{"id":63234,"date":"2023-09-12T18:42:00","date_gmt":"2023-09-12T13:12:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=63234"},"modified":"2023-11-09T11:25:29","modified_gmt":"2023-11-09T05:55:29","slug":"rd-sharma-solutions-class-11-maths-chapter-3-functions","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/","title":{"rendered":"RD Sharma Solutions Class 11 Maths Chapter 3 &#8211; Functions (Updated For 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-119084\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-3.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 3 \" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-3.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-3-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><b>RD Sharma Solutions Class 11 Maths Chapter 3 &#8211; Functions:\u00a0<\/b><span style=\"font-weight: 400;\">Though the chapter is always small when it comes to Functions it doesn\u2019t mean that you can leave it or take it for granted. Understand all the concepts nicely and you will be able to solve any question that appears in your examination with the help of <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths<\/a> Chapter 3.<\/span><\/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-69db36114c7db\" 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-69db36114c7db\"><\/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-11-maths-chapter-3-functions\/#download-rd-sharma-solutions-class-11-maths-chapter-3-functions-pdf\" title=\"Download RD Sharma Solutions Class 11 Maths Chapter 3 Functions PDF\">Download RD Sharma Solutions Class 11 Maths Chapter 3 Functions PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#exercise-wise-rd-sharma-solutions-class-11-maths-chapter-3-functions-pdf\" title=\"Exercise-Wise RD Sharma Solutions Class 11 Maths Chapter 3 Functions PDF\">Exercise-Wise RD Sharma Solutions Class 11 Maths Chapter 3 Functions PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#detailed-exercise-wise-rd-sharma-solutions-class-11-maths-chapter-3-important-topics\" title=\"Detailed Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 3 &amp; Important Topics\">Detailed Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 3 &amp; Important Topics<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#rd-sharma-solutions-class-11-maths-chapter-3-exercise-31\" title=\"RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.1\">RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#rd-sharma-solutions-class-11-maths-chapter-3-exercise-32\" title=\"RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.2\">RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#rd-sharma-solutions-class-11-maths-chapter-3-exercise-33\" title=\"RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.3\">RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.3<\/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-11-maths-chapter-3-functions\/#rd-sharma-solutions-class-11-maths-chapter-3-exercise-34\" title=\"RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.4\">RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.4<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#access-rd-sharma-solutions-class-11-maths-chapter-3\" title=\"Access RD Sharma Solutions Class 11 Maths Chapter 3\">Access RD Sharma Solutions Class 11 Maths Chapter 3<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#exercise-31-page-no-37\" title=\"EXERCISE 3.1 PAGE NO: 3.7\">EXERCISE 3.1 PAGE NO: 3.7<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#1-define-a-function-as-a-set-of-ordered-pairs\" title=\"1. Define a function as a set of ordered pairs.\">1. Define a function as a set of ordered pairs.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#2-define-a-function-as-a-correspondence-between-two-sets\" title=\"2. Define a function as a correspondence between two sets.\">2. Define a function as a correspondence between two sets.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#3-what-is-the-fundamental-difference-between-a-relation-and-a-function-is-every-relation-a-function\" title=\"3. What is the fundamental difference between a relation and a function? Is every relation a function?\">3. What is the fundamental difference between a relation and a function? Is every relation a function?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#4-let-a-%e2%80%932-%e2%80%931-0-1-2-and-f-a-%e2%86%92-z-be-a-function-defined-by-fx-x2-%e2%80%93-2x-%e2%80%93-3-find-i-range-of-f-ie-f-a-ii-pre-images-of-6-%e2%80%933-and-5\" title=\"4. Let A = {\u20132, \u20131, 0, 1, 2} and f: A\u00a0\u2192\u00a0Z be a function defined by f(x) = x2\u00a0\u2013 2x \u2013 3. Find: (i) range of f i.e. f (A) (ii) pre-images of 6, \u20133 and 5\">4. Let A = {\u20132, \u20131, 0, 1, 2} and f: A\u00a0\u2192\u00a0Z be a function defined by f(x) = x2\u00a0\u2013 2x \u2013 3. Find: (i) range of f i.e. f (A) (ii) pre-images of 6, \u20133 and 5<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#5-if-a-function-f-r-%e2%86%92-r-be-defined-by\" title=\"5. If a function f: R\u00a0\u2192\u00a0R be defined by \">5. If a function f: R\u00a0\u2192\u00a0R be defined by <\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#6-a-function-f-r-%e2%86%92-r-is-defined-by-fx-x2-determine-i-range-of-f-ii-x-fx-4-iii-y-fy-%e2%80%931\" title=\"6. A function f: R\u00a0\u2192\u00a0R is defined by f(x) = x2. Determine (i) range of f (ii) {x: f(x) = 4} (iii) {y: f(y) = \u20131}\">6. A function f: R\u00a0\u2192\u00a0R is defined by f(x) = x2. Determine (i) range of f (ii) {x: f(x) = 4} (iii) {y: f(y) = \u20131}<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#7-let-f-r%e2%86%92-r-where-r-is-the-set-of-all-positive-real-numbers-be-such-that-fx-loge-x-determine-i-the-image-set-of-the-domain-of-f-ii-x-f-x-%e2%80%932-iii-whether-f-xy-f-x-f-y-holds\" title=\"7.\u00a0Let f: R+\u2192\u00a0R, where R+\u00a0is the set of all positive real numbers, be such that f(x) = loge\u00a0x. Determine (i) the image set of the domain of f (ii) {x: f (x) = \u20132} (iii) whether f (xy) = f (x) + f (y) holds.\">7.\u00a0Let f: R+\u2192\u00a0R, where R+\u00a0is the set of all positive real numbers, be such that f(x) = loge\u00a0x. Determine (i) the image set of the domain of f (ii) {x: f (x) = \u20132} (iii) whether f (xy) = f (x) + f (y) holds.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-17\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#8-write-the-following-relations-as-sets-of-ordered-pairs-and-find-which-of-them-are-functions-i-x-y-y-3x-x-%e2%88%88-1-2-3-y-%e2%88%88-3-6-9-12-ii-x-y-y-%3e-x-1-x-1-2-and-y-2-4-6-iii-x-y-x-y-3-x-y-%e2%88%88-0-1-2-3\" title=\"8. Write the following relations as sets of ordered pairs and find which of them are functions: (i) {(x, y): y = 3x, x\u00a0\u2208\u00a0{1, 2, 3}, y\u00a0\u2208\u00a0{3, 6, 9, 12}} (ii) {(x, y): y &gt; x + 1, x = 1, 2 and y = 2, 4, 6} (iii) {(x, y): x + y = 3, x, y\u00a0\u2208\u00a0{0, 1, 2, 3}}\">8. Write the following relations as sets of ordered pairs and find which of them are functions: (i) {(x, y): y = 3x, x\u00a0\u2208\u00a0{1, 2, 3}, y\u00a0\u2208\u00a0{3, 6, 9, 12}} (ii) {(x, y): y &gt; x + 1, x = 1, 2 and y = 2, 4, 6} (iii) {(x, y): x + y = 3, x, y\u00a0\u2208\u00a0{0, 1, 2, 3}}<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-18\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#9-let-f-r-%e2%86%92-r-and-g-c-%e2%86%92-c-be-two-functions-defined-as-fx-x2-and-gx-x2-are-they-equal-functions\" title=\"9. Let f: R\u00a0\u2192\u00a0R and g: C\u00a0\u2192\u00a0C be two functions defined as f(x) = x2\u00a0and g(x) = x2. Are they equal functions?\">9. Let f: R\u00a0\u2192\u00a0R and g: C\u00a0\u2192\u00a0C be two functions defined as f(x) = x2\u00a0and g(x) = x2. Are they equal functions?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-19\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#exercise-32-page-no-311\" title=\"EXERCISE 3.2 PAGE NO: 3.11\">EXERCISE 3.2 PAGE NO: 3.11<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-20\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#1-if-f-x-x2-%e2%80%93-3x-4-then-find-the-values-of-x-satisfying-the-equation-f-x-f-2x-1\" title=\"1. If f (x) = x2\u00a0\u2013 3x + 4, then find the values of x satisfying the equation f (x) = f (2x + 1).\">1. If f (x) = x2\u00a0\u2013 3x + 4, then find the values of x satisfying the equation f (x) = f (2x + 1).<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-21\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#2-if-f-x-x-%e2%80%93-a2-x-%e2%80%93-b2-find-f-a-b\" title=\"2. If f (x) = (x \u2013 a)2\u00a0(x \u2013 b)2, find f (a + b).\">2. If f (x) = (x \u2013 a)2\u00a0(x \u2013 b)2, find f (a + b).<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-22\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#3-if-y-f-x-ax-%e2%80%93-b-bx-%e2%80%93-a-show-that-x-f-y\" title=\"3. If y = f (x) = (ax \u2013 b) \/ (bx \u2013 a), show that x = f (y).\">3. If y = f (x) = (ax \u2013 b) \/ (bx \u2013 a), show that x = f (y).<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-23\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#4-if-f-x-1-1-%e2%80%93-x-show-that-f-f-f-x-x\" title=\"4. If f (x) = 1 \/ (1 \u2013 x), show that f [f {f (x)}] = x.\">4. If f (x) = 1 \/ (1 \u2013 x), show that f [f {f (x)}] = x.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-24\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#5-if-f-x-x-1-x-%e2%80%93-1-show-that-f-f-x-x\" title=\"5. If f (x) = (x + 1) \/ (x \u2013 1), show that f [f (x)] = x.\">5. If f (x) = (x + 1) \/ (x \u2013 1), show that f [f (x)] = x.