{"id":69075,"date":"2023-09-11T19:42:00","date_gmt":"2023-09-11T14:12:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=69075"},"modified":"2023-12-01T10:57:43","modified_gmt":"2023-12-01T05:27:43","slug":"rd-sharma-solutions-class-12-maths-chapter-17-exercise-17-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-17-exercise-17-2\/","title":{"rendered":"RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-120796\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-17-Exercise-17.2.jpg\" alt=\"RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-17-Exercise-17.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-17-Exercise-17.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2<\/strong>: RD Sharma&#8217;s book provides a variety of options and types of short answer questions that students can use to improve their analytical thinking and time management skills. Facts have shown that they are essential for studying the course and developing the confidence needed to take the exam. Encourage students to pass RD Sharma solutions, as these solutions become more repetitive in competitive exams and exams.<\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-12-solutions\/\" target=\"_blank\" rel=\"noopener\"><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Solutions for Class 12&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:16897,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;17&quot;:1}\">RD Sharma Solutions for Class 12 Maths All Chapters\u00a0<\/span><\/a>\u00a0<\/li>\n<\/ul>\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-69e750d98e221\" 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id=\"item-69e750d98e221\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-17-exercise-17-2\/#download-rd-sharma-solutions-class-12-maths-chapter-17-exercise-172-pdf\" title=\"Download RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 PDF:\">Download RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 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-12-maths-chapter-17-exercise-17-2\/#access-rd-sharma-solutions-class-12-maths-chapter-17-exercise-172\" title=\"Access RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2\">Access RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2<\/a><\/li><li 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ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-17-exercise-17-2\/#is-rd-sharma-enough-for-the-class-12-exams\" title=\"Is RD Sharma enough for the Class 12 Exams?\">Is RD Sharma enough for the Class 12 Exams?<\/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-12-maths-chapter-17-exercise-17-2\/#what-are-the-important-topics-included-in-rd-sharma-solutions-class-12-maths-chapter-17-exercise-172\" title=\"What are the important topics included in RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2?\">What are the important topics included in RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-17-exercise-17-2\/#can-i-download-rd-sharma-solutions-class-12-maths-chapter-17-exercise-172-pdf-free\" title=\"Can I download RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 PDF free?\">Can I download RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 PDF free?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-12-maths-chapter-17-exercise-172-pdf\"><\/span>Download RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 PDF:<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 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\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-17-Ex-17.2-1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-17-Ex-17.2-1.pdf\">RD-Sharma-Solutions-Class-12-Maths-Chapter-17-Ex-17.