{"id":68979,"date":"2021-08-23T15:54:00","date_gmt":"2021-08-23T10:24:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=68979"},"modified":"2021-08-31T18:31:06","modified_gmt":"2021-08-31T13:01:06","slug":"rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-1\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Exercise 2.1 Chapter 2 Function (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-117938\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.1.png\" alt=\"RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.1.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-2-Exercise-2.1-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1:<\/strong> Here you will find the PDF of <strong>RD Sharma Solutions<\/strong> for Class 12 Maths Chapter 2 Function Exercise 2.1. <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-functions\/\"><strong>RD Sharma Maths Solutions for Class 12 Chapter 2<\/strong><\/a> can help students prepare for their exams and deliver their answers without making any mistakes. Students can improve their problem-solving speed by adopting short-cut strategies on a regular basis, resulting in higher exam scores. The pdf of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-12-solutions\/\"><strong>RD Sharma Solutions Class 12 Maths<\/strong><\/a> Chapter 2 Exercise 2.1 can be downloaded by students using the links provided below. This exercise focuses on the fundamentals of functions.<\/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-69d00870a9d0b\" 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\" 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class=\"ez-toc-section-end\"><\/span><\/h4>\n<p><strong>1. Give an example of a function\u00a0<\/strong><\/p>\n<p><strong>(i) Which is one-one but not onto.<\/strong><\/p>\n<p><strong>(ii) Which is not one-one but onto.<\/strong><\/p>\n<p><strong>(iii) Which is neither one-one nor onto.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Let f:\u00a0Z\u00a0\u2192\u00a0Z\u00a0given by\u00a0f(x) = 3x\u00a0+ 2<\/p>\n<p>Let us check one-one condition on f(x) = 3x\u00a0+ 2<\/p>\n<p>Injectivity:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (Z), such that\u00a0f(x) =\u00a0f(y).<\/p>\n<p>\u00a0f\u00a0(x) =\u00a0f(y)<\/p>\n<p>\u21d2 3x\u00a0+ 2 =3y\u00a0+ 2<\/p>\n<p>\u21d2\u00a03x\u00a0= 3y<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0y<\/p>\n<p>\u21d2\u00a0f(x) =\u00a0f(y)\u00a0<\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is one-one.<\/p>\n<p>Surjectivity:<br \/><br \/>Let\u00a0y\u00a0be any element in the co-domain (Z), such that\u00a0f(x) =\u00a0y\u00a0for some element\u00a0x\u00a0in\u00a0Z(domain).<\/p>\n<p>Let f(x) =\u00a0y<\/p>\n<p>\u21d2 3x\u00a0+ 2 =\u00a0y<\/p>\n<p>\u21d2\u00a03x\u00a0=\u00a0y\u00a0\u2013 2<\/p>\n<p>\u21d2\u00a0x =\u00a0(y \u2013 2)\/3.\u00a0It\u00a0may\u00a0not\u00a0be\u00a0in\u00a0the\u00a0domain\u00a0(Z)<\/p>\n<p>Because\u00a0if\u00a0we\u00a0take\u00a0y\u00a0=\u00a03,<\/p>\n<p>x = (y \u2013 2)\/3 = (3-2)\/3 = 1\/3 \u2209\u00a0domain\u00a0Z.<\/p>\n<p>So, for every element in the co domain there need not be any element in the domain such that\u00a0f(x) =\u00a0y.<br \/><br \/>Thus,\u00a0f\u00a0is not onto.<\/p>\n<p>(ii) Example for the function which is not one-one but onto<\/p>\n<p>Let f:\u00a0Z\u00a0\u2192\u00a0N\u00a0\u222a {0} given by\u00a0f(x) = |x|<\/p>\n<p>Injectivity:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (Z),<\/p>\n<p>Such that\u00a0f(x) =\u00a0f(y).<\/p>\n<p>\u21d2 |x| = |y|<\/p>\n<p>\u21d2\u00a0x =\u00a0\u00b1 y<\/p>\n<p>So, different elements of domain\u00a0f\u00a0may give the same image.<br \/><br \/>So,\u00a0f\u00a0is not one-one.<\/p>\n<p>Surjectivity:<br \/><br \/>Let\u00a0y\u00a0be any element in the co domain (Z), such that\u00a0f(x) =\u00a0y\u00a0for some element\u00a0x\u00a0in\u00a0Z (domain).<\/p>\n<p>f(x) =\u00a0y<\/p>\n<p>\u21d2 |x| =\u00a0y<\/p>\n<p>\u21d2 x\u00a0= \u00b1 y<\/p>\n<p>Which is an element in\u00a0Z\u00a0(domain).<br \/><br \/>So, for every element in the co-domain, there exists a pre-image in the domain.<br \/><br \/>Thus,\u00a0f\u00a0is onto.<\/p>\n<p>(iii) Example for the function which is neither one-one nor onto.<\/p>\n<p>Let f:\u00a0Z\u00a0\u2192\u00a0Z\u00a0given by\u00a0f(x) = 2x<sup>2<\/sup>\u00a0+ 1<\/p>\n<p>Injectivity:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (Z), such that\u00a0f(x) =\u00a0f(y).<\/p>\n<p>f(x)\u00a0=\u00a0f(y)<\/p>\n<p>\u21d2\u00a02x<sup>2<\/sup>+1\u00a0=\u00a02y<sup>2<\/sup>+1<\/p>\n<p>\u21d2\u00a02x<sup>2<\/sup>\u00a0=\u00a02y<sup>2<\/sup><\/p>\n<p>\u21d2\u00a0x<sup>2\u00a0<\/sup>=\u00a0y<sup>2<\/sup><\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0\u00b1\u00a0y<\/p>\n<p>So, different elements of domain\u00a0f\u00a0may give the same image.<br \/><br \/>Thus,\u00a0f\u00a0is not one-one.<\/p>\n<p>Surjectivity:<br \/><br \/>Let\u00a0y\u00a0be any element in the co-domain (Z), such that\u00a0f(x) =\u00a0y\u00a0for some element\u00a0x\u00a0in\u00a0Z (domain).