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-25\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#6-if\" title=\"6. If\">6. If<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-26\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#find\" title=\"Find:\">Find:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-27\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#i-f-12\" title=\"(i) f (1\/2)\">(i) f (1\/2)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-28\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#ii-f-2\" title=\"(ii) f (-2)\">(ii) f (-2)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-29\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#iii-f-1\" title=\"(iii) f (1)\">(iii) f (1)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-30\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#iv-f-%e2%88%9a3\" title=\"(iv) f (\u221a3)\">(iv) f (\u221a3)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-31\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#v-f-%e2%88%9a-3\" title=\"(v) f (\u221a-3)\">(v) f (\u221a-3)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-32\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#exercise-33-page-no-318\" title=\"EXERCISE 3.3 PAGE NO: 3.18\">EXERCISE 3.3 PAGE NO: 3.18<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-33\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#1-find-the-domain-of-each-of-the-following-real-valued-functions-of-real-variable\" title=\"1. Find the domain of each of the following real valued functions of real variable:\">1. Find the domain of each of the following real valued functions of real variable:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-34\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#i-f-x-1x\" title=\"(i) f (x) = 1\/x\">(i) f (x) = 1\/x<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-35\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#ii-f-x-1x-7\" title=\"(ii) f (x) = 1\/(x-7)\">(ii) f (x) = 1\/(x-7)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-36\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#iii-f-x-3x-2x1\" title=\"(iii) f (x) = (3x-2)\/(x+1)\">(iii) f (x) = (3x-2)\/(x+1)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-37\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#iv-f-x-2x1x2-9\" title=\"(iv) f (x) = (2x+1)\/(x2-9)\">(iv) f (x) = (2x+1)\/(x2-9)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-38\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#v-f-x-x22x1x2-8x12\" title=\"(v) f (x) = (x2+2x+1)\/(x2-8x+12)\">(v) f (x) = (x2+2x+1)\/(x2-8x+12)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-39\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#2-find-the-domain-of-each-of-the-following-real-valued-functions-of-real-variable\" title=\"2. Find the domain of each of the following real valued functions of real variable:\">2. Find the domain of each of the following real valued functions of real variable:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-40\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#i-f-x-%e2%88%9ax-2\" title=\"(i) f (x) = \u221a(x-2)\">(i) f (x) = \u221a(x-2)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-41\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#ii-f-x-1%e2%88%9ax2-1\" title=\"(ii) f (x) = 1\/(\u221a(x2-1))\">(ii) f (x) = 1\/(\u221a(x2-1))<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-42\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#iii-f-x-%e2%88%9a9-x2\" title=\"(iii) f (x) = \u221a(9-x2)\">(iii) f (x) = \u221a(9-x2)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-43\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#iv-f-x-%e2%88%9ax-23-x\" title=\"(iv) f (x) = \u221a(x-2)\/(3-x)\">(iv) f (x) = \u221a(x-2)\/(3-x)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-44\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#3-find-the-domain-and-range-of-each-of-the-following-real-valued-functions\" title=\"3. Find the domain and range of each of the following real valued functions:\">3. Find the domain and range of each of the following real valued functions:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-45\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#i-f-x-axbbx-a\" title=\"(i) f (x) = (ax+b)\/(bx-a)\">(i) f (x) = (ax+b)\/(bx-a)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-46\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#ii-f-x-ax-bcx-d\" title=\"(ii) f (x) = (ax-b)\/(cx-d)\">(ii) f (x) = (ax-b)\/(cx-d)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-47\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#iii-f-x-%e2%88%9ax-1\" title=\"(iii) f (x) = \u221a(x-1)\">(iii) f (x) = \u221a(x-1)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-48\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#iv-f-x-%e2%88%9ax-3\" title=\"(iv) f (x) = \u221a(x-3)\">(iv) f (x) = \u221a(x-3)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-49\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#v-f-x-x-22-x\" title=\"(v) f (x) = (x-2)\/(2-x)\">(v) f (x) = (x-2)\/(2-x)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-50\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#vi-f-x-x-1\" title=\"(vi) f (x) = |x-1|\">(vi) f (x) = |x-1|<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-51\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#vii-f-x-x\" title=\"(vii) f (x) = -|x|\">(vii) f (x) = -|x|<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-52\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#viii-f-x-%e2%88%9a9-x2\" title=\"(viii) f (x) = \u221a(9-x2)\">(viii) f (x) = \u221a(9-x2)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-53\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#exercise-34-page-no-338\" title=\"EXERCISE 3.4 PAGE NO: 3.38\">EXERCISE 3.4 PAGE NO: 3.38<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-54\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#1-find-f-g-f-%e2%80%93-g-cf-c-%e2%88%88-r-c-%e2%89%a0-0-fg-1f-and-fg-in-each-of-the-following-i-f-x-x3-1-and-g-x-x-1\" title=\"1. Find f + g, f \u2013 g, cf (c\u00a0\u2208\u00a0R, c \u2260 0), fg, 1\/f and f\/g in each of the following: (i) f (x) = x3\u00a0+ 1 and g (x) = x + 1\">1. Find f + g, f \u2013 g, cf (c\u00a0\u2208\u00a0R, c \u2260 0), fg, 1\/f and f\/g in each of the following: (i) f (x) = x3\u00a0+ 1 and g (x) = x + 1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-55\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#ii-f-x-%e2%88%9ax-1-and-g-x-%e2%88%9ax1\" title=\"(ii) f (x) =\u00a0\u221a(x-1) and g (x) = \u221a(x+1)\">(ii) f (x) =\u00a0\u221a(x-1) and g (x) = \u221a(x+1)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-56\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#2-let-fx-2x-5-and-gx-x2-x-describe\" title=\"2.\u00a0Let f(x) = 2x + 5 and g(x) = x2\u00a0+ x. Describe\">2.\u00a0Let f(x) = 2x + 5 and g(x) = x2\u00a0+ x. Describe<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-57\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#i-f-g-ii-f-%e2%80%93-g-iii-fg-iv-fg-find-the-domain-in-each-case\" title=\"(i) f + g (ii) f \u2013 g (iii) fg (iv)\u00a0f\/g Find the domain in each case.\">(i) f + g (ii) f \u2013 g (iii) fg (iv)\u00a0f\/g Find the domain in each case.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-58\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#3-if-fx-be-defined-on-%e2%80%932-2-and-is-given-by\" title=\"3.\u00a0If f(x) be defined on [\u20132, 2] and is given by\">3.\u00a0If f(x) be defined on [\u20132, 2] and is given by<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-59\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#4-let-f-g-be-two-real-functions-defined-by-fx-%e2%88%9ax1-and-gx-%e2%88%9a9-x2-then-describe-each-of-the-following-functions-i-f-g-ii-g-%e2%80%93-f-iii-fg-iv-fg-v-gf-vi-2f-%e2%80%93-%e2%88%9a5g-vii-f2-7f-viii-5g\" title=\"4. Let f, g be two real functions defined by f(x) = \u221a(x+1) and\u00a0g(x) = \u221a(9-x2). Then, describe each of the following functions. (i) f + g (ii) g \u2013 f (iii) fg (iv) f\/g (v) g\/f (vi) 2f \u2013 \u221a5g (vii) f2\u00a0+ 7f (viii) 5\/g\u00a0\">4. Let f, g be two real functions defined by f(x) = \u221a(x+1) and\u00a0g(x) = \u221a(9-x2). Then, describe each of the following functions. (i) f + g (ii) g \u2013 f (iii) fg (iv) f\/g (v) g\/f (vi) 2f \u2013 \u221a5g (vii) f2\u00a0+ 7f (viii) 5\/g\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-60\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#5-if-fx-loge-1-%e2%80%93-x-and-gx-x-then-determine-each-of-the-following-functions-i-f-g-ii-fg\" title=\"5. If f(x) = loge\u00a0(1 \u2013 x) and g(x) = [x], then determine each of the following functions: (i) f + g (ii) fg\">5. If f(x) = loge\u00a0(1 \u2013 x) and g(x) = [x], then determine each of the following functions: (i) f + g (ii) fg<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-61\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#iii-fg-iv-gf\" title=\"(iii)\u00a0f\/g (iv) g\/f\u00a0\">(iii)\u00a0f\/g (iv) g\/f\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-62\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#also-find-f-g-%e2%80%931-fg-0-fg-12-and-gf-12\" title=\"Also, find (f + g) (\u20131), (fg) (0), (f\/g) (1\/2) and (g\/f) (1\/2).\">Also, find (f + g) (\u20131), (fg) (0), (f\/g) (1\/2) and (g\/f) (1\/2).<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-63\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#important-topics-from-rd-sharma-solutions-class-11-maths\" title=\"Important Topics from RD Sharma Solutions Class 11 Maths\">Important Topics from RD Sharma Solutions Class 11 Maths<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-64\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#access-other-important-chapters-of-rd-sharma-solutions-class-11-maths\" title=\"Access Other Important Chapters of RD Sharma Solutions Class 11 Maths\">Access Other Important Chapters of RD Sharma Solutions Class 11 Maths<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-65\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#faqs-on-rd-sharma-solutions-class-11-maths-chapter-3\" title=\"FAQs on RD Sharma Solutions Class 11 Maths Chapter 3\">FAQs on RD Sharma Solutions Class 11 Maths Chapter 3<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-66\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#from-where-can-i-download-the-rd-sharma-solutions-class-11-maths-chapter-3-pdf\" title=\"From where can I download the RD Sharma Solutions Class 11 Maths Chapter 3 PDF?\">From where can I download the RD Sharma Solutions Class 11 Maths Chapter 3 PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-67\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#can-i-access-the-rd-sharma-solutions-class-11-maths-chapter-3-pdf-offline\" title=\"Can I access the RD Sharma Solutions Class 11 Maths Chapter 3 PDF offline?\">Can I access the RD Sharma Solutions Class 11 Maths Chapter 3 PDF offline?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-68\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/#how-much-does-it-cost-to-download-the-rd-sharma-solutions-class-11-maths-chapter-3-pdf\" title=\"How much does it cost to download the RD Sharma Solutions Class 11 Maths Chapter 3 PDF?