Partial Fractions (SAQs)
Maths-2A | 7. Partial Fractions – SAQs:
Welcome to SAQs in Chapter 7: Partial Fractions. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.
SAQ-1 : Resolve (3x+7)/(x2-3x+2) into partial fractions
G.E = 3x + \frac{7}{x^2} – 3x + 2 = 3x + \frac{7}{(x – 1)(x – 2)}
Let \frac{3x + 7}{(x – 1)(x – 2)} = \frac{A}{x- 1} + \frac{B}{x – 2} = \frac{A(x – 2) + B(x – 1)}{(x – 1)(x – 2)}
A(x-2) + B(x-1) = 3x + 7…(1)
Putting x = 2 in (1) we get A(2-2) + B(2-1) = 3(2) + 7 \Rightarrow B = 13
Putting x = 1 in (1) we get A(1-2) + B(1-1) = 3(1) + 7 \Rightarrow -A = 10 \Rightarrow A = -10
3x + \frac{7}{x^2} – 3x + 2 = \frac{A}{(x-1)} + \frac{B}{(x-2)} = \frac{-10}{(x-1)} + \frac{13}{(x-2)} = \frac{10}{(1-x)} + \frac{13}{(x-2)}
SAQ-2 : Resolve (x+4)/(x2-4)(x+1) into partial fractions
G.E = x + \frac{4}{(x^2 – 4)(x + 1)} = x + \frac{4}{(x+ 2)(x – 2)(x + 1)}
Let \frac{x + 4}{(x + 2)(x – 2)(x + 1)} = \frac{A}{x + 2} + \frac{B}{x – 2} + \frac{C}{x + 1}
= \frac{A(x-2)(x+1) + B(x+2)(x+1) + C(x^2-4)}{(x+2)(x-2)(x+1)}
A(x-2)(x+1) + B(x+2)(x+1) + C(x^2-4) = x+4…(1)
Putting x = −2 in (1) we get A(-2-2)(-2+1) + B(0) + C(0) = -2 + 4 \Rightarrow 4A = 2 \Rightarrow A = 1/2
Putting x = 2 in (1) we get A(0) + B(2+2)(2+1) + C(0) = 2 + 4 \Rightarrow 12B = 6 \Rightarrow B = 1/2
Putting x = −1 in (1) we get A(0) + B(0) + C(1-4) = -1 + 4 \Rightarrow -3C = 3 \Rightarrow C = -1
x + \frac{4}{(x^2 – 4)(x + 1)} = \frac{A}{(x+2)} + \frac{B}{(x-2)} + \frac{C}{(x+1)} = \frac{1}{2}(x+2) + \frac{1}{2}(x-2) – \frac{1}{(x+1)}
SAQ-3 : Resolve (x2-x+1)/(x+1)(x-1)2 into partial fractions
Let x^2 – x + \frac{1}{(x + 1)(x – 1)^2} = \frac{A}{x + 1} + \frac{B}{x – 1} + \frac{C}{(x – 1)^2}
x^2 – x + \frac{1}{(x+1)(x-1)^2} = \frac{A(x-1)^2 + B(x+1)(x-1) + C(x+1)}{(x+1)(x-1)^2}
A(x-1)^2 + B(x+1)(x-1) + C(x+1) = x^2-x+1…(1)
Putting x = 1 in (1) we get 1 = A(1-1)^2 + B(2)(0) + C(1+1) \Rightarrow 2C = 1 \Rightarrow C = 1/2
Putting x = −1 in (1) we get = A(1-(-1))^2 + B(-1+1)(-1-1) + C(-1+1) = 3 \Rightarrow 4A = 3 \Rightarrow A = 3/4
Equating the coefficients of x2 we get A+B = 1 \Rightarrow B = 1 – A = 1 – 3/4 = 1/4
x^2 – x + \frac{1}{(x + 1)(x – 1)^2} = \frac{A}{x + 1} + \frac{B}{x – 1} + \frac{C}{(x – 1)^2} = \frac{3}{4}(x + 1) + \frac{1}{4}(x – 1) + \frac{1}{2}(x – 1)^2
SAQ-4 : Resolve 2x2+2x+1/x3+x2 into partial fractions
Let 2x^2 + 2x + \frac{1}{x^3} + x^2 = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1}
= \frac{Ax(x+1) + B(x+1) + Cx^2}{x^2(x+1)}
Ax(x+1) + B(x+1) + Cx^2 = 2x^2 + 2x + 1…(1)
Putting x = 0 in (1) we get: A(0) + B(1) + C(0) = 1 \Rightarrow B = 1
Putting x = −1 in (1) we get: A(0) + B(0) + C(-1)^2 = 2(-1)^2 + 2(-1) + 1 \Rightarrow C = 1
Equating the coefficients of x2 we get: 2 = A + C \Rightarrow A = 2 – C = 2 – 1 = 1
2x^2 + 2x + \frac{1}{x^3} + x^2 = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x + 1}
