:<br>(i) fog<br>(ii)&nbsp;gof<br>(iii)&nbsp;fof<br>(iv)&nbsp;f<sup>2<\/sup><br>Also, show that&nbsp;fof&nbsp;\u2260&nbsp;f<sup>2<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>f(x)&nbsp;and&nbsp;g(x)&nbsp;are polynomials.<\/p>\n<p>\u21d2 f:&nbsp;R&nbsp;\u2192&nbsp;R&nbsp;and&nbsp;g:&nbsp;R&nbsp;\u2192&nbsp;R.<\/p>\n<p>So,&nbsp;fog:&nbsp;R&nbsp;\u2192&nbsp;R&nbsp;and&nbsp;gof:&nbsp;R&nbsp;\u2192&nbsp;R.<\/p>\n<p>(i) (fog) (x) = f (g (x))<\/p>\n<p>= f (x<sup>2<\/sup>&nbsp;+ 1)<\/p>\n<p>= 2 (x<sup>2&nbsp;<\/sup>+ 1) + 5<\/p>\n<p>=2x<sup>2<\/sup>&nbsp;+ 2 + 5<\/p>\n<p>= 2x<sup>2<\/sup>&nbsp;+7<\/p>\n<p>(ii) (gof) (x) = g (f (x))<\/p>\n<p>= g (2x +5)<\/p>\n<p>&nbsp;= (2x + 5)<sup>2<\/sup>&nbsp;+ 1<\/p>\n<p>= 4x<sup>2<\/sup>&nbsp;+ 20x + 26<\/p>\n<p>(iii) (fof) (x) = f (f (x))<\/p>\n<p>= f (2x +5)<\/p>\n<p>= 2 (2x + 5) + 5<\/p>\n<p>= 4x + 10 + 5<\/p>\n<p>= 4x + 15<\/p>\n<p>(iv) f<sup>2<\/sup>&nbsp;(x) = f (x) x f (x)<\/p>\n<p>= (2x + 5) (2x + 5)&nbsp;<\/p>\n<p>= (2x + 5)<sup>2<\/sup><\/p>\n<p>= 4x<sup>2<\/sup>&nbsp;+ 20x +25<\/p>\n<p>Hence, from (iii) and (iv) clearly fof&nbsp;\u2260&nbsp;f<sup>2<\/sup><\/p>\n<p><strong>5. If&nbsp;f(x) = sin&nbsp;x&nbsp;and&nbsp;g(x) = 2x be two real functions, then describe&nbsp;gof&nbsp;and&nbsp;fog.&nbsp;Are these equal functions?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f(x) = sin&nbsp;x&nbsp;and&nbsp;g(x) = 2x<\/p>\n<p>We&nbsp;know&nbsp;that<\/p>\n<p>f:&nbsp;R\u2192&nbsp;[\u22121,&nbsp;1]&nbsp;and&nbsp;g:&nbsp;R\u2192&nbsp;R<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;f&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;g.<\/p>\n<p>gof:&nbsp;R\u2192&nbsp;R<\/p>\n<p>(gof)&nbsp;(x)&nbsp;=&nbsp;g&nbsp;(f&nbsp;(x))<\/p>\n<p>=&nbsp;g&nbsp;(sin&nbsp;x)<\/p>\n<p>=&nbsp;2&nbsp;sin&nbsp;x<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>fog:&nbsp;R&nbsp;\u2192&nbsp;R<\/p>\n<p>So,&nbsp;(fog)&nbsp;(x)&nbsp;=&nbsp;f&nbsp;(g&nbsp;(x))<\/p>\n<p>=&nbsp;f&nbsp;(2x)<\/p>\n<p>=&nbsp;sin&nbsp;(2x)<\/p>\n<p>Clearly,&nbsp;fog \u2260 gof<\/p>\n<p>Hence they are not equal functions.<\/p>\n<p><strong>6. Let&nbsp;f,&nbsp;g,&nbsp;h&nbsp;be real functions given by&nbsp;f(x) = sin&nbsp;x,&nbsp;g&nbsp;(x) = 2x&nbsp;and&nbsp;h&nbsp;(x) = cos&nbsp;x. Prove that&nbsp;fog&nbsp;=&nbsp;go&nbsp;(f h).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that f(x) = sin&nbsp;x,&nbsp;g&nbsp;(x) = 2x&nbsp;and&nbsp;h&nbsp;(x) = cos&nbsp;x<\/p>\n<p>We&nbsp;know&nbsp;that&nbsp;f:&nbsp;R\u2192 [\u22121,&nbsp;1]&nbsp;and&nbsp;g:&nbsp;R\u2192 R<\/p>\n<p>Clearly,&nbsp;the&nbsp;range&nbsp;of&nbsp;g&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;the&nbsp;domain&nbsp;of&nbsp;f.<\/p>\n<p>fog:&nbsp;R&nbsp;\u2192&nbsp;R<\/p>\n<p>Now,&nbsp;(f h)&nbsp;(x) = f&nbsp;(x) h&nbsp;(x)&nbsp;=&nbsp;(sin&nbsp;x)&nbsp;(cos&nbsp;x)&nbsp;= \u00bd sin&nbsp;(2x)<\/p>\n<p>Domain&nbsp;of&nbsp;f h&nbsp;is&nbsp;R.<\/p>\n<p>Since&nbsp;range&nbsp;of&nbsp;sin&nbsp;x&nbsp;is&nbsp;[-1, 1], \u22121&nbsp;\u2264&nbsp;sin&nbsp;2x&nbsp;\u2264&nbsp;1<\/p>\n<p>\u21d2 -1\/2 \u2264 sin x\/2 \u2264 1\/2<\/p>\n<p>Range&nbsp;of&nbsp;f h&nbsp;= [-1\/2, 1\/2]<\/p>\n<p>So,&nbsp;(f h):&nbsp;R&nbsp;\u2192 [(-1)\/2, 1\/2]<\/p>\n<p>Clearly,&nbsp;range&nbsp;of&nbsp;f h&nbsp;is&nbsp;a&nbsp;subset&nbsp;of&nbsp;g.