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-12-maths-chapter-17-exercise-172\"><\/span>Access RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Exercise 17.2 Page No: 17.33<\/p>\n<p><strong>1. Find the intervals in which the following functions are increasing or decreasing.<\/strong><\/p>\n<p><strong>(i) f (x) = 10 \u2013 6x \u2013 2x<sup>2<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-3.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 3\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-4.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 4\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-5.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 5\" \/><\/p>\n<p><strong>(ii) f (x) = x<sup>2<\/sup>\u00a0+ 2x \u2013 5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-6.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 6\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-7.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 7\" \/><\/p>\n<p><strong>(iii) f (x) = 6 \u2013 9x \u2013 x<sup>2<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-8.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 8\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-9.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 9\" \/><\/p>\n<p><strong>(iv) f(x) = 2x<sup>3<\/sup>\u00a0\u2013 12x<sup>2<\/sup>\u00a0+ 18x + 15<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-10.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 10\" \/><\/p>\n<p><strong>(v) f (x) = 5 + 36x + 3x<sup>2<\/sup>\u00a0\u2013 2x<sup>3<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = 5 + 36x + 3x<sup>2<\/sup>\u00a0\u2013 2x<sup>3<\/sup><\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-11.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 11\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 36 + 6x \u2013 6x<sup>2<\/sup><\/p>\n<p>For f(x), now we have to find critical point, we must have<\/p>\n<p>\u21d2\u00a0f\u2019(x) = 0<\/p>\n<p>\u21d2\u00a036 + 6x \u2013 6x<sup>2<\/sup>\u00a0= 0<\/p>\n<p>\u21d2\u00a06(\u2013x<sup>2<\/sup>\u00a0+ x + 6) = 0<\/p>\n<p>\u21d2\u00a06(\u2013x<sup>2<\/sup>\u00a0+ 3x \u2013 2x + 6) = 0<\/p>\n<p>\u21d2\u00a0\u2013x<sup>2<\/sup>\u00a0+ 3x \u2013 2x + 6 = 0<\/p>\n<p>\u21d2\u00a0x<sup>2<\/sup>\u00a0\u2013 3x + 2x \u2013 6 = 0<\/p>\n<p>\u21d2\u00a0(x \u2013 3) (x + 2) = 0<\/p>\n<p>\u21d2\u00a0x = 3, \u2013 2<\/p>\n<p>Clearly, f\u2019(x) &gt; 0 if \u20132&lt; x &lt; 3 and f\u2019(x) &lt; 0 if x &lt; \u20132 and x &gt; 3<\/p>\n<p>Thus, f(x) increases on x\u00a0\u2208\u00a0(\u20132, 3) and f(x) is decreasing on interval (\u2013\u221e, \u20132)\u00a0\u222a\u00a0(3, \u221e)<\/p>\n<p><strong>(vi) f (x) = 8 + 36x + 3x<sup>2<\/sup>\u00a0\u2013 2x<sup>3<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = 8 + 36x + 3x<sup>2<\/sup>\u00a0\u2013 2x<sup>3<\/sup><\/p>\n<p>Now differentiating with respect to x<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-12.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 12\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 36 + 6x \u2013 6x<sup>2<\/sup><\/p>\n<p>For f(x), we have to find critical point, we must have<\/p>\n<p>\u21d2\u00a0f\u2019(x) = 0<\/p>\n<p>\u21d2\u00a036 + 6x \u2013 6x<sup>2<\/sup>\u00a0= 0<\/p>\n<p>\u21d2\u00a06(\u2013x<sup>2<\/sup>\u00a0+ x + 6) = 0<\/p>\n<p>\u21d2\u00a06(\u2013x<sup>2<\/sup>\u00a0+ 3x \u2013 2x + 6) = 0<\/p>\n<p>\u21d2\u00a0\u2013x<sup>2<\/sup>\u00a0+ 3x \u2013 2x + 6 = 0<\/p>\n<p>\u21d2\u00a0x<sup>2<\/sup>\u00a0\u2013 3x + 2x \u2013 6 = 0<\/p>\n<p>\u21d2\u00a0(x \u2013 3) (x + 2) = 0<\/p>\n<p>\u21d2\u00a0x = 3, \u2013 2<\/p>\n<p>Clearly, f\u2019(x) &gt; 0 if \u20132 &lt; x &lt; 3 and f\u2019(x) &lt; 0 if x &lt; \u20132 and x &gt; 3<\/p>\n<p>Thus, f(x) increases on x\u00a0\u2208\u00a0(\u20132, 3) and f(x) is decreasing on interval (\u2013\u221e, 2)\u00a0\u222a\u00a0(3, \u221e)<\/p>\n<p><strong>(vii) f(x) = 5x<sup>3<\/sup>\u00a0\u2013 15x<sup>2<\/sup>\u00a0\u2013 120x + 3<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f(x) = 5x<sup>3<\/sup>\u00a0\u2013 15x<sup>2<\/sup>\u00a0\u2013 120x + 3<\/p>\n<p>Now by differentiating the above equation with respect x, we get<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-13.