\">How much does it cost to download the RD Sharma Solutions Class 11 Maths Chapter 3 PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-3-functions-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 3 Functions PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-3-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 3<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<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\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-3-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-11-maths-chapter-3-functions-pdf\"><\/span>Exercise-Wise RD Sharma Solutions Class 11 Maths Chapter 3 Functions PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-exercise-3-1\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma Solutions Exercise-3.1<\/span><\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-exercise-3-2\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma Solutions Exercise-3.2<\/span><\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-exercise-3-3\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma Solutions Exercise-3.3<\/span><\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-exercise-3-4\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma Solutions Exercise-3.4<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-rd-sharma-solutions-class-11-maths-chapter-3-important-topics\"><\/span><span style=\"font-size: 30px; background-color: initial;\">Detailed Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 3 &amp; Important Topics<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-11-maths-chapter-3-exercise-31\"><\/span>RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">The exercise-3.1 in <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\">RD Sharma Solution<\/a> is one of the most basic introductory exercises of chapter 3-Functions. The exercise starts with an introduction of functions, description of functions, further explaining the meaning of domain, co-domain, and the range of functions with the help of some illustrations (examples). Exercise solutions also cover the concept of equal functions.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Go through these basic terminologies. Try to understand them with the help of your teachers or your friends and then move towards the exercise.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">In this exercise, you will encounter questions like these- <\/span><i><span style=\"font-weight: 400;\">A function f: R -&gt; R is defined by f(x) =. Determine the range off.<\/span><\/i><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-11-maths-chapter-3-exercise-32\"><\/span>RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">Though the chapters in RD Sharma Solutions are small and exercises are way too small but it doesn\u2019t mean that you can skip any exercise. Every concept is inter-linked to others. Make sure to have at least basic knowledge about it in case a question appears in the examination.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The exercise-3.2 is very small covering only a small topic of real functions or real-valued functions.<\/span><\/p>\n<p><b><i>Real-Valued Function: <\/i><\/b><i><span style=\"font-weight: 400;\">A function is said to be a real-valued function if a function f: A -&gt; B if B is a subset of R (set of all real numbers).<\/span><\/i><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-11-maths-chapter-3-exercise-33\"><\/span>RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.3<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">The exercise-3.3 is all about finding the domain of real functions. RD Sharma Solutions provide you with a detailed step-by-step for every question you have in the RD Sharma Class 11 Mathematics.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The topic covers all the possible questions related to finding a domain of real functions. Practice questions carefully and take the help of the solutions booklet in case you feel stuck.<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-11-maths-chapter-3-exercise-34\"><\/span>RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">Exercise-3.4 is the last exercise of this chapter. You have covered a large portion and now you can solve questions of exercise-3.4. This exercise a small concept of modulus and the properties of modulus function.<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-3\"><\/span>Access RD Sharma Solutions Class 11 Maths Chapter 3<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"exercise-31-page-no-37\"><\/span>EXERCISE 3.1 PAGE NO: 3.7<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"1-define-a-function-as-a-set-of-ordered-pairs\"><\/span><strong>1. Define a function as a set of ordered pairs.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let A and B be two non-empty sets. A relation from A to B, i.e., a subset of A\u00d7B, is called a function (or a mapping) from A to B, if<\/p>\n<p>(i) for each a \u2208 A there exists b \u2208 B such that (a, b) \u2208 f<\/p>\n<p>(ii) (a, b) \u2208 f and (a, c) \u2208 f \u21d2 b = c<\/p>\n<h3><span class=\"ez-toc-section\" id=\"2-define-a-function-as-a-correspondence-between-two-sets\"><\/span>2. Define a function as a correspondence between two sets.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let A and B be two non-empty sets. Then a function \u2018f\u2019 from set A to B is a rule or method or correspondence which associates elements of set A to elements of set B such that:<\/p>\n<p>(i) all elements of set A are associated to elements in set B.<\/p>\n<p>(ii) an element of set A is associated to a unique element in set B.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"3-what-is-the-fundamental-difference-between-a-relation-and-a-function-is-every-relation-a-function\"><\/span>3. What is the fundamental difference between a relation and a function? Is every relation a function?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let \u2018f\u2019 be a function and R be a relation defined from set X to set Y.<\/p>\n<p>The domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to X. This is because each element of the domain of a function must have an element associated with it, whereas this is not necessary for a relation.<\/p>\n<p>In relation, one element of X might be associated with one or more elements of Y, while it must be associated with only one element of Y in a function.<\/p>\n<p>Thus, not every relation is a function. However, every function is necessarily a relation.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"4-let-a-%e2%80%932-%e2%80%931-0-1-2-and-f-a-%e2%86%92-z-be-a-function-defined-by-fx-x2-%e2%80%93-2x-%e2%80%93-3-find-i-range-of-f-ie-f-a-ii-pre-images-of-6-%e2%80%933-and-5\"><\/span>4. Let A = {\u20132, \u20131, 0, 1, 2} and f: A\u00a0\u2192\u00a0Z be a function defined by f(x) = x<sup>2<\/sup>\u00a0\u2013 2x \u2013 3. Find:<br \/>(i) range of f i.e. f (A)<br \/>(ii) pre-images of 6, \u20133 and 5<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>A = {\u20132, \u20131, 0, 1, 2}<\/p>\n<p>f : A\u00a0\u2192\u00a0Z such that f(x) = x<sup>2<\/sup>\u00a0\u2013 2x \u2013 3<\/p>\n<p><strong>(i)<\/strong>\u00a0Range of f i.e. f (A)<\/p>\n<p>A is the domain of the function f. Hence, range is the set of elements f(x) for all x\u00a0\u2208\u00a0A.<\/p>\n<p>Substituting x = \u20132 in f(x), we get<\/p>\n<p>f(\u20132) = (\u20132)<sup>2<\/sup>\u00a0\u2013 2(\u20132) \u2013 3<\/p>\n<p>= 4 + 4 \u2013 3<\/p>\n<p>= 5<\/p>\n<p>Substituting x = \u20131 in f(x), we get<\/p>\n<p>f(\u20131) = (\u20131)<sup>2<\/sup>\u00a0\u2013 2(\u20131) \u2013 3<\/p>\n<p>= 1 + 2 \u2013 3<\/p>\n<p>= 0<\/p>\n<p>Substituting x = 0 in f(x), we get<\/p>\n<p>f(0) = (0)<sup>2<\/sup>\u00a0\u2013 2(0) \u2013 3<\/p>\n<p>= 0 \u2013 0 \u2013 3<\/p>\n<p>= \u2013 3<\/p>\n<p>Substituting x = 1 in f(x), we get<\/p>\n<p>f(1) = 1<sup>2<\/sup>\u00a0\u2013 2(1) \u2013 3<\/p>\n<p>= 1 \u2013 2 \u2013 3<\/p>\n<p>= \u2013 4<\/p>\n<p>Substituting x = 2 in f(x), we get<\/p>\n<p>f(2) = 2<sup>2<\/sup>\u00a0\u2013 2(2) \u2013 3<\/p>\n<p>= 4 \u2013 4 \u2013 3<\/p>\n<p>= \u20133<\/p>\n<p>Thus, the range of f is {-4, -3, 0, 5}.<\/p>\n<p><strong>(ii)<\/strong>\u00a0pre-images of 6, \u20133 and 5<\/p>\n<p>Let x be the pre-image of 6\u00a0\u21d2\u00a0f(x) = 6<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2x \u2013 3 = 6<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2x \u2013 9 = 0<\/p>\n<p>x = [-(-2) \u00b1\u00a0<strong>\u221a<\/strong>\u00a0((-2)<sup>2<\/sup>\u00a0\u2013 4(1) (-9))] \/ 2(1)<\/p>\n<p>= [2 \u00b1\u00a0<strong>\u221a<\/strong>\u00a0(4+36)] \/ 2<\/p>\n<p>= [2 \u00b1\u00a0<strong>\u221a<\/strong>40] \/ 2<\/p>\n<p>= 1 \u00b1\u00a0<strong>\u221a<\/strong>10<\/p>\n<p>However,\u00a01 \u00b1\u00a0<strong>\u221a<\/strong>10 \u2209 A<\/p>\n<p>Thus, there exists no pre-image of 6.<\/p>\n<p>Now, let x be the pre-image of \u20133\u00a0\u21d2\u00a0f(x) = \u20133<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2x \u2013 3 = \u20133<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2x = 0<\/p>\n<p>x(x \u2013 2) = 0<\/p>\n<p>x = 0 or 2<\/p>\n<p>Clearly, both 0 and 2 are elements of A.<\/p>\n<p>Thus, 0 and 2 are the pre-images of \u20133.<\/p>\n<p>Now, let x be the pre-image of 5\u00a0\u21d2\u00a0f(x) = 5<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2x \u2013 3 = 5<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2x \u2013 8= 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 4x + 2x \u2013 8= 0<\/p>\n<p>x(x \u2013 4) + 2(x \u2013 4) = 0<\/p>\n<p>(x + 2)(x \u2013 4) = 0<\/p>\n<p>x = \u20132 or 4<\/p>\n<p>However, 4\u00a0\u2209\u00a0A but\u00a0\u20132\u00a0\u2208\u00a0A<\/p>\n<p>Thus, \u20132 is the pre-images of 5.<\/p>\n<p>\u2234 \u00d8, {0, 2}, -2 are the pre-images of 6, -3, 5<\/p>\n<h3><span class=\"ez-toc-section\" id=\"5-if-a-function-f-r-%e2%86%92-r-be-defined-by\"><\/span>5. If a function f: R\u00a0\u2192\u00a0R be defined by<strong><br \/><\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-1-Ex-1.1-2.pdf\"><img class=\"alignnone size-full wp-image-117814\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-11-maths-chapter-3.png\" alt=\"\" width=\"168\" height=\"80\"><\/a><\/p>\n<p><strong>Find: f (1), f (\u20131), f (0), f (2).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Let us find f (1), f (\u20131), f (0) and f (2).<\/p>\n<p>When x &gt; 0, f (x) = 4x + 1<\/p>\n<p>Substituting x = 1 in the above equation, we get<\/p>\n<p>f (1) = 4(1) + 1<\/p>\n<p>= 4 + 1<\/p>\n<p>= 5<\/p>\n<p>When x &lt; 0, f(x) = 3x \u2013 2<\/p>\n<p>Substituting x = \u20131 in the above equation, we get<\/p>\n<p>f (\u20131) = 3(\u20131) \u2013 2<\/p>\n<p>= \u20133 \u2013 2<\/p>\n<p>= \u20135<\/p>\n<p>When x = 0, f(x) = 1<\/p>\n<p>Substituting x = 0 in the above equation, we get<\/p>\n<p>f (0) = 1<\/p>\n<p>When x &gt; 0, f(x) = 4x + 1<\/p>\n<p>Substituting x = 2 in the above equation, we get<\/p>\n<p>f (2) = 4(2) + 1<\/p>\n<p>= 8 + 1<\/p>\n<p>= 9<\/p>\n<p>\u2234 f (1) = 5, f (\u20131) = \u20135, f (0) = 1 and f (2) = 9.