<\/p>\n<p>\u21d2&nbsp;go&nbsp;(f h):&nbsp;R&nbsp;\u2192&nbsp;R<\/p>\n<p>\u21d2 Domains&nbsp;of&nbsp;fog&nbsp;and&nbsp;go&nbsp;(f h)&nbsp;are&nbsp;the&nbsp;same.<\/p>\n<p>So,&nbsp;(fog)&nbsp;(x) = f&nbsp;(g&nbsp;(x))&nbsp;<\/p>\n<p>=&nbsp;f&nbsp;(2x)&nbsp;<\/p>\n<p>=&nbsp;sin&nbsp;(2x)<\/p>\n<p>And&nbsp;(go&nbsp;(f h))&nbsp;(x)&nbsp;=&nbsp;g&nbsp;((f(x). h(x))&nbsp;<\/p>\n<p>=&nbsp;g&nbsp;(sin x&nbsp;cos&nbsp;x)&nbsp;<\/p>\n<p>=&nbsp;2sin&nbsp;x&nbsp;cos&nbsp;x&nbsp;<\/p>\n<p>=&nbsp;sin&nbsp;(2x)<\/p>\n<p>\u21d2&nbsp;(fog)&nbsp;(x)&nbsp;=&nbsp;(go (f h))&nbsp;(x),&nbsp;\u2200x&nbsp;\u2208&nbsp;R<\/p>\n<p>Hence,&nbsp;fog&nbsp;=&nbsp;go&nbsp;(f h)<\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-12-maths-chapter-2-exercise-23-important-topics\"><\/span>RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3: 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>The 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 2.3. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>&nbsp;Class 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-23\"><\/span>FAQs on RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3<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-1629725571185\" 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-23\"><\/span>How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 6 questions in RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629725641006\" 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-23-for-free\"><\/span>Is RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3 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 2.3 for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629725653757\" 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-23-free-pdf\"><\/span>Where can I download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3 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 2.3 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 2.3:&nbsp;In Chapter 2, Exercise 2.3 RD Sharma Class 12 Maths Solutions, you will solve problems concerning the composition of real functions. Class 12 is a turning point in a student&#8217;s life. It mainly prepares students to make important decisions about their future education and aspirations.&nbsp;You can &#8230; <a title=\"RD Sharma Solutions for Class 12 Maths Exercise 2.3 Chapter 2 Function (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-3\/\" aria-label=\"More on RD Sharma Solutions for Class 12 Maths Exercise 2.3 Chapter 2 Function (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":117944,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3429,73223,61598,73664,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68981"}],"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=68981"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68981\/revisions"}],"predecessor-version":[{"id":505764,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68981\/revisions\/505764"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/117944"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=68981"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=68981"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=68981"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}