{"id":35212,"date":"2022-08-25T11:20:27","date_gmt":"2022-08-25T05:50:27","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35212"},"modified":"2022-08-25T11:20:27","modified_gmt":"2022-08-25T05:50:27","slug":"inter-2nd-year-maths-2a-partial-fractions-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2a-partial-fractions-important-questions\/","title":{"rendered":"Inter 2nd Year Maths 2A Partial Fractions Important Questions"},"content":{"rendered":"

Students get through Maths 2A Important Questions<\/a> Inter 2nd Year Maths 2A Partial Fractions Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 2nd Year Maths 2A Partial Fractions Important Questions<\/h2>\n

Question 1.
\n\\(\\frac{x+4}{\\left(x^{2}-4\\right)(x+1)}\\) (Mar. 14)
\nSolution:
\n\\(\\frac{x+4}{\\left(x^{2}-4\\right)(x+1)}\\) = \\(\\frac{A}{x+1}\\) + \\(\\frac{B}{x+2}\\) + \\(\\frac{c}{x-2}\\)
\nMultiplying with (x2<\/sup> – 4) (x + 1)
\nx + 4 = A(x2<\/sup> – 4) + B(x + 1) (x – 2) + C (x + 1) (x + 2)
\nx = -1 \u21d2 3 = A(1 – 4) = -3A \u21d2 A = -1
\nx = -2 \u21d2 2 = B(-2 + 1) (-2 – 2)
\n= 4B \u21d2 B = + \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\)
\nx = 2 \u21d2 6 = C(2 + 1)(2 + 2)
\n= 12C \u21d2 C = \\(\\frac{1}{2}\\)<\/p>\n

Question 2.
\n\\(\\frac{x^{2}-x+1}{(x+1)(x-1)^{2}}\\) (TS Mar. 15)
\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}}\\)
\nMultiplying with (x + 1) (x – 1)2<\/sup>
\nx2<\/sup> – x + 1 = A(x – 1)2<\/sup> + B(x + 1) (x – 1) + C(x + 1)
\nPut x = -1, 1 + 1 + 1 = A(4)
\n\u21d2 A = \\(\\frac{3}{4}\\)
\nPut x = 1, 1 – 1 + 1 = C(2)
\n\u21d2 C = + \\(\\frac{1}{2}\\)
\nEquating the coefficients of x2<\/sup>,
\nA + B = 1
\n\u21d2 B = 1 – A = 1 – \\(\\frac{3}{4}\\) = \\(\\frac{1}{4}\\)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 3.
\nResolve the \\(\\frac{2 x^{2}+3 x+4}{(x-1)\\left(x^{2}+2\\right)}\\) into partial fractions. (AP Mar. ’15, ’11; May ’11) (TS Mar. ’17)
\nSolution:
\nLet \\(\\frac{2 x^{2}+3 x+4}{(x-1)\\left(x^{2}+2\\right)}\\) = \\(\\frac{A}{x-1}\\) + \\(\\frac{\\mathrm{Bx}+\\mathrm{C}}{\\mathrm{x}^{2}+2}\\)
\nMultiplying with (x – 1) (x2<\/sup> + 2)
\n2x2<\/sup> + 3x + 4 = A(x2<\/sup> + 2) + (Bx + C)(x – 1)
\nx = 1 \u21d2 2 + 3 + 4 = A(1 + 2)
\n9 = 3A \u21d2 A = 3
\nEquating the coefficients of x2<\/sup>
\n2 = A + B \u21d2 B = 2 – A = 2 – 3 = -1
\nEquating constants
\n4 = 2A – C \u21d2 C = 2A – 4 = 6 – 4 = 2
\n\\(\\frac{2 x^{2}+3 x+4}{(x-1)\\left(x^{2}+2\\right)}\\) = \\(\\frac{3}{x-1}\\) + \\(\\frac{-x+2}{x^{2}+2}\\)<\/p>\n

Question 4.
\nResolve \\(\\frac{x^{4}}{(x-1)(x-2)}\\) into partial fractions.
\nSolution:
\n\"Inter
\nEquating the coefficients of (x – 1) (x – 2),
\n15x – 14 = A(x – 2) + B(x – 1)
\nPut x = 1, 15 – 14 \u21d2 A(-1) = A = -1
\nPut x = 2, 30 – 14 = B(1) \u21d2 B = 16
\n\u2234 \\(\\frac{x^{4}}{(x-1)(x-2)}\\) = x2<\/sup> + 3x + 7 – \\(\\frac{1}{x-1}\\) + \\(\\frac{16}{x-2}\\)<\/p>\n