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 13\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 15x<sup>2<\/sup>\u00a0\u2013 30x \u2013 120<\/p>\n<p>For f(x) we have to find critical point, we must have<\/p>\n<p>\u21d2\u00a0f\u2019(x) = 0<\/p>\n<p>\u21d2\u00a015x<sup>2<\/sup>\u00a0\u2013 30x \u2013 120 = 0<\/p>\n<p>\u21d2\u00a015(x<sup>2<\/sup>\u00a0\u2013 2x \u2013 8) = 0<\/p>\n<p>\u21d2\u00a015(x<sup>2<\/sup>\u00a0\u2013 4x + 2x \u2013 8) = 0<\/p>\n<p>\u21d2\u00a0x<sup>2<\/sup>\u00a0\u2013 4x + 2x \u2013 8 = 0<\/p>\n<p>\u21d2\u00a0(x \u2013 4) (x + 2) = 0<\/p>\n<p>\u21d2\u00a0x = 4, \u2013 2<\/p>\n<p>Clearly, f\u2019(x) &gt; 0 if x &lt; \u20132 and x &gt; 4 and f\u2019(x) &lt; 0 if \u20132 &lt; x &lt; 4<\/p>\n<p>Thus, f(x) increases on (\u2013\u221e,\u20132)\u00a0\u222a\u00a0(4, \u221e) and f(x) is decreasing on interval x\u00a0\u2208\u00a0(\u20132, 4)<\/p>\n<p><strong>(viii) f(x) = x<sup>3<\/sup>\u00a0\u2013 6x<sup>2<\/sup>\u00a0\u2013 36x + 2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = x<sup>3<\/sup>\u00a0\u2013 6x<sup>2<\/sup>\u00a0\u2013 36x + 2<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-14.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 14\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 3x<sup>2<\/sup>\u00a0\u2013 12x \u2013 36<\/p>\n<p>For f(x), we have to find critical point, we must have<\/p>\n<p>\u21d2\u00a0f\u2019(x) = 0<\/p>\n<p>\u21d2\u00a03x<sup>2<\/sup>\u00a0\u2013 12x \u2013 36 = 0<\/p>\n<p>\u21d2\u00a03(x<sup>2<\/sup>\u00a0\u2013 4x \u2013 12) = 0<\/p>\n<p>\u21d2\u00a03(x<sup>2<\/sup>\u00a0\u2013 6x + 2x \u2013 12) = 0<\/p>\n<p>\u21d2\u00a0x<sup>2<\/sup>\u00a0\u2013 6x + 2x \u2013 12 = 0<\/p>\n<p>\u21d2\u00a0(x \u2013 6) (x + 2) = 0<\/p>\n<p>\u21d2\u00a0x = 6, \u2013 2<\/p>\n<p>Clearly, f\u2019(x) &gt; 0 if x &lt; \u20132 and x &gt; 6 and f\u2019(x) &lt; 0 if \u20132&lt; x &lt; 6<\/p>\n<p>Thus, f(x) increases on (\u2013\u221e,\u20132)\u00a0\u222a\u00a0(6, \u221e) and f(x) is decreasing on interval x\u00a0\u2208\u00a0(\u20132, 6)<\/p>\n<p><strong>(ix) f(x) = 2x<sup>3<\/sup>\u00a0\u2013 15x<sup>2<\/sup>\u00a0+ 36x + 1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = 2x<sup>3<\/sup>\u00a0\u2013 15x<sup>2<\/sup>\u00a0+ 36x + 1<\/p>\n<p>Now by differentiating the above equation with respect x, we get<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-15.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 15\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 6x<sup>2<\/sup>\u00a0\u2013 30x + 36<\/p>\n<p>For f(x), we have to find critical point, we must have<\/p>\n<p>\u21d2\u00a0f\u2019(x) = 0<\/p>\n<p>\u21d2\u00a06x<sup>2<\/sup>\u00a0\u2013 30x + 36 = 0<\/p>\n<p>\u21d2\u00a06 (x<sup>2<\/sup>\u00a0\u2013 5x + 6) = 0<\/p>\n<p>\u21d2\u00a06(x<sup>2<\/sup>\u00a0\u2013 3x \u2013 2x + 6) = 0<\/p>\n<p>\u21d2\u00a0x<sup>2<\/sup>\u00a0\u2013 3x \u2013 2x + 6 = 0<\/p>\n<p>\u21d2\u00a0(x \u2013 3) (x \u2013 2) = 0<\/p>\n<p>\u21d2\u00a0x = 3, 2<\/p>\n<p>Clearly, f\u2019(x) &gt; 0 if x &lt; 2 and x &gt; 3 and f\u2019(x) &lt; 0 if 2 &lt; x &lt; 3<\/p>\n<p>Thus, f(x) increases on (\u2013\u221e, 2)\u00a0\u222a\u00a0(3, \u221e) and f(x) is decreasing on interval x\u00a0\u2208\u00a0(2, 3)<\/p>\n<p><strong>(x) f (x) = 2x<sup>3<\/sup>\u00a0+ 9x<sup>2<\/sup>\u00a0+ 12x + 20<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = 2x<sup>3<\/sup>\u00a0+ 9x<sup>2<\/sup>\u00a0+ 12x + 20<\/p>\n<p>Differentiating the above equation we get<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-16.