<\/p>\n<p>f (x) =\u00a0y<\/p>\n<p>\u21d2\u00a02x<sup>2<\/sup>+1=y<\/p>\n<p>\u21d2\u00a02x<sup>2<\/sup>=\u00a0y\u00a0\u2212\u00a01<\/p>\n<p>\u21d2\u00a0x<sup>2<\/sup>\u00a0= (y-1)\/2<\/p>\n<p>\u21d2\u00a0x = \u221a ((y-1)\/2)\u00a0\u2209 Z always.<\/p>\n<p>For\u00a0example,\u00a0if\u00a0we\u00a0take,\u00a0y\u00a0=\u00a04,<\/p>\n<p>x = \u00b1 \u221a ((y-1)\/2)<\/p>\n<p>= \u00b1 \u221a ((4-1)\/2)<\/p>\n<p>= \u00b1 \u221a (3\/2) \u2209 Z<\/p>\n<p>So,\u00a0x\u00a0may\u00a0not\u00a0be\u00a0in\u00a0Z\u00a0(domain).<\/p>\n<p>Thus,\u00a0f\u00a0is not onto.<\/p>\n<p><strong>2. Which of the following functions from\u00a0A\u00a0to\u00a0B\u00a0are one-one and onto?<br \/>(i) f<sub>1<\/sub>\u00a0= {(1, 3), (2, 5), (3, 7)};\u00a0A\u00a0= {1, 2, 3},\u00a0B\u00a0= {3, 5, 7}<\/strong><\/p>\n<p><strong>(ii) f<sub>2<\/sub>\u00a0= {(2,\u00a0a), (3,\u00a0b), (4,\u00a0c)};\u00a0A\u00a0= {2, 3, 4},\u00a0B\u00a0= {a,\u00a0b,\u00a0c}<\/strong><\/p>\n<p><strong>(iii) f<sub>3<\/sub>\u00a0= {(a,\u00a0x), (b,\u00a0x), (c,\u00a0z), (d,\u00a0z)};\u00a0A\u00a0= {a,\u00a0b,\u00a0c,\u00a0d,},\u00a0B\u00a0= {x,\u00a0y,\u00a0z}.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Consider f<sub>1<\/sub>\u00a0= {(1, 3), (2, 5), (3, 7)};\u00a0A\u00a0= {1, 2, 3},\u00a0B\u00a0= {3, 5, 7}<\/p>\n<p>Injectivity:<br \/><br \/>f<sub>1<\/sub>\u00a0(1) = 3<br \/><br \/>f<sub>1\u00a0<\/sub>(2) = 5<br \/><br \/>f<sub>1<\/sub>\u00a0(3) = 7<\/p>\n<p>\u21d2 Every element of\u00a0A\u00a0has different images in\u00a0B.<br \/><br \/>So,\u00a0f<sub>1<\/sub>\u00a0is one-one.<\/p>\n<p>Surjectivity:<br \/><br \/>Co-domain of\u00a0f<sub>1<\/sub>\u00a0= {3, 5, 7}<br \/><br \/>Range of\u00a0f<sub>1<\/sub>\u00a0=set of images = {3, 5, 7}<\/p>\n<p>\u21d2 Co-domain = range<br \/><br \/>So,\u00a0f<sub>1<\/sub>\u00a0is onto.<\/p>\n<p>(ii) Consider f<sub>2<\/sub>\u00a0= {(2,\u00a0a), (3,\u00a0b), (4,\u00a0c)};\u00a0A\u00a0= {2, 3, 4},\u00a0B\u00a0= {a,\u00a0b,\u00a0c}<\/p>\n<p>Injectivity:<br \/><br \/>f<sub>2<\/sub>\u00a0(2)\u00a0= a<br \/><br \/>f<sub>2<\/sub>\u00a0(3)\u00a0= b<br \/><br \/>f<sub>2<\/sub>\u00a0(4)\u00a0= c<\/p>\n<p>\u21d2 Every element of\u00a0A\u00a0has different images in\u00a0B.<br \/><br \/>So,\u00a0f<sub>2<\/sub>\u00a0is one-one.<\/p>\n<p>Surjectivity:<br \/><br \/>Co-domain of\u00a0f<sub>2<\/sub>\u00a0= {a,\u00a0b,\u00a0c}<\/p>\n<p>Range of\u00a0f<sub>2<\/sub>\u00a0= set of images = {a,\u00a0b,\u00a0c}<\/p>\n<p>\u21d2 Co-domain = range<\/p>\n<p>So,\u00a0f<sub>2<\/sub>\u00a0is onto.<\/p>\n<p>(iii) Consider f<sub>3<\/sub>\u00a0= {(a,\u00a0x), (b,\u00a0x), (c,\u00a0z), (d,\u00a0z)} ;\u00a0A\u00a0= {a,\u00a0b,\u00a0c,\u00a0d,},\u00a0B\u00a0= {x,\u00a0y,\u00a0z}<\/p>\n<p>Injectivity:<br \/><br \/>f<sub>3<\/sub>\u00a0(a) = x<br \/><br \/>f<sub>3<\/sub>\u00a0(b) = x<br \/><br \/>f<sub>3<\/sub>\u00a0(c) = z<br \/><br \/>f<sub>3<\/sub>\u00a0(d) = z<\/p>\n<p>\u21d2 a\u00a0and\u00a0b\u00a0have the same image\u00a0x.<\/p>\n<p>Also\u00a0c\u00a0and\u00a0d\u00a0have the same image\u00a0z<br \/><br \/>So,\u00a0f<sub>3<\/sub>\u00a0is not one-one.<\/p>\n<p>Surjectivity:<br \/><br \/>Co-domain of\u00a0f<sub>3<\/sub>\u00a0={x, y, z}\u00a0<br \/><br \/>Range of\u00a0f<sub>3<\/sub>\u00a0=set of images =\u00a0{x, z}<br \/><br \/>So, the co-domain is not same as the range.<br \/><br \/>So,\u00a0f<sub>3<\/sub>\u00a0is not onto.<\/p>\n<p><strong>3. Prove that the function\u00a0f:\u00a0N\u00a0\u2192\u00a0N, defined by\u00a0f(x) =\u00a0x<sup>2<\/sup>\u00a0+\u00a0x\u00a0+ 1, is one-one but not onto<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f:\u00a0N\u00a0\u2192\u00a0N, defined by\u00a0f(x) =\u00a0x<sup>2<\/sup>\u00a0+\u00a0x\u00a0+ 1<\/p>\n<p>Now we have to prove that given function is one-one<\/p>\n<p>Injectivity:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (N), such that\u00a0f(x) = f(y).\u00a0<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0+ x + 1 = y<sup>2<\/sup>\u00a0+ y + 1<\/p>\n<p>\u21d2 (x<sup>2<\/sup>\u00a0\u2013 y<sup>2<\/sup>) + (x \u2013 y) = 0 `<\/p>\n<p>\u21d2 (x + y) (x- y ) + (x \u2013 y ) = 0<\/p>\n<p>\u21d2 (x \u2013 y) (x + y + 1) = 0\u00a0<\/p>\n<p>\u21d2 x \u2013 y = 0 [x + y + 1\u00a0cannot be zero because x and y are natural numbers\u00a0<\/p>\n<p>\u21d2 x = y<\/p>\n<p>So,\u00a0f\u00a0is one-one.<\/p>\n<p>Surjectivity:\u00a0<\/p>\n<p>When x = 1<\/p>\n<p>x<sup>2<\/sup>\u00a0+ x + 1 = 1 + 1 + 1 = 3<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0+ x +1 \u2265 3,\u00a0for every x in N.<\/p>\n<p>\u21d2\u00a0f(x) will not assume the values 1 and 2.\u00a0<\/p>\n<p>So, f is not onto.<\/p>\n<p><strong>4. Let\u00a0A\u00a0= {\u22121, 0, 1} and\u00a0f\u00a0= {(x,\u00a0x<sup>2<\/sup>) :\u00a0x\u00a0\u2208\u00a0A}. Show that\u00a0f\u00a0:\u00a0A\u00a0\u2192\u00a0A\u00a0is neither one-one nor onto.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given A\u00a0= {\u22121, 0, 1} and\u00a0f\u00a0= {(x,\u00a0x<sup>2<\/sup>):\u00a0x\u00a0\u2208\u00a0A}<br \/><br \/>Also given that,\u00a0f(x) = x<sup>2<\/sup><\/p>\n<p>Now we have to prove that given function neither one-one or nor onto.