Question 5.
\nResolve \\(\\frac{x^{2}-3}{(x+2)\\left(x^{2}+1\\right)}\\) into partial fractions. (AP Mar. ’17, ’16)
\nSolution:
\nLet \\(\\frac{x^{2}-3}{(x+2)\\left(x^{2}+1\\right)}\\) = \\(\\frac{A}{x+2}\\) + \\(\\frac{B x+C}{x^{2}+1}\\)
\nMultiplying with (x + 2) (x2<\/sup> + 1)
\nx2<\/sup> – 3 = A(x2<\/sup> + 1) + (Bx + C)(x + 2)
\nx = -2 \u21d2 4 – 3 = A(4 + 1)
\n1 = 5A \u21d2 A = \\(\\frac{1}{5}\\)
\nEquating the coefficients of x2<\/sup>
\n1 = A + B \u21d2 B = 1 – A
\n= 1 – \\(\\frac{1}{5}\\) = \\(\\frac{4}{5}\\)
\nEquating the constants – 3 = A + 2C
\n2C = -3 – A = -3 – \\(\\frac{1}{5}\\) = – \\(\\frac{16}{5}\\)
\nC = –\\(\\frac{8}{5}\\)
\n\\(\\frac{x^{2}-3}{(x+2)\\left(x^{2}+1\\right)}\\) = \\(\\frac{1}{5(x+2)}\\) + \\(\\frac{4 x-8}{5\\left(x^{2}+1\\right)}\\)<\/p>\n

\"Inter<\/p>\n

Question 6.
\nResolve \\(\\frac{1}{(x-1)^{2}(x-2)}\\) into partial fractions. (May ’13)
\nSolution:
\n\"Inter
\nPut x = 1 in (1)
\n1 = A(0) + B(1 – 2) + C(0)
\n\u21d2 -B = 1 \u21d2 B = -1
\nPut x = 2 in (1)
\n\u21d2 1 = A(0) + B(0) + C (2 – 1)2<\/sup>
\n\u21d2 C = 1
\nEquating the coefficients of x2<\/sup> in (1)
\n0 = A + C \u21d2 A = -C = -1
\nA = -1
\n\u2234 \\(\\frac{1}{(x-1)^{2}(x-2)}\\) = \\(\\frac{-1}{x-1}\\) – \\(\\frac{1}{(x-1)^{2}}\\) + \\(\\frac{1}{x-2}\\)<\/p>\n

Question 7.
\nResolve \\(\\frac{5 x+1}{(x+2)(x-1)}\\) into Partial fractions.
\nSolution:
\n\"Inter
\n\u21d2 5x + 1 = A(x – 1) + B(x + 2) —– (1)
\nPut x = 1 in(1)
\n5(1) + 1,= A(0) + B(1 + 2)
\n\u21d2 3B = 6 \u21d2 B = 2
\nPut x = -2 in (1)
\n5(-2) + 1 = A (-2 – 1) + B(0)
\n\u21d2 -9 = -3A \u21d2 A = 3
\n\u2234 \\(\\frac{5 x+1}{(x+2)(x-1)}\\) = \\(\\frac{3}{x+2}\\) + \\(\\frac{2}{x-1}\\)<\/p>\n

\"Inter<\/p>\n

Question 8.
\nResolve \\(\\frac{2 x+3}{5(x+2)(2 x+1)}\\) into Partial fractions.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 9.
\nResolve \\(\\frac{13 x+43}{2 x^{2}+17 x+30}\\) into partial fractions.
\nSolution:
\n2x2<\/sup> + 17x + 30 = 2x2<\/sup> + 12x + 5x + 30
\n= 2x(x + 6) + 5(x + 6)
\n= (x + 6) (2x + 5)
\nLet \\(\\frac{13 x+43}{(x+6)(2 x+5)}\\) = \\(\\frac{A}{x+6}\\) + \\(\\frac{B}{2x+5}\\)
\n\u2234 13x + 43 = A(2x + 5) + B(x + 6)
\nputting x = -6
\n-78 + 43 = A(-12 + 5)
\n-35 = -7A
\n\u21d2 A = 5
\nPutting x = \\(\\frac{-5}{2}\\)
\n13\\(\\left(\\frac{-5}{2}\\right)\\) + 43 = \\(B\\left(\\frac{-5}{2}+6\\right)\\)
\n\u21d2 -65 + 86 = B(\u20145 + 12)
\n\u21d2 21 = 7B
\n\u21d2 B = 3
\n\u2234 \\(\\frac{13 x+43}{2 x^{2}+17 x+30}\\) = \\(\\frac{5}{x+6}\\) + \\(\\frac{3}{2 x+5}\\)<\/p>\n