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 16\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 6x<sup>2<\/sup>\u00a0+ 18x + 12<\/p>\n<p>For f(x), we have to find critical point, we must have<\/p>\n<p>\u21d2\u00a0f\u2019(x) = 0<\/p>\n<p>\u21d2\u00a06x<sup>2<\/sup>\u00a0+ 18x + 12 = 0<\/p>\n<p>\u21d2\u00a06(x<sup>2<\/sup>\u00a0+ 3x + 2) = 0<\/p>\n<p>\u21d2\u00a06(x<sup>2<\/sup>\u00a0+ 2x + x + 2) = 0<\/p>\n<p>\u21d2\u00a0x<sup>2<\/sup>\u00a0+ 2x + x + 2 = 0<\/p>\n<p>\u21d2\u00a0(x + 2) (x + 1) = 0<\/p>\n<p>\u21d2\u00a0x = \u20131, \u20132<\/p>\n<p>Clearly, f\u2019(x) &gt; 0 if \u20132 &lt; x &lt; \u20131 and f\u2019(x) &lt; 0 if x &lt; \u20131 and x &gt; \u20132<\/p>\n<p>Thus, f(x) increases on x\u00a0\u2208\u00a0(\u20132,\u20131) and f(x) is decreasing on interval (\u2013\u221e, \u20132)\u00a0\u222a\u00a0(\u20132, \u221e)<\/p>\n<p><strong>2. Determine the values of x for which the function f(x) = x<sup>2<\/sup>\u00a0\u2013 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x<sup>2<\/sup>\u00a0\u2013 6x + 9 where the normal is parallel to the line y = x + 5.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f(x) = x<sup>2<\/sup>\u00a0\u2013 6x + 9<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-17.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 17\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 2x \u2013 6<\/p>\n<p>\u21d2\u00a0f\u2019(x) = 2(x \u2013 3)<\/p>\n<p>For f(x), let us find critical point, we must have<\/p>\n<p>\u21d2\u00a0f\u2019(x) = 0<\/p>\n<p>\u21d2\u00a02(x \u2013 3) = 0<\/p>\n<p>\u21d2\u00a0(x \u2013 3) = 0<\/p>\n<p>\u21d2\u00a0x = 3<\/p>\n<p>Clearly, f\u2019(x) &gt; 0 if x &gt; 3 and f\u2019(x) &lt; 0 if x &lt; 3<\/p>\n<p>Thus, f(x) increases on (3, \u221e) and f(x) is decreasing on interval x\u00a0\u2208\u00a0(\u2013\u221e, 3)<\/p>\n<p>Now, let us find the coordinates of the point<\/p>\n<p>Equation\u00a0of curve is f(x) = x<sup>2<\/sup>\u00a0\u2013 6x + 9<\/p>\n<p>The slope of this curve is given by<\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-18.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 18\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-18-1.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 18\" \/><\/p>\n<p><strong>3. Find the intervals in which f(x) = sin x \u2013 cos x, where 0 &lt; x &lt; 2\u03c0 is increasing or decreasing.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-19.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 19\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-20.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 20\" \/><\/p>\n<p><strong>4. Show that f(x) = e<sup>2x<\/sup>\u00a0is increasing on R.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = e<sup>2x<\/sup><\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-21.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 21\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 2e<sup>2x<\/sup><\/p>\n<p>For f(x) to be increasing, we must have<\/p>\n<p>\u21d2\u00a0f\u2019(x) &gt; 0<\/p>\n<p>\u21d2\u00a02e<sup>2x<\/sup>\u00a0&gt; 0<\/p>\n<p>\u21d2\u00a0e<sup>2x<\/sup>\u00a0&gt; 0<\/p>\n<p>Since the value of e lies between 2 and 3<\/p>\n<p>So, whatever be the power of e (that is x in domain R) will be greater than zero.<\/p>\n<p>Thus f(x) is increasing on interval R<\/p>\n<p><strong>5. Show that f (x) = e<sup>1\/x<\/sup>, x \u2260 0 is a decreasing function for all x \u2260 0.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-22.