<\/p>\n<p>Injectivity:<br \/><br \/>Let x = 1<\/p>\n<p>Therefore f(1) = 1<sup>2<\/sup>=1 and<br \/><br \/>f(-1)=(-1)<sup>2<\/sup>=1<\/p>\n<p>\u21d2 1 and -1 have the same images.<br \/><br \/>So,\u00a0f\u00a0is not one-one.<\/p>\n<p>Surjectivity:<br \/><br \/>Co-domain of\u00a0f =\u00a0{-1, 0, 1}<\/p>\n<p>f(1) = 1<sup>2<\/sup>\u00a0= 1,<br \/><br \/>f(-1) = (-1)<sup>2<\/sup>\u00a0= 1 and<br \/><br \/>f(0) = 0<br \/><br \/>\u21d2 Range of\u00a0f = {0, 1}<br \/><br \/>So, both are not same.<br \/><br \/>Hence,\u00a0f\u00a0is not onto<\/p>\n<p><strong>5. Classify the following function as injection, surjection or bijection:<\/strong><\/p>\n<p><strong>(i)\u00a0f:\u00a0N\u00a0\u2192\u00a0N\u00a0given by\u00a0f(x) =\u00a0x<sup>2<\/sup><\/strong><\/p>\n<p><strong>(ii)\u00a0f:\u00a0Z\u00a0\u2192\u00a0Z\u00a0given by\u00a0f(x) =\u00a0x<sup>2<\/sup><\/strong><\/p>\n<p><strong>(iii) f:\u00a0N\u00a0\u2192\u00a0N\u00a0given by\u00a0f(x) =\u00a0x<sup>3<\/sup><\/strong><\/p>\n<p><strong>(iv) f:\u00a0Z\u00a0\u2192\u00a0Z\u00a0given by\u00a0f(x) =\u00a0x<sup>3<\/sup><\/strong><\/p>\n<p><strong>(v) f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = |x|<\/strong><\/p>\n<p><strong>(vi) f:\u00a0Z\u00a0\u2192\u00a0Z, defined by\u00a0f(x) =\u00a0x<sup>2<\/sup>\u00a0+\u00a0x<\/strong><\/p>\n<p><strong>(vii) f:\u00a0Z\u00a0\u2192\u00a0Z, defined by\u00a0f(x) =\u00a0x\u00a0\u2212 5<\/strong><\/p>\n<p><strong>(viii) f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = sin x<\/strong><\/p>\n<p><strong>(ix) f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) =\u00a0x<sup>3<\/sup>\u00a0+ 1<\/strong><\/p>\n<p><strong>(x) f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) =\u00a0x<sup>3<\/sup>\u00a0\u2212\u00a0x<\/strong><\/p>\n<p><strong>(xi) f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = sin<sup>2<\/sup>x\u00a0+ cos<sup>2<\/sup>x<\/strong><\/p>\n<p><strong>(xii) f:\u00a0Q\u00a0\u2212 {3} \u2192\u00a0Q, defined by f (x) = (2x +3)\/(x-3)<\/strong><\/p>\n<p><strong>(xiii) f:\u00a0Q\u00a0\u2192\u00a0Q, defined by\u00a0f(x) =\u00a0x<sup>3<\/sup>\u00a0+ 1<\/strong><\/p>\n<p><strong>(xiv) f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = 5x<sup>3<\/sup>\u00a0+ 4<\/strong><\/p>\n<p><strong>(xv) f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = 5x<sup>3<\/sup>\u00a0+ 4<\/strong><\/p>\n<p><strong>(xvi) f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = 1 +\u00a0x<sup>2<\/sup><\/strong><\/p>\n<p><strong>(xvii) f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = x\/(x<sup>2\u00a0<\/sup>+ 1)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given f:\u00a0N\u00a0\u2192\u00a0N,\u00a0given by\u00a0f(x) =\u00a0x<sup>2<\/sup><\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection condition:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (N), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x) = f(y)<\/p>\n<p>x<sup>2\u00a0<\/sup>= y<sup>2<\/sup><\/p>\n<p>x = y\u00a0(We\u00a0do\u00a0not\u00a0get\u00a0\u00b1\u00a0because\u00a0x\u00a0and\u00a0y\u00a0are\u00a0in\u00a0N that is natural numbers)<\/p>\n<p>So,\u00a0f\u00a0is an injection.<\/p>\n<p>Surjection condition:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (N),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0N (domain).<\/p>\n<p>f(x)\u00a0= y<\/p>\n<p>x<sup>2<\/sup>=\u00a0y<\/p>\n<p>x = \u221ay,\u00a0which\u00a0may\u00a0not\u00a0be\u00a0in\u00a0N.<\/p>\n<p>For\u00a0example,\u00a0if\u00a0y\u00a0= 3,<\/p>\n<p>x = \u221a3\u00a0is\u00a0not\u00a0in\u00a0N.<\/p>\n<p>So,\u00a0f\u00a0is not a surjection.<\/p>\n<p>Also f is not a bijection.<\/p>\n<p>(ii) Given f:\u00a0Z\u00a0\u2192\u00a0Z,\u00a0given by\u00a0f(x) =\u00a0x<sup>2<\/sup><\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection condition:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (Z), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>x<sup>2\u00a0<\/sup>= y<sup>2<\/sup><\/p>\n<p>x\u00a0=\u00a0\u00b1y<\/p>\n<p>So,\u00a0f\u00a0is not an injection.<\/p>\n<p>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (Z),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0Z (domain).<\/p>\n<p>f(x)\u00a0= y<\/p>\n<p>x<sup>2\u00a0<\/sup>=\u00a0y<\/p>\n<p>x = \u00b1 \u221ay\u00a0which\u00a0may\u00a0not\u00a0be\u00a0in\u00a0Z.<\/p>\n<p>For\u00a0example,\u00a0if\u00a0y\u00a0= 3,<\/p>\n<p>x\u00a0=\u00a0\u00b1 \u221a 3 is\u00a0not\u00a0in\u00a0Z.<\/p>\n<p>So,\u00a0f\u00a0is not a surjection.<\/p>\n<p>Also f is not bijection.<\/p>\n<p>(iii) Given f:\u00a0N\u00a0\u2192\u00a0N\u00a0given by\u00a0f(x) =\u00a0x<sup>3<\/sup><\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection condition:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (N), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>x<sup>3<\/sup>\u00a0=\u00a0y<sup>3<\/sup><\/p>\n<p>x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is an injection\u00a0<\/p>\n<p>Surjection condition:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (N),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0N\u00a0(domain).<\/p>\n<p>f(x)\u00a0= y<\/p>\n<p>x<sup>3<\/sup>=\u00a0y<\/p>\n<p>x = \u221by which\u00a0may\u00a0not\u00a0be\u00a0in\u00a0N.