{"id":11415,"date":"2024-03-09T10:04:45","date_gmt":"2024-03-09T04:34:45","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=11415"},"modified":"2024-03-12T17:48:35","modified_gmt":"2024-03-12T12:18:35","slug":"ts-inter-2nd-year-maths-2a-solutions-chapter-7-ex-7a","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2a-solutions-chapter-7-ex-7a\/","title":{"rendered":"TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a)"},"content":{"rendered":"

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

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

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

I.
\nQuestion 1.
\n\\(\\frac{2 x+3}{(x+1)(x-3)}\\)
\nSolution:
\nLet \\(\\frac{2 x+3}{(x+1)(x-3)}=\\frac{A}{x+1}+\\frac{B}{x-3}\\)
\n\u21d2 A (x – 3) + B (x + 1) = 2x – 3 …………..(1)
\nSubstituting x = 3 in (1),
\nweget 4B = 9 .
\n\u21d2 B = \\(\\frac{9}{4}\\)
\nSubstituting x = – 1 in (1),
\nwe get – 4A = 1
\n\u21d2 A = \\(\\frac{-1}{4}\\)
\n\u2234 \\(\\frac{2 x+3}{(x+1)(x-3)}=\\frac{9}{4(x-3)}-\\frac{1}{4(x+1)}\\).<\/p>\n

Question 2.
\n\\(\\frac{5 x+6}{(2+x)(1-x)}\\)
\nSolution:
\nLet \\(\\frac{5 x+6}{(2+x)(1-x)}=\\frac{A}{2+x}+\\frac{B}{1-x}\\)
\n\u21d2 A (1 – x) + B (2 + x) = 5x + 6 ……………..(1)
\nSubstituting x = 1 in (I),
\nweget 3B = 11
\n\u21d2 B = \\(\\frac{11}{3}\\)
\nSubstituting x = – 2 in (1),
\nwe get 3A = – 4
\n\u21d2 A = \\(\\frac{-4}{3}\\)
\n\u2234 \\(\\frac{5 x+6}{(2+x)(1-x)}=\\frac{11}{3(1-x)}-\\frac{4}{3(2+x)}\\).<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\n\\(\\frac{3 x+7}{x^2-3 x+2}\\)
\nSolution:
\nWe know that
\n\\(\\frac{3 x+7}{x^2-3 x+2}=\\frac{3 x+7}{(x-2)(x-1)}\\)
\nLet \\(\\frac{3 x+7}{(x-2)(x-1)}=\\frac{A}{x-2}+\\frac{B}{x-1}\\)
\n\u21d2 A (x – 1) + B(x – 2) = 3x + 7 …………..(1)
\nSubstitutIng x = 2 in (1)
\nwe get A = 13
\nSubstituting x = 1 in (1)
\nwe get – B = 10 i.e., B = – 10
\n\u2234 \\(\\frac{3 x+7}{x^2-3 x+2}=\\frac{13}{x-2}-\\frac{10}{x-1}\\)<\/p>\n

Question 2.
\n\\(\\frac{x+4}{\\left(x^2-4\\right)(x+1)}\\)
\nSolution:
\nWe know that
\n\\(\\frac{x+4}{\\left(x^2-4\\right)(x+1)}=\\frac{x+4}{(x-2)(x+2)(x+1)}\\)
\nLet \\(\\frac{x+4}{(x-2)(x+2)(x+1)}\\) = \\(\\frac{A}{x-2}+\\frac{B}{x+2}+\\frac{C}{x+1}\\)
\nA (x + 2) (x + 1) + B (x – 2) (x + 1) + C (x – 2) (x + 2) = x + 4 …………..(1)
\nSubstituting x = 2 in (1), we have
\n12A = 6
\nA = \\(\\frac{1}{2}\\)
\nSubstituting x = – 2 in (1), we have
\n4B = 2
\n\u21d2 B = \\(\\frac{1}{2}\\)
\nSubstituting x = – 1 in (1), we have
\n– 3C = 3
\n\u21d2 C = – 1
\n\u2234 \\(\\frac{x+4}{\\left(x^2-4\\right)(x+1)}\\) = \\(\\frac{1}{2(x-2)}+\\frac{1}{2(x+2)}-\\frac{1}{x+1}\\)<\/p>\n

