{"id":520,"date":"2023-11-01T09:27:04","date_gmt":"2023-11-01T03:57:04","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=520"},"modified":"2023-11-02T10:28:59","modified_gmt":"2023-11-02T04:58:59","slug":"kseeb-solutions-class-9-maths-chapter-4-ex-4-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-4-ex-4-2\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.2.<\/p>\n

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.2<\/h2>\n

Question 1.
\nFind the value of the polynomial 5x – 4x2<\/sup> + 3 at
\ni) x = 0
\nii) x = -1
\niii) x = 2
\nAnswer:
\ni) f(x) = 5x – 4x2<\/sup> + 3 x = 0 then,
\nf(0) = 5(0) – 4(0)2<\/sup> + 3
\n=0 – 0 + 3
\nf(0) = 3<\/p>\n

ii) f(x) = 5x – 4x2<\/sup> + 3 x = -1 then,
\nf(-1) = 5(-1) – 4(-1)2 + 3
\n= -5 – 4(+1) + 3
\n= -5 – 4 + 3
\n= -9 + 3
\nf(-1) = -6<\/p>\n

iii) f(x) = 5x – 4x2<\/sup> + 3 x = 2 then,
\nf(2) = 5(2) – 4(2)2<\/sup> + 3
\n= 5(2) – 4(4) + 3
\n= 10 – 16 + 3
\n= 13 – 16
\nf(2) = -3<\/p>\n

Question 2.
\nFind p(0), p(1) and p(2) for each of the following polynomials :
\ni) p(y) = y2<\/sup> – y + 1
\nii) p(t) = 2 + t + 2t2<\/sup> – t3<\/sup>
\niii) p(x) = x3<\/sup>
\niv) p(x) = (x – 1) (x + 1)
\nAnswer:
\ni) (a) p(y) = y2<\/sup> – y + 1
\np(0) = (0)2<\/sup> – 0 + 1
\n= 0 – 0 + 1
\n\u2234 p(0) = =1<\/p>\n

(b) p(y) = y2<\/sup> – y + 1
\np(1) = (1)2<\/sup> – 1 + 1
\n= 1 – 1 + 1
\n\u2234 p(1) = 1<\/p>\n

(c) p(y) = y2<\/sup> – y +
\np(2) = (2)2<\/sup> – 2 + 1
\n= 4 – 2 + 1
\n= 5 – 2
\n\u2234 p(2) = 3<\/p>\n

ii) (a) p(t) = 2 + t + 2t2<\/sup> – t3<\/sup>
\np(0) = 2 + 0 + 22<\/sup> – (0)3<\/sup>
\n= 2 + 0 + 0 – 0
\n\u2234 p(0) = 2<\/p>\n

(b) p(t) = 2 + t + 2t2<\/sup> – t3<\/sup>
\np(1) = 2 + 1 + 2(1)2<\/sup> – (1)3<\/sup>
\n= 2 + 1 + 2(1) – 1
\n= 2 + 1 + 2 – 1
\n\u2234 p(1) = 4<\/p>\n

(c) p(t) – 2 + t + 2t2<\/sup> – t3<\/sup>
\np(2) = 2 + 2 + 2(2)2<\/sup> – (2)3<\/sup>
\n= 2 + 2 + 2(4) – 8
\n= 2 + 2 + 8 – 8
\n\u2234 p(2) = 4<\/p>\n

iii) (a) p(x) = x3<\/sup>
\np(0) = (0)3<\/sup>
\np(0) = 0<\/p>\n

(b) p(x) = x3<\/sup>
\np(1) = (1)3<\/sup>
\n\u2234 p(1) = 1<\/p>\n

(c) p(x) = x3<\/sup>
\np(2) = (2)3<\/sup>
\n\u2234 p(2) = 8<\/p>\n

iv) (a) p(x) = (x – 1)(x + 1)
\np(x) = x2<\/sup> – 1 [\u2235 (a + b)(a – b) = a2<\/sup> – b]
\np(0) = (0)2<\/sup> – 1
\n= 0 – 1
\n\u2234 p(o) = -1<\/p>\n

(b) p(x) = (x – 1)(x + 1)
\np(x) = x2<\/sup> – a [\u2235 (a + b)(a – b) = a2<\/sup> – b2<\/sup>]
\np(1) = (1)2<\/sup> – 1
\n= 1 – 1
\n\u2234 p(1) = 0<\/p>\n

(c) p(x) = (x – 1)(x + 1)
\np(x) = x2<\/sup> – 1 [\u2235 (a + b)(a – b) = a2<\/sup> – b2<\/sup>]
\np(2) = (2)2<\/sup> – 1
\n= 4 – 1
\n\u2234 p(0) = 3<\/p>\n