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"6-a-function-f-r-%e2%86%92-r-is-defined-by-fx-x2-determine-i-range-of-f-ii-x-fx-4-iii-y-fy-%e2%80%931\"><\/span>6. A function f: R\u00a0\u2192\u00a0R is defined by f(x) = x<sup>2<\/sup>. Determine<br \/>(i) range of f<br \/>(ii) {x: f(x) = 4}<br \/>(iii) {y: f(y) = \u20131}<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>f : R\u00a0\u2192\u00a0R and f(x) = x<sup>2<\/sup>.<\/p>\n<p><strong>(i)<\/strong>\u00a0range of f<\/p>\n<p>Domain of f = R (set of real numbers)<\/p>\n<p>We know that the square of a real number is always positive or equal to zero.<\/p>\n<p>\u2234 range of f = R<sup>+<\/sup>\u222a\u00a0{0}<\/p>\n<p><strong>(ii)<\/strong>\u00a0{x: f(x) = 4}<\/p>\n<p>Given:<\/p>\n<p>f(x) = 4<\/p>\n<p>we know, x<sup>2<\/sup>\u00a0= 4<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 4 = 0<\/p>\n<p>(x \u2013 2)(x + 2) = 0<\/p>\n<p>\u2234\u00a0x = \u00b1 2<\/p>\n<p>\u2234 {x: f(x) = 4} = {\u20132, 2}<\/p>\n<p><strong>(iii)<\/strong>\u00a0{y: f(y) = \u20131}<\/p>\n<p>Given:<\/p>\n<p>f(y) = \u20131<\/p>\n<p>y<sup>2<\/sup>\u00a0= \u20131<\/p>\n<p>However, the domain of f is R, and for every real number y, the value of y<sup>2<\/sup>\u00a0is non-negative.<\/p>\n<p>Hence, there exists no real y for which y<sup>2<\/sup>\u00a0= \u20131.<\/p>\n<p>\u2234{y: f(y) = \u20131} =\u00a0\u2205<\/p>\n<h3><span class=\"ez-toc-section\" id=\"7-let-f-r%e2%86%92-r-where-r-is-the-set-of-all-positive-real-numbers-be-such-that-fx-loge-x-determine-i-the-image-set-of-the-domain-of-f-ii-x-f-x-%e2%80%932-iii-whether-f-xy-f-x-f-y-holds\"><\/span>7.\u00a0Let f: R<sup>+<\/sup>\u2192\u00a0R, where R<sup>+<\/sup>\u00a0is the set of all positive real numbers, be such that f(x) = log<sub>e\u00a0<\/sub>x. Determine<br \/>(i) the image set of the domain of f<br \/>(ii) {x: f (x) = \u20132}<br \/>(iii) whether f (xy) = f (x) + f (y) holds.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f: R<sup>+<\/sup>\u2192\u00a0R and f(x) = log<sub>e\u00a0<\/sub>x.<\/p>\n<p><strong>(i)<\/strong>\u00a0the image set of the domain of f<\/p>\n<p>Domain of f = R<sup>+<\/sup>\u00a0(set of positive real numbers)<\/p>\n<p>We know the value of logarithm to the base e (natural logarithm) can take all possible real values.<\/p>\n<p>\u2234 The image set of f = R<\/p>\n<p><strong>(ii)<\/strong>\u00a0{x: f(x) = \u20132}<\/p>\n<p>Given f(x) = \u20132<\/p>\n<p>log<sub>e\u00a0<\/sub>x = \u20132<\/p>\n<p>\u2234\u00a0x = e<sup>-2<\/sup>\u00a0[since,\u00a0log<sub>b\u00a0<\/sub>a = c\u00a0\u21d2\u00a0a = b<sup>c<\/sup>]<\/p>\n<p>\u2234 {x: f(x) = \u20132} = {e<sup>\u20132<\/sup>}<\/p>\n<p><strong>(iii)<\/strong>\u00a0Whether f (xy) = f (x) + f (y) holds.<\/p>\n<p>We have f (x) = log<sub>e\u00a0<\/sub>x\u00a0\u21d2\u00a0f (y) = log<sub>e\u00a0<\/sub>y<\/p>\n<p>Now, let us consider f (xy)<\/p>\n<p>F (xy) = log<sub>e\u00a0<\/sub>(xy)<\/p>\n<p>f (xy) = log<sub>e\u00a0<\/sub>(x \u00d7 y) [since,\u00a0log<sub>b\u00a0<\/sub>(a\u00d7c) = log<sub>b\u00a0<\/sub>a + log<sub>b\u00a0<\/sub>c]<\/p>\n<p>f (xy) = log<sub>e\u00a0<\/sub>x + log<sub>e\u00a0<\/sub>y<\/p>\n<p>f (xy) = f (x) + f (y)<\/p>\n<p>\u2234 the equation f (xy) = f (x) + f (y) holds.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"8-write-the-following-relations-as-sets-of-ordered-pairs-and-find-which-of-them-are-functions-i-x-y-y-3x-x-%e2%88%88-1-2-3-y-%e2%88%88-3-6-9-12-ii-x-y-y-%3e-x-1-x-1-2-and-y-2-4-6-iii-x-y-x-y-3-x-y-%e2%88%88-0-1-2-3\"><\/span>8. Write the following relations as sets of ordered pairs and find which of them are functions:<br \/>(i) {(x, y): y = 3x, x\u00a0\u2208\u00a0{1, 2, 3}, y\u00a0\u2208\u00a0{3, 6, 9, 12}}<br \/>(ii) {(x, y): y &gt; x + 1, x = 1, 2 and y = 2, 4, 6}<br \/>(iii) {(x, y): x + y = 3, x, y\u00a0\u2208\u00a0{0, 1, 2, 3}}<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a0{(x, y): y = 3x, x\u00a0\u2208\u00a0{1, 2, 3}, y\u00a0\u2208\u00a0{3, 6, 9, 12}}<\/p>\n<p>When x = 1, y = 3(1) = 3<\/p>\n<p>When x = 2, y = 3(2) = 6<\/p>\n<p>When x = 3, y = 3(3) = 9<\/p>\n<p>\u2234 R = {(1, 3), (2, 6), (3, 9)}<\/p>\n<p>Hence, the given relation R is a function.<\/p>\n<p><strong>(ii)<\/strong>\u00a0{(x, y): y &gt; x + 1, x = 1, 2 and y = 2, 4, 6}<\/p>\n<p>When x = 1, y &gt; 1 + 1 or y &gt; 2\u00a0\u21d2\u00a0y = {4, 6}<\/p>\n<p>When x = 2, y &gt; 2 + 1 or y &gt; 3\u00a0\u21d2\u00a0y = {4, 6}<\/p>\n<p>\u2234 R = {(1, 4), (1, 6), (2, 4), (2, 6)}<\/p>\n<p>Hence, the given relation R is not a function.<\/p>\n<p><strong>(iii)<\/strong>\u00a0{(x, y): x + y = 3, x, y\u00a0\u2208\u00a0{0, 1, 2, 3}}<\/p>\n<p>When x = 0, 0 + y = 3\u00a0\u21d2\u00a0y = 3<\/p>\n<p>When x = 1, 1 + y = 3\u00a0\u21d2\u00a0y = 2<\/p>\n<p>When x = 2, 2 + y = 3\u00a0\u21d2\u00a0y = 1<\/p>\n<p>When x = 3, 3 + y = 3\u00a0\u21d2\u00a0y = 0<\/p>\n<p>\u2234 R = {(0, 3), (1, 2), (2, 1), (3, 0)}<\/p>\n<p>Hence, the given relation R is a function.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"9-let-f-r-%e2%86%92-r-and-g-c-%e2%86%92-c-be-two-functions-defined-as-fx-x2-and-gx-x2-are-they-equal-functions\"><\/span>9. Let f: R\u00a0\u2192\u00a0R and g: C\u00a0\u2192\u00a0C be two functions defined as f(x) = x<sup>2<\/sup>\u00a0and g(x) = x<sup>2<\/sup>. Are they equal functions?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>f: R\u00a0\u2192\u00a0R\u00a0\u2208\u00a0f(x) = x<sup>2<\/sup>\u00a0and g : R\u00a0\u2192\u00a0R\u00a0\u2208\u00a0g(x) = x<sup>2<\/sup><\/p>\n<p>f is defined from R to R, the domain of f = R.<\/p>\n<p>g is defined from C to C, the domain of g = C.<\/p>\n<p>Two functions are equal only when the domain and codomain of both the functions are equal.<\/p>\n<p>In this case, the domain of f \u2260 domain of g.<\/p>\n<p>\u2234 f and g are not equal functions.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"exercise-32-page-no-311\"><\/span>EXERCISE 3.2 PAGE NO: 3.11<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"1-if-f-x-x2-%e2%80%93-3x-4-then-find-the-values-of-x-satisfying-the-equation-f-x-f-2x-1\"><\/span>1. If f (x) = x<sup>2<\/sup>\u00a0\u2013 3x + 4, then find the values of x satisfying the equation f (x) = f (2x + 1).<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>f(x) = x<sup>2<\/sup>\u00a0\u2013 3x + 4.<\/p>\n<p>Let us find x satisfying f (x) = f (2x + 1).<\/p>\n<p>We have,<\/p>\n<p>f (2x + 1) = (2x + 1)<sup>2<\/sup>\u00a0\u2013 3(2x + 1) + 4<\/p>\n<p>= (2x)<sup>\u00a02<\/sup>\u00a0+ 2(2x) (1) + 1<sup>2<\/sup>\u00a0\u2013 6x \u2013 3 + 4<\/p>\n<p>= 4x<sup>2<\/sup>\u00a0+ 4x + 1 \u2013 6x + 1<\/p>\n<p>= 4x<sup>2<\/sup>\u00a0\u2013 2x + 2<\/p>\n<p>Now, f (x) = f (2x + 1)<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 3x + 4 = 4x<sup>2<\/sup>\u00a0\u2013 2x + 2<\/p>\n<p>4x<sup>2<\/sup>\u00a0\u2013 2x + 2 \u2013 x<sup>2<\/sup>\u00a0+ 3x \u2013 4 = 0<\/p>\n<p>3x<sup>2<\/sup>\u00a0+ x \u2013 2 = 0<\/p>\n<p>3x<sup>2<\/sup>\u00a0+ 3x \u2013 2x \u2013 2 = 0<\/p>\n<p>3x(x + 1) \u2013 2(x + 1) = 0<\/p>\n<p>(x + 1)(3x \u2013 2) = 0<\/p>\n<p>x + 1 = 0 or 3x \u2013 2 = 0<\/p>\n<p>x = \u20131 or 3x = 2<\/p>\n<p>x = \u20131 or\u00a02\/3<\/p>\n<p>\u2234 The values of x are \u20131 and 2\/3.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"2-if-f-x-x-%e2%80%93-a2-x-%e2%80%93-b2-find-f-a-b\"><\/span>2. If f (x) = (x \u2013 a)<sup>2\u00a0<\/sup>(x \u2013 b)<sup>2<\/sup>, find f (a + b).<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>F (x) = (x \u2013 a)<sup>2<\/sup>(x \u2013 b)<sup>2<\/sup><\/p>\n<p>Let us find f (a + b).<\/p>\n<p>We have,<\/p>\n<p>f (a + b) = (a + b \u2013 a)<sup>2\u00a0<\/sup>(a + b \u2013 b)<sup>2<\/sup><\/p>\n<p>f (a + b) = (b)<sup>2<\/sup>\u00a0(a)<sup>2<\/sup><\/p>\n<p>\u2234\u00a0f (a + b) = a<sup>2<\/sup>b<sup>2<\/sup><\/p>\n<h3><span class=\"ez-toc-section\" id=\"3-if-y-f-x-ax-%e2%80%93-b-bx-%e2%80%93-a-show-that-x-f-y\"><\/span>3. If y = f (x) = (ax \u2013 b) \/ (bx \u2013 a), show that x = f (y).<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>y = f (x) = (ax \u2013 b) \/ (bx \u2013 a) \u21d2 f (y) = (ay \u2013 b) \/ (by \u2013 a)<\/p>\n<p>Let us prove that x = f (y).<\/p>\n<p>We have,<\/p>\n<p>y = (ax \u2013 b) \/ (bx \u2013 a)<\/p>\n<p>By cross-multiplying,<\/p>\n<p>y(bx \u2013 a) = ax \u2013 b<\/p>\n<p>bxy \u2013 ay = ax \u2013 b<\/p>\n<p>bxy \u2013 ax = ay \u2013 b<\/p>\n<p>x(by \u2013 a) = ay \u2013 b<\/p>\n<p>x = (ay \u2013 b) \/ (by \u2013 a) = f (y)<\/p>\n<p>\u2234\u00a0x = f (y)<\/p>\n<p>Hence proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"4-if-f-x-1-1-%e2%80%93-x-show-that-f-f-f-x-x\"><\/span>4. If f (x) = 1 \/ (1 \u2013 x), show that f [f {f (x)}] = x.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>f (x) = 1 \/ (1 \u2013 x)<\/p>\n<p>Let us prove that f [f {f (x)}] = x.<\/p>\n<p>Firstly, let us solve for f {f (x)}.<\/p>\n<p>f {f (x)} = f {1\/(1 \u2013 x)}<\/p>\n<p>= 1 \/ 1 \u2013 (1\/(1 \u2013 x))<\/p>\n<p>= 1 \/ [(1 \u2013 x \u2013 1)\/(1 \u2013 x)]<\/p>\n<p>= 1 \/ (-x\/(1 \u2013 x))<\/p>\n<p>= (1 \u2013 x) \/ -x<\/p>\n<p>= (x \u2013 1) \/ x<\/p>\n<p>\u2234 f {f (x)} = (x \u2013 1) \/ x<\/p>\n<p>Now, we shall solve for f [f {f (x)}]<\/p>\n<p>f [f {f (x)}] = f [(x-1)\/x]<\/p>\n<p>= 1 \/ [1 \u2013 (x-1)\/x]<\/p>\n<p>= 1 \/ [(x \u2013 (x-1))\/x]<\/p>\n<p>= 1 \/ [(x \u2013 x + 1)\/x]<\/p>\n<p>= 1 \/ (1\/x)<\/p>\n<p>\u2234\u00a0f [f {f (x)}] = x<\/p>\n<p>Hence proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"5-if-f-x-x-1-x-%e2%80%93-1-show-that-f-f-x-x\"><\/span>5. If f (x) = (x + 1) \/ (x \u2013 1), show that f [f (x)] = x.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>f (x) = (x + 1) \/ (x \u2013 1)<\/p>\n<p>Let us prove that f [f (x)] = x.<\/p>\n<p>f [f (x)] = f [(x+1)\/(x-1)]<\/p>\n<p>= [(x+1)\/(x-1) + 1] \/ [(x+1)\/(x-1) \u2013 1]<\/p>\n<p>= [[(x+1) + (x-1)]\/(x-1)] \/ [[(x+1) \u2013 (x-1)]\/(x-1)]<\/p>\n<p>= [(x+1) + (x-1)] \/ [(x+1) \u2013 (x-1)]<\/p>\n<p>= (x+1+x-1)\/(x+1-x+1)<\/p>\n<p>= 2x\/2<\/p>\n<p>= x<\/p>\n<p>\u2234\u00a0f [f (x)] = x<\/p>\n<p>Hence proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"6-if\"><\/span>6. If<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-3-1.