Question 3.
\n\\(\\frac{2 x^2+2 x+1}{x^3+x^2}\\)
\nSolution:
\nWe know that
\n\\(\\frac{2 x^2+2 x+1}{x^3+x^2}=\\frac{2 x^2+2 x+1}{x^2(x+1)}\\)
\nLet \\(\\frac{2 x^2+2 x+1}{x^2(x+1)}=\\frac{A}{x}+\\frac{B}{x^2}+\\frac{C}{x+1}\\)
\n\u21d2 Ax (x + 1) + B (x + 1) + Cx2 = 2x + 2x + 1
\nSubstituting x = 0 in (1), we have B = 1
\nSubstituting x = – 1 in (1), we have C = 1
\nEquating coefficient of x2<\/sup> on both sides in (1), we have
\nA + C = 2
\n\u21d2 A = 1
\n\u2234 \\(\\frac{2 x^2+2 x+1}{x^3+x^2}=\\frac{1}{x}+\\frac{1}{x^2}+\\frac{1}{x+1}\\).<\/p>\n

\"TS<\/p>\n

Question 4.
\n\\(\\frac{2 x+3}{(x-1)^3}\\)
\nSolution:
\nLet \\(\\frac{2 x+3}{(x-1)^3}\\) = \\(\\frac{A}{x-1}+\\frac{B}{(x-1)^2}+\\frac{C}{(x-1)^3}\\)
\n\u21d2 A(x – 1)2<\/sup> + B(x – 1) + C = 2x + 3 ……………..(1)
\nSubstituting x = 1 in (1).
\nwe get C = 5
\nEquating coefficient of x2<\/sup> on both sides in (1)
\nWe get A = 0
\nEquating coefficient of x on both sides in (1)
\nWe get – 2A + B = 2
\n\u21d2 B = 2.<\/p>\n

Alternate method:
\nLet x – 1 = y<\/p>\n

\"TS<\/p>\n

Question 5.
\n\\(\\frac{x^2-2 x+6}{(x-2)^3}\\)
\nSolution:
\nLet x – 2 = y<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

III.
\nQuestion 1.
\n\\(\\frac{x^2-x+1}{(x+1)(x-1)^2}\\)
\nSolution:
\nLet \\(\\frac{x^2-x+1}{(x+1)(x-1)^2}\\) = \\(\\frac{A}{x+1}+\\frac{B}{x-1}+\\frac{C}{(x-1)^2}\\)
\nA (x – 1)2<\/sup> + B (x + 1) (x – 1) + C (x + 1) = x2<\/sup> – x + 1 ………..(1)
\nSubstituting x = 1 in (1), we get
\n2C = 1
\n\u21d2 C = \\(\\frac{1}{2}\\)
\nSubstituting x = – 1 in (1), we get
\n4A = 3
\n\u21d2 A = \\(\\frac{3}{4}\\)
\nEquating coefficient of x2<\/sup> on both sides in (1)
\nWe get A + B = 1
\n\\(\\frac{3}{4}\\) + B = 1
\n\u21d2 B = \\(\\frac{1}{4}\\)
\n\u2234 \\(\\frac{x^2-x+1}{(x+1)(x-1)^2}\\) = \\(\\frac{3}{4(x+1)}+\\frac{1}{4(x-1)}+\\frac{1}{2(x-1)^2}\\)<\/p>\n

