{"id":124995,"date":"2023-07-09T13:31:00","date_gmt":"2023-07-09T08:01:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=124995"},"modified":"2023-11-23T10:17:54","modified_gmt":"2023-11-23T04:47:54","slug":"rd-sharma-class-9-solutions-chapter-4-exercise-4-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-1\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Exercise 4.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone wp-image-125235 size-full\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-4-Exercise-4.1.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-4-Exercise-4.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-4-Exercise-4.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1:<\/strong> Practice questions from <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> and ace your Maths exam this year. You don&#8217;t have to study too many books as <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-4-algebraic-identities\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 4 <\/a> Exercise 4.1 has everything you might need. To know more, you have to read the whole blog.<\/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-69ea1ecace18d\" 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-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-1\/#access-rd-sharma-class-9-solutions-chapter-4-exercise-41\" title=\"Access RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1\">Access RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1<\/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-class-9-solutions-chapter-4-exercise-4-1\/#exercise-61-page-no-62\" title=\"Exercise 6.1 Page No: 6.2\">Exercise 6.1 Page No: 6.2<\/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-class-9-solutions-chapter-4-exercise-4-1\/#exercise-62-page-no-68\" title=\"Exercise 6.2 Page No: 6.8\">Exercise 6.2 Page No: 6.8<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-1\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-4-exercise-41\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.1\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.1<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-1\/#how-many-questions-are-there-in-rd-sharma-solutions-for-class-9-maths-chapter-4-exercise-41\" title=\"How many questions are there in RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1?\">How many questions are there in RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1?<\/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-class-9-solutions-chapter-4-exercise-4-1\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-4-exercise-41\" title=\"From where can I download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1\">From where can I download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-1\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-4-exercise-41\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-4-exercise-41-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-4-Ex-4.1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions for Chapter 4 Exercise 4.1 Class 9 PDF<\/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\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-4-Ex-4.1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-4-exercise-41\"><\/span><strong>Access RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"exercise-61-page-no-62\"><\/span>Exercise 6.1 Page No: 6.