{"id":11521,"date":"2024-03-08T14:21:38","date_gmt":"2024-03-08T08:51:38","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=11521"},"modified":"2024-03-11T17:54:20","modified_gmt":"2024-03-11T12:24:20","slug":"ts-inter-2nd-year-maths-2a-solutions-chapter-7-ex-7c","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2a-solutions-chapter-7-ex-7c\/","title":{"rendered":"TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(c)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions<\/a> Chapter 7 Partial Fractions Ex 7(c) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(c)<\/h2>\n

Resolve the following fractions into partial fractions.<\/p>\n

Question 1.
\n\\(\\frac{x^2}{(x-1)(x-2)}\\)
\nSolution:
\nThe given rational fraction \\(\\frac{x^2}{(x-1)(x-2)}\\) is improper with degree of numerater is equal
\nto degree of denominator.
\n\u2234 \\(\\frac{x^2}{(x-1)(x-2)}\\) = 1 + \\(\\frac{r(x)}{(x-1)(x-2)}\\)
\nLet \\(\\frac{x^2}{(x-1)(x-2)}\\) = 1 + \\(\\frac{A}{x-1}+\\frac{B}{x-2}\\)
\n\u21d2 (x – 1) (x – 2) + A (x – 2) + B (x – 1) = x2<\/sup> …………..(1)
\nSubstituting x = 1 in (1), we get
\n– A = 1
\n\u21d2 A = – 1
\nSubstituting x = 2 in (1), we get
\nB = 4
\n\u2234 \\(\\frac{x^2}{(x-1)(x-2)}\\) = 1 – \\(\\frac{1}{x-1}+\\frac{4}{x-2}\\).<\/p>\n

\"TS<\/p>\n

Question 2.
\n\\(\\frac{x^3}{(x-1)(x+2)}\\)
\nSolution:
\nThe given rational fraction \\(\\frac{x^3}{(x-1)(x+2)}\\) is improper with degree of numerator is greater than degree of denominator.
\nClearly
\n\\(\\frac{x^3}{(x-1)(x+2)}\\) = (x – 1) + \\(\\frac{3 x-2}{(x-1)(x+2)}\\)
\nLet \\(\\frac{3 x-2}{(x-1)(x+2)}\\) = \\(\\frac{A}{(x-1)}+\\frac{B}{x+2}\\)
\n\u21d2 A (x + 2) + B(x – 1) = 3x – 2 …………..(1)
\nSubstituting x = 1 in (1), we get
\n3A = 1
\n\u21d2 A = \\(\\frac{1}{3}\\)
\nSubstituting x = – 2 in (1), we get
\n– 3B = – 8
\n\u21d2 B = \\(\\frac{8}{3}\\)
\n\u2234 \\(\\frac{x^3}{(x-1)(x+2)}\\) = x – 1 + \\(\\frac{1}{3(x-1)}+\\frac{8}{3(x+2)}\\).<\/p>\n

\"TS<\/p>\n

Question 3.
\n\\(\\frac{x^3}{(2 x-1)(x-1)^2}\\)
\nSolution:
\nThe given rational fraction \\(\\frac{x^3}{(2 x-1)(x-1)^2}\\) is improper as degree of numerator is equal to degree of denominator.
\nClearly \\(\\frac{x^3}{(2 x-1)(x-1)^2}=\\frac{1}{2}+\\frac{r(x)}{(2 x-1)(x-1)^2}\\)
\nLet \\(\\frac{x^3}{(2 x-1)(x-1)^2}\\) = \\(\\frac{1}{2}\\) + \\(\\frac{A}{(2 x-1)}+\\frac{B}{(x-1)}+\\frac{C}{(x-1)^2}\\)
\n\u21d2 (2x – 1) (x – 1)2<\/sup> + 2A (x – 1)2<\/sup> + 2B (2x – 1)(x – 1) + 2C (2x – 1) = 2x3<\/sup> ………..(1)
\nSubstituting x = 1 in (1), we get C = 1
\nSubstituting x = \\(\\frac{1}{2}\\) in (1), we get
\n\\(\\frac{\\mathrm{A}}{2}=\\frac{1}{4}\\)
\n\u21d2 A = \\(\\frac{1}{2}\\)
\nSubstituting x = 0 in (I), we get
\n– 1 + 2A + 2B – C = 0
\n\u21d2 – 1 + 1 + 2B – 2 = 0
\n\u21d2 B = 1
\n\u2234 \\(\\frac{x^3}{(2 x-1)(x-1)^2}\\) = \\(\\frac{1}{2}+\\frac{1}{2(2 x-1)}+\\frac{1}{x-1}+\\frac{1}{(x-1)^2}\\).<\/p>\n

\"TS<\/p>\n

Question 4.
\n\\(\\frac{x^3}{(x-a)(x-b)(x-c)}\\)
\nSolution:
\nThe given rational fraction \\(\\frac{x^3}{(x-a)(x-b)(x-c)}\\) is improper as degree of numerator is equal to degree of denominator.
\nLet \\(\\frac{x^3}{(x-a)(x-b)(x-c)}\\) = 1 + \\(\\frac{A}{x-a}+\\frac{B}{x-b}+\\frac{C}{x-c}\\)
\n\u21d2 (x – a) (x – b) (x – c) + A (x – b) (x – c) + B (x – a) (x – c) + C (x – a) (x – b) = x3<\/sup> ………..(1)
\nSubstituting x = a in (1), we get,
\nA (a – b) (a – c) = a3<\/sup>
\n\u21d2 A = \\(\\frac{a^3}{(a-b)(a-c)}\\)
\nSubstituting x = b in (1), we get.
\nB = \\(\\frac{b^3}{(b-c)(b-a)}\\)
\nSubstituting x= c in (1), we get<\/p>\n

\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(c) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(c) Resolve the following fractions into partial fractions. Question 1. Solution: The given rational fraction is improper with degree … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11521"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=11521"}],"version-history":[{"count":2,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11521\/revisions"}],"predecessor-version":[{"id":11524,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11521\/revisions\/11524"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=11521"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=11521"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=11521"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}