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 22\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-23.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 23\" \/><\/p>\n<p><strong>6. Show that f(x) = log<sub>a<\/sub>\u00a0x, 0 &lt; a &lt; 1 is a decreasing function for all x &gt; 0.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-24.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 24\" \/><\/p>\n<p><strong>7. Show that f(x) = sin x is increasing on (0, \u03c0\/2) and decreasing on (\u03c0\/2, \u03c0) and neither increasing nor decreasing in (0, \u03c0).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-25.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 25\" \/><\/p>\n<p><strong>8. Show that f(x) = log sin x is increasing on (0, \u03c0\/2) and decreasing on (\u03c0\/2, \u03c0).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-26.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 26\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-27.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 27\" \/><\/p>\n<p><strong>9. Show that f(x) = x \u2013 sin x is increasing for all x\u00a0\u03f5\u00a0R.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = x \u2013 sin x<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-28.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 28\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 1 \u2013 cos x<\/p>\n<p>Now, as given x\u00a0\u03f5\u00a0R<\/p>\n<p>\u21d2\u00a0\u20131 &lt; cos x &lt; 1<\/p>\n<p>\u21d2\u00a0\u20131 &gt; cos x &gt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &gt; 0<\/p>\n<p>Hence, the condition for f(x) to be increasing<\/p>\n<p>Thus f(x) is increasing on interval x\u00a0\u2208\u00a0R<\/p>\n<p><strong>10. Show that f(x) = x<sup>3<\/sup>\u00a0\u2013 15x<sup>2<\/sup>\u00a0+ 75x \u2013 50 is an increasing function for all x\u00a0\u03f5\u00a0R.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f(x) = x<sup>3<\/sup>\u00a0\u2013 15x<sup>2<\/sup>\u00a0+ 75x \u2013 50<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-29.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 29\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 3x<sup>2<\/sup>\u00a0\u2013 30x + 75<\/p>\n<p>\u21d2\u00a0f\u2019(x) = 3(x<sup>2<\/sup>\u00a0\u2013 10x + 25)<\/p>\n<p>\u21d2\u00a0f\u2019(x) = 3(x \u2013 5)<sup>2<\/sup><\/p>\n<p>Now, as given x\u00a0\u03f5\u00a0R<\/p>\n<p>\u21d2\u00a0(x \u2013 5)<sup>2<\/sup>\u00a0&gt; 0<\/p>\n<p>\u21d2\u00a03(x \u2013 5)<sup>2<\/sup>\u00a0&gt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &gt; 0<\/p>\n<p>Hence, the condition for f(x) to be increasing<\/p>\n<p>Thus f(x) is increasing on interval x\u00a0\u2208\u00a0R<\/p>\n<p><strong>11. Show that f(x) = cos<sup>2<\/sup>\u00a0x is a decreasing function on (0, \u03c0\/2).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = cos<sup>2<\/sup>\u00a0x<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-30.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 30\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 2 cos x (\u2013sin x)<\/p>\n<p>\u21d2\u00a0f\u2019(x) = \u20132 sin (x) cos (x)<\/p>\n<p>\u21d2\u00a0f\u2019(x) = \u2013sin2x<\/p>\n<p>Now, as given x belongs to (0, \u03c0\/2).<\/p>\n<p>\u21d2\u00a02x\u00a0\u2208\u00a0(0,<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/https-gradeup-question-images-grdp-co-livedata-proj23971-1543903130748996-png.png\" alt=\"https:\/\/gradeup-question-images.grdp.co\/liveData\/PROJ23971\/1543903130748996.png\" \/>\u03c0)<\/p>\n<p>\u21d2\u00a0Sin (2x)&gt; 0<\/p>\n<p>\u21d2\u00a0\u2013Sin (2x) &lt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &lt; 0<\/p>\n<p>Hence, the condition for f(x) to be decreasing<\/p>\n<p>Thus f(x) is decreasing on interval\u00a0(0, \u03c0\/2).