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Learn/ 11th Class Mathematics/ Unit 5 /Numericals

Unit 5: Partial Fractions — Exercise 5 2

11th Class Mathematics · Unit 5: Partial Fractions

1.Resolve into partial fractions: $\frac{2x^2+3x+3}{(x+1)(x^2+1)}$
Formula
Assume form 2x2+3x+3(x+1)(x2+1) = Ax+1 + Bx+Cx2+1
Multiply by $(x+1)(x^2+1)$ 2x2+3x+3 = A(x2+1) + (Bx+C)(x+1)
Put $x = -1$ 2(1) - 3 + 3 = A(1+1) ⇒ 2 = 2A ⇒ A = 1
Comparing coefficients of $x^2$ A + B = 2 ⇒ 1 + B = 2 ⇒ B = 1
Comparing coefficients of $x$ B + C = 3 ⇒ 1 + C = 3 ⇒ C = 2
Result
Answer 1x+1 + x+2x2+1
2.Resolve into partial fractions: $\frac{2x+1}{(x-2)(x^2+3x+5)}$
Formula
Assume form 2x+1(x-2)(x2+3x+5) = Ax-2 + Bx+Cx2+3x+5
Multiply by $(x-2)(x^2+3x+5)$ 2x+1 = A(x2+3x+5) + (Bx+C)(x-2)
Put $x = 2$ 5 = A(4+6+5) ⇒ 5 = 15A ⇒ A = 13
Comparing coefficients of $x^2$ A + B = 0 ⇒ B = -13
Comparing coefficients of $x$ 3A - 2B + C = 2 ⇒ C = 2 - 1 + 23 = 13
Result
Answer 13(x-2) - x-13(x2+3x+5)
3.Resolve into partial fractions: $\frac{2x+32}{(x-2)(x^2+2)^2}$
Formula
Assume form 2x+32(x-2)(x2+2)2 = Ax-2 + Bx+Cx2+2 + Ex+D(x2+2)2
Multiply by $(x-2)(x^2+2)^2$ 2x+32 = A(x2+2)2 + (Bx+C)(x-2)(x2+2) + (Ex+D)(x-2)
Put $x = 2$ 36 = A(4+2)2 = 36A ⇒ A = 1
Comparing $x^3$ coefficient A + B = 0 ⇒ B = -1
Comparing $x^2$ coefficient -2B + C = 0 ⇒ C = -2
Comparing $x$ coefficient 4A + 2B - 2C + E = 0 ⇒ 4 - 2 + 4 + E = 0 ⇒ E = -6
Comparing constant term 4A - 4C - 2D = 32 ⇒ D = -10
Result
Answer 1x-2 + x+2x2+2 + 2(3x+5)(x2+2)2
4.Resolve into partial fractions: $\frac{3x^2+3}{x^3+1}$
Factor denominator x3+1 = (x+1)(x2-x+1)
Formula
Assume form 3x2+3(x+1)(x2-x+1) = Ax+1 + Bx+Cx2-x+1
Multiply by $(x+1)(x^2-x+1)$ 3x2+3 = A(x2-x+1) + (Bx+C)(x+1)
Put $x = -1$ 6 = A(1+1+1) ⇒ A = 2
Comparing coefficient of $x^2$ A + B = 3 ⇒ B = 1
Comparing coefficient of $x$ -A + B + C = 0 ⇒ C = 1
Result
Answer 2x+1 + x+1x2-x+1
5.Resolve into partial fractions: $\frac{x^4}{x^4+2x^2+1}$
Recognize improper fraction, perform long division x4x4+2x2+1 = 1 + -2x2-1x4+2x2+1
Factor denominator x4+2x2+1 = (x2+1)2
Formula
Decompose remainder 2x2+1(x2+1)2 = Ax+Bx2+1 + Cx+D(x2+1)2
Comparing coefficients A = 0, B = 2, C = 0, D = -1
Result
Answer 1 - 2x2+1 + 1(x2+1)2
6.Resolve into partial fractions: $\frac{6x^4+40x^2}{(4-x^2)(x^2+4)^2}$
Factor denominator 4-x2 = (2-x)(2+x)
Formula
Assume form 6x4+40x2(2-x)(2+x)(x2+4)2 = A2-x + B2+x + Cx+Dx2+4 + Ex+F(x2+4)2
Put $x = 2$ 256 = 256A ⇒ A = 1
Put $x = -2$ 256 = 256B ⇒ B = 1
Expand and compare coefficients of $x^5, x^4, x^3, x$ C = 0, D = -2, E = 0, F = -8
Result
Answer 12-x + 12+x + 2x2+4 + 8(x2+4)2