{"id":24648,"date":"2021-06-19T16:32:28","date_gmt":"2021-06-19T11:02:28","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=24648"},"modified":"2022-03-02T10:38:14","modified_gmt":"2022-03-02T05:08:14","slug":"ncert-solutions-for-class-9-maths-chapter-2-ex-2-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2<\/h2>\n

Question 1.
\nFind the value of the polynomial 5x – 4x2<\/sup> + 3 at
\n(i) x = 0
\n(ii) x = -1
\n(iii) x = 2
\nSolution:
\n(i) Let P(x) = 5x – 4x2<\/sup> + 3
\nTherefore, P(0) = 5(0) – 4(0)2<\/sup> + 3
\n= 0 – 0 + 3
\n= 3
\nSo, the value of P(x) at x = 0 is 3.<\/p>\n

(ii) Let P(x) = 5x – 4x2<\/sup> + 3
\nTherefore, P(-1) = 5(-1) – 4(-1)2<\/sup> + 3
\n= -5 – 4 + 3
\n= -6
\nSo, the value of P(x) at x = -1 is -6.<\/p>\n

(iii) Let P(x) = 5x – 4x2<\/sup> + 3
\nTherefore, P(2) = 5(2) – 4(2)2<\/sup> + 3
\n= 10 – 16 + 3
\n= -3
\nSo, the value of P(x) at x = 2 is -3.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nFind p(0), p(1) and p(2) for each of the following polynomials:
\n(i) p(y) = y2<\/sup> – y + 1
\n(ii) p(t) = 2 + t + 2t2<\/sup> – t3<\/sup>
\n(iii) p(x) = x3<\/sup>
\n(iv) p(x) = (x – 1) (x + 1)
\nSolution:
\n(i) We have given
\np(y) = y2<\/sup> – y + 1
\nTherefore, the value of polynomial p(y) at y = 0 is
\np(0) = 02<\/sup> – 0 + 1 = 1
\nAgain, the value of polynomial p(y) at y = 1 is
\np(1) = 12<\/sup> – 1 + 1 = 1
\nAgain, the value of polynomial p(y) at y = 2 is
\np(2) = 22<\/sup> – 2 + 1 = 3<\/p>\n

(ii) We have given,
\np(t) = 2 + t + 2t2<\/sup> – t3<\/sup>
\nTherefore, the value of polynomial p(t) at t = 0 is
\np(0) = 2 + 0 + 2(0)2<\/sup> – (0)3<\/sup> = 2
\nAgain, the value of polynomial p(t) at t = 1 is
\np(1) = 2 + 1 + 2(1)2<\/sup> – (1)3<\/sup> = 4
\nAgain, the value of polynomial p(t) at t = 2 is
\np(2) = 2 + 2 + 2(2)2<\/sup> – (2)3<\/sup> = 4<\/p>\n

(iii) We have given,
\nP(x) = x3<\/sup>,
\nTherefore, the value of polynomial p(x) at x = 0 is
\np(0) = (0)3<\/sup> = 0
\nAgain the value of polynomial p(x) at x = 1 is
\np(1) = (1)3<\/sup> = 1
\nAgain, the value of polynomial p(x) at x = 2 is
\nP(2) = (2)3<\/sup> = 8<\/p>\n