<\/p>\n<p>Hence proved<\/p>\n<p><strong>12. Show that f(x) = sin x is an increasing function on (\u2013\u03c0\/2, \u03c0\/2).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = sin x<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-31.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 31\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = cos x<\/p>\n<p>Now, as given x \u2208 (\u2013\u03c0\/2, \u03c0\/2).<\/p>\n<p>That is 4<sup>th<\/sup>\u00a0quadrant, where<\/p>\n<p>\u21d2\u00a0Cos x&gt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &gt; 0<\/p>\n<p>Hence, the condition for f(x) to be increasing<\/p>\n<p>Thus f(x) is increasing on interval\u00a0(\u2013\u03c0\/2, \u03c0\/2).<\/p>\n<p><strong>13. Show that f(x) = cos x is a decreasing function on (0, \u03c0), increasing in (\u2013\u03c0, 0) and neither increasing nor decreasing in (\u2013\u03c0, \u03c0).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f(x) = cos x<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-32.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 32\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = \u2013sin x<\/p>\n<p>Taking different regions from 0 to 2\u03c0<\/p>\n<p>Let\u00a0x \u2208 (0, \u03c0).<\/p>\n<p>\u21d2\u00a0Sin(x) &gt; 0<\/p>\n<p>\u21d2\u00a0\u2013sin x &lt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &lt; 0<\/p>\n<p>Thus f(x) is decreasing in\u00a0(0, \u03c0)<\/p>\n<p>Let x \u2208 (\u2013\u03c0, o).<\/p>\n<p>\u21d2\u00a0Sin (x) &lt; 0<\/p>\n<p>\u21d2\u00a0\u2013sin x &gt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &gt; 0<\/p>\n<p>Thus f(x) is increasing in\u00a0(\u2013\u03c0, 0).<\/p>\n<p>Therefore, from the above condition, we find that<\/p>\n<p>\u21d2\u00a0f (x) is decreasing in (0, \u03c0) and increasing in\u00a0(\u2013\u03c0, 0).<\/p>\n<p>Hence, condition for f(x) neither increasing nor decreasing in (\u2013\u03c0, \u03c0)<\/p>\n<p><strong>14. Show that f(x) = tan x is an increasing function on (\u2013\u03c0\/2, \u03c0\/2).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = tan x<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-33.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 33\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = sec<sup>2<\/sup>x<\/p>\n<p>Now, as given<\/p>\n<p>x \u2208 (\u2013\u03c0\/2, \u03c0\/2).<\/p>\n<p>That is 4<sup>th<\/sup>\u00a0quadrant, where<\/p>\n<p>\u21d2\u00a0sec<sup>2<\/sup>x &gt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &gt; 0<\/p>\n<p>Hence, the condition for f(x) to be increasing<\/p>\n<p>Thus f(x) is increasing on interval\u00a0(\u2013\u03c0\/2, \u03c0\/2).<\/p>\n<p><strong>15. Show that f(x) = tan<sup>\u20131<\/sup>\u00a0(sin x + cos x) is a decreasing function on the interval (\u03c0\/4, \u03c0 \/2).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-34.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 34\" \/><\/p>\n<p><strong>16. Show that the function f (x) = sin (2x + \u03c0\/4) is decreasing on (3\u03c0\/8, 5\u03c0\/8).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-35.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 35\" \/><\/p>\n<p>Thus f (x) is decreasing on the interval (3\u03c0\/8, 5\u03c0\/8).<\/p>\n<p><strong>17. Show that the function f(x) = cot<sup>\u20131<\/sup>\u00a0(sin x + cos x) is decreasing on (0, \u03c0\/4) and increasing on (\u03c0\/4, \u03c0\/2).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f(x) = cot<sup>\u20131<\/sup>\u00a0(sin x + cos x)<\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-36.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 36\" \/><\/p>\n<p><strong>18. Show that f(x) = (x \u2013 1) e<sup>x<\/sup>\u00a0+ 1 is an increasing function for all x &gt; 0.