Question 10.
\nResolve \\(\\frac{x^{2}+5 x+7}{(x-3)^{3}}\\) into partial fractions.
\nSolution:
\nLet x – 3 = y \u21d2 x = y + 3
\n\\(\\frac{x^{2}+5 x+7}{(x-3)^{3}}\\) = \\(\\frac{(y+3)^{2}+5(y+3)+7}{y^{3}}\\)
\n\"Inter<\/p>\n

Question 11.
\nResolve \\(\\frac{x^{2}+13 x+15}{(2 x+3)(x+3)^{2}}\\) into partial fractions.
\nSolution:
\n\"Inter<\/p>\n

Question 12.
\nResolve \\(\\frac{1}{(x-1)^{2}(x-2)}\\) into Partial fractions.
\nSolution:
\n\"Inter
\nPut x = 1 in (1)
\n1 = A (0) + B(1 – 2) + C(0)
\n\u21d2 -B = 1 \u21d2 B = -1
\nPut x = 2 in (1)
\n1 = A(0) + B(0) + C(2 – 1)2<\/sup>
\n\u21d2 C = 1
\nEquating the coefficients of x2<\/sup> in (1)
\n0 = A + C \u21d2 A = -C = -1
\nA = -1
\n\u2234 \\(\\frac{1}{(x-1)^{2}(x-2)}\\) = \\(\\frac{-1}{x-1}\\) – \\(\\frac{1}{(x-1)^{2}}\\) + \\(\\frac{1}{x-2}\\)<\/p>\n

\"Inter<\/p>\n

Question 13.
\nResolve \\(\\frac{3 x-18}{x^{3}(x+3)}\\) into Partial fractions.
\nSolution:
\n\"Inter
\nPut x = -3 in(1)
\n3(-3) – 18 = A(0) + B(0) + C(O) + D (-3)3<\/sup>
\n\u21d2 -27D = -27 \u21d2 D = 1
\nPut x = 0 in (1)
\n3(0) – 18 = A (0) + B(0) + C(0 + 3) + D(0)
\n\u21d2 3C = -18 \u21d2 C = -6
\nEquating the coefficients of x3<\/sup> in (1)
\n0 = A + D
\n\u21d2 A = -D = -1 \u21d2 A = -1
\nEquating the coefficients of x2<\/sup> in (1)
\n0 = 3A + B
\n\u21d2 B = -3A = -3(-1) = 3 \u21d2 B = 31
\n\u2234 \\(\\frac{3 x-18}{x^{3}(x+3)}\\) = \\(\\frac{-1}{x}\\) + \\(\\frac{3}{x^{2}}\\) – \\(\\frac{6}{x^{3}}\\) + \\(\\frac{1}{x+3}\\)<\/p>\n

Question 14.
\nResolve \\(\\frac{x-1}{(x+1)(x-2)^{2}}\\) into Partial fractions.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 15.
\nResolve \\(\\frac{2 x^{2}+1}{x^{3}-1}\\) into partial fractions.
\nSolution:
\n\"Inter
\nPut x = 1 in (1)
\n2(1) + 1 = A(1 + 1 + 1) + (B + C)(0)
\n\u21d2 3A = 3 \u21d2 A = 1
\nPut x = 0 in (1)
\n0 + 1 = A(1) + (0 + C)(0 – 1)
\n\u21d2 1 = A – C
\n\u21d2 1 = 1 – C \u21d2 C = 0
\nEquating the coefficients of x2<\/sup> in (1)
\n2 = A + B
\n\u21d2 2 = 1 + B \u21d2 B = 1
\n\"Inter<\/p>\n

Question 16.
\nResolve \\(\\frac{x^{3}+x^{2}+1}{\\left(x^{2}+2\\right)\\left(x^{2}+3\\right)}\\) into partial fractions.
\nSolution:
\n\"Inter
\nComparing the coefficients of x3<\/sup>, x2<\/sup>, x and constant terms
\nA + C = 1, B + D = 1, 3A + 2C = 0, 3B + 2D = 1
\nSolve
\n\"Inter<\/p>\n

Question 17.
\nResolve \\(\\frac{3 x^{3}-2 x^{2}-1}{x^{4}+x^{2}+1}\\) into Partial fractions.
\nSolution:
\n\"Inter
\nA – B + C + D = 0 —— (3)
\nB + D = -1 ——- (4) D = -1 – B
\nSubstitute C, D in (2)
\n-A + B + 3 – A – 1 – B = -2
\n\u21d2 -2A = -4 \u21d2 A = 2
\nSubstitute C, D in (3)
\nA – B + 3 – A – 1 – B = 0 \u21d2 2 = 2B \u21d2 B = 1
\n\u2234 C = 3 – 2 = 1, D = -1 – 1 = -2
\nAx + B = 2x + 1, Cx + D = x – 2
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 18.
\nResolve \\(\\frac{x^{4}+24 x^{2}+28}{\\left(x^{2}+1\\right)^{3}}\\) into partial fractions.
\nSolution:
\n\"Inter<\/p>\n