(iv) We have given
\np(x) = (x – 1)(x + 1)
\nTherefore, the value of polynomial p(x) at x = 0 is
\nP(0) = (0 – 1)(0 + 1)
\n= (-1) \u00d7 (+1)
\n= -1
\nAgain, the value of polynomial p(x) at x = 1 is
\np(1) = (1 – 1) \u00d7 (1 + 1)
\n= 0 \u00d7 2
\n= 0
\nAgain, the value of polynomial p(x), at x = 2 is
\np(2) = (2 – 1)(2 + 1)
\n= 1 \u00d7 3
\n= 3<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nVerify whether the following are zeros of the polynomial indicated against them:
\n(i) p(x) = 3x + 1, x = \\(-\\frac{1}{3}\\)
\n(ii) p(x) = 5x – p, x = \\(\\frac{4}{5}\\)
\n(iii) p(x) = x2<\/sup> – 1, x = 1, -1
\n(iv) p(x) = (x + 1) (x – 2), x = -1, 2
\n(v) p(x) = x2<\/sup>, x = 0
\n(vi) p(x) = 1x + m, x = \\(-\\frac{m}{1}\\)
\n(vii) p(x) = 3x2<\/sup> – 1, x = \\(-\\frac{1}{\\sqrt{3}}, \\frac{2}{\\sqrt{3}}\\)
\n(viii) p(x) = 2x + 1, x = \\(\\frac{1}{2}\\)
\nSolution:
\n(i) We have given that, p(x) = 3x + 1.
\nTherefore, the value of polynomial p(x) at x = \\(-\\frac{1}{3}\\) is
\n\\(P\\left(-\\frac{1}{3}\\right)=3 \\times \\frac{(-1)}{3}+1\\) = 0
\nyes x = \\(-\\frac{1}{3}\\) is the zero of polynomial p(x).<\/p>\n

(ii) We have given that p(x) = 5x – p
\nTherefore, the value of polynomial p(x) at x = \\(\\frac{4}{5}\\) is
\n\\(P\\left(\\frac{4}{5}\\right)=5 \\times \\frac{4}{5}-\\frac{22}{7}\\) (\u2235 \u03c0 = \\(\\frac {22}{7}\\))
\n= 4 – \\(\\frac{22}{7}\\)
\n\\(P\\left(\\frac{4}{5}\\right)=\\frac{6}{7}\\)
\nNo, x = \\(\\frac{4}{5}\\) is not the zero of p(x) = 5x – \u03c0.<\/p>\n

(iii) We have given that p(x) = x2<\/sup> – 1
\nTherefore, the value of polynomial p(x) at x = 1 is
\np(1) = 12<\/sup> – 1 = 0
\nAgain, the value of polynomial p(x) at x = -1 is
\np(-1) = (-1)2<\/sup> – 1 = 0
\nyes x = 1, -1 are the zero of polynomial p(x) = x2<\/sup> – 1<\/p>\n

(iv) We have given that
\np(x) = (x + 1)(x – 2)
\nTherefore, the value of polynomial p(x) at x = -1 is
\np(-1) = (-1 + 1)(-1 – 2)
\n= 0 \u00d7 (-3)
\n= 0
\nAgain, the value of polynomial p(x) at x = 2 is
\nP(2) = (2 + 1)(2 – 2)
\n= 3 \u00d7 0
\n= 0
\nyes x = -1, 2 are the zero of polynomial p(x) = (x + 1)(x – 2)<\/p>\n

\"NCERT<\/p>\n

(v) We have given that p(x) = x2<\/sup>
\nTherefore, the value of polynomial p(x) at x = 0 is
\np(0) = 02<\/sup> = 0
\nyes x = 0 is the zero of polynomial p(x) = x2<\/sup>.<\/p>\n

(vi) We have given that, p(x) = lx + m
\nTherefore, the value of polynomial p(x) at x = \\(-\\frac{m}{\\ell}\\) is
\n\\(P\\left(-\\frac{m}{\\ell}\\right)=\\ell\\left(-\\frac{m}{\\ell}\\right)+m\\) = 0
\nyes x = \\(-\\frac{m}{\\ell}\\) is the zero of polynomial p(x) = lx + m.<\/p>\n

(vii) We have given that, p(x) = 3x2<\/sup> – 1
\nTherefore, the value of polynomial p(x) at x = \\(-\\frac{1}{\\sqrt{3}}\\) is
\n\\(P\\left(-\\frac{1}{\\sqrt{3}}\\right)=3 \\times\\left(-\\frac{1}{\\sqrt{3}}\\right)^{2}-1\\)
\n= \\(3 \\times \\frac{1}{3}-1\\)
\n= 0
\nAgain, the value of polynomial p(x) at x = \\(\\frac{2}{\\sqrt{3}}\\) is
\n\\(P\\left(\\frac{2}{\\sqrt{3}}\\right)=3 \\times\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}-1\\)
\n= \\(3 \\times \\frac{4}{3}-1\\)
\n= 4 – 1
\n= 3
\nTherefore, x = \\(-\\frac{1}{\\sqrt{3}}\\) is the zero of polynomial p(x) and x = \\(\\frac{2}{\\sqrt{3}}\\) is not the zero of polynomial p(x).<\/p>\n