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = (x \u2013 1) e<sup>x<\/sup>\u00a0+ 1<\/p>\n<p>Now differentiating the given equation with respect to x, we get<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-37.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 37\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = e<sup>x<\/sup>\u00a0+ (x \u2013 1) e<sup>x<\/sup><\/p>\n<p>\u21d2\u00a0f\u2019(x) = e<sup>x<\/sup>(1+ x \u2013 1)<\/p>\n<p>\u21d2\u00a0f\u2019(x) = x e<sup>x<\/sup><\/p>\n<p>As given x &gt; 0<\/p>\n<p>\u21d2\u00a0e<sup>x<\/sup>\u00a0&gt; 0<\/p>\n<p>\u21d2\u00a0x e<sup>x<\/sup>\u00a0&gt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &gt; 0<\/p>\n<p>Hence, the condition for f(x) to be increasing<\/p>\n<p>Thus f(x) is increasing on interval x &gt; 0<\/p>\n<p><strong>19. Show that the function x<sup>2<\/sup>\u00a0\u2013 x + 1 is neither increasing nor decreasing on (0, 1).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f(x) = x<sup>2<\/sup>\u00a0\u2013 x + 1<\/p>\n<p>Now by differentiating the given equation with respect to x, we get<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-38.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 38\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 2x \u2013 1<\/p>\n<p>Taking different regions from (0, 1)<\/p>\n<p>Let x \u2208 (0, \u00bd)<\/p>\n<p>\u21d2\u00a02x \u2013 1 &lt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &lt; 0<\/p>\n<p>Thus f(x) is decreasing in (0, \u00bd)<\/p>\n<p>Let x \u2208 (\u00bd, 1)<\/p>\n<p>\u21d2\u00a02x \u2013 1 &gt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &gt; 0<\/p>\n<p>Thus f(x) is increasing in (\u00bd, 1)<\/p>\n<p>Therefore, from the above condition, we find that<\/p>\n<p>\u21d2\u00a0f (x) is decreasing in\u00a0(0, \u00bd) \u00a0and increasing in\u00a0(\u00bd, 1)<\/p>\n<p>Hence, the condition for f(x) is neither increasing nor decreasing in (0, 1)<\/p>\n<p><strong>20. Show that f(x) = x<sup>9<\/sup>\u00a0+ 4x<sup>7<\/sup>\u00a0+ 11 is an increasing function for all x\u00a0\u03f5\u00a0R.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given f (x) = x<sup>9<\/sup>\u00a0+ 4x<sup>7<\/sup>\u00a0+ 11<\/p>\n<p>Now by differentiating the above equation with respect to x, we get<\/p>\n<p>\u21d2<br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-17-increaing-and-decreasing-functions-image-39.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 39\" \/><\/p>\n<p>\u21d2\u00a0f\u2019(x) = 9x<sup>8<\/sup>\u00a0+ 28x<sup>6<\/sup><\/p>\n<p>\u21d2\u00a0f\u2019(x) = x<sup>6<\/sup>(9x<sup>2<\/sup>\u00a0+ 28)<\/p>\n<p>As given x\u00a0\u03f5\u00a0R<\/p>\n<p>\u21d2\u00a0x<sup>6<\/sup>\u00a0&gt; 0 and 9x<sup>2<\/sup>\u00a0+ 28 &gt; 0<\/p>\n<p>\u21d2\u00a0x<sup>6\u00a0<\/sup>(9x<sup>2<\/sup>\u00a0+ 28) &gt; 0<\/p>\n<p>\u21d2\u00a0f\u2019(x) &gt; 0<\/p>\n<p>Hence, the condition for f(x) to be increasing<\/p>\n<p>Thus f(x) is increasing on interval x\u00a0\u2208\u00a0R<\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-12-maths-chapter-17-exercise-172-important-topics-from-the-chapter\"><\/span>RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2: Important Topics From The Chapter<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Some of the essential topics of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-17-increasing-and-decreasing-functions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 12 Maths Chapter 17<\/a> Exercise 17.2 are listed below.