SAQ-5 : Resolve (3x-18)/(x3(x+3) into partial fractions
Let 3x – \frac{18}{x^3(x + 3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x + 3}
\Rightarrow 3x – \frac{18}{x^3(x + 3)} = \frac{Ax^2(x + 3) + Bx(x + 3) + C(x + 3) + Dx^3}{x^3(x + 3)}
Ax^2(x + 3) + Bx(x + 3) + C(x + 3) + Dx^3 = 3x – 18….(1)
Putting x = −3 in (1) we get: A(0) + B(0) + C(0) + D(-3)^3 = 3(-3) – 18 \Rightarrow -27D = -27 \Rightarrow D = 1
Putting x = 0 in (1) we get: A(0) + B(0) + C(0+3) + D(0) = 3(0) – 18 \Rightarrow 3C = -18 \Rightarrow C = -6
Equating the coefficient of x3 we get: A + D = 0 \Rightarrow A = -D \Rightarrow A = -1
Equating the coefficient of x2 we get: 3A + B = 0 \Rightarrow B = -3A = -3(-1) \Rightarrow B = 3
3x – \frac{18}{x^3(x + 3)} = \frac{-1}{x} + \frac{3}{x^2} – \frac{6}{x^3} + \frac{1}{x + 3}
SAQ-6 : Resolve (x2+5x+7)/(x-3)3 into partial fractions
Put x – 3 = y
\frac{x^2 + 5x + 7}{(x – 3)^3} = \frac{(y + 3)^2 + 5(y + 3) + 7}{y^3} = \frac{y^2 + 6y + 9 + 5y + 15 + 7}{y^3} = \frac{y^2 + 11y + 31}{y^3}
= \frac{1}{y} + \frac{11}{y^2} + \frac{31}{y^3} = \frac{1}{x – 3} + \frac{11}{(x – 3)^2} + \frac{31}{(x – 3)^3}
SAQ-7 : Resolve 2x2+1/x3-1into partial fractions
Let 2x^2 + \frac{1}{x^3 – 1} = 2x^2 + \frac{1}{(x – 1)(x^2 + x + 1)}
2x^2 + \frac{1}{(x – 1)(x^2 + x + 1)} = \frac{A}{x – 1} + \frac{Bx + C}{x^2 + x + 1} = \frac{A(x^2 + x + 1) + (Bx + C)(x – 1)}{(x – 1)(x^2 + x + 1)}
A(x^2 + x + 1) + (Bx + C)(x – 1) = 2x^2 + 1…(1)
Putting x = 1 in (1) we get A(1 + 1 + 1) + (B + C)(0) = 2(1)^2 + 1 \Rightarrow 3A = 3 \Rightarrow A = 1
Comparing the coefficients of x2 in (1) we get A + B = 2 \Rightarrow B = 2 – A = 2 – 1 \Rightarrow B = 1
Comparing the constant terms in (1) we get A + C = 1 \Rightarrow C = 1 – A = 1 – 1 \Rightarrow C = 0
2x^2 + \frac{1}{x^3 – 1} = \frac{A}{x – 1} + \frac{Bx + C}{x^2 + x + 1} = \frac{1}{x – 1} + \frac{x}{x^2 + x + 1}
SAQ-8 : Resolve 2x2+3x+4/(x-1)(x2+2) into partial fractions
Let 2x^2 + 3x + 4/(x – 1)(x^2 + 2) = \frac{A}{x – 1} + \frac{Bx + C}{x^2 + 2} = \frac{A(x^2 + 2) + (Bx + C)(x – 1)}{(x – 1)(x^2 + 2)}
A(x^2 + 2) + (Bx + C)(x – 1) = 2x^2 + 3x + 4…(1)
Putting x = 1 in (1) we get A(1^2 + 2) + (B + C)(0) = 2(1)^2 + 3(1) + 4
\Rightarrow 3A = 9 \Rightarrow A = 3
Putting x = 0 in (1) we get A(0^2 + 2) + (0 + C)(0 – 1) = 2(0)^2 + 3(0) + 4 \Rightarrow 2A – C = 4
\Rightarrow C = 2A – 4 = 2(3) – 4 = 2
Comparing the coefficient of x2 in (1) we get A + B = 2 \Rightarrow 3 + B = 2 \Rightarrow B = -1
2x^2 + 3x + 4/(x – 1)(x^2 + 2) = \frac{3}{x – 1} + \frac{-1x + 2}{x^2 + 2} = \frac{3}{x – 1} + \frac{x}{x^2 + 2} + \frac{2}{x^2 + 2}
SAQ-9 : Resolve x2-3/(x+2)(x2+1) into partial fractions
Let x^2 – \frac{3}{(x + 2)(x^2 + 1)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1} = \frac{A(x^2+1) + (Bx + C)(x + 2)}{(x+2)(x^2+1)}
\Rightarrow A(x^2+1) + (Bx+C)(x+2) = x^2-3…(1)