Question 19.
\nResolve \\(\\frac{x+3}{(1-x)^{2}\\left(1+x^{2}\\right)}\\) into partial fractions.
\nSolution:
\nLet \\(\\frac{x+3}{(1-x)^{2}\\left(1+x^{2}\\right)}\\)
\n= \\(\\frac{\\mathrm{A}}{1-x}\\) + \\(\\frac{B}{(1-x)^{2}}\\) + \\(\\frac{C x+D}{1+x^{2}}\\)
\n= x + 3 = A(1 – x) (1 – x2<\/sup>) + B(1 + x2<\/sup>) + (Cx + D)(1 – x2<\/sup>)
\nComparing the coefficients of like powers of x, we get
\nA + B + D = 3 \u2192 (1)
\n-A + C – 2D = 1 \u2192 (2)
\nA + B – 2C + D = 0 \u2192 (3)
\n-A + C = 0 \u2192 (4)
\nSolving these equations, we get
\n\"Inter<\/p>\n

Question 20.
\nResolve \\(\\frac{x^{3}}{(2 x-1)(x+2)(x-3)}\\) into partial fractions.
\nSolution:
\n\\(\\frac{x^{3}}{(2 x-1)(x+2)(x-3)}\\)
\n= \\(\\frac{1}{2}\\) + \\(\\frac{A}{2 x-1}\\) + \\(\\frac{B}{x+2}\\) + \\(\\frac{c}{x-3}\\)
\nMultiplying with 2(2x – 1) (x + 2) (x – 3)
\n2x3<\/sup> = (2x – 1) (x + 2) (x – 3) + 2A(x + 2)
\n(x – 3) + 2B (2x – 1) (x – 3) + 2C (2x -1) (x + 2)
\n\"Inter<\/p>\n

Question 21.
\nResolve \\(\\frac{x^{4}}{(x-1)(x-2)}\\) into partial fractions. (T.S. Mar. ’16, March – 2013)
\nSolution:
\n\"Inter
\nEquating the coefficients of (x – 1) (x – 2),
\n15x – 14 = A(x – 2) + B(x – 1)
\nPut x = 1, 15 – 14 = A(-1) \u21d2 A = -1
\nPut x = 2, 30 – 14 = B(1) \u21d2 B = 16
\n\u2234 \\(\\frac{x^{4}}{(x-1)(x-2)}\\) = x2<\/sup> + 3x + 7 – \\(\\frac{1}{x-1}\\) + \\(\\frac{16}{x-2}\\)<\/p>\n

Question 22.
\nFind the coefficient of x4<\/sup> in the expansion of \\(\\frac{3 x}{(x-2)(x+1)}\\) in powers of x specifying the interval in which the expansion is valid.
\nSolution:
\n\\(\\frac{3 x}{(x-2)(x+1)}\\) = \\(\\frac{A}{x-2}\\) + \\(\\frac{B}{x+1}\\)
\nMultiplying with (x – 2) (x + 1)
\n3x = A(x + 1) + B(x – 2)
\nPut x = -1, -3 = B(-3) \u21d2 B = 1
\nPut x = 2, 6 = A(3) \u21d2 A = 2
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 23.
\nFind the coefficient of xn<\/sup> in the power series expansion of \\(\\frac{x}{(x-1)^{2}(x-2)}\\) specifying the region in which the expansion is valid.
\nSolution:
\n\\(\\frac{x}{(x-1)^{2}(x-2)}\\) = \\(\\frac{A}{x-1}\\) + \\(\\frac{B}{(x-1)^{2}}\\) + \\(\\frac{C}{x-2}\\)
\nMultiplying with (x – 1)2<\/sup> (x – 2)
\nx = A(x – 1) (x – 2) + B(x – 2) + C(x – 1)2<\/sup>
\nPut x = 1, 1 = B(-1) \u21d2 B = -1
\nPut x = 2, 2 = C(1) \u21d2 C = 2
\nPut x = 0, 0 = 2A – 2B + C \u21d2 2A = 2B – C
\n= -2 – 2 = -4 \u21d2 A = -2
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Partial Fractions Important Questions which are most likely to be asked in the exam. Intermediate 2nd Year Maths 2A Partial Fractions Important Questions Question 1. (Mar. 14) Solution: = + + Multiplying with (x2 – 4) (x + 1) x + 4 = …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35212"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35212"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35212\/revisions"}],"predecessor-version":[{"id":35213,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35212\/revisions\/35213"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35212"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35212"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35212"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}