<\/p>\n<ul>\n<li>The solution of rational algebraic inequations<\/li>\n<li>Strictly increasing functions<\/li>\n<li>Strictly decreasing functions<\/li>\n<li>Monotonic functions<\/li>\n<li>Monotonically increasing function<\/li>\n<li>Monotonically decreasing functions<\/li>\n<li>Necessary and sufficient conditions for monotonicity<\/li>\n<li>Finding the intervals in which a function is increasing or decreasing<\/li>\n<li>Proving the monotonicity of a function on a given interval<\/li>\n<li>Finding the interval in which a function is increasing or decreasing<\/li>\n<\/ul>\n<p>We have included all the information regarding\u00a0<a href=\"http:\/\/cbseacademic.nic.in\/\" target=\"_blank\" rel=\"noopener nofollow\">CBSE<\/a> RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2. If you have any queries feel free to ask in the comment section.\u00a0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-solutions-class-12-maths-chapter-17-exercise-172\"><\/span>FAQ: RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630073321994\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-the-class-12-exams\"><\/span>Is RD Sharma enough for the Class 12 Exams?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma has a variety of questions for class 12 board exams.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630073357612\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-important-topics-included-in-rd-sharma-solutions-class-12-maths-chapter-17-exercise-172\"><\/span>What are the important topics included in RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can refer to the above article to know important topics included in RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630073358488\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-open-rd-sharma-solutions-class-12-maths-chapter-17-exercise-172-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 PDF on my smartphone?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can open RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 on any device.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630073359529\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-from-rd-sharma-solutions-class-12\"><\/span>What are the benefits of studying from RD Sharma Solutions Class 12?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing these solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630073363281\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-rd-sharma-solutions-class-12-maths-chapter-17-exercise-172-pdf-free\"><\/span>Can I download RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 PDF free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can download RRD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2: RD Sharma&#8217;s book provides a variety of options and types of short answer questions that students can use to improve their analytical thinking and time management skills. Facts have shown that they are essential for studying the course and developing the confidence needed to take &#8230; <a title=\"RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-17-exercise-17-2\/\" aria-label=\"More on RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":120796,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2917,73716,73411,2985,73410],"tags":[3429,73223,73664,76768],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/69075"}],"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\/244"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=69075"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/69075\/revisions"}],"predecessor-version":[{"id":515184,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/69075\/revisions\/515184"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/120796"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=69075"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=69075"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=69075"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}