Putting x = −2 in (1) we get A(4+1) + (B(-2)+C)(0) = 4 – 3 \Rightarrow 5A = 1 \Rightarrow A = 1/5
Putting x = 0 in (1) we get A(0^2 + 1) + (0 + C)(2) = 0^2 – 3 \Rightarrow 1A + 2C = -3 \Rightarrow C = -\frac{8}{5}
Comparing the coefficients of x2 we get A + B = 1 \Rightarrow B = 1 – A = 1 – \frac{1}{5} = \frac{4}{5}
x^2 – \frac{3}{(x+2)(x^2+1)} = \frac{1}{5}(x+2) + \left(\frac{4}{5}x – \frac{8}{5}\right)/(x^2+1)
SAQ-10 : Resolve x3/(x-1)(x+2) into partial fractions
Here the degree of numerator 3 ≥ degree of denominator 2 So it is an improper function
Also (x-1)(x+2) = x^2 + x – 2
Now on dividing x3 by x2 + x − 2 we have:
x^3/(x-1)(x+2) = (x-1) + \frac{3x-2}{x^2+x-2}
Now \frac{3x-2}{x^2+x-2} = \frac{3x-2}{(x-1)(x+2)}
= \frac{A}{x-1} + \frac{B}{x+2} = \frac{A(x+2)+B(x-1)}{(x-1)(x+2)}
\Rightarrow A(x+2) + B(x-1) = 3x – 2…(1)
Putting x = 1 in (1) we get A(3) + B(0) = 3 – 2 \Rightarrow 3A = 1 \Rightarrow A = 1/3
Putting x = −2 in (1) we get A(0) + B(-3) = 3(-2)-2 \Rightarrow -3B = -8 \Rightarrow B = 8/3
\Rightarrow \frac{3x-2}{x^2+x-2} = \frac{A}{x-1} + \frac{B}{x+2} = \frac{1}{3}(x-1) + \frac{8}{3}(x+2)
x^3/(x-1)(x+2) = x-1 + \frac{1}{3}(x-1) + \frac{8}{3}(x+2)
SAQ-11 : Resolve x4/(x-1)(x-2) into partial fractions
(x-1)(x-2) = x^2-3x+2
x^4/(x-1)(x-2)
= (x^2 + 3x + 7) + \frac{15x-14}{x^2-3x+2}
Now, \frac{15x-14}{x^2-3x+2} = \frac{15x-14}{(x-1)(x-2)}
= \frac{A}{(x-1)} + \frac{B}{(x-2)} = \frac{A(x-2) + B(x-1)}{(x-1)(x-2)}
\Rightarrow A(x-2) + B(x-1) = 15x-14…(1)
Putting x = 1 in (1) we get A(1-2) + B(0) = 15(1) – 14 = 1 \Rightarrow A = -1
Putting x = 2 in (1) we get A(0) + B(2-1) = 15(2) – 14 = 16 \Rightarrow B = 16
x^4/(x-1)(x-2) = x^2 + 3x + 7 + \frac{-1}{x-1} + \frac{16}{x-2}
SAQ-12 : Resolve x3/(2x-1)(x-1)2 into partial fractions
Let x^3/(2x-1)(x-1)^2 = \frac{1}{2} + \frac{A}{2x-1} + \frac{B}{x-1} + \frac{C}{(x-1)^2}
= \frac{(2x-1)(x-1)^2+2A(x-1)^2+2B(2x-1)(x-1)+2C(2x-1)}{2(2x-1)(x-1)^2}
\Rightarrow (2x-1)(x-1)^2+2A(x-1)^2+2B(2x-1)(x-1)+2C(2x-1)=2x^3…(1)
Putting x=\frac{1}{2}
Putting x = 1 in (1) we get 2C(2\cdot1-1) = 2(1)^3 \Rightarrow C = 1
Putting x = 0 in (1) gives:
(-1)(-1)^2+2A(-1)^2+2B(2\cdot0-1)(-1)+2C(2\cdot0-1)=0
2A+2B-2C=0
Given A = \frac{1}{2}
2B=2-2A
2B=2-1
B=1/2
x^3/(2x-1)(x-1)^2 = \frac{1}{2} + \frac{1}{2}(2x-1) + \frac{1}{x-1} + \frac{1}{(x-1)^2}
SAQ-13 : Resolve x3/(x-a)(x-b)(x-c) into partial fractions
Let x^3/(x-a)(x-b)(x-c) = 1 + \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}
= \frac{(x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)}{(x-a)(x-b)(x-c)}
\Rightarrow (x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b) = x^3…(1)
Putting x = a in (1) we get 0+A(a-b)(a-c)+0+0 = a^3 \Rightarrow A = \frac{a^3}{(a-b)(a-c)}
Similarly, by putting x=b
x^3/(x-a)(x-b)(x-c) = 1 + \frac{a^3}{(a-b)(a-c)}(x-a) + \frac{b^3}{(b-c)(b-a)}(x-b) + \frac{c^3}{(c-a)(c-b)}(x-c)