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: Which of the following expressions are polynomials in one variable, and which are not? State reasons for your answer:<\/strong><\/p>\n<p><strong>(i) 3x<sup>2<\/sup>\u00a0\u2013 4x + 15<\/strong><\/p>\n<p><strong>(ii) y<sup>2<\/sup>\u00a0+ 2\u221a3<\/strong><\/p>\n<p><strong>(iii) 3\u221ax + \u221a2x<\/strong><\/p>\n<p><strong>(iv) x \u2013 4\/x<\/strong><\/p>\n<p><strong>(v) x<sup>12<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ t<sup>50<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a03x<sup>2<\/sup>\u00a0\u2013 4x + 15<\/p>\n<p><em>It is a polynomial of x.<\/em><\/p>\n<p><strong>(ii)<\/strong>\u00a0y<sup>2<\/sup>\u00a0+ 2\u221a3<\/p>\n<p><em>It is a polynomial of y.<\/em><\/p>\n<p><strong>(iii)<\/strong>\u00a03\u221ax + \u221a2x<\/p>\n<p><em>It is not a polynomial since the exponent of 3\u221ax is a rational term.<\/em><\/p>\n<p><strong>(iv)<\/strong>\u00a0x \u2013 4\/x<\/p>\n<p><em>It is not a polynomial since the exponent of \u2013 4\/x is not a positive term.<\/em><\/p>\n<p><strong>(v)<\/strong>\u00a0x<sup>12<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ t<sup>50<\/sup><\/p>\n<p><em>It is a three-variable polynomial, x, y and t.<\/em><\/p>\n<p><strong>Question 2: Write the coefficient of x<sup>2<\/sup>\u00a0in each of the following:<\/strong><\/p>\n<p><strong>(i) 17 \u2013 2x + 7x<sup>2<\/sup><\/strong><\/p>\n<p><strong>(ii) 9 \u2013 12x + x<sup>3<\/sup><\/strong><\/p>\n<p><strong>(iii) \u220f\/6 x<sup>2<\/sup>\u00a0\u2013 3x + 4<\/strong><\/p>\n<p><strong>(iv) \u221a3x \u2013 7<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a017 \u2013 2x + 7x<sup>2<\/sup><\/p>\n<p><em>Coefficient of x<sup>2<\/sup>\u00a0= 7<\/em><\/p>\n<p><strong>(ii)<\/strong>\u00a09 \u2013 12x + x<sup>3<\/sup><\/p>\n<p><em>Coefficient of x<sup>2\u00a0<\/sup>=0<\/em><\/p>\n<p><strong>(iii)<\/strong>\u00a0\u220f\/6 x<sup>2<\/sup>\u00a0\u2013 3x + 4<\/p>\n<p><em>Coefficient of x<sup>2\u00a0<\/sup>= \u220f\/6<\/em><\/p>\n<p><strong>(iv)<\/strong>\u00a0\u221a3x \u2013 7<\/p>\n<p><em>Coefficient of x<sup>2\u00a0<\/sup>= 0<\/em><\/p>\n<p><strong>Question 3: Write the degrees of each of the following polynomials:<\/strong><\/p>\n<p><strong>(i) 7x<sup>3<\/sup>\u00a0+ 4x<sup>2<\/sup>\u00a0\u2013 3x + 12<\/strong><\/p>\n<p><strong>(ii) 12 \u2013 x + 2x<sup>3<\/sup><\/strong><\/p>\n<p><strong>(iii) 5y \u2013 \u221a2<\/strong><\/p>\n<p><strong>(iv) 7<\/strong><\/p>\n<p><strong>(v) 0<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>As we know, degree is the highest power in the polynomial<\/p>\n<p><strong>(i)<\/strong>\u00a0Degree of the polynomial 7x<sup>3<\/sup>\u00a0+ 4x<sup>2<\/sup>\u00a0\u2013 3x + 12 is\u00a0<em>3<\/em><\/p>\n<p><strong>(ii)<\/strong>\u00a0Degree of the polynomial 12 \u2013 x + 2x<sup>3<\/sup>\u00a0is\u00a0<em>3<\/em><\/p>\n<p><strong>(iii)<\/strong>\u00a0Degree of the polynomial 5y \u2013 \u221a2 is\u00a0<em>1<\/em><\/p>\n<p><strong>(iv)<\/strong>\u00a0Degree of the polynomial 7 is\u00a0<em>0<\/em><\/p>\n<p><strong>(v)<\/strong>\u00a0Degree of the polynomial 0 is\u00a0<em>undefined.<\/em><\/p>\n<p><strong>Question 4: Classify the following polynomials as linear, quadratic, cubic and biquadratic polynomials:<\/strong><\/p>\n<p><strong>(i) x + x<sup>2<\/sup>\u00a0+ 4<\/strong><\/p>\n<p><strong>(ii) 3x \u2013 2<\/strong><\/p>\n<p><strong>(iii) 2x + x<sup>2<\/sup><\/strong><\/p>\n<p><strong>(iv) 3y<\/strong><\/p>\n<p><strong>(v) t<sup>2<\/sup>\u00a0+ 1<\/strong><\/p>\n<p><strong>(vi) 7t<sup>4<\/sup>\u00a0+ 4t<sup>3<\/sup>\u00a0+ 3t \u2013 2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a0x + x<sup>2<\/sup>\u00a0+ 4: It is a quadratic polynomial as its degree is\u00a0<em>2<\/em>.<\/p>\n<p><strong>(ii)<\/strong>\u00a03x \u2013 2 : It is a linear polynomial as its degree is\u00a0<em>1<\/em>.<\/p>\n<p><strong>(iii)<\/strong>\u00a02x + x<sup>2<\/sup>: It is a quadratic polynomial as its degree is\u00a0<em>2<\/em>.