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 2\" \/><\/strong><\/p>\n<h3><span class=\"ez-toc-section\" id=\"find\"><\/span>Find:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"i-f-12\"><\/span>(i) f (1\/2)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"ii-f-2\"><\/span>(ii) f (-2)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iii-f-1\"><\/span>(iii) f (1)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iv-f-%e2%88%9a3\"><\/span>(iv) f (\u221a3)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"v-f-%e2%88%9a-3\"><\/span>(v) f (\u221a-3)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a0f (1\/2)<\/p>\n<p>When, 0 \u2264 x \u2264 1, f(x) = x<\/p>\n<p>\u2234 f (1\/2) = \u00bd<\/p>\n<p><strong>(ii)<\/strong>\u00a0f (-2)<\/p>\n<p>When, x &lt; 0, f(x) = x<sup>2<\/sup><\/p>\n<p>f (\u20132) = (\u20132)<sup>2<\/sup><\/p>\n<p>= 4<\/p>\n<p>\u2234\u00a0f (\u20132) = 4<\/p>\n<p><strong>(iii)<\/strong>\u00a0f (1)<\/p>\n<p>When, x \u2265 1, f (x) = 1\/x<\/p>\n<p>f (1) = 1\/1<\/p>\n<p>\u2234\u00a0f(1) = 1<\/p>\n<p><strong>(iv)<\/strong>\u00a0f (\u221a3)<\/p>\n<p>We have \u221a3 = 1.732 &gt; 1<\/p>\n<p>When, x \u2265 1, f (x) = 1\/x<\/p>\n<p>\u2234 f (\u221a3) = 1\/\u221a3<\/p>\n<p><strong>(v)<\/strong>\u00a0f (\u221a-3)<\/p>\n<p>We know\u00a0\u221a-3\u00a0is not a real number and the function f(x) is defined only when x\u00a0\u2208\u00a0R.<\/p>\n<p>\u2234\u00a0f (\u221a-3)\u00a0does not exist.<\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-33-page-no-318\"><\/span>EXERCISE 3.3 PAGE NO: 3.18<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"1-find-the-domain-of-each-of-the-following-real-valued-functions-of-real-variable\"><\/span>1. Find the domain of each of the following real valued functions of real variable:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"i-f-x-1x\"><\/span>(i) f (x) = 1\/x<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"ii-f-x-1x-7\"><\/span>(ii) f (x) = 1\/(x-7)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iii-f-x-3x-2x1\"><\/span>(iii) f (x) = (3x-2)\/(x+1)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iv-f-x-2x1x2-9\"><\/span>(iv) f (x) = (2x+1)\/(x<sup>2<\/sup>-9)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"v-f-x-x22x1x2-8x12\"><\/span>(v) f (x) = (x<sup>2<\/sup>+2x+1)\/(x<sup>2<\/sup>-8x+12)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>f (x) = 1\/x<\/p>\n<p>We know, f (x) is defined for all real values of x, except for the case when x = 0.<\/p>\n<p>\u2234 Domain of f = R \u2013 {0}<\/p>\n<p><strong>(ii)<\/strong>\u00a0f (x) = 1\/(x-7)<\/p>\n<p>We know, f (x) is defined for all real values of x, except for the case when x \u2013 7 = 0 or x = 7.<\/p>\n<p>\u2234 Domain of f = R \u2013 {7}<\/p>\n<p><strong>(iii)<\/strong>\u00a0f (x) = (3x-2)\/(x+1)<\/p>\n<p>We know, f(x) is defined for all real values of x, except for the case when x + 1 = 0 or x = \u20131.<\/p>\n<p>\u2234 Domain of f = R \u2013 {\u20131}<\/p>\n<p><strong>(iv)\u00a0<\/strong>f (x) = (2x+1)\/(x<sup>2<\/sup>-9)<\/p>\n<p>We know, f (x) is defined for all real values of x, except for the case when x<sup>2<\/sup>\u00a0\u2013 9 = 0.<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 9 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 3<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(x + 3)(x \u2013 3) = 0<\/p>\n<p>x + 3 = 0 or x \u2013 3 = 0<\/p>\n<p>x = \u00b1 3<\/p>\n<p>\u2234 Domain of f = R \u2013 {\u20133, 3}<\/p>\n<p><strong>(v)<\/strong>\u00a0f (x) = (x<sup>2<\/sup>+2x+1)\/(x<sup>2<\/sup>-8x+12)<\/p>\n<p>We know, f(x) is defined for all real values of x, except for the case when x<sup>2<\/sup>\u00a0\u2013 8x + 12 = 0.<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 8x + 12 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2x \u2013 6x + 12 = 0<\/p>\n<p>x(x \u2013 2) \u2013 6(x \u2013 2) = 0<\/p>\n<p>(x \u2013 2)(x \u2013 6) = 0<\/p>\n<p>x \u2013 2 = 0 or x \u2013 6 = 0<\/p>\n<p>x = 2 or 6<\/p>\n<p>\u2234 Domain of f = R \u2013 {2, 6}<\/p>\n<h3><span class=\"ez-toc-section\" id=\"2-find-the-domain-of-each-of-the-following-real-valued-functions-of-real-variable\"><\/span>2. Find the domain of each of the following real valued functions of real variable:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"i-f-x-%e2%88%9ax-2\"><\/span>(i) f (x) = \u221a(x-2)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"ii-f-x-1%e2%88%9ax2-1\"><\/span>(ii) f (x) = 1\/(\u221a(x<sup>2<\/sup>-1))<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iii-f-x-%e2%88%9a9-x2\"><\/span>(iii) f (x) = \u221a(9-x<sup>2<\/sup>)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iv-f-x-%e2%88%9ax-23-x\"><\/span>(iv) f (x) = \u221a(x-2)\/(3-x)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>f (x) = \u221a(x-2)<\/p>\n<p>We know the square of a real number is never negative.<\/p>\n<p>f (x) takes real values only when x \u2013 2 \u2265 0<\/p>\n<p>x \u2265 2<\/p>\n<p>\u2234\u00a0x\u00a0\u2208\u00a0[2, \u221e)<\/p>\n<p>\u2234 Domain (f) = [2, \u221e)<\/p>\n<p><strong>(ii)<\/strong>\u00a0f (x) = 1\/(\u221a(x<sup>2<\/sup>-1))<\/p>\n<p>We know the square of a real number is never negative.<\/p>\n<p>f (x) takes real values only when x<sup>2<\/sup>\u00a0\u2013 1 \u2265 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 1<sup>2<\/sup>\u00a0\u2265 0<\/p>\n<p>(x + 1) (x \u2013 1) \u2265 0<\/p>\n<p>x \u2264 \u20131 or x \u2265 1<\/p>\n<p>\u2234\u00a0x\u00a0\u2208\u00a0(\u2013\u221e, \u20131]\u00a0\u222a\u00a0[1, \u221e)<\/p>\n<p>In addition, f (x) is also undefined when x<sup>2<\/sup>\u00a0\u2013 1 = 0 because denominator will be zero and the result will be indeterminate.<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 1 = 0\u00a0\u21d2\u00a0x = \u00b1 1<\/p>\n<p>So, x\u00a0\u2208\u00a0(\u2013\u221e, \u20131]\u00a0\u222a\u00a0[1, \u221e) \u2013 {\u20131, 1}<\/p>\n<p>x\u00a0\u2208\u00a0(\u2013\u221e, \u20131)\u00a0\u222a\u00a0(1, \u221e)<\/p>\n<p>\u2234 Domain (f) = (\u2013\u221e, \u20131)\u00a0\u222a\u00a0(1, \u221e)<\/p>\n<p><strong>(iii)<\/strong>\u00a0f (x) = \u221a(9-x<sup>2<\/sup>)<\/p>\n<p>We know the square of a real number is never negative.<\/p>\n<p>f (x) takes real values only when 9 \u2013 x<sup>2<\/sup>\u00a0\u2265 0<\/p>\n<p>9 \u2265 x<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0\u2264 9<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 9 \u2264 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 3<sup>2<\/sup>\u00a0\u2264 0<\/p>\n<p>(x + 3)(x \u2013 3) \u2264 0<\/p>\n<p>x \u2265 \u20133 and x \u2264 3<\/p>\n<p>x\u00a0\u2208\u00a0[\u20133, 3]<\/p>\n<p>\u2234 Domain (f) = [\u20133, 3]<\/p>\n<p><strong>(iv)<\/strong>\u00a0f (x) = \u221a(x-2)\/(3-x)<\/p>\n<p>We know the square root of a real number is never negative.<\/p>\n<p>f (x) takes real values only when x \u2013 2 and 3 \u2013 x are both positive and negative.<\/p>\n<p><strong>(a)<\/strong>\u00a0Both x \u2013 2 and 3 \u2013 x are positive<\/p>\n<p>x \u2013 2 \u2265 0<\/p>\n<p>x \u2265 2<\/p>\n<p>3 \u2013 x \u2265 0<\/p>\n<p>x \u2264 3<\/p>\n<p>Hence, x \u2265 2 and x \u2264 3<\/p>\n<p>\u2234\u00a0x\u00a0\u2208\u00a0[2, 3]<\/p>\n<p><strong>(b)<\/strong>\u00a0Both x \u2013 2 and 3 \u2013 x are negative<\/p>\n<p>x \u2013 2 \u2264 0<\/p>\n<p>x \u2264 2<\/p>\n<p>3 \u2013 x \u2264 0<\/p>\n<p>x \u2265 3<\/p>\n<p>Hence, x \u2264 2 and x \u2265 3<\/p>\n<p>However, the intersection of these sets is null set. Thus, this case is not possible.<\/p>\n<p>Hence, x\u00a0\u2208\u00a0[2, 3] \u2013 {3}<\/p>\n<p>x\u00a0\u2208\u00a0[2, 3]<\/p>\n<p>\u2234 Domain (f) = [2, 3]<\/p>\n<h3><span class=\"ez-toc-section\" id=\"3-find-the-domain-and-range-of-each-of-the-following-real-valued-functions\"><\/span>3. Find the domain and range of each of the following real valued functions:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"i-f-x-axbbx-a\"><\/span>(i) f (x) = (ax+b)\/(bx-a)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"ii-f-x-ax-bcx-d\"><\/span>(ii) f (x) = (ax-b)\/(cx-d)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iii-f-x-%e2%88%9ax-1\"><\/span>(iii) f (x) = \u221a(x-1)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iv-f-x-%e2%88%9ax-3\"><\/span>(iv) f (x) = \u221a(x-3)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"v-f-x-x-22-x\"><\/span>(v) f (x) = (x-2)\/(2-x)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"vi-f-x-x-1\"><\/span>(vi) f (x) = |x-1|<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"vii-f-x-x\"><\/span>(vii) f (x) = -|x|<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"viii-f-x-%e2%88%9a9-x2\"><\/span>(viii) f (x) = \u221a(9-x<sup>2<\/sup>)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>f (x) = (ax+b)\/(bx-a)<\/p>\n<p>f(x) is defined for all real values of x, except for the case when bx \u2013 a = 0 or x = a\/b.<\/p>\n<p>Domain (f) = R \u2013\u00a0(a\/b)<\/p>\n<p>Let f (x) = y<\/p>\n<p>(ax+b)\/(bx-a) = y<\/p>\n<p>ax + b = y(bx\u00a0\u2013 a)<\/p>\n<p>ax + b = bxy\u00a0\u2013 ay<\/p>\n<p>ax \u2013 bxy =\u00a0\u2013ay \u2013 b<\/p>\n<p>x(a \u2013 by) =\u00a0\u2013(ay + b)<\/p>\n<p>\u2234 x = \u2013 (ay+b)\/(a-by)<\/p>\n<p>When a \u2013 by = 0 or y = a\/b<\/p>\n<p>Hence, f(x) cannot take the value a\/b.<\/p>\n<p>\u2234 Range (f) = R \u2013 (a\/b)<\/p>\n<p><strong>(ii)\u00a0<\/strong>f (x) = (ax-b)\/(cx-d)<\/p>\n<p>f(x) is defined for all real values of x, except for the case when cx \u2013 d = 0 or x = d\/c. Domain (f) = R \u2013 (d\/c)<\/p>\n<p>Let f (x) = y<\/p>\n<p>(ax-b)\/(cx-d) = y<\/p>\n<p>ax \u2013 b = y(cx \u2013 d)<\/p>\n<p>ax \u2013 b = cxy \u2013 dy<\/p>\n<p>ax \u2013 cxy = b \u2013 dy<\/p>\n<p>x(a \u2013 cy) = b \u2013 dy<\/p>\n<p>\u2234 x = (b-dy)\/(a-cy)<\/p>\n<p>When a \u2013 cy = 0 or y = a\/c,<\/p>\n<p>Hence, f(x) cannot take the value a\/c.<\/p>\n<p>\u2234 Range (f) = R \u2013\u00a0(a\/c)<\/p>\n<p><strong>(iii)<\/strong>\u00a0f (x) = \u221a(x-1)<\/p>\n<p>We know the square of a real number is never negative.<\/p>\n<p>f(x) takes real values only when x \u2013 1 \u2265 0<\/p>\n<p>x \u2265 1<\/p>\n<p>\u2234\u00a0x\u00a0\u2208\u00a0[1, \u221e)<\/p>\n<p>Thus, domain (f) = [1, \u221e)<\/p>\n<p>When x \u2265 1, we have x \u2013 1 \u2265 0<\/p>\n<p>Hence, \u221a(x-1) \u2265 0 \u21d2 f (x) \u2265 0<\/p>\n<p>f(x)\u00a0\u2208\u00a0[0, \u221e)<\/p>\n<p>\u2234 Range (f) = [0, \u221e)<\/p>\n<p><strong>(iv)<\/strong>\u00a0f (x) = \u221a(x-3)<\/p>\n<p>We know the square of a real number is never negative.<\/p>\n<p>f (x) takes real values only when x \u2013 3 \u2265 0<\/p>\n<p>x \u2265 3<\/p>\n<p>\u2234\u00a0x\u00a0\u2208\u00a0[3, \u221e)<\/p>\n<p>Domain (f) = [3, \u221e)<\/p>\n<p>When x \u2265 3, we have x \u2013 3 \u2265 0<\/p>\n<p>Hence, \u221a(x-3) \u2265 0 \u21d2 f (x) \u2265 0<\/p>\n<p>f(x)\u00a0\u2208\u00a0[0, \u221e)<\/p>\n<p>\u2234 Range (f) = [0, \u221e)<\/p>\n<p><strong>(v)<\/strong>\u00a0f (x) = (x-2)\/(2-x)<\/p>\n<p>f(x) is defined for all real values of x, except for the case when 2 \u2013 x = 0 or x = 2.