<\/p>\n<p>For\u00a0example,\u00a0if\u00a0y\u00a0= 3,<\/p>\n<p>X = \u221b3\u00a0is\u00a0not\u00a0in\u00a0N.<\/p>\n<p>So,\u00a0f\u00a0is not a surjection and\u00a0f\u00a0is not a bijection.<\/p>\n<p>(iv) Given f:\u00a0Z\u00a0\u2192\u00a0Z\u00a0given by\u00a0f(x) =\u00a0x<sup>3<\/sup><\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection condition:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (Z), such that\u00a0f(x)\u00a0= f(y)<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>x<sup>3<\/sup>\u00a0=\u00a0y<sup>3<\/sup><\/p>\n<p>x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is an injection.<\/p>\n<p>Surjection condition:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (Z),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0Z (domain).<\/p>\n<p>f(x) = y<\/p>\n<p>x<sup>3<\/sup>\u00a0=\u00a0y<\/p>\n<p>x = \u221by\u00a0which\u00a0may\u00a0not\u00a0be\u00a0in\u00a0Z.<\/p>\n<p>For\u00a0example,\u00a0if\u00a0y\u00a0= 3,<\/p>\n<p>x = \u221b3 is\u00a0not\u00a0in\u00a0Z.<\/p>\n<p>So,\u00a0f\u00a0is not a surjection and\u00a0f\u00a0is not a bijection.<\/p>\n<p>(v) Given f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = |x|<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection test:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (R), such that\u00a0f(x)\u00a0= f(y)<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>|x|=|y|<\/p>\n<p>x = \u00b1y<\/p>\n<p>So,\u00a0f\u00a0is not an injection.<br \/><br \/>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (R), such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0R\u00a0(domain).<\/p>\n<p>f(x)\u00a0= y<\/p>\n<p>|x|=y<\/p>\n<p>x\u00a0=\u00a0\u00b1\u00a0y\u00a0\u2208\u00a0Z<\/p>\n<p>So,\u00a0f\u00a0is a surjection and\u00a0f\u00a0is not a bijection.<\/p>\n<p>(vi) Given f:\u00a0Z\u00a0\u2192\u00a0Z, defined by\u00a0f(x) =\u00a0x<sup>2<\/sup>\u00a0+\u00a0x<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection test:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (Z), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>x<sup>2<\/sup>+\u00a0x\u00a0=\u00a0y<sup>2\u00a0<\/sup>+\u00a0y<\/p>\n<p>Here,\u00a0we\u00a0cannot\u00a0say\u00a0that\u00a0x\u00a0=\u00a0y.<\/p>\n<p>For\u00a0example,\u00a0x\u00a0=\u00a02\u00a0and\u00a0y\u00a0=\u00a0\u2013\u00a03<\/p>\n<p>Then,<\/p>\n<p>x<sup>2\u00a0<\/sup>+ x = 2<sup>2\u00a0<\/sup>+ 2 =\u00a06<\/p>\n<p>y<sup>2\u00a0<\/sup>+ y = (\u22123)<sup>2\u00a0<\/sup>\u2013 3 =\u00a06<\/p>\n<p>So,\u00a0we\u00a0have\u00a0two\u00a0numbers\u00a02\u00a0and\u00a0-3\u00a0in\u00a0the\u00a0domain\u00a0Z\u00a0whose\u00a0image\u00a0is\u00a0same\u00a0as\u00a06.<\/p>\n<p>So,\u00a0f\u00a0is not an injection.<br \/><br \/>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (Z),<\/p>\n<p>such that\u00a0f(x)\u00a0= y for some element\u00a0x\u00a0in\u00a0Z\u00a0(domain).<\/p>\n<p>f(x)\u00a0= y<\/p>\n<p>x<sup>2<\/sup>\u00a0+\u00a0x\u00a0=\u00a0y<\/p>\n<p>Here,\u00a0we\u00a0cannot\u00a0say\u00a0x\u00a0\u2208\u00a0Z.<\/p>\n<p>For\u00a0example,\u00a0y\u00a0= \u2013 4.<\/p>\n<p>x<sup>2<\/sup>\u00a0+\u00a0x\u00a0=\u00a0\u2212\u00a04<\/p>\n<p>x<sup>2\u00a0<\/sup>+\u00a0x\u00a0+\u00a04\u00a0=\u00a00<\/p>\n<p>x\u00a0= (-1 \u00b1 \u221a-5)\/2 = (-1 \u00b1 i \u221a5)\/2\u00a0which\u00a0is\u00a0not\u00a0in\u00a0Z.<\/p>\n<p>So,\u00a0f\u00a0is not a surjection and\u00a0f\u00a0is not a bijection.<\/p>\n<p>(vii) Given f:\u00a0Z\u00a0\u2192\u00a0Z, defined by\u00a0f(x) =\u00a0x\u00a0\u2013 5<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection test:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (Z), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>x \u2013\u00a05 =\u00a0y \u2013\u00a05<\/p>\n<p>x = y<br \/><br \/>So,\u00a0f\u00a0is an injection.<br \/><br \/>Surjection test:<br \/><br \/>Let\u00a0y\u00a0be any element in the co-domain (Z),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0Z\u00a0(domain).<\/p>\n<p>f(x)\u00a0= y<br \/><br \/>x \u2013\u00a05 =\u00a0y<\/p>\n<p>x = y\u00a0+ 5, which is in\u00a0Z.<\/p>\n<p>So,\u00a0f\u00a0is a surjection and\u00a0f\u00a0is a bijection<\/p>\n<p>(viii) Given f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = sin x<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection test:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (R), such that\u00a0f(x)\u00a0= f(y).<br \/><br \/>f(x)\u00a0= f(y)<\/p>\n<p>Sin\u00a0x\u00a0=\u00a0sin\u00a0y<\/p>\n<p>Here,\u00a0x\u00a0may\u00a0not\u00a0be\u00a0equal\u00a0to\u00a0y\u00a0because\u00a0sin 0 = sin \u03c0.<\/p>\n<p>So,\u00a00\u00a0and\u00a0\u03c0\u00a0have\u00a0the\u00a0same\u00a0image\u00a00.<\/p>\n<p>So,\u00a0f\u00a0is not an injection.<br \/><br \/>Surjection test:<\/p>\n<p>Range of\u00a0f\u00a0= [-1, 1]<\/p>\n<p>Co-domain of\u00a0f\u00a0=\u00a0R<br \/><br \/>Both are not same.<br \/><br \/>So,\u00a0f\u00a0is not a surjection and\u00a0f\u00a0is not a bijection.