<\/p>\n<p><strong>(iv)<\/strong>\u00a03y: It is a linear polynomial as its degree is\u00a0<em>1<\/em>.<\/p>\n<p><strong>(v)<\/strong>\u00a0t<sup>2<\/sup>+ 1: It is a quadratic polynomial as its degree is\u00a0<em>2<\/em>.<\/p>\n<p><strong>(vi)<\/strong>\u00a07t<sup>4<\/sup>\u00a0+ 4t<sup>3<\/sup>\u00a0+ 3t \u2013 2: It is a biquadratic polynomial as its degree is\u00a0<em>4<\/em>.<\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-62-page-no-68\"><\/span>Exercise 6.2 Page No: 6.8<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: If f(x) = 2x<sup>3<\/sup>\u00a0\u2013 13x<sup>2<\/sup>\u00a0+ 17x + 12, find<\/strong><\/p>\n<p><strong>(i) f (2)<\/strong><\/p>\n<p><strong>(ii) f (-3)<\/strong><\/p>\n<p><strong>(iii) f(0)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>f(x) = 2x<sup>3<\/sup>\u00a0\u2013 13x<sup>2<\/sup>\u00a0+ 17x + 12<\/p>\n<p><strong>(i)<\/strong>\u00a0f(2) = 2(2)<sup>3<\/sup>\u00a0\u2013 13(2)<sup>\u00a02<\/sup>\u00a0+ 17(2) + 12<\/p>\n<p>= 2 x 8 \u2013 13 x 4 + 17 x 2 + 12<\/p>\n<p>= 16 \u2013 52 + 34 + 12<\/p>\n<p>= 62 \u2013 52<\/p>\n<p><em>= 10<\/em><\/p>\n<p><strong>(ii)<\/strong>\u00a0f(-3) = 2(-3)<sup>3<\/sup>\u00a0\u2013 13(-3)<sup>\u00a02<\/sup>\u00a0+ 17 x (-3) + 12<\/p>\n<p>= 2 x (-27) \u2013 13 x 9 + 17 x (-3) + 12<\/p>\n<p>= -54 \u2013 117 -51 + 12<\/p>\n<p>= -222 + 12<\/p>\n<p><em>= -210<\/em><\/p>\n<p><strong>(iii)<\/strong>\u00a0f(0) = 2 x (0)<sup>3<\/sup>\u00a0\u2013 13(0)<sup>\u00a02<\/sup>\u00a0+ 17 x 0 + 12<\/p>\n<p>= 0-0 + 0+ 12<\/p>\n<p><em>= 12<\/em><\/p>\n<p><strong>Question 2: Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:<\/strong><\/p>\n<p><strong>(i) f(x) = 3x + 1, x = \u22121\/3<\/strong><\/p>\n<p><strong>(ii) f(x) = x<sup>2<\/sup>\u00a0\u2013 1, x = 1,\u22121<\/strong><\/p>\n<p><strong>(iii) g(x) = 3x<sup>2<\/sup>\u00a0\u2013 2 , x = 2\/\u221a3 , \u22122\/\u221a3<\/strong><\/p>\n<p><strong>(iv) p(x) = x<sup>3<\/sup>\u00a0\u2013 6x<sup>2<\/sup>\u00a0+ 11x \u2013 6 , x = 1, 2, 3<\/strong><\/p>\n<p><strong>(v) f(x) = 5x \u2013 \u03c0, x = 4\/5<\/strong><\/p>\n<p><strong>(vi) f(x) = x<sup>2<\/sup>\u00a0, x = 0<\/strong><\/p>\n<p><strong>(vii) f(x) = lx + m, x = \u2212m\/l<\/strong><\/p>\n<p><strong>(viii) f(x) = 2x + 1, x = 1\/2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a0f(x) = 3x + 1, x = \u22121\/3<\/p>\n<p>f(x) = 3x + 1<\/p>\n<p>Substitute x = \u22121\/3 in f(x)<\/p>\n<p>f( \u22121\/3) = 3(\u22121\/3) + 1<\/p>\n<p>= -1 + 1<\/p>\n<p>= 0<\/p>\n<p><em>Since, the result is 0, so x = \u22121\/3 is the root of 3x + 1<\/em><\/p>\n<p><strong>(ii)<\/strong>\u00a0f(x) = x<sup>2<\/sup>\u00a0\u2013 1, x = 1,\u22121<\/p>\n<p>f(x) = x<sup>2<\/sup>\u00a0\u2013 1<\/p>\n<p>Given that x = (1 , -1)<\/p>\n<p>Substitute x = 1 in f(x)<\/p>\n<p>f(1) = 1<sup>2<\/sup>\u00a0\u2013 1<\/p>\n<p>= 1 \u2013 1<\/p>\n<p>= 0<\/p>\n<p>Now, substitute x = (-1) in f(x)<\/p>\n<p>f(-1) = (\u22121)<sup>2<\/sup>\u00a0\u2013 1<\/p>\n<p>= 1 \u2013 1<\/p>\n<p>= 0<\/p>\n<p><em>Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x<sup>2\u00a0<\/sup>\u2013 1<\/em><\/p>\n<p><strong>(iii)<\/strong>\u00a0g(x) = 3x<sup>2<\/sup>\u00a0\u2013 2 , x = 2\/\u221a3 , \u22122\/\u221a3<\/p>\n<p>g(x) = 3x<sup>2<\/sup>\u00a0\u2013 2<\/p>\n<p>Substitute x = 2\/\u221a3 in g(x)<\/p>\n<p>g(2\/\u221a3) = 3(2\/\u221a3)<sup>2<\/sup>\u00a0\u2013 2<\/p>\n<p>= 3(4\/3) \u2013 2<\/p>\n<p>= 4 \u2013 2<\/p>\n<p>= 2 \u2260 0<\/p>\n<p>Now, Substitute x = \u22122\/\u221a3 in g(x)<\/p>\n<p>g(2\/\u221a3) = 3(-2\/\u221a3)<sup>2<\/sup>\u00a0\u2013 2<\/p>\n<p>= 3(4\/3) \u2013 2<\/p>\n<p>= 4 \u2013 2<\/p>\n<p>= 2 \u2260 0<\/p>\n<p><em>The results when x = 2\/\u221a3 and x = \u22122\/\u221a3) are not 0. Therefore, (2\/\u221a3 , \u22122\/\u221a3 ) are not zeros of 3x<sup>2<\/sup>\u20132.