<\/p>\n<p>Domain (f) = R \u2013 {2}<\/p>\n<p>We have, f (x) = (x-2)\/(2-x)<\/p>\n<p>f (x) = -(2-x)\/(2-x)<\/p>\n<p>= \u20131<\/p>\n<p>When x \u2260 2, f(x) = \u20131<\/p>\n<p>\u2234 Range (f) = {\u20131}<\/p>\n<p><strong>(vi)<\/strong>\u00a0f (x) = |x-1|<\/p>\n<p><img class=\"alignnone size-full wp-image-117844\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-11-maths-chapter-3.gif\" alt=\"\" width=\"210\" height=\"45\"><\/p>\n<p><img class=\"alignnone size-full wp-image-117847\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-11-maths-chapter-3-1.gif\" alt=\"\" width=\"240\" height=\"45\"><\/p>\n<p><img class=\"alignnone size-full wp-image-117849\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-11-maths-chapter-3-2.gif\" alt=\"\" width=\"258\" height=\"45\"><\/p>\n<p>Hence, f(x) is defined for all real numbers x.<\/p>\n<p>Domain (f) = R<\/p>\n<p>When, x &lt; 1, we have x \u2013 1 &lt; 0 or 1 \u2013 x &gt; 0.<\/p>\n<p>|x \u2013 1| &gt; 0\u00a0\u21d2\u00a0f(x) &gt; 0<\/p>\n<p>When, x \u2265 1, we have x \u2013 1 \u2265 0.<\/p>\n<p>|x \u2013 1| \u2265 0\u00a0\u21d2\u00a0f(x) \u2265 0<\/p>\n<p>\u2234\u00a0f(x) \u2265 0 or f(x)\u00a0\u2208\u00a0[0, \u221e)<\/p>\n<p>Range (f) = [0, \u221e)<\/p>\n<p><strong>(vii)<\/strong>\u00a0f (x) = -|x|<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-3-3.gif\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 6\" \/><\/p>\n<p>Now we have,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 7\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-3-4.gif\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 7\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 8\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-3-5.gif\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 8\" \/><\/p>\n<p>Hence, f(x) is defined for all real numbers x.<\/p>\n<p>Domain (f) = R<\/p>\n<p>When, x &lt; 0, we have \u2013|x| &lt; 0<\/p>\n<p>f (x) &lt; 0<\/p>\n<p>When, x \u2265 0, we have \u2013x \u2264 0.<\/p>\n<p>\u2013|x| \u2264 0\u00a0\u21d2\u00a0f (x) \u2264 0<\/p>\n<p>\u2234\u00a0f (x) \u2264 0 or f (x)\u00a0\u2208\u00a0(\u2013\u221e, 0]<\/p>\n<p>Range (f) = (\u2013\u221e, 0]<\/p>\n<p><strong>(viii)\u00a0<\/strong>f (x) = \u221a(9-x<sup>2<\/sup>)<\/p>\n<p>We know the square of a real number is never negative.<\/p>\n<p>f(x) takes real values only when 9 \u2013 x<sup>2<\/sup>\u00a0\u2265 0<\/p>\n<p>9 \u2265 x<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0\u2264 9<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 9 \u2264 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 3<sup>2<\/sup>\u00a0\u2264 0<\/p>\n<p>(x + 3)(x \u2013 3) \u2264 0<\/p>\n<p>x \u2265 \u20133 and x \u2264 3<\/p>\n<p>\u2234\u00a0x\u00a0\u2208\u00a0[\u20133, 3]<\/p>\n<p>Domain (f) = [\u20133, 3]<\/p>\n<p>When, x\u00a0\u2208\u00a0[\u20133, 3], we have 0 \u2264 9 \u2013 x<sup>2<\/sup>\u00a0\u2264 9<\/p>\n<p>0 \u2264 \u221a(9-x<sup>2<\/sup>) \u2264 3 \u21d2 0 \u2264 f (x) \u2264 3<\/p>\n<p>\u2234\u00a0f(x)\u00a0\u2208\u00a0[0, 3]<\/p>\n<p>Range (f) = [0, 3]<\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-34-page-no-338\"><\/span>EXERCISE 3.4 PAGE NO: 3.38<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"1-find-f-g-f-%e2%80%93-g-cf-c-%e2%88%88-r-c-%e2%89%a0-0-fg-1f-and-fg-in-each-of-the-following-i-f-x-x3-1-and-g-x-x-1\"><\/span><strong>1. Find f + g, f \u2013 g, cf (c\u00a0\u2208\u00a0R, c \u2260 0), fg, 1\/f and f\/g in each of the following:<br \/>(i) f (x) = x<sup>3<\/sup>\u00a0+ 1 and g (x) = x + 1<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"ii-f-x-%e2%88%9ax-1-and-g-x-%e2%88%9ax1\"><\/span><strong>(ii) f (x) =\u00a0<\/strong>\u221a(x-1) and g (x) = \u221a(x+1)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>f (x) = x<sup>3<\/sup>\u00a0+ 1 and g(x) = x + 1<\/p>\n<p>We have f(x): R\u00a0\u2192\u00a0R and g(x): R\u00a0\u2192\u00a0R<\/p>\n<p>(a) f + g<\/p>\n<p>We know, (f + g) (x) = f(x) + g(x)<\/p>\n<p>(f + g) (x) = x<sup>3<\/sup>\u00a0+ 1 + x + 1<\/p>\n<p>= x<sup>3<\/sup>\u00a0+ x + 2<\/p>\n<p>So, (f + g) (x): R\u00a0\u2192\u00a0R<\/p>\n<p>\u2234 f + g: R\u00a0\u2192\u00a0R is given by (f + g) (x) = x<sup>3<\/sup>\u00a0+ x + 2<\/p>\n<p>(b) f \u2013 g<\/p>\n<p>We know, (f \u2013 g) (x) = f(x) \u2013 g(x)<\/p>\n<p>(f \u2013 g) (x) = x<sup>3<\/sup>\u00a0+ 1 \u2013 (x + 1)<\/p>\n<p>= x<sup>3<\/sup>\u00a0+ 1 \u2013 x \u2013 1<\/p>\n<p>= x<sup>3<\/sup>\u00a0\u2013 x<\/p>\n<p>So, (f \u2013 g) (x): R\u00a0\u2192\u00a0R<\/p>\n<p>\u2234 f \u2013 g: R\u00a0\u2192\u00a0R is given by (f \u2013 g) (x) = x<sup>3<\/sup>\u00a0\u2013 x<\/p>\n<p>\u00a0<\/p>\n<p>(c) cf (c\u00a0\u2208\u00a0R, c \u2260 0)<\/p>\n<p>We know, (cf) (x) = c \u00d7 f(x)<\/p>\n<p>(cf)(x) = c(x<sup>3<\/sup>\u00a0+ 1)<\/p>\n<p>= cx<sup>3<\/sup>\u00a0+ c<\/p>\n<p>So, (cf) (x) : R\u00a0\u2192\u00a0R<\/p>\n<p>\u2234 cf: R\u00a0\u2192\u00a0R is given by (cf) (x) = cx<sup>3<\/sup>\u00a0+ c<\/p>\n<p>(d) fg<\/p>\n<p>We know, (fg) (x) = f(x) g(x)<\/p>\n<p>(fg) (x) = (x<sup>3<\/sup>\u00a0+ 1) (x + 1)<\/p>\n<p>= (x + 1) (x<sup>2<\/sup>\u00a0\u2013 x + 1) (x + 1)<\/p>\n<p>= (x + 1)<sup>2\u00a0<\/sup>(x<sup>2<\/sup>\u00a0\u2013 x + 1)<\/p>\n<p>So, (fg) (x): R\u00a0\u2192\u00a0R<\/p>\n<p>\u2234 fg: R\u00a0\u2192\u00a0R is given by (fg) (x) = (x + 1)<sup>2<\/sup>(x<sup>2<\/sup>\u00a0\u2013 x + 1)<\/p>\n<p>(e)\u00a01\/f<\/p>\n<p>We know, (1\/f) (x) = 1\/f (x)<\/p>\n<p>1\/f (x)\u00a0= 1 \/ (x<sup>3<\/sup>\u00a0+ 1)<\/p>\n<p>Observe that 1\/f(x)\u00a0is undefined when f(x) = 0 or when x = \u2013 1.<\/p>\n<p>So, 1\/f: R \u2013 {\u20131}\u00a0\u2192\u00a0R is given by 1\/f (x) = 1 \/ (x<sup>3<\/sup>\u00a0+ 1)<\/p>\n<p>(f)\u00a0f\/g<\/p>\n<p>We know, (f\/g) (x) = f(x)\/g(x)<\/p>\n<p>(f\/g) (x) = (x<sup>3<\/sup>\u00a0+ 1) \/ (x + 1)<\/p>\n<p>Observe that (x<sup>3<\/sup>\u00a0+ 1) \/ (x + 1)\u00a0is undefined when g(x) = 0 or when x = \u20131.<\/p>\n<p>Using x<sup>3<\/sup>\u00a0+ 1 = (x + 1) (x<sup>2<\/sup>\u00a0\u2013 x + 1), we have<\/p>\n<p>(f\/g) (x) = [(x+1) (x<sup>2<\/sup>\u2013 x+1)\/(x+1)]<\/p>\n<p>= x<sup>2<\/sup>\u00a0\u2013 x + 1<\/p>\n<p>\u2234 f\/g: R \u2013 {\u20131}\u00a0\u2192\u00a0R is given by (f\/g) (x) = x<sup>2<\/sup>\u00a0\u2013 x + 1<\/p>\n<p><strong>(ii)\u00a0<\/strong>f (x) = \u221a(x-1) and g (x) = \u221a(x+1)<\/p>\n<p>We have f(x): [1, \u221e)\u00a0\u2192\u00a0R<sup>+<\/sup>\u00a0and g(x): [\u20131, \u221e)\u00a0\u2192\u00a0R<sup>+<\/sup>\u00a0as real square root is defined only for non-negative numbers.<\/p>\n<p>(a) f + g<\/p>\n<p>We know, (f + g) (x) = f(x) + g(x)<\/p>\n<p>(f+g) (x) = \u221a(x-1) + \u221a(x+1)<\/p>\n<p>Domain of (f + g) = Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>Domain of (f + g) = [1, \u221e)\u00a0\u2229\u00a0[\u20131, \u221e)<\/p>\n<p>Domain of (f + g) = [1, \u221e)<\/p>\n<p>\u2234 f + g: [1, \u221e)\u00a0\u2192\u00a0R is given by (f+g) (x) = \u221a(x-1) + \u221a(x+1)<\/p>\n<p>(b) f \u2013 g<\/p>\n<p>We know, (f \u2013 g) (x) = f(x) \u2013 g(x)<\/p>\n<p>(f-g) (x) = \u221a(x-1) \u2013 \u221a(x+1)<\/p>\n<p>Domain of (f \u2013 g) = Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>Domain of (f \u2013 g) = [1, \u221e)\u00a0\u2229\u00a0[\u20131, \u221e)<\/p>\n<p>Domain of (f \u2013 g) = [1, \u221e)<\/p>\n<p>\u2234 f \u2013 g: [1, \u221e)\u00a0\u2192\u00a0R is given by (f-g) (x) = \u221a(x-1) \u2013 \u221a(x+1)<\/p>\n<p>(c) cf (c\u00a0\u2208\u00a0R, c \u2260 0)<\/p>\n<p>We know, (cf) (x) = c \u00d7 f(x)<\/p>\n<p>(cf) (x) = c\u221a(x-1)<\/p>\n<p>Domain of (cf) = Domain of f<\/p>\n<p>Domain of (cf) = [1, \u221e)<\/p>\n<p>\u2234 cf: [1, \u221e)\u00a0\u2192\u00a0R is given by (cf) (x) = c\u221a(x-1)<\/p>\n<p>(d) fg<\/p>\n<p>We know, (fg) (x) = f(x) g(x)<\/p>\n<p>(fg) (x) = \u221a(x-1) \u221a(x+1)<\/p>\n<p>= \u221a(x<sup>2<\/sup>\u00a0-1)<\/p>\n<p>Domain of (fg) = Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>Domain of (fg) = [1, \u221e)\u00a0\u2229\u00a0[\u20131, \u221e)<\/p>\n<p>Domain of (fg) = [1, \u221e)<\/p>\n<p>\u2234 fg: [1, \u221e)\u00a0\u2192\u00a0R is given by (fg) (x) = \u221a(x<sup>2<\/sup>\u00a0-1)<\/p>\n<p>(e)\u00a01\/f<\/p>\n<p>We know, (1\/f) (x) = 1\/f(x)<\/p>\n<p>(1\/f) (x) = 1\/\u221a(x-1)<\/p>\n<p>Domain of (1\/f) =\u00a0Domain of f<\/p>\n<p>Domain of\u00a0(1\/f)\u00a0= [1, \u221e)<\/p>\n<p>Observe that 1\/\u221a(x-1)\u00a0is also undefined when x \u2013 1 = 0 or x = 1.<\/p>\n<p>\u2234 1\/f: (1, \u221e)\u00a0\u2192\u00a0R is given by (1\/f) (x) = 1\/\u221a(x-1)<\/p>\n<p>(f)\u00a0f\/g<\/p>\n<p>We know, (f\/g) (x) = f(x)\/g(x)<\/p>\n<p>(f\/g) (x) = \u221a(x-1)\/\u221a(x+1)<\/p>\n<p>(f\/g) (x) = \u221a[(x-1)\/(x+1)]<\/p>\n<p>Domain of\u00a0(f\/g)\u00a0= Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>Domain of\u00a0(f\/g)\u00a0= [1, \u221e)\u00a0\u2229\u00a0[\u20131, \u221e)<\/p>\n<p>Domain of\u00a0(f\/g)\u00a0= [1, \u221e)<\/p>\n<p>\u2234\u00a0f\/g: [1, \u221e)\u00a0\u2192\u00a0R is given by (f\/g) (x) = \u221a[(x-1)\/(x+1)]<\/p>\n<h3><span class=\"ez-toc-section\" id=\"2-let-fx-2x-5-and-gx-x2-x-describe\"><\/span>2.\u00a0Let f(x) = 2x + 5 and g(x) = x<sup>2<\/sup>\u00a0+ x. Describe<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"i-f-g-ii-f-%e2%80%93-g-iii-fg-iv-fg-find-the-domain-in-each-case\"><\/span>(i) f + g<br \/>(ii) f \u2013 g<br \/>(iii) fg<br \/>(iv)\u00a0f\/g<br \/>Find the domain in each case.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>f(x) = 2x + 5 and g(x) = x<sup>2<\/sup>\u00a0+ x<\/p>\n<p>Both f(x) and g(x) are defined for all x\u00a0\u2208\u00a0R.<\/p>\n<p>So, domain of f = domain of g = R<\/p>\n<p><strong>(i)<\/strong>\u00a0f + g<\/p>\n<p>We know, (f + g)(x) = f(x) + g(x)<\/p>\n<p>(f + g)(x) = 2x + 5 + x<sup>2<\/sup>\u00a0+ x<\/p>\n<p>= x<sup>2<\/sup>\u00a0+ 3x + 5<\/p>\n<p>(f + g)(x) Is defined for all real numbers x.<\/p>\n<p>\u2234\u00a0The domain of (f + g) is R<\/p>\n<p><strong>(ii)<\/strong>\u00a0f \u2013 g<\/p>\n<p>We know, (f \u2013 g)(x) = f(x) \u2013 g(x)<\/p>\n<p>(f \u2013 g)(x) = 2x + 5 \u2013 (x<sup>2<\/sup>\u00a0+ x)<\/p>\n<p>= 2x + 5 \u2013 x<sup>2<\/sup>\u00a0\u2013 x<\/p>\n<p>= 5 + x \u2013 x<sup>2<\/sup><\/p>\n<p>(f \u2013 g)(x) is defined for all real numbers x.<\/p>\n<p>\u2234\u00a0The domain of (f \u2013 g) is R<\/p>\n<p><strong>(iii)<\/strong>\u00a0fg<\/p>\n<p>We know, (fg)(x) = f(x)g(x)<\/p>\n<p>(fg)(x) = (2x + 5)(x<sup>2<\/sup>\u00a0+ x)<\/p>\n<p>= 2x(x<sup>2<\/sup>\u00a0+ x) + 5(x<sup>2<\/sup>\u00a0+ x)<\/p>\n<p>= 2x<sup>3<\/sup>\u00a0+ 2x<sup>2<\/sup>\u00a0+ 5x<sup>2<\/sup>\u00a0+ 5x<\/p>\n<p>= 2x<sup>3<\/sup>\u00a0+ 7x<sup>2<\/sup>\u00a0+ 5x<\/p>\n<p>(fg)(x) is defined for all real numbers x.<\/p>\n<p>\u2234\u00a0The domain of fg is R<\/p>\n<p><strong>(iv)<\/strong>\u00a0f\/g<\/p>\n<p>We know, (f\/g) (x) = f(x)\/g(x)<\/p>\n<p>(f\/g) (x) = (2x+5)\/(x<sup>2<\/sup>+x)<\/p>\n<p>(f\/g) (x)\u00a0is defined for all real values of x, except for the case when x<sup>2<\/sup>\u00a0+ x = 0.<\/p>\n<p>x<sup>2<\/sup>\u00a0+ x = 0<\/p>\n<p>x(x + 1) = 0<\/p>\n<p>x = 0 or x + 1 = 0<\/p>\n<p>x = 0 or \u20131<\/p>\n<p>When x = 0 or \u20131,\u00a0(f\/g) (x)\u00a0will be undefined as the division result will be indeterminate.