<\/p>\n<p>(ix) Given f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) =\u00a0x<sup>3<\/sup>\u00a0+ 1<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection test:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (R), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>x<sup>3<\/sup>+1\u00a0=\u00a0y<sup>3<\/sup>+\u00a01<\/p>\n<p>x<sup>3<\/sup>= y<sup>3<\/sup><\/p>\n<p>x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is an injection.<\/p>\n<p>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (R),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0R\u00a0(domain).<\/p>\n<p>f(x) = y<\/p>\n<p>x<sup>3<\/sup>+1=y<\/p>\n<p>x = \u221b (y \u2013 1) \u2208 R<\/p>\n<p>So,\u00a0f\u00a0is a surjection.<\/p>\n<p>So,\u00a0f\u00a0is a bijection.<\/p>\n<p>(x) Given f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) =\u00a0x<sup>3<\/sup>\u00a0\u2212\u00a0x<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection test:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (R), such that\u00a0f(x)\u00a0= f(y).<br \/><br \/>f(x)\u00a0= f(y)<\/p>\n<p>x<sup>3\u00a0<\/sup>\u2013 x = y<sup>3\u00a0<\/sup>\u2212 y<\/p>\n<p>Here,\u00a0we\u00a0cannot\u00a0say\u00a0x = y.<\/p>\n<p>For\u00a0example,\u00a0x = 1\u00a0and\u00a0y = -1<\/p>\n<p>x<sup>3\u00a0<\/sup>\u2212 x\u00a0= 1 \u2212 1 =\u00a00<\/p>\n<p>y<sup>3\u00a0<\/sup>\u2013 y = (\u22121)<sup>3<\/sup>\u2212 (\u22121) \u2013 1 + 1 = 0<\/p>\n<p>So,\u00a01\u00a0and\u00a0-1\u00a0have\u00a0the\u00a0same\u00a0image\u00a00.<\/p>\n<p>So,\u00a0f\u00a0is not an injection.<br \/><br \/>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (R),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0R\u00a0(domain).<\/p>\n<p>f(x) =\u00a0y<\/p>\n<p>x<sup>3<\/sup>\u00a0\u2212\u00a0x\u00a0=\u00a0y<\/p>\n<p>By\u00a0observation\u00a0we\u00a0can\u00a0say\u00a0that\u00a0there\u00a0exist\u00a0some\u00a0x\u00a0in\u00a0R,\u00a0such\u00a0that\u00a0x<sup>3\u00a0<\/sup>\u2013 x = y.<\/p>\n<p>So,\u00a0f\u00a0is a surjection and\u00a0f\u00a0is not a bijection.<\/p>\n<p>(xi) Given f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = sin<sup>2<\/sup>x\u00a0+ cos<sup>2<\/sup>x<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection condition:<\/p>\n<p>f(x) = sin<sup>2<\/sup>x\u00a0+ cos<sup>2<\/sup>x\u00a0<\/p>\n<p>We know that sin<sup>2<\/sup>x\u00a0+ cos<sup>2<\/sup>x\u00a0= 1<\/p>\n<p>So,\u00a0f(x) = 1 for every\u00a0x\u00a0in R.<\/p>\n<p>So, for all elements in the domain, the image is 1.<\/p>\n<p>So,\u00a0f\u00a0is not an injection.<br \/><br \/>Surjection condition:<br \/><br \/>Range of\u00a0f\u00a0= {1}<br \/><br \/>Co-domain of\u00a0f\u00a0=\u00a0R<br \/><br \/>Both are not same.<br \/><br \/>So,\u00a0f\u00a0is not a surjection and\u00a0f\u00a0is not a bijection.<\/p>\n<p>(xii) Given f:\u00a0Q\u00a0\u2212 {3} \u2192\u00a0Q, defined by f (x) = (2x +3)\/(x-3)<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection test:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (Q\u00a0\u2212\u00a0{3}), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0=\u00a0f(y)<\/p>\n<p>(2x + 3)\/(x \u2013 3) = (2y + 3)\/(y \u2013 3)<\/p>\n<p>(2x + 3)\u00a0(y \u2212 3) =\u00a0(2y + 3)\u00a0(x \u2212 3)<\/p>\n<p>2xy\u00a0\u2212\u00a06x +\u00a03y \u2212 9\u00a0=\u00a02xy\u00a0\u2212\u00a06y + 3x \u2212 9<\/p>\n<p>9x\u00a0=\u00a09y\u00a0<\/p>\n<p>x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is an injection.<\/p>\n<p>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (Q\u00a0\u2212\u00a0{3}),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0Q\u00a0(domain).<\/p>\n<p>f(x) =\u00a0y<\/p>\n<p>(2x + 3)\/(x \u2013 3) = y<\/p>\n<p>2x + 3 = x y \u2212 3y<\/p>\n<p>2x \u2013 x y = \u22123y \u2212 3<\/p>\n<p>x\u00a0(2\u2212y) = \u22123\u00a0(y + 1)<\/p>\n<p>x = -3(y + 1)\/(2 \u2013 y)\u00a0which\u00a0is\u00a0not\u00a0defined\u00a0at\u00a0y = 2.<\/p>\n<p>So,\u00a0f\u00a0is not a surjection and\u00a0f\u00a0is not a bijection.<\/p>\n<p>(xiii) Given f:\u00a0Q\u00a0\u2192\u00a0Q, defined by\u00a0f(x) =\u00a0x<sup>3<\/sup>\u00a0+ 1<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection test:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (Q), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>x<sup>3\u00a0<\/sup>+ 1\u00a0=\u00a0y<sup>3\u00a0<\/sup>+\u00a01<\/p>\n<p>x<sup>3<\/sup>\u00a0=\u00a0y<sup>3<\/sup><\/p>\n<p>x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is an injection.<br \/><br \/>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (Q),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0Q (domain).<\/p>\n<p>f(x)\u00a0= y<\/p>\n<p>x<sup>3<\/sup>+\u00a01\u00a0=\u00a0y<\/p>\n<p>x = \u221b(y-1), which\u00a0may\u00a0not\u00a0be\u00a0in\u00a0Q.<\/p>\n<p>For\u00a0example,\u00a0if\u00a0y=\u00a08,<\/p>\n<p>x<sup>3<\/sup>+\u00a01\u00a0= 8<\/p>\n<p>x<sup>3<\/sup>=\u00a07<\/p>\n<p>x = \u221b7, which\u00a0is\u00a0not\u00a0in\u00a0Q.<\/p>\n<p>So,\u00a0f\u00a0is not a surjection and\u00a0f\u00a0is not a bijection.<\/p>\n<p>(xiv) Given f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = 5x<sup>3<\/sup>\u00a0+ 4<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection test:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (R), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>5x<sup>3\u00a0<\/sup>+ 4\u00a0=\u00a05y<sup>3\u00a0<\/sup>+ 4<\/p>\n<p>5x<sup>3<\/sup>=\u00a05y<sup>3<\/sup><\/p>\n<p>x<sup>3\u00a0<\/sup>=\u00a0y<sup>3<\/sup><\/p>\n<p>x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is an injection.