<\/em><\/p>\n<p><strong>(iv)<\/strong>\u00a0p(x) = x<sup>3<\/sup>\u00a0\u2013 6x<sup>2<\/sup>\u00a0+ 11x \u2013 6 , x = 1, 2, 3<\/p>\n<p>p(1) = 1<sup>3<\/sup>\u00a0\u2013 6(1)<sup>2<\/sup>\u00a0+ 11x 1 \u2013 6 = 1 \u2013 6 + 11 \u2013 6 = 0<\/p>\n<p>p(2) = 2<sup>3<\/sup>\u00a0\u2013 6(2)<sup>2<\/sup>\u00a0+ 11\u00d72 \u2013 6 = 8 \u2013 24 + 22 \u2013 6 = 0<\/p>\n<p>p(3) = 3<sup>3<\/sup>\u00a0\u2013 6(3)<sup>2<\/sup>\u00a0+ 11\u00d73 \u2013 6 = 27 \u2013 54 + 33 \u2013 6 = 0<\/p>\n<p><em>Therefore, x = 1, 2, 3 are zeros of p(x).<\/em><\/p>\n<p><strong>(v)<\/strong>\u00a0f(x) = 5x \u2013 \u03c0, x = 4\/5<\/p>\n<p>f(4\/5) = 5 x 4\/5 \u2013 \u03c0 = 4 \u2013 \u03c0 \u2260 0<\/p>\n<p><em>Therefore, x = 4\/5 is not a zeros of f(x).<\/em><\/p>\n<p><strong>(vi)<\/strong>\u00a0f(x) = x<sup>2<\/sup>\u00a0, x = 0<\/p>\n<p>f(0) = 0<sup>2<\/sup>\u00a0= 0<\/p>\n<p><em>Therefore, x = 0 is a zero of f(x).<\/em><\/p>\n<p><strong>(vii)<\/strong>\u00a0f(x) = lx + m, x = \u2212m\/l<\/p>\n<p>f(\u2212m\/l) = l x \u2212m\/l + m = -m + m = 0<\/p>\n<p><em>Therefore, x = \u2212m\/l is a zero of f(x).<\/em><\/p>\n<p><strong>(viii)<\/strong>\u00a0f(x) = 2x + 1, x = \u00bd<\/p>\n<p>f(1\/2) = 2x 1\/2 + 1 = 1 + 1 = 2 \u2260 0<\/p>\n<p><em>Therefore, x = \u00bd is not a zero of f(x).<\/em><\/p>\n<p>This is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.1. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 9 Maths exam, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-9-maths-chapter-4-exercise-41\"><\/span><strong>FAQs on RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.1<\/strong><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-1631002597091\" 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-for-class-9-maths-chapter-4-exercise-41\"><\/span>How many questions are there in RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 14 questions in\u00a0RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.1.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631002617013\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-4-exercise-41\"><\/span>From where can I download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631002645155\" 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-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-4-exercise-41\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1: Practice questions from RD Sharma Solutions Class 9 Maths and ace your Maths exam this year. You don&#8217;t have to study too many books as RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1 has everything you might need. To know more, you have to read &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Exercise 4.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-1\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Exercise 4.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":125235,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411],"tags":[3086,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124995"}],"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=124995"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124995\/revisions"}],"predecessor-version":[{"id":511194,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124995\/revisions\/511194"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125235"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=124995"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=124995"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=124995"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}