<\/p>\n<p>\u2234 The domain of\u00a0f\/g\u00a0= R \u2013 {\u20131, 0}<\/p>\n<h3><span class=\"ez-toc-section\" id=\"3-if-fx-be-defined-on-%e2%80%932-2-and-is-given-by\"><\/span>3.\u00a0If f(x) be defined on [\u20132, 2] and is given by<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img class=\"alignnone size-full wp-image-117862\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-11-maths-chapter-3-6.gif\" alt=\"\" width=\"200\" height=\"45\"> <strong>and g(x) = f(|x|) + |f(x)|. Find g(x).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p><img class=\"alignnone size-full wp-image-117867\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-11-maths-chapter-3-6-1.gif\" alt=\"\" width=\"200\" height=\"45\"> <strong>and g(x) = f(|x|) + |f(x)|. Find g(x).<\/strong><\/p>\n<p>Now we have,<\/p>\n<p><img class=\"alignnone size-full wp-image-117867\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-11-maths-chapter-3-6-1.gif\" alt=\"\" width=\"200\" height=\"45\"><\/p>\n<p>However, |x| \u2265 0 \u21d2 f (|x|) = |x| &#8211; 1 when 0 &lt; |x| \u2264 2<\/p>\n<p>We also have,\u00a0<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-3-2.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 10\" \/><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-3-3.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 11\" \/><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-3-4.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 3 \u2013 Functions image - 12\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"4-let-f-g-be-two-real-functions-defined-by-fx-%e2%88%9ax1-and-gx-%e2%88%9a9-x2-then-describe-each-of-the-following-functions-i-f-g-ii-g-%e2%80%93-f-iii-fg-iv-fg-v-gf-vi-2f-%e2%80%93-%e2%88%9a5g-vii-f2-7f-viii-5g\"><\/span>4. Let f, g be two real functions defined by f(x) = \u221a(x+1) and\u00a0g(x) = \u221a(9-x<sup>2<\/sup>). Then, describe each of the following functions.<br \/>(i) f + g<br \/>(ii) g \u2013 f<br \/>(iii) fg<br \/>(iv) f\/g<br \/>(v) g\/f<br \/>(vi) 2f \u2013 \u221a5g<br \/>(vii) f<sup>2<\/sup>\u00a0+ 7f<strong><br \/><\/strong>(viii) 5\/g\u00a0<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>f(x) = \u221a(x+1) and\u00a0g(x) = \u221a(9-x<sup>2<\/sup>)<\/p>\n<p>We know the square of a real number is never negative.<\/p>\n<p>So, f(x) takes real values only when x + 1 \u2265 0<\/p>\n<p>x \u2265 \u20131, x\u00a0\u2208\u00a0[\u20131, \u221e)<\/p>\n<p>Domain of f = [\u20131, \u221e)<\/p>\n<p>Similarly, g(x) takes real values only when 9 \u2013 x<sup>2<\/sup>\u00a0\u2265 0<\/p>\n<p>9 \u2265 x<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0\u2264 9<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 9 \u2264 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 3<sup>2<\/sup>\u00a0\u2264 0<\/p>\n<p>(x + 3)(x \u2013 3) \u2264 0<\/p>\n<p>x \u2265 \u20133 and x \u2264 3<\/p>\n<p>\u2234\u00a0x\u00a0\u2208\u00a0[\u20133, 3]<\/p>\n<p>Domain of g = [\u20133, 3]<\/p>\n<p><strong>(i)<\/strong>\u00a0f + g<\/p>\n<p>We know, (f + g)(x) = f(x) + g(x)<\/p>\n<p>(f + g) (x) = \u221a(x+1) + \u221a(9-x<sup>2<\/sup>)<\/p>\n<p>Domain of f + g = Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>= [\u20131, \u221e)\u00a0\u2229\u00a0[\u20133, 3]<\/p>\n<p>= [\u20131, 3]<\/p>\n<p>\u2234 f + g: [\u20131, 3]\u00a0\u2192\u00a0R is given by (f + g) (x) = f(x) + g(x) = \u221a(x+1) + \u221a(9-x<sup>2<\/sup>)<\/p>\n<p><strong>(ii)<\/strong>\u00a0g \u2013 f<\/p>\n<p>We know, (g \u2013 f)(x) = g(x) \u2013 f(x)<\/p>\n<p>(g \u2013 f) (x) = \u221a(9-x<sup>2<\/sup>) \u2013 \u221a(x+1)<\/p>\n<p>Domain of g \u2013 f = Domain of g\u00a0\u2229\u00a0Domain of f<\/p>\n<p>= [\u20133, 3] \u2229\u00a0[\u20131, \u221e)<\/p>\n<p>= [\u20131, 3]<\/p>\n<p>\u2234 g \u2013 f: [\u20131, 3]\u00a0\u2192\u00a0R is given by\u00a0(g \u2013 f) (x) = g(x) \u2013 f(x) = \u221a(9-x<sup>2<\/sup>) \u2013 \u221a(x+1)<\/p>\n<p><strong>(iii)<\/strong>\u00a0fg<\/p>\n<p>We know, (fg) (x) = f(x)g(x)<\/p>\n<p>(fg) (x) = \u221a(x+1) \u221a(9-x<sup>2<\/sup>)<\/p>\n<p>= \u221a[(x+1) (9-x<sup>2<\/sup>)]<\/p>\n<p>= \u221a[x(9-x<sup>2<\/sup>) + (9-x<sup>2<\/sup>)]<\/p>\n<p>= \u221a(9x-x<sup>3<\/sup>+9-x<sup>2<\/sup>)<\/p>\n<p>= \u221a(9+9x-x<sup>2<\/sup>-x<sup>3<\/sup>)<\/p>\n<p>Domain of fg = Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>= [\u20131, \u221e)\u00a0\u2229\u00a0[\u20133, 3]<\/p>\n<p>= [\u20131, 3]<\/p>\n<p>\u2234 fg: [\u20131, 3]\u00a0\u2192\u00a0R is given by\u00a0(fg) (x) = f(x) g(x) = \u221a(x+1) \u221a(9-x<sup>2<\/sup>) = \u221a(9+9x-x<sup>2<\/sup>-x<sup>3<\/sup>)<\/p>\n<p><strong>(iv)<\/strong>\u00a0f\/g<\/p>\n<p>We know, (f\/g) (x) = f(x)\/g(x)<\/p>\n<p>(f\/g) (x) = \u221a(x+1) \/ \u221a(9-x<sup>2<\/sup>)<\/p>\n<p>= \u221a[(x+1) \/ (9-x<sup>2<\/sup>)]<\/p>\n<p>Domain of f\/g = Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>= [\u20131, \u221e)\u00a0\u2229\u00a0[\u20133, 3]<\/p>\n<p>= [\u20131, 3]<\/p>\n<p>However, (f\/g) (x)\u00a0is defined for all real values of x\u00a0\u2208\u00a0[\u20131, 3], except for the case when 9 \u2013 x<sup>2<\/sup>\u00a0= 0 or x = \u00b1 3<\/p>\n<p>When x = \u00b13,\u00a0(f\/g) (x)\u00a0will be undefined as the division result will be indeterminate.<\/p>\n<p>Domain of\u00a0f\/g\u00a0= [\u20131, 3] \u2013 {\u20133, 3}<\/p>\n<p>Domain of\u00a0f\/g\u00a0= [\u20131, 3)<\/p>\n<p>\u2234 f\/g: [\u20131, 3)\u00a0\u2192\u00a0R is given by\u00a0(f\/g) (x) = f(x)\/g(x)\u00a0= \u221a(x+1) \/ \u221a(9-x<sup>2<\/sup>)<\/p>\n<p><strong>(v)<\/strong>\u00a0g\/f<\/p>\n<p>We know, (g\/f) (x) = g(x)\/f(x)<\/p>\n<p>(g\/f) (x) = \u221a(9-x<sup>2<\/sup>) \/ \u221a(x+1)<\/p>\n<p>= \u221a[(9-x<sup>2<\/sup>) \/ (x+1)]<\/p>\n<p>Domain of g\/f = Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>= [\u20131, \u221e)\u00a0\u2229\u00a0[\u20133, 3]<\/p>\n<p>= [\u20131, 3]<\/p>\n<p>However, (g\/f) (x)\u00a0is defined for all real values of x\u00a0\u2208\u00a0[\u20131, 3], except for the case when x + 1 = 0 or x = \u20131<\/p>\n<p>When x = \u20131,\u00a0(g\/f) (x)\u00a0will be undefined as the division result will be indeterminate.<\/p>\n<p>Domain of\u00a0g\/f\u00a0= [\u20131, 3] \u2013 {\u20131}<\/p>\n<p>Domain of\u00a0g\/f\u00a0= (\u20131, 3]<\/p>\n<p>\u2234 g\/f: (\u20131, 3]\u00a0\u2192\u00a0R is given by\u00a0(g\/f) (x) = g(x)\/f(x)\u00a0= \u221a(9-x<sup>2<\/sup>) \/ \u221a(x+1)<\/p>\n<p><strong>(vi)<\/strong>\u00a02f \u2013 \u221a5g<strong>\u00a0<\/strong><\/p>\n<p>We know, (2f \u2013 \u221a5g)<strong>\u00a0<\/strong>(x) = 2f(x) \u2013 \u221a5g(x)<\/p>\n<p>(2f \u2013 \u221a5g)\u00a0(x) = 2f (x) \u2013 \u221a5g (x)<\/p>\n<p>= 2\u221a(x+1) \u2013 \u221a5\u221a(9-x<sup>2<\/sup>)<\/p>\n<p>= 2\u221a(x+1) \u2013 \u221a(45- 5x<sup>2<\/sup>)<\/p>\n<p>Domain of 2f \u2013 \u221a5g<strong>\u00a0<\/strong>= Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>= [\u20131, \u221e)\u00a0\u2229\u00a0[\u20133, 3]<\/p>\n<p>= [\u20131, 3]<\/p>\n<p>\u2234 2f \u2013 \u221a5g: [\u20131, 3]\u00a0\u2192\u00a0R is given by (2f \u2013 \u221a5g)\u00a0(x) = 2f (x) \u2013 \u221a5g (x) = 2\u221a(x+1) \u2013 \u221a(45- 5x<sup>2<\/sup>)<\/p>\n<p><strong>(vii)<\/strong>\u00a0f<sup>2<\/sup>\u00a0+ 7f<\/p>\n<p>We know, (f<sup>2<\/sup>\u00a0+ 7f) (x) = f<sup>2<\/sup>(x) + (7f)(x)<\/p>\n<p>(f<sup>2<\/sup>\u00a0+ 7f) (x) = f(x) f(x) + 7f(x)<\/p>\n<p>= \u221a(x+1) \u221a(x+1) + 7\u221a(x+1)<\/p>\n<p>= x + 1 + 7\u221a(x+1)<\/p>\n<p>Domain of f<sup>2<\/sup>\u00a0+ 7f is same as domain of f.<\/p>\n<p>Domain of f<sup>2<\/sup>\u00a0+ 7f = [\u20131, \u221e)<\/p>\n<p>\u2234 f<sup>2<\/sup>\u00a0+ 7f: [\u20131, \u221e)\u00a0\u2192\u00a0R is given by (f<sup>2<\/sup>\u00a0+ 7f) (x) = f(x) f(x) + 7f(x) = x + 1 + 7\u221a(x+1)<\/p>\n<p><strong>(viii)<\/strong>\u00a05\/g<\/p>\n<p>We know, (5\/g) (x) = 5\/g(x)<\/p>\n<p>(5\/g) (x) = 5\/\u221a(9-x<sup>2<\/sup>)<\/p>\n<p>Domain of\u00a05\/g\u00a0= Domain of g = [\u20133, 3]<\/p>\n<p>However, (5\/g) (x)\u00a0is defined for all real values of x\u00a0\u2208\u00a0[\u20133, 3], except for the case when 9 \u2013 x<sup>2<\/sup>\u00a0= 0 or x = \u00b1 3<\/p>\n<p>When x = \u00b13,\u00a0(5\/g) (x)\u00a0will be undefined as the division result will be indeterminate.<\/p>\n<p>Domain of\u00a05\/g\u00a0= [\u20133, 3] \u2013 {\u20133, 3}<\/p>\n<p>= (\u20133, 3)<\/p>\n<p>\u2234 5\/g: (\u20133, 3)\u00a0\u2192\u00a0R is given by (5\/g) (x) = 5\/g(x) = 5\/\u221a(9-x<sup>2<\/sup>)<\/p>\n<h3><span class=\"ez-toc-section\" id=\"5-if-fx-loge-1-%e2%80%93-x-and-gx-x-then-determine-each-of-the-following-functions-i-f-g-ii-fg\"><\/span>5. If f(x) = log<sub>e\u00a0<\/sub>(1 \u2013 x) and g(x) = [x], then determine each of the following functions:<br \/>(i) f + g<br \/>(ii) fg<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iii-fg-iv-gf\"><\/span>(iii)\u00a0f\/g<br \/>(iv) g\/f\u00a0<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"also-find-f-g-%e2%80%931-fg-0-fg-12-and-gf-12\"><\/span>Also, find (f + g) (\u20131), (fg) (0), (f\/g) (1\/2) and (g\/f) (1\/2).<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>f(x) = log<sub>e\u00a0<\/sub>(1 \u2013 x) and g(x) = [x]<\/p>\n<p>We know, f(x) takes real values only when 1 \u2013 x &gt; 0<\/p>\n<p>1 &gt; x<\/p>\n<p>x &lt; 1, \u2234\u00a0x\u00a0\u2208\u00a0(\u2013\u221e, 1)<\/p>\n<p>Domain of f = (\u2013\u221e, 1)<\/p>\n<p>Similarly, g(x) is defined for all real numbers x.<\/p>\n<p>Domain of g = [x], x \u2208 R<\/p>\n<p>= R<\/p>\n<p><strong>(i)<\/strong>\u00a0f + g<\/p>\n<p>We know, (f + g) (x) = f(x) + g(x)<\/p>\n<p>(f + g) (x) = log<sub>e\u00a0<\/sub>(1 \u2013 x) + [x]<\/p>\n<p>Domain of f + g = Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>Domain of f + g = (\u2013\u221e, 1)\u00a0\u2229\u00a0R<\/p>\n<p>= (\u2013\u221e, 1)<\/p>\n<p>\u2234 f + g: (\u2013\u221e, 1)\u00a0\u2192\u00a0R is given by (f + g) (x) = log<sub>e\u00a0<\/sub>(1 \u2013 x) + [x]<\/p>\n<p><strong>(ii)<\/strong>\u00a0fg<\/p>\n<p>We know, (fg) (x) = f(x) g(x)<\/p>\n<p>(fg) (x) = log<sub>e\u00a0<\/sub>(1 \u2013 x) \u00d7 [x]<\/p>\n<p>= [x] log<sub>e\u00a0<\/sub>(1 \u2013 x)<\/p>\n<p>Domain of fg = Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>= (\u2013\u221e, 1)\u00a0\u2229\u00a0R<\/p>\n<p>= (\u2013\u221e, 1)<\/p>\n<p>\u2234 fg: (\u2013\u221e, 1)\u00a0\u2192\u00a0R is given by (fg) (x) = [x] log<sub>e\u00a0<\/sub>(1 \u2013 x)<\/p>\n<p><strong>(iii)<\/strong>\u00a0f\/g<\/p>\n<p>We know, (f\/g) (x) = f(x)\/g(x)<\/p>\n<p>(f\/g) (x) = log<sub>e\u00a0<\/sub>(1 \u2013 x) \/ [x]<\/p>\n<p>Domain of f\/g = Domain of f\u00a0\u2229\u00a0Domain of g<\/p>\n<p>= (\u2013\u221e, 1)\u00a0\u2229\u00a0R<\/p>\n<p>= (\u2013\u221e, 1)<\/p>\n<p>However, (f\/g) (x)\u00a0is defined for all real values of x\u00a0\u2208\u00a0(\u2013\u221e, 1), except for the case when [x] = 0.