<\/p>\n<p>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (R),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0R (domain).<\/p>\n<p>f(x) = y<\/p>\n<p>5x<sup>3<\/sup>+ 4\u00a0=\u00a0y<\/p>\n<p>x<sup>3<\/sup>\u00a0= (y \u2013 4)\/5 \u2208 R<\/p>\n<p>So,\u00a0f\u00a0is a surjection and\u00a0f\u00a0is a bijection.<\/p>\n<p>(xv) Given f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = 5x<sup>3<\/sup>\u00a0+ 4<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection condition:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (R), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>5x<sup>3\u00a0<\/sup>+ 4\u00a0=\u00a05y<sup>3\u00a0<\/sup>+ 4<\/p>\n<p>5x<sup>3\u00a0<\/sup>=\u00a05y<sup>3<\/sup><\/p>\n<p>x<sup>3\u00a0<\/sup>=\u00a0y<sup>3<\/sup><\/p>\n<p>x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is an injection.<\/p>\n<p>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (R),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0R (domain).<\/p>\n<p>f(x) = y<\/p>\n<p>5x<sup>3\u00a0<\/sup>+ 4\u00a0=\u00a0y<\/p>\n<p>x<sup>3<\/sup>\u00a0= (y \u2013 4)\/5 \u2208 R<\/p>\n<p>So,\u00a0f\u00a0is a surjection and\u00a0f\u00a0is a bijection.<\/p>\n<p>(xvi) Given f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = 1 +\u00a0x<sup>2<\/sup><\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection condition:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (R), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>1\u00a0+\u00a0x<sup>2\u00a0<\/sup>= 1\u00a0+\u00a0y<sup>2<\/sup><\/p>\n<p>x<sup>2\u00a0<\/sup>=\u00a0y<sup>2<\/sup><\/p>\n<p>x\u00a0=\u00a0\u00b1\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is not an injection.<br \/><br \/>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (R),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0R (domain).<\/p>\n<p>f(x)\u00a0= y<\/p>\n<p>1\u00a0+\u00a0x<sup>2\u00a0<\/sup>=\u00a0y<\/p>\n<p>x<sup>2\u00a0<\/sup>=\u00a0y\u00a0\u2212\u00a01<\/p>\n<p>x = \u00b1 \u221a-1 = \u00b1 i`\u00a0is\u00a0not\u00a0in\u00a0R.<\/p>\n<p>So,\u00a0f\u00a0is not a surjection and\u00a0f\u00a0is not a bijection.<\/p>\n<p>(xvii) Given f:\u00a0R\u00a0\u2192\u00a0R, defined by\u00a0f(x) = x\/(x<sup>2\u00a0<\/sup>+ 1)<\/p>\n<p>Now we have to check for the given function is injection, surjection and bijection condition.<\/p>\n<p>Injection condition:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (R), such that\u00a0f(x)\u00a0= f(y).<\/p>\n<p>f(x)\u00a0= f(y)<\/p>\n<p>x \/(x<sup>2\u00a0<\/sup>+ 1) = y \/(y<sup>2<\/sup>\u00a0+ 1)<\/p>\n<p>x y<sup>2<\/sup>+\u00a0x\u00a0=\u00a0x<sup>2<\/sup>y\u00a0+\u00a0y<\/p>\n<p>xy<sup>2\u00a0<\/sup>\u2212 x<sup>2<\/sup>y\u00a0+\u00a0x\u00a0\u2212 y\u00a0=\u00a00<\/p>\n<p>\u2212x y\u00a0(\u2212y + x) +\u00a01\u00a0(x \u2212 y)\u00a0=\u00a00<\/p>\n<p>(x \u2212 y)\u00a0(1 \u2013 x y)\u00a0=\u00a00<\/p>\n<p>x\u00a0=\u00a0y\u00a0or\u00a0x\u00a0= 1\/y<\/p>\n<p>So,\u00a0f\u00a0is not an injection.<br \/><br \/>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain\u00a0(R),\u00a0such that\u00a0f(x)\u00a0= y\u00a0for some element\u00a0x\u00a0in\u00a0R (domain).<\/p>\n<p>f(x)\u00a0= y<\/p>\n<p>x \/(x<sup>2<\/sup>\u00a0+ 1) = y<\/p>\n<p>y x<sup>2\u00a0<\/sup>\u2013 x + y = 0<\/p>\n<p>x = (-(-1) \u00b1 \u221a (1-4y<sup>2<\/sup>))\/(2y)\u00a0if\u00a0y\u00a0\u2260\u00a00<\/p>\n<p>= (1 \u00b1 \u221a (1-4y<sup>2<\/sup>))\/ (2y), which\u00a0may\u00a0not\u00a0be\u00a0in\u00a0R<\/p>\n<p>For\u00a0example,\u00a0if\u00a0y=1,\u00a0then<\/p>\n<p>(1 \u00b1 \u221a (1-4)) \/ (2y) = (1 \u00b1 i \u221a3)\/2, which\u00a0is\u00a0not\u00a0in\u00a0R<\/p>\n<p>So,\u00a0f\u00a0is\u00a0not\u00a0surjection\u00a0and\u00a0f\u00a0is\u00a0not\u00a0bijection.<\/p>\n<p><strong>6. If\u00a0f:\u00a0A\u00a0\u2192\u00a0B\u00a0is an injection, such that range of\u00a0f\u00a0= {a}, determine the number of elements in\u00a0A.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f:\u00a0A\u00a0\u2192\u00a0B\u00a0is an injection<\/p>\n<p>And also given that range of\u00a0f\u00a0= {a}<br \/><br \/>So, the number of images of f\u00a0= 1<br \/><br \/>Since,\u00a0f is an injection, there will be exactly one image for each element of\u00a0f\u00a0.<br \/><br \/>So, number of elements in\u00a0A\u00a0= 1.<\/p>\n<p><strong>7. Show that the function\u00a0f:\u00a0R\u00a0\u2212 {3} \u2192\u00a0R\u00a0\u2212 {2} given by\u00a0f(x) = (x-2)\/(x-3)\u00a0is a bijection.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that f:\u00a0R\u00a0\u2212 {3} \u2192\u00a0R\u00a0\u2212 {2} given by f (x) = (x-2)\/(x-3)<\/p>\n<p>Now we have to show that the given function is one-one and on-to<\/p>\n<p>Injectivity:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (R\u00a0\u2212 {3}), such that\u00a0f(x) = f(y).