Question 3.
\nVerify whether the following are zeroes of the polynomial, induced against them.
\ni) p(x) = 3x + 1; \\(x=-\\frac{1}{3}\\)
\nii) p(x) = 5x – \u03c0; \\(x=\\frac{4}{5}\\)
\niii) p(x) = x2<\/sup> – 1; x = 1, -1
\niv) p(x) = (x + 1) (x – 2); x = -1, 2
\nv) p(x) = x2<\/sup>; x = 0
\nvi) p(x) = lx + m; \\(x=-\\frac{m}{l}\\)
\nvii) p(x) = 3x2<\/sup> – 1; \\(x=-\\frac{1}{\\sqrt{3}}, \\frac{2}{\\sqrt{3}}\\)
\nviii) p(x) = 2x + 1; \\(x=\\frac{1}{2}\\)
\nAnswer:
\ni) p(x) = 3x + 1; \\(x=-\\frac{1}{3}\\)
\n\"KSEEB
\nHere value of polynomial is zero.
\n\\(x=-\\frac{1}{3}\\) is not zero of the polynomial<\/p>\n

ii) p(x) = 5x – \u03c0; \\( x=\\frac{4}{5}\\)
\n\"KSEEB
\nHere value of polynomial is not zero.
\n\\(x=\\frac{4}{5}\\) is not zero of the polynomial<\/p>\n

iii) p(x) = x2<\/sup> – 1; x = 1, -1
\np(1) = (1)2<\/sup> – 1
\n= 1 – 1
\np(1) = 0
\nHere value of p(x) is zero.
\nhence its zero is 1.
\np(x) = x2<\/sup> – 1; x = -1
\np(-1) = (-1)2<\/sup> – 1
\n= 1 – 1
\np(-1) = 0
\nHere value of p(x) is zero.
\n\u2234 -1 is zero.<\/p>\n

iv) p(x) = (x – 1)(x – 2); x = -1, 2
\np(x) = x2<\/sup> – 2x + x – 2
\np(x) = x2<\/sup> – x + 2 x = -1
\np(-1) = (-1)2<\/sup> – (-1)2<\/sup> + 2
\n= 1 + 1 + 2
\np(-1) = 4
\nHere value of polynomila is not zero.
\n\u2234 -1 is not zero.
\np(x) = x2<\/sup> – x + 2 x = 2
\np(2) = (2)2<\/sup> – (2) + 2
\n= 4 – 2 + 2
\np(-1) = 4
\nHere value of polynomial is not zero.
\n\u2234 2 is not zero.<\/p>\n

(v) p(x) = x2<\/sup>; x = 0
\np(0) = (0)2<\/sup>
\np(0) = 0
\nHere value of p(x) is zero.
\n\u2234 0 is its zero.<\/p>\n

vi) p(x) = lx + m; \\(x=-\\frac{m}{l}\\)
\n\\(\\mathrm{p}\\left(-\\frac{\\mathrm{m}}{l}\\right)=l\\left(-\\frac{\\mathrm{m}}{l}\\right)+\\mathrm{m}\\)
\n= -m + m
\n= 0
\nHere p(x) is zero.
\n\u2234 \\(-\\frac{\\mathrm{m}}{l}\\) is its zero.<\/p>\n

(vii) p(x) = 3x2<\/sup> – 1; \\(x=-\\frac{1}{\\sqrt{3}}, \\frac{2}{\\sqrt{3}}\\)
\n\"KSEEB<\/p>\n

viii) p(x) = 2x + 1; \\(x=\\frac{1}{2}\\)
\n\"KSEEB
\nHere value of p(x) is not zero.
\n\u2234 \\(\\frac{1}{2}\\) is not its zero.<\/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.
\nAnswer:
\ni) p(x) = x + 5
\nLet p(x) =0, then,
\np(x) = x + 5 = 0
\nx = 0 – 5
\n\u2234 x = -5
\n-5 is zero of p(x).<\/p>\n

ii) p(x) = x – 5
\nIf p(x) = 0, then
\np(x) = x – 5 = 0
\nx = 0 + 5
\n\u2234 x = 5
\n5 is the zero of p(x).<\/p>\n

iii) p(x) = 2x + 5
\nIf p(x)= 0, then
\np(x) = 2x + 5 = 0
\n2x = – 5
\n\u2234 \\(x=\\frac{-5}{2}\\)
\n\\(\\frac{-5}{2}\\) is the zero of p(x).<\/p>\n

iv) p(x) = 3x – 2
\nIf p(x)= 0, then
\np(x) = 3x – 2 = 0
\n3x = 2
\n\u2234 \\(x=\\frac{2}{3}\\)
\n\\(\\frac{2}{3}\\) is the zero of p(x).<\/p>\n

v) p(x) = 3x
\nIf p(x) = 0, then
\np(x) = 3x = 0
\n\u2234 \\(x=\\frac{0}{3}\\)
\n\\(\\frac{0}{3}\\) is the zero of p(x)<\/p>\n

vi) p(x) = ax, a \u2260 0
\nIf p(x)= 0, then
\np(x) = ax = 0
\n\u2234 \\(x=\\frac{0}{a}\\) \u2234 x = \u221e(infinity)
\n\u221e is the zero of p(x).<\/p>\n

vii) p(x) = cx + d, c \u2260 0, c, d are real numbers
\nIf p(x)= 0, then
\np(x) = cx + d = 0
\ncx = 0 – d
\ncx = -d
\n\u2234 \\(x=-\\frac{d}{c}\\)
\n\\(-\\frac{\\mathrm{d}}{\\mathrm{c}}\\) is the zero of p(x).<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.2, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.2. Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.2 Question 1. Find the value of the polynomial 5x – 4×2 …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/520"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=520"}],"version-history":[{"count":1,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/520\/revisions"}],"predecessor-version":[{"id":90874,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/520\/revisions\/90874"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=520"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=520"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=520"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}