<\/p>\n<p>We have, [x] = 0 when 0 \u2264 x &lt; 1 or x\u00a0\u2208\u00a0[0, 1)<\/p>\n<p>When 0 \u2264 x &lt; 1,\u00a0(f\/g) (x)\u00a0will be undefined as the division result will be indeterminate.<\/p>\n<p>Domain of\u00a0f\/g\u00a0= (\u2013\u221e, 1) \u2013 [0, 1)<\/p>\n<p>= (\u2013\u221e, 0)<\/p>\n<p>\u2234 f\/g: (\u2013\u221e, 0)\u00a0\u2192\u00a0R is given by (f\/g) (x) = log<sub>e\u00a0<\/sub>(1 \u2013 x) \/ [x]<\/p>\n<p><strong>(iv)<\/strong>\u00a0g\/f<\/p>\n<p>We know, (g\/f) (x) = g(x)\/f(x)<\/p>\n<p>(g\/f) (x) = [x] \/ log<sub>e\u00a0<\/sub>(1 \u2013 x)<\/p>\n<p>However, (g\/f) (x)\u00a0is defined for all real values of x\u00a0\u2208\u00a0(\u2013\u221e, 1), except for the case when log<sub>e\u00a0<\/sub>(1 \u2013 x) = 0.<\/p>\n<p>log<sub>e\u00a0<\/sub>(1 \u2013 x) = 0\u00a0\u21d2\u00a01 \u2013 x = 1 or x = 0<\/p>\n<p>When x = 0,\u00a0(g\/f) (x)\u00a0will be undefined as the division result will be indeterminate.<\/p>\n<p>Domain of\u00a0g\/f\u00a0= (\u2013\u221e, 1) \u2013 {0}<\/p>\n<p>= (\u2013\u221e, 0)\u00a0\u222a\u00a0(0, 1)<\/p>\n<p>\u2234 g\/f: (\u2013\u221e, 0)\u00a0\u222a\u00a0(0, 1)\u00a0\u2192\u00a0R is given by (g\/f) (x) = [x] \/ log<sub>e\u00a0<\/sub>(1 \u2013 x)<\/p>\n<p>(a) We need to find (f + g) (\u20131).<\/p>\n<p>We have, (f + g) (x) = log<sub>e\u00a0<\/sub>(1 \u2013 x) + [x], x\u00a0\u2208\u00a0(\u2013\u221e, 1)<\/p>\n<p>Substituting x = \u20131 in the above equation, we get<\/p>\n<p>(f + g)(\u20131) = log<sub>e\u00a0<\/sub>(1 \u2013 (\u20131)) + [\u20131]<\/p>\n<p>= log<sub>e\u00a0<\/sub>(1 + 1) + (\u20131)<\/p>\n<p>= log<sub>e<\/sub>2 \u2013 1<\/p>\n<p>\u2234 (f + g) (\u20131) = log<sub>e<\/sub>2 \u2013 1<\/p>\n<p>(b) We need to find (fg) (0).<\/p>\n<p>We have, (fg) (x) = [x] log<sub>e\u00a0<\/sub>(1 \u2013 x), x\u00a0\u2208\u00a0(\u2013\u221e, 1)<\/p>\n<p>Substituting x = 0 in the above equation, we get<\/p>\n<p>(fg) (0) = [0] log<sub>e\u00a0<\/sub>(1 \u2013 0)<\/p>\n<p>= 0 \u00d7 log<sub>e<\/sub>1<\/p>\n<p>\u2234\u00a0(fg) (0) = 0<\/p>\n<p>(c) We need to find (f\/g) (1\/2)<\/p>\n<p>We have, (f\/g) (x) = log<sub>e\u00a0<\/sub>(1 \u2013 x) \/ [x], x\u00a0\u2208\u00a0(\u2013\u221e, 0)<\/p>\n<p>However,\u00a01\/2\u00a0is not in the domain of f\/g.<\/p>\n<p>\u2234 (f\/g) (1\/2)\u00a0does not exist.<\/p>\n<p>\u00a0<\/p>\n<p>(d) We need to find (g\/f) (1\/2)<\/p>\n<p>We have, (g\/f) (x) = [x] \/ log<sub>e\u00a0<\/sub>(1 \u2013 x), x\u00a0\u2208\u00a0(\u2013\u221e, 0)\u00a0\u222a\u00a0(0, \u221e)<\/p>\n<p>Substituting x=1\/2\u00a0in the above equation, we get<\/p>\n<p>(g\/f) (1\/2)\u00a0= [x] \/ log<sub>e\u00a0<\/sub>(1 \u2013 x)<\/p>\n<p>= (1\/2)\/ log<sub>e\u00a0<\/sub>(1 \u2013 1\/2)<\/p>\n<p>= 0.5\/ log<sub>e\u00a0<\/sub>(1\/2)<\/p>\n<p>= 0 \/ log<sub>e\u00a0<\/sub>(1\/2)<\/p>\n<p>= 0<\/p>\n<p>\u2234 (g\/f) (1\/2) = 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"important-topics-from-rd-sharma-solutions-class-11-maths\"><\/span>Important Topics from RD Sharma Solutions Class 11 Maths<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">Every chapter has something that you can\u2019t afford to miss. The same is the case with Functions as well. The important topics of the chapter are:<\/span><\/p>\n<ul>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Finding the domain of the function<\/span><\/li>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Real functions<\/span><\/li>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Finding the range of the function<\/span><\/li>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Modulus function<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"access-other-important-chapters-of-rd-sharma-solutions-class-11-maths\"><\/span>Access Other Important Chapters of RD Sharma Solutions Class 11 Maths<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 1 &#8211; Sets<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 2 &#8211; Relations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 4 &#8211; Measurement of Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 5 &#8211; Trigonometric Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 6 &#8211; Graphs of Trigonometric Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 7 &#8211; Trigonometric Ratios of Compound Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 8 &#8211; Transformation Formulae<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 9 &#8211; Trigonometric Ratios of Multiple and Sub Multiple Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 10 &#8211; Sine and Cosine Formulae and Their Applications<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 11 &#8211; Trigonometric Equations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 12 &#8211; Mathematical Induction<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 13 &#8211; Complex Numbers<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 14 &#8211; Quadratic Equations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 15 &#8211; Linear Inequations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-16-permutations\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 16 &#8211; Permutations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-17-combinations\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 17 &#8211; Combinations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18-binomial-theorem\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 18 &#8211; Binomial Theorem<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-19-arithmetic-progressions\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 19 &#8211; Arithmetic Progressions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-20-geometric-progressions\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 20 &#8211; Geometric Progressions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-21-some-special-series\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 21 &#8211; Some Special Series<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-22-brief-review-of-cartesian-system-of-rectangular-coordinates\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 22 &#8211; Brief Review of Cartesian System of Rectangular Coordinates<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-23-the-straight-lines\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 23 &#8211; The Straight Lines<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 24 &#8211; The Circle<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-25-parabola\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 25 &#8211; Parabola<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-26-ellipse\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 26 &#8211; Ellipse<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-27-hyperbola\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 27 &#8211; Hyperbola<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-28-introduction-to-3d-coordinate-geometry\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 28 &#8211; Introduction To 3D Coordinate Geometry<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-29-limits\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 29 &#8211; Limits<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-30-derivatives\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 30 &#8211; Derivatives<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-31-mathematical-reasoning\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 31 &#8211; Mathematical Reasoning<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-32-statistics\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 32 &#8211; Statistics<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-33-probability\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 33 &#8211; Probability<\/a><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">In case you have any queries related to the chapter, do let us know by commenting below. We would love to answer all your queries. <\/span><span style=\"font-weight: 400;\">Keep Practicing for you <a href=\"http:\/\/cbse.nic.in\" target=\"_blank\" rel=\"noopener noreferrer\">CBSE<\/a> class 11 mathematics exam!<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-11-maths-chapter-3\"><\/span>FAQs on RD Sharma Solutions Class 11 Maths Chapter 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-1629723920460\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-rd-sharma-solutions-class-11-maths-chapter-3-pdf\"><\/span>From where can I download the RD Sharma Solutions Class 11 Maths Chapter 3 PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the PDF link in the above blog. <\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629724003432\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-class-11-maths-chapter-3-pdf-offline\"><\/span>Can I access the RD Sharma Solutions Class 11 Maths Chapter 3 PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You should download the PDF online first and then you can access it offline, whenever required. <\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629724198036\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-rd-sharma-solutions-class-11-maths-chapter-3-pdf\"><\/span>How much does it cost to download the RD Sharma Solutions Class 11 Maths Chapter 3 PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download the PDF for free. <\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 3 &#8211; Functions:\u00a0Though the chapter is always small when it comes to Functions it doesn\u2019t mean that you can leave it or take it for granted. Understand all the concepts nicely and you will be able to solve any question that appears in your examination with the help &#8230; <a title=\"RD Sharma Solutions Class 11 Maths Chapter 3 &#8211; Functions (Updated For 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-3-functions\/\" aria-label=\"More on RD Sharma Solutions Class 11 Maths Chapter 3 &#8211; Functions (Updated For 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":119084,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63234"}],"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=63234"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63234\/revisions"}],"predecessor-version":[{"id":504777,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63234\/revisions\/504777"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/119084"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=63234"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=63234"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=63234"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}