<\/p>\n<p>f(x) = f(y)<\/p>\n<p>\u21d2 (x \u2013 2) \/(x \u2013 3) = (y \u2013 2) \/(y \u2013 3)<\/p>\n<p>\u21d2 (x \u2013 2) (y \u2013 3) =\u00a0(y \u2013 2) (x \u2013 3)<\/p>\n<p>\u21d2 x y \u2013 3 x \u2013 2 y +\u00a06 = x y \u2013 3y \u2013 2x + 6<\/p>\n<p>\u21d2 x = y<\/p>\n<p>So,\u00a0f\u00a0is one-one.<\/p>\n<p>Surjectivity:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (R\u00a0\u2212 {2}), such that\u00a0f(x) = y\u00a0for some element\u00a0x in\u00a0R\u00a0\u2212 {3}\u00a0(domain).<\/p>\n<p>f(x) = y<\/p>\n<p>\u21d2 (x \u2013 2) \/(x \u2013 3) = y<\/p>\n<p>\u21d2 x \u2013 2 = x y \u2013 3y<\/p>\n<p>\u21d2 x y \u2013 x = 3y \u2013 2<\/p>\n<p>\u21d2 x ( y \u2013 1 ) = 3y \u2013 2<\/p>\n<p>\u21d2\u00a0x = (3y \u2013 2)\/ (y \u2013 1), which\u00a0is\u00a0in\u00a0R \u2013 {3}<\/p>\n<p>So, for every element in the co-domain, there exists some pre-image in the domain.\u00a0<\/p>\n<p>\u21d2\u00a0f\u00a0is onto.<\/p>\n<p>Since,\u00a0f\u00a0is both one-one and onto, it is a bijection.<\/p>\n<p><strong>8. Let\u00a0<em>A<\/em>\u00a0= [-1, 1]. Then, discuss whether the following function from\u00a0<em>A<\/em>\u00a0to itself is one-one,\u00a0onto or bijective:<\/strong><\/p>\n<p><strong>(i) f (x) = x\/2<\/strong><\/p>\n<p><strong>(ii) g (x) = |x|<\/strong><\/p>\n<p><strong>(iii) h (x) = x<sup>2<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given f: A\u00a0\u2192 A, given\u00a0by f (x) = x\/2<\/p>\n<p>Now we have to show that the given function is one-one and on-to<\/p>\n<p>Injection test:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (A), such that\u00a0f(x) =\u00a0f(y).<\/p>\n<p>f(x) =\u00a0f(y)<\/p>\n<p>x\/2 = y\/2<\/p>\n<p>x = y<\/p>\n<p>So,\u00a0f\u00a0is one-one.<br \/><br \/>Surjection test:<\/p>\n<p>Let\u00a0y\u00a0be any element in the co-domain (A), such that\u00a0f(x) =\u00a0y\u00a0for some element\u00a0x\u00a0in\u00a0A (domain)<\/p>\n<p>f(x) =\u00a0y<\/p>\n<p>x\/2 = y<\/p>\n<p>x\u00a0= 2y, which may not be in\u00a0A.<\/p>\n<p>For example, if\u00a0y\u00a0= 1, then<br \/><br \/>x\u00a0= 2, which is not in\u00a0A.<br \/><br \/>So,\u00a0f\u00a0is not onto.<br \/><br \/>So,\u00a0f\u00a0is not bijective.<\/p>\n<p>(ii) Given g: A\u00a0\u2192 A, given\u00a0by g (x) = |x|<\/p>\n<p>Now we have to show that the given function is one-one and on-to<\/p>\n<p>Injection test:<br \/><br \/>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (A), such that\u00a0f(x) =\u00a0f(y).<\/p>\n<p>g(x) =\u00a0g(y)<\/p>\n<p>|x| = |y|<\/p>\n<p>x\u00a0=\u00a0\u00b1 y<\/p>\n<p>So,\u00a0f\u00a0is not one-one.<br \/><br \/>Surjection test:<\/p>\n<p>For\u00a0y\u00a0= -1, there is no value of\u00a0x\u00a0in\u00a0A.<\/p>\n<p>So,\u00a0g\u00a0is not onto.<br \/><br \/>So,\u00a0g\u00a0is not bijective.<\/p>\n<p>(iii) Given h: A\u00a0\u2192 A, given\u00a0by h (x) = x<sup>2<\/sup><\/p>\n<p>Now we have to show that the given function is one-one and on-to<\/p>\n<p>Injection test:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (A), such that\u00a0h(x) =\u00a0h(y).<\/p>\n<p>h(x) =\u00a0h(y)<br \/><br \/>x<sup>2<\/sup>\u00a0=\u00a0y<sup>2<\/sup><\/p>\n<p>x\u00a0= \u00b1y<\/p>\n<p>So,\u00a0f\u00a0is not one-one.<br \/><br \/>Surjection test:<\/p>\n<p>For\u00a0y\u00a0= \u2013 1, there is no value of\u00a0x\u00a0in\u00a0A.<br \/><br \/>So,\u00a0h\u00a0is not onto.<br \/><br \/>So,\u00a0h\u00a0is not bijective.<\/p>\n<p><strong>9. Are the following set of ordered pair of\u00a0a function? If so, examine whether the mapping is\u00a0injective or surjective:<\/strong><\/p>\n<p><strong>(i) {(x,\u00a0y):\u00a0x\u00a0is a person,\u00a0y\u00a0is the mother of\u00a0x}<\/strong><\/p>\n<p><strong>(ii) {(a,\u00a0b):\u00a0a\u00a0is a person,\u00a0b\u00a0is an ancestor of\u00a0a}\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let f\u00a0= {(x,\u00a0y):\u00a0x\u00a0is a person,\u00a0y\u00a0is the mother of\u00a0x}<\/p>\n<p>As, for each element\u00a0x\u00a0in domain set, there is a unique related element\u00a0y\u00a0in co-domain set.<\/p>\n<p>So,\u00a0f\u00a0is the function.<\/p>\n<p>Injection test:<br \/><br \/>As,\u00a0y\u00a0can be mother of two or more persons<br \/><br \/>So,\u00a0f\u00a0is not injective.<\/p>\n<p>Surjection test:<\/p>\n<p>For every mother\u00a0y\u00a0defined by (x,\u00a0y), there exists a person\u00a0x\u00a0for whom\u00a0y\u00a0is mother.<br \/><br \/>So,\u00a0f\u00a0is surjective.<br \/><br \/>Therefore,\u00a0f\u00a0is surjective function.<\/p>\n<p>(ii) Let g\u00a0= {(a,\u00a0b):\u00a0a\u00a0is a person,\u00a0b\u00a0is an ancestor of\u00a0a}<br \/><br \/>Since, the ordered map (a,\u00a0b) does not map \u2018a\u2019 \u2013 a person to a living person.<br \/><br \/>So,\u00a0g\u00a0is not a function.<\/p>\n<p><strong>10. Let\u00a0<em>A<\/em>\u00a0= {1, 2, 3}. Write all one-one from\u00a0<em>A<\/em>\u00a0to itself.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given A = {1, 2, 3}<br \/><br \/>Number of elements in A\u00a0= 3<br \/><br \/>Number of one-one functions = number of ways of arranging 3 elements = 3! = 6<\/p>\n<p>(i) {(1, 1), (2, 2), (3, 3)}<br \/><br \/>(ii) {(1, 1), (2, 3), (3, 2)}<br \/><br \/>(iii) {(1, 2 ), (2, 2), (3, 3 )}<br \/><br \/>(iv) {(1, 2), (2, 1), (3, 3)}<br \/><br \/>(v) {(1, 3), (2, 2), (3, 1)}<br \/><br \/>(vi) {(1, 3), (2, 1), (3,2 )}<\/p>\n<p><strong>11. If\u00a0f:\u00a0R\u00a0\u2192\u00a0R\u00a0be the function defined by\u00a0f(x) = 4x<sup>3<\/sup>\u00a0+ 7, show that\u00a0f\u00a0is a bijection.<\/strong><\/p>\n<p>\u00a0<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f:\u00a0R\u00a0\u2192\u00a0R\u00a0is a function defined by\u00a0f(x) = 4x<sup>3<\/sup>\u00a0+ 7<br \/><br \/>Injectivity:<\/p>\n<p>Let\u00a0x\u00a0and\u00a0y\u00a0be any two elements in the domain (R), such that\u00a0f(x) = f(y)<\/p>\n<p>\u21d2\u00a04x<sup>3\u00a0<\/sup>+ 7\u00a0=\u00a04y<sup>3\u00a0<\/sup>+\u00a07<\/p>\n<p>\u21d2\u00a04x<sup>3\u00a0<\/sup>=\u00a04y<sup>3<\/sup><\/p>\n<p>\u21d2\u00a0x<sup>3\u00a0<\/sup>=\u00a0y<sup>3<\/sup><\/p>\n<p>\u21d2\u00a0x\u00a0=\u00a0y<\/p>\n<p>So,\u00a0f\u00a0is one-one.<\/p>\n<p>Surjectivity:<br \/><br \/>Let\u00a0y\u00a0be any element in the co-domain\u00a0(R),\u00a0such that\u00a0f(x) = y\u00a0for some element\u00a0x\u00a0in\u00a0R (domain)<\/p>\n<p>f(x) = y<\/p>\n<p>\u21d2\u00a04x<sup>3\u00a0<\/sup>+ 7\u00a0=\u00a0y<\/p>\n<p>\u21d2\u00a04x<sup>3\u00a0<\/sup>=\u00a0y\u00a0\u2212 7<\/p>\n<p>\u21d2 x<sup>3<\/sup>\u00a0= (y \u2013 7)\/4<\/p>\n<p>\u21d2 x = \u221b(y-7)\/4 in R<\/p>\n<p>So, for every element in the co-domain, there exists some pre-image in the domain. f\u00a0is onto.<br \/><br \/>Since,\u00a0f\u00a0is both one-to-one and onto, it is a bijection.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-12-maths-chapter-2-exercise-21-important-topics\"><\/span>RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1: Important Topics<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Let us have a look at some of the important concepts that are discussed in this chapter.<\/p>\n<ul>\n<li>Classification of functions\n<ul>\n<li>Types of functions\n<ul>\n<li>Constant function<\/li>\n<li>Identity function<\/li>\n<li>Modulus function<\/li>\n<li>Integer function<\/li>\n<li>Exponential function<\/li>\n<li>Logarithmic function<\/li>\n<li>Reciprocal function<\/li>\n<li>Square root function<\/li>\n<\/ul>\n<\/li>\n<li>Operations on real functions<\/li>\n<li>Kinds of functions\n<ul>\n<li>One-one function<\/li>\n<li>On-to function<\/li>\n<li>Many one function<\/li>\n<li>In to function<\/li>\n<li>Bijection<\/li>\n<\/ul>\n<\/li>\n<li>Composition of functions<\/li>\n<li>Properties of the composition of functions<\/li>\n<li>Composition of real function<\/li>\n<li>Inverse of a function<\/li>\n<li>Inverse of an element<\/li>\n<li>Relation between graphs of a function and its inverse<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>We have provided complete details of RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 12 Exam, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-12-maths-chapter-2-exercise-21\"><\/span>FAQs on RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1<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-1629723031214\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-class-12-maths-chapter-2-exercise-21\"><\/span>How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 11 questions in RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629723072134\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-solutions-class-12-maths-chapter-2-exercise-21-for-free\"><\/span>Is RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1 for free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, You can get RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1 for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629723102065\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-solutions-class-12-maths-chapter-2-exercise-21-free-pdf\"><\/span>Where can I download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.1: Here you will find the PDF of RD Sharma Solutions for Class 12 Maths Chapter 2 Function Exercise 2.1. RD Sharma Maths Solutions for Class 12 Chapter 2 can help students prepare for their exams and deliver their answers without making any mistakes. Students can &#8230; <a title=\"RD Sharma Solutions for Class 12 Maths Exercise 2.1 Chapter 2 Function (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-2-exercise-2-1\/\" aria-label=\"More on RD Sharma Solutions for Class 12 Maths Exercise 2.1 Chapter 2 Function (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":117938,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2917,2985,73410],"tags":[3429,73223,73664],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68979"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=68979"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68979\/revisions"}],"predecessor-version":[{"id":121958,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68979\/revisions\/121958"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/117938"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=68979"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=68979"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=68979"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}