Question 2.
\n\\(\\frac{9}{(x-1)(x+2)^2}\\)
\nSolution:
\nLet \\(\\frac{9}{(x-1)(x+2)^2}\\) = \\(\\frac{A}{x-1}+\\frac{B}{x+2}+\\frac{C}{(x+2)^2}\\)
\n\u21d2 A (x + 2)2<\/sup> + B (x – 1) (x + 2) + C (x – 1) = 9 …………….(1)
\nSubstituting x = 1 in (1), we get
\n9A = 9
\n\u21d2 A = 1
\nSubstituting x = – 2 in (1), we get
\n– 3C = 9
\n\u21d2 C = – 3
\nEquating coefficient of x2<\/sup> on both sides in (1),
\nwe get A + B = 0
\n\u21d2 B = – 1
\n\u2234 \\(\\frac{9}{(x-1)(x+2)^2}\\) = \\(\\frac{1}{x-1}-\\frac{1}{(x+2)}-\\frac{3}{(x+2)^2}\\).<\/p>\n

\"TS<\/p>\n

Question 3.
\n\\(\\frac{1}{(1-2 x)^2(1-3 x)}\\)
\nSolution:
\nLet \\(\\frac{1}{(1-2 x)^2(1-3 x)}\\) = \\(\\frac{A}{(1-3 x)}+\\frac{B}{(1-2 x)}+\\frac{C}{(1-2 x)^2}\\)
\n\u21d2 A (1 – 2x)2<\/sup> + B (1 – 3x) (1 – 2x) + C (1 – 3x) = 1 …………..(1)
\nSubstituting x = \\(\\frac{1}{2}\\) in (1),
\nwe get \\(\\frac{-C}{2}\\) = 1
\n\u21d2 C = – 2
\nSubstituting x = \\(\\frac{1}{3}\\) in (1),
\nwe get \\(\\frac{\\mathrm{A}}{9}\\) = 1
\n\u21d2 A = 9
\nSubstituting x = 0 in (1),
\nWe get A + B + C = 1
\n\u21d2 9 + B – 2 = 1
\n\u21d2 B = – 6
\n\u2234 \\(\\frac{1}{(1-2 x)^2(1-3 x)}\\) = \\(\\frac{9}{1-3 x}-\\frac{6}{1-2 x}-\\frac{2}{(1-2 x)^2}\\)<\/p>\n

Question 4.
\n\\(\\frac{1}{x^3(x+a)}\\)
\nSol.
\nLet \\(\\frac{1}{x^3(x+a)}=\\frac{A}{x+a}+\\frac{B}{x}+\\frac{C}{x^2}+\\frac{D}{x^3}\\)
\n\u21d2 Ax3<\/sup> + Bx2<\/sup> (x + a) + Cx (x + a) + D (x + a) = 1 …………(1)
\nSubstituting x = – a in (1),
\nwe get – a3<\/sup>A = 1
\n\u21d2 A = \\(\\frac{-1}{a^3}\\)
\nEquating coefficient of x3<\/sup> on both sides,
\nwe get A + B = 0
\n\u21d2 B = \\(\\frac{-1}{a^3}\\)
\nSubstituting x = 0 in (1),
\nwe get aD = 1
\nEquating coefficient of x on both sides,
\nwe get aC + D = 0
\n\u21d2 aC + \\(\\frac{1}{a}\\) = 0
\n\u21d2 C = \\(\\frac{-1}{a^2}\\)
\n\u2234 \\(\\frac{1}{x^3(x+a)}\\) = \\(\\frac{-1}{a^3(x+a)}+\\frac{1}{a^3 x}-\\frac{1}{a^2 x^2}+\\frac{1}{a x^3}\\).<\/p>\n

\"TS<\/p>\n

Question 5.
\n\\(\\frac{x^2+5 x+7}{(x-3)^3}\\)
\nSolution:
\nLet x – 3 = y<\/p>\n

\"TS<\/p>\n

Question 6.
\n\\(\\frac{3 x^3-8 x^2+10}{(x-1)^4}\\)
\nSolution:
\nLet x – 1 = y<\/p>\n

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

Students must practice this TS Intermediate Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a) Resolve the following fractions into partial fractions. I. Question 1. Solution: Let \u21d2 A (x – 3) + … 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\/11415"}],"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=11415"}],"version-history":[{"count":1,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11415\/revisions"}],"predecessor-version":[{"id":11421,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11415\/revisions\/11421"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=11415"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=11415"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=11415"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}