(viii) We have given that p(x) = 2x + 1
\nTherefore, the value of polynomial p(x) at x = \\(\\frac{1}{2}\\) is
\n\\(\\mathrm{P}\\left(\\frac{1}{2}\\right)=2 \\times \\frac{1}{2}+1\\) = 2
\nTherefore, x = \\(\\frac{1}{2}\\) is not the zero of polynomial p(x).<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nFind the zero of the polynomial in each of the following cases:
\n(i) p(x) = x + 5
\n(ii) p(x) = x – 5
\n(iii) p(x) = 2x + 5
\n(iv) p(x) = 3x – 2
\n(v) p(x) = 3x
\n(vi) p(x) = ax, a \u2260 0
\n(vii) p(x) = cx + d, c \u2260 0; c, d are real numbers.
\nSolution:
\n(i) We have given that
\np(x) = x + 5 ……..(i)
\nTo find the zero of polynomial p(x), we can take
\nP(x) = 0 ………(ii)
\nfrom equation (i) and (ii)
\nx + 5 = 0
\n\u2234 x = -5
\nTherefore, x = -5 is the zero of polynomial p(x) = x + 5.<\/p>\n

(ii) We have given that,
\np(x) = x – 5 ……..(i)
\nTo find the zero of polynomial p(x) we can take
\np(x) = 0 ……..(ii)
\nfrom equation (i) and (ii)
\nx – 5 = 0
\n\u2234 x = 5
\nTherefore x = 5 is the zero of polynomial p(x) = x – 5.<\/p>\n

(iii) We have given that
\np(x) = 2x + 5 ………(i)
\nTo find the zero of polynomial p(x), we can take
\np(x) = 0 …….(ii)
\nfrom equation (i) and (ii)
\n2x + 5 = 0
\nor x = \\(-\\frac{5}{2}\\)
\nTherefore x = \\(-\\frac{5}{2}\\) is the zero of polynomial p(x) = 2x + 5.<\/p>\n

\"NCERT<\/p>\n

(iv) We have given that
\np(x) = 3x – 2 …….(i)
\nTo find the zero of polynomial p(x) we can take
\np(x) = 0 ……..(ii)
\n3x – 2 = 0
\nx = \\(\\frac{2}{3}\\)
\nTherefore, x = \\(\\frac{2}{3}\\) is the zero of polynomial p(x) = 3x – 2.<\/p>\n

(v) We have given that
\np(x) = 3x ……..(i)
\nTo find the zero of polynomial p(x) we can take
\nP(x) = 0 ……..(ii)
\nfrom equation (i) and (ii)
\n3x = 0
\nx = 0
\nTherefore, x = 0 is the zero of polynomial p(x) = 3x.<\/p>\n

(vi) We have given that,
\np(x) = ax, a \u2260 0 ……(i)
\nTo find the zero of polynomial p(x) we can take
\np(x) = 0 ………(ii)
\nfrom equation (i) and (ii)
\nax = 0
\nx = \\(\\frac{0}{a}\\) = 0
\nTherefore, x = 0 is the zero of polynomial p(x) = ax, a \u2260 0.<\/p>\n

\"NCERT<\/p>\n

(vii) We have given
\np(x) = cx + d …….(i)
\nc \u2260 0, c, d are real numbers.
\nTo find the zero of polynomial p(x) we can take
\np(x) = 0 ………(ii)
\nfrom equation (i) and (ii)
\ncx + d = 0
\nx = \\(-\\frac{d}{c}\\)
\nwhere c \u2260 0 and c, d are real numbers.
\nTherefore, x = \\(\\frac{d}{c}\\), where c \u2260 0 and c, d are real numbers is the solution of polynomial p(x) = cx + d.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Question 1. Find the value of the polynomial 5x – 4